Post by: jakesilove on January 28, 2016, 07:58:56 pm
HSC PHYSICS Q&A THREAD
To go straight to posts for the new syllabus, click here.
What is this thread for?
If you have general questions about the HSC Physics course or how to improve in certain areas, this is the place to ask! 👌
Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.
Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!
There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.
To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!
OTHER PHYSICS RESOURCES
CLICK ME!
* Free Physics notes
* HSC Physics Question Thread
* How to Get a Band 6 in HSC Physics
* Physics MEGA THREAD
* Physics News and Cool Stuff
* Foolproof Method for Maths Questions in HSC Physics
* Physics: 8 Mistakes Every Student Makes!
* A Quick Guide to Motion and Kinematics
* A Guide to Understanding Special Relativity
* A Guide to Motors and the Motor Effect
* A Guide to Generators and Induction
* A Guide to Transformers and AC Power
* A Guide to Cathode Rays
* A Guide to The Photoelectric Effect
* 60 Hours Until Physics
* Quantum Physics in 10 Minutes
* How to Go About Studying for Any Science
* HSC Physics Question Thread
* How to Get a Band 6 in HSC Physics
* Physics MEGA THREAD
* Physics News and Cool Stuff
* Foolproof Method for Maths Questions in HSC Physics
* Physics: 8 Mistakes Every Student Makes!
* A Quick Guide to Motion and Kinematics
* A Guide to Understanding Special Relativity
* A Guide to Motors and the Motor Effect
* A Guide to Generators and Induction
* A Guide to Transformers and AC Power
* A Guide to Cathode Rays
* A Guide to The Photoelectric Effect
* 60 Hours Until Physics
* Quantum Physics in 10 Minutes
* How to Go About Studying for Any Science
Original post.
Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Hey everyone!
A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!
This is a community, so we want you to feel like you can post any type of Physics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Physics is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.
Remember that Physics can be a difficult course. There will be lots of answers to the same questions, and I'll try give you the best or easiest to remember ones.
I got an ATAR of 99.80, and a mark of 93 in the the Physics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Hey everyone!
A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!
This is a community, so we want you to feel like you can post any type of Physics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Physics is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.
Remember that Physics can be a difficult course. There will be lots of answers to the same questions, and I'll try give you the best or easiest to remember ones.
I got an ATAR of 99.80, and a mark of 93 in the the Physics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Post by: chloe9756 on January 30, 2016, 10:09:45 am
The truck driver wishes to pass the car but notices an oncoming car which he estimates to be 1.0 km away.
The truck driver assumes that the oncoming car is also travelling at 30 ms-1. If the car is 3.0 m long, the truck 15
m long and two car lengths are allowed as clearance before and after passing:
(a) Assuming that the truck driver’s assumptions are correct, does the truck make it? Give reasons for your
answer.
(b) The truck driver’s distance estimate is correct but the oncoming car is travelling at a speed greater than 30
ms-1. What is the maximum speed that the oncoming car can travel and still avoid a collision?
ans:
a) yes, by 100m
b) 36.7 ms-1
Post by: jamonwindeyer on January 30, 2016, 12:00:26 pm
A truck is travelling along a straight highway at a speed of 30 ms-1. Ahead of it is a car travelling at 28 ms-1.
The truck driver wishes to pass the car but notices an oncoming car which he estimates to be 1.0 km away.
The truck driver assumes that the oncoming car is also travelling at 30 ms-1. If the car is 3.0 m long, the truck 15
m long and two car lengths are allowed as clearance before and after passing:
(a) Assuming that the truck driver’s assumptions are correct, does the truck make it? Give reasons for your
answer.
(b) The truck driver’s distance estimate is correct but the oncoming car is travelling at a speed greater than 30
ms-1. What is the maximum speed that the oncoming car can travel and still avoid a collision?
ans:
a) yes, by 100m
b) 36.7 ms-1
Hi Chloe! I'm not Jake (obviously ;)) but let me field this one for you! I've written up a solution below. The quality is a little blurry, I'm not quite sure why. I'd normally write up the solution in the response itself, but I'm having some issues with my LaTex equation formatting, so this will hopefully be a good substitute!
Have a read, maybe twice, it's a little tricky to wrap your head around, and of course let me know if anything doesn't quite make sense. This is a question on relative velocities, which can be difficult, but if you are struggling, picture driving in a car on the motorway. When a car overtakes you, they are moving at say, 120 kilometres an hour. But you don't see them whiz past, you see them slowly edge in front of you. Relative to you (your frame of reference, if you remember from Space topics), they are only moving 10 kilometres an hour, since you are travelling at 110km/h. If a police officer was sitting on the side of the freeway, in their frame of reference, they see the overtaker moving at 120km/h (and would likely go and fine them for speeding).
Thanks so much for the question! I guarantee a whole bunch of people will benefit from you asking, be sure to keep sending questions to Jake and myself! We are happy to assist :D enjoy your weekend!
(http://i.imgur.com/6Wcdks1.png)
Post by: chloe9756 on January 30, 2016, 01:52:52 pm
a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?
these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.
2. a small car (500kg) pulls on a caravan (200kg). the car's engine supplies a driving force of 18 000N. The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.
while the car starts off from a red light, is the force it moves itself with greater, equal or less than the force it uses to move the caravan? give a qualitative and quantitative explanation.
thank you so much
Post by: Happy Physics Land on January 30, 2016, 08:01:20 pm
1. a sailor hoists a heavy weight up by pulling down a rope slung over a pulley:
a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?
these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.
thank you so much
Hey Chloe:
I am a year 12 student currently undertaking physics, and in regards to question 1, I believe that I have a valid explanation to these questions however I cannot be 100% certain. I will post my solutions and explanations here and if jake happens to scroll through this chat he can confirm whether or not my reasonings are correct. Tension questions as such as perhaps the most difficult part of Moving About, and to be honest some teachers simply skip this section of the textbook because its not frequently asked in HSC exams, but more so in engineering studied exams.
a) The tension in the rope when the mass is at rest should be the same as the tension of the rope when the mass is moving upwards with constant velocity. This would seem very counter-intuitive at first because in real life when we pull up an object away from the ground, we would always feel more exhausted than simply holding the object statically in one position. However, keep in mind that we are dealing with FORCE here. In your response, chloe, you have made a very valid point that the upwards acceleration of the cradle would require more tensile force. However your answer is a little inaccurate in saying that the point of a higher tensile force is to prevent the object from falling. The correct explanation to your answer of having a greater tensile force should be to accelerated the object to our desired "constant speed" value, because as we all know, according to Newton's second law, when we have an acceleration, we have a force (F=ma).
Ok, you have truly state that a greater tensile force is required to accelerate the mass. But when the mass begins to travel at a constant velocity, as phrased in the question, it would no longer be accelerating. Therefore, the value of acceleration would be 0, and if we substitute a=0 into F=ma, then we would discover that the extra tensile force required for a constant velocity is indeed 0N. The weight force of the mass is still equal but opposite to the tension on the rope (Newton's 3rd Law) and there are no other forces acting in the system.Hence the tensile force in the rope should be the same in both scenarios.
The reason why a lot of students would be confused in this case is that in real life, when we pull an object up at constant velocity, we feel exhausted after a while not because of the extra force we apply, but because as the object is pulled up, we are required to exert more kinetic energy to the mass as the object increases in its gravitational potential energy.
b) I feel like there is a very sneaky trap in the second question, your answer seems very logical and it does make sense. Let's first look at how you would have structured your answer better. Something you should mention is as the mass speeds up, it possesses an acceleration and hence according to Newton's 2nd law, F=ma, a tensile force would be required to speed up the mass. This reference to Newton's law would make your response stronger and there would be a higher chance for you to score full marks if you refer to an established scientific law.
But heres the problem. Lets say the crate is accelerating at 10ms^-2, then F = ma will give a net force of 10m Newtons. However, consider the case where the crate is decelerating at 10ms^-2, meaning that a= -10ms^-2, F=ma will give a net force of -10m Newtons. If we compare their magnitude, the force in both scenarios would be the same, since the negative sign is only indicative of a direction. So personally I would say that when the crate is speeding up at a larger rate than its slowing down, then your answer would be correct. If the rate of speeding up is the same as slowing down, then the tension in the rope would be the same.
I hope my answers make sense to you and if you remain dubious about question b), perhaps Jake can confirm my argument when he comes around. But these are very good mechanical questions to think about.
Best Regards
Happy Physics Land
Post by: jamonwindeyer on January 31, 2016, 12:18:25 am
1. a sailor hoists a heavy weight up by pulling down a rope slung over a pulley:
a) is the tension in the rope greater when the mass is at rest or moving upwards with constant speed? justify your answer.
b) if the mass is travelling upwards, is the tension in the rope greatest when the crate is speeding up or when it is slowing down?
these are my answers, but i have no idea if they're correct or not.
a) the tension is greater when the mass is moving upwards. An acceleration of the crate would require more tension to prevent it from falling.
b) the tension is greater as the crate speeds up, as there is more force required to pull a mass up as it accelerates.
2. a small car (500kg) pulls on a caravan (200kg). the car's engine supplies a driving force of 18 000N. The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.
while the car starts off from a red light, is the force it moves itself with greater, equal or less than the force it uses to move the caravan? give a qualitative and quantitative explanation.
thank you so much
Hey Chloe! Happy Physics Land, you did an awesome job answering the question, spot on!
The important thing to remember is that the tension in the rope is a force which corresponds directly to the force that the rope is applying to the crate. In this way, it is a reactive force.
To confirm the answer, Chloe, your reasoning that an acceleration of the crate would result in greater tension in the rope. But Happy has correctly identified that, when moving at a constant speed, there is no acceleration in the crate. Therefore, the tension is identical to if it was stationary. This is assuming the crate isn't on the ground, in which case, there would be no tension in the rope.
For the second part however, I think my interpretation is different. Consider the crate travelling upward at a constant speed, with some level of tension. To accelerate it upwards would require an additional upwards force, provided by the rope, and thus tension would increase. If the crate then accelerated downward (still moving upwards, remember), this downwards acceleration would in fact be caused by gravity, no tension involved. I would therefore say that tension is higher in the rope when accelerating upwards.
Your second question is very interesting indeed. The key is in this phrase:
The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.
What this tells us is that acceleration is identical for the car and trailer. So, if acceleration is equal, but the masses are different, then the forces must also be different. The resultant calculation is an application of Newton's 2nd Law:
Total Force = Car Force + Trailer Force
Car Force = 500a
Trailer Force = 200a
So Total Force = 700a
18000 = 700a
a = 25.7...
So Car Force = 12850, Trailer Force = 5140 (there is a rounding error there), and so the car uses more force pulling itself than the trailer! :D
I hope this explanation helps Chloe, and Happy Physics Land, thanks heaps for tackling that doozy of a first question! I hope it is also helping you on your HSC journey, it is certainly helping a lot of other people ;D
Post by: Happy Physics Land on January 31, 2016, 12:49:06 pm
Thank you very much for your kind words, I believe that I have over-complicated 1b) after seeing your solution, but yes thank you so much for your approval and your encouragement. It means a lot to me!
Best Regards
Happy Physics Land
Post by: chloe9756 on January 31, 2016, 04:16:25 pm
1. a car of mass 2.0 x 10^3kg cruises North down the high way at 100 km/h, with a driving force of 1.2 x 10^4N.
a) calculate the retarding force of friction acting on the car.
b) the car speeds up to 110m/s in 5 seconds. Calculate the acceleration of the car, and thus determine the new driving force of the car.
2. a dog pulls a 80kg sled along the ground with 500N, which encounters a friction force of 150N. On the sled is a 20kg box.
a) calculate the acceleration of the sled.
b) calculate the friction force of the sled on the box which moves the box forwards along with the sled.
c) calculate the net force on the sled.
these are some of my answers but i have no idea whether they're correct:
2. a) a = (500-150)/(80+20) = 3.5 m/s/s forwards
b) F = 3.5 x 20 = 70N forwards
Post by: brenden on January 31, 2016, 04:53:11 pm
sorry for all the questions but i have no idea how to do some of theseWell, I know nothing about Physics, but I know that people asking questions is the foundation for everything we do - so don't be sorry for that! Keep asking them! Thanks Chloe :)
Post by: cajama on January 31, 2016, 05:15:18 pm
A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.
a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)
Thanks in advance :D
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.
Post by: jakesilove on January 31, 2016, 06:38:06 pm
I have a question from motors and gens (from the Surfing book) :)
A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.
a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)
Thanks in advance :D
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.
Hey cajama!
Really really great questions, and definitely one of the most difficult of its type. It requires a good knowledge of multiple sections of the course, which is where the most difficult questions in your HSC will lie. I hope you can follow my explanation; once you've covered even one of these questions, and gone over it a few times, you'll be totally fine with anything even similar to it. My biggest piece of advice for M&G questions where you're just like WHAT THE HELL DO I EVEN DO??!??!?!! is write down the information you have, and just try sub them into a formula.
(http://i.imgur.com/nLqIUWz.png?1)
(http://i.imgur.com/FU1gakp.png?1)
(http://i.imgur.com/zpQIYFI.png?1)
Post by: Happy Physics Land on January 31, 2016, 07:25:23 pm
I have a question from motors and gens (from the Surfing book) :)
A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.
a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)
Thanks in advance :D
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.
Hello Cajama:
I am a year 12 student currently undertaking physics and I am more than happy to help you out here. In regards to questions like these, a lot of people get stuck on it because of how abstract the question is (after all, we cant really see the magnetic field and whoever made this question is so time-poor that they wouldnt even bother giving us a diagram). A lot of my friends are in the same situation and what I recommend to do is to draw a simple 2D diagram just to make life easier. In a HSC exam, questions like these would often be accompanied by a 3D diagram, and similarly what I would recommend for you to do is to draw a 2D diagram which allows you to observe whats happening much more clearly.
Ok so recommendations aside, lets get into the question.
(http://i.imgur.com/QbNXVpg.png?1)
N.B. In the 2D diagram I have attached, I have assumed the direction of the magnetic field vectors and the negative and positive terminals on the voltage supplier. Now, you will soon discover that even if the magnetic field direction reverses, or if the negative and positive terminals reverse, the current will still be travelling in a direction that is perpendicular to the direction of the magnetic field. Hence in this case I am allowed to assume these directions. The importance of this will be discussed later on in part B.
a) This question is rather simple. After you manage to draw out your 2D diagram, you will see thats it's simply an application of Ohm's Law, something we learnt back in year 11 in the "Electricity at homes" module.
Ohm's law states that R = V/I. In this question, our R = 4 ohms, V = 36V and hence when we substitute these values into the formula, we get an answer of 9A currents (R = V/I, I = V/R, I = 36/4, I = 9A)
b) Usually in a question that is divided into several parts, the outcome of the previous part will usually relate to solving the second part. In part a), we were demanded to calculate the current flowing through the metal bar. Recall all the formulae you have learnt in Motors and Generators module, there are two formulae that heavily relate to the concept of force: F = BILsin(theta) and F=qvBsin(theta). If we think about F=qvBsin(theta), we will soon discover that we dont exactly have a value for q, and nor do we have a value for v. Hence F=qvB would be unsuitable for solving part b). Let's consider F =BIL sin(theta). We calculated the current flowing through the bar in part a), which we found out to be 9A. We are also provided with the length of the bar which is 0.5m and the magnitude of the magnetic field which is 0.3T.
But, what about theta? Clearly we are NOT provided with the value of theta. BUT THROUGH DRAWING A 2D DIAGRAM, we can find out what theta is. Lets go back to the first principles and define what this theta is, in the equation F = BILsin(theta). This theta is defined as the angle between the direction of the current and the magnetic field vector. So, when you consider the 2D diagram, it becomes apparent that the current is travelling in a direction that is perpendicular to the magnetic field direction. Therefore, the value of theta is 90 degrees.
Now let's substitute everything into the equation F= BILsin(theta). B = 0.3T, I = 9A, L = 0.5m and theta = 90 degrees. Hence
F = 0.3 x 0.5 x 9 x sin(90) = 1.35 N which is the magnitude of the force. The direction of the force here cannot be calculated because we are not provided with the direction of the magnetic field, we only know that the direction of the magnetic field, whether to the right or to the left, will be perpendicular to the current flow within the metal bar.
c) Ok so this is perhaps the most tricky part of the question because it combined the knowledge from TWO MODULES!! The way I figured out this question is through observing which part of the question I havent used yet. So far, we have used resistance, voltage, length and magnetic field values provided by the question. So now we are left with 0.04kg mass, and like what I said beforehand, the previous part in a question will usually relate to its subsequent part. In part b) we found out the magnitude of force which is 1.35N, hence using Newton's 2nd law F=ma, we discover that 1.35 = 0.04(a), hence a=33.75 ms^-2. Sweet, now we have acceleration, but we were asked to find the displacement of the bar. A convenient way to figure out what to do is to just skim through the formula sheet, finding out the equation that relate acceleration and displacement together. Hence in this case it would be appropriate to use the formula y = ut + 1/2 at^2, because we have all the necessary details to figure out displacement (y). Initially, velocity of the bar is 0, hence u= 0 ms^-1. The acceleration we have worked out is 33.75 ms^-2 and the time in this case is provided (t = 0.25s). Everything after this step now is easy, substitute in all the values, we will obtain that y = 0(0.25) + 1/2 (33.75) (0.25)^2 and throw this into the calculator will will obtain an answer of 1.05m (2.d.p.).
Post by: Happy Physics Land on January 31, 2016, 07:54:06 pm
I have a question from motors and gens (from the Surfing book) :)
A metal bar, mass 0.04kg, resistance 4 ohms, and length 0.5m, is placed on two frictionless conducting rails in a magnetic field of 0.3T. The bar is connected to a 36 V power source.
a) Calculate the current in the bar (ans: 9A)
b) Calculate the magnitude of the force which acts on the bar (ans: 1.35 N)
c) If the bar is free to move, and the force is in the appropriate direction, calculate how far it will move along the rails in 0.25 s. (ans: 1.05m)
Thanks in advance :D
P.s. I can't say for sure if the answers are correct on the book because some of the answers we get from questions can be really dodgy.
I am really sorry I cant upload my 2D diagram because of some technical issues that state that my image is "unwritable" but I hope my explanation was still helpful and clear to you!
Post by: brenden on January 31, 2016, 08:03:41 pm
I am really sorry I cant upload my 2D diagram because of some technical issues that state that my image is "unwritable" but I hope my explanation was still helpful and clear to you!That issue should be fixed tomorrow. For now, you can go to Imgur and upload the image, then once the image is uploaded, right click it, "open image in new tab", then take the URL of that and put it in between the code [img ][/img]
Post by: Happy Physics Land on January 31, 2016, 08:17:44 pm
That issue should be fixed tomorrow. For now, you can go to Imgur and upload the image, then once the image is uploaded, right click it, "open image in new tab", then take the URL of that and put it in between the code [img ][/img]
Noice Brenden!!! Thank you so much you are a lifesaver!!!!! :D
Post by: jakesilove on January 31, 2016, 08:59:32 pm
Hello Cajama:
I am a year 12 student currently undertaking physics and I am more than happy to help you out here. In regards to questions like these, a lot of people get stuck on it because of how abstract the question is (after all, we cant really see the magnetic field and whoever made this question is so time-poor that they wouldnt even bother giving us a diagram). A lot of my friends are in the same situation and what I recommend to do is to draw a simple 2D diagram just to make life easier. In a HSC exam, questions like these would often be accompanied by a 3D diagram, and similarly what I would recommend for you to do is to draw a 2D diagram which allows you to observe whats happening much more clearly.
Ok so recommendations aside, lets get into the question.
(http://i.imgur.com/QbNXVpg.png?1)
N.B. In the 2D diagram I have attached, I have assumed the direction of the magnetic field vectors and the negative and positive terminals on the voltage supplier. Now, you will soon discover that even if the magnetic field direction reverses, or if the negative and positive terminals reverse, the current will still be travelling in a direction that is perpendicular to the direction of the magnetic field. Hence in this case I am allowed to assume these directions. The importance of this will be discussed later on in part B.
a) This question is rather simple. After you manage to draw out your 2D diagram, you will see thats it's simply an application of Ohm's Law, something we learnt back in year 11 in the "Electricity at homes" module.
Ohm's law states that R = V/I. In this question, our R = 4 ohms, V = 36V and hence when we substitute these values into the formula, we get an answer of 9A currents (R = V/I, I = V/R, I = 36/4, I = 9A)
b) Usually in a question that is divided into several parts, the outcome of the previous part will usually relate to solving the second part. In part a), we were demanded to calculate the current flowing through the metal bar. Recall all the formulae you have learnt in Motors and Generators module, there are two formulae that heavily relate to the concept of force: F = BILsin(theta) and F=qvBsin(theta). If we think about F=qvBsin(theta), we will soon discover that we dont exactly have a value for q, and nor do we have a value for v. Hence F=qvB would be unsuitable for solving part b). Let's consider F =BIL sin(theta). We calculated the current flowing through the bar in part a), which we found out to be 9A. We are also provided with the length of the bar which is 0.5m and the magnitude of the magnetic field which is 0.3T.
But, what about theta? Clearly we are NOT provided with the value of theta. BUT THROUGH DRAWING A 2D DIAGRAM, we can find out what theta is. Lets go back to the first principles and define what this theta is, in the equation F = BILsin(theta). This theta is defined as the angle between the direction of the current and the magnetic field vector. So, when you consider the 2D diagram, it becomes apparent that the current is travelling in a direction that is perpendicular to the magnetic field direction. Therefore, the value of theta is 90 degrees.
Now let's substitute everything into the equation F= BILsin(theta). B = 0.3T, I = 9A, L = 0.5m and theta = 90 degrees. Hence
F = 0.3 x 0.5 x 9 x sin(90) = 1.35 N which is the magnitude of the force. The direction of the force here cannot be calculated because we are not provided with the direction of the magnetic field, we only know that the direction of the magnetic field, whether to the right or to the left, will be perpendicular to the current flow within the metal bar.
c) Ok so this is perhaps the most tricky part of the question because it combined the knowledge from TWO MODULES!! The way I figured out this question is through observing which part of the question I havent used yet. So far, we have used resistance, voltage, length and magnetic field values provided by the question. So now we are left with 0.04kg mass, and like what I said beforehand, the previous part in a question will usually relate to its subsequent part. In part b) we found out the magnitude of force which is 1.35N, hence using Newton's 2nd law F=ma, we discover that 1.35 = 0.04(a), hence a=33.75 ms^-2. Sweet, now we have acceleration, but we were asked to find the displacement of the bar. A convenient way to figure out what to do is to just skim through the formula sheet, finding out the equation that relate acceleration and displacement together. Hence in this case it would be appropriate to use the formula y = ut + 1/2 at^2, because we have all the necessary details to figure out displacement (y). Initially, velocity of the bar is 0, hence u= 0 ms^-1. The acceleration we have worked out is 33.75 ms^-2 and the time in this case is provided (t = 0.25s). Everything after this step now is easy, substitute in all the values, we will obtain that y = 0(0.25) + 1/2 (33.75) (0.25)^2 and throw this into the calculator will will obtain an answer of 1.05m (2.d.p.).
Happy Physics Land, this is a great answer to Cajama's questions! Again, I'd recommend anyone who wants to ask a question, or answer a question, please does so.
Remember that before you can ask a question, or answer one, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
I think that my answer essentially covers most of what Happy Physics Land has included, with one key area lacking: I didn't draw a diagram. I think that's extremely, extremely important and so thank you HPL for doing so. When you get a complicated question like this, ALWAYS draw a diagram trying to figure out what's actually going on. Great one!
Jake :)
Post by: Happy Physics Land on January 31, 2016, 09:37:39 pm
Happy Physics Land, this is a great answer to Cajama's questions! Again, I'd recommend anyone who wants to ask a question, or answer a question, please does so.
Remember that before you can ask a question, or answer one, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Thank you jake for your recognition, I do believe that your answer was much more succinctly stated than mine and its a lot more easier to understand actually. I hope I can soon get on your level jake!!! :D
I think that my answer essentially covers most of what Happy Physics Land has included, with one key area lacking: I didn't draw a diagram. I think that's extremely, extremely important and so thank you HPL for doing so. When you get a complicated question like this, ALWAYS draw a diagram trying to figure out what's actually going on. Great one!
Jake :)
Thank you very much jake!!! I think that your response was much clearer and succint than mine actually and I really understood every part of your explanation very easily. I HOPE I CAN GET ON YOUR LEVEL SOON JAKE! :D (I posted this comment half an hour ago ldk why it wasnt there damn o.O)
Post by: cajama on January 31, 2016, 10:51:50 pm
Post by: jamonwindeyer on January 31, 2016, 11:45:14 pm
sorry for all the questions but i have no idea how to do some of these
1. a car of mass 2.0 x 10^3kg cruises North down the high way at 100 km/h, with a driving force of 1.2 x 10^4N.
a) calculate the retarding force of friction acting on the car.
b) the car speeds up to 110m/s in 5 seconds. Calculate the acceleration of the car, and thus determine the new driving force of the car.
2. a dog pulls a 80kg sled along the ground with 500N, which encounters a friction force of 150N. On the sled is a 20kg box.
a) calculate the acceleration of the sled.
b) calculate the friction force of the sled on the box which moves the box forwards along with the sled.
c) calculate the net force on the sled.
these are some of my answers but i have no idea whether they're correct:
2. a) a = (500-150)/(80+20) = 3.5 m/s/s forwards
b) F = 3.5 x 20 = 70N forwards
Hello again Chloe! Don't be sorry, it's what we are here for! Let's begin with your first question (I apologise in advance that I am not using formatted formulae, this is a brief technical issue which will be fixed tomorrow!).
The car is cruising at 100 km/h. What this means (though perhaps not stated explicitly) is that the car is travelling at a constant speed. Thus, it has an acceleration of 0! What does this mean? Well, consider Newton's 2nd Law:
Net Force = Mass x Acceleration
If the acceleration is zero, then the net force must also be zero. But how can this be? The answer is friction. When a car is cruising at constant speed, the driving force of the motor is perfectly balanced by the retarding force of friction, caused by the friction of the road on the tyres, and the friction caused by air resistance.
Therefore, the answer to your first question is 12 000N, the retarding frictional force (retarding simply means de-accelerating, or acting against) is equal to the driving force, since speed is constant.
For your second question, remember that acceleration is simply the rate of change of velocity. Ie - velocity increasing by 1 metre per second, in 1 second, corresponds to an acceleration of 1ms^-2. In this question, the velocity increases by 10 metres per second, in 5 seconds. This corresponds to an acceleration of:
Acceleration = Change in Velocity/Time = 2 ms^-2
So, that's the acceleration. To get the force, we need to be a bit tricky. We will calculate the additional force, which causes this acceleration, by plugging this value and the mass into Newtons 2nd Law.
F = Mass x Acceleration = 2 x 2000 = 4000N additional force to speed the car up.
We then add this to the original driving force, to obtain an answer of 16 000 Newtons . Note, this question does require the assumption that the retarding force does not change based on speed, which in real life, it certainly does. But Physics at this level is about keeping things simple ;D
Right, second question:
For part A, you are spot on. Acceleration is just obtained by plugging in to Newton's 2nd Law, and you've made all the correct inclusions (box + sled weight, and net force). Awesome!
For part B, again, spot on. The force of friction "drags" the box along with the sled, and this friction must provide the same acceleration as in Part A. Therefore, your answer is correct again, well done!
Part C requires a little clever thinking. Let's think about all the forces on the sled, we have:
- The 500N driving force (or dog force, if you will) FORWARD
- The 150N of friction from the ground BACKWARD
- 70 N from the box BACKWARD
But where does this third force come from? Just as the sled is dragging along the box (due to friction), the box is dragging the sled backwards. This is a consequence of Newton's 3rd Law. Therefore, the answer is:
F = 500 - 150 - 70 = 280 Newtons in the direction of motion.
You could also consider the 350N net force on the sled and box together, and that since 70N of this must be spent accelerating the box, the rest must be spent accelerating the sled. Both are valid interpretations which lead to the same answer.
I hope these answers helped Chloe! And never apologise for asking a question, ask as many as you can, as much as you want, we and your peers want you to improve and succeed (the way Physics scales is entirely dependent on how well Physics students do in the HSC, so it is in everyones best interests to help each other out ;)) .
Post by: sire123 on February 04, 2016, 06:00:12 pm
Post by: Happy Physics Land on February 04, 2016, 07:25:50 pm
Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!
Hello Sire 123:
Here are ALL the formulae you will need to know for the Topic of Space in HSC Physics. Everything that I have underlined using a red colour is compulsory for you to know, everything l underlined using a yellow-ish colour is everything that I would recommend you to know because you can refer to them in your questions. Everything else that are not underlined are just beneficial for you to know but you wouldnt need to worry too much about them. I will post another one for motors and generators soon and I will organise it in such a way that it would be easier for you to see. Sorry about the layout of this formula sheet which wouldnt be the most user-friendly.
(http://i.imgur.com/XfGtXRX.jpg)
If you have any queries dont hesitate to pop up and ask!!! :)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on February 04, 2016, 08:15:19 pm
Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!
Hello Sire 123:
Here is the complete set of notes on all the formulae you are required to know for the topic Motors and Generators, again, if you have any queries on what the letters represent or how to derive any formula, please dont hesitate to pop up and ask, because others may have the same question as you and you are actually helping everyone by proposing questions. Likewise to my previous formulae for the space topic, the red ticks suggest those formulae that you must know by heart, the yellow ticks indicate those formulaw that is strongly recommended for you to know. Those formulae without a tick would be beneficial for you to know however they are not too necessary for HSC exams (you can still refer to them). Just a sidenote, it is really important when you are answering motors and generators questions to refer to one of these formulae or refer to Faraday or Lenz's law wherever appropriate. These will enhance your reasoning and your argument. Also, by doing so, the markers can easily recognise that you have a solid understanding of these knowledge.
(http://i.imgur.com/KdqWBQU.jpg)
In regards to the formulae on Ideas into Implementation, Jake will be able to help you out, since I have not done the topic yet. But anyways, I hope you a successful HSC year and all the best in Physics!
Best Regards
Happy Physics Land
Post by: jakesilove on February 04, 2016, 08:46:55 pm
Yo guys, does anyone have a summary of just the formulas in Physics? The core topics I mean (not option) cos I feel like the one I wrote doesn't cover all of them. It'll be really great if u can send it via email, or just pics of anything. Cheers!
Hey Sire!
Firstly, let's just give HPL a massive round of applause for the massive effort he's put into answering this question. A stellar summary, and honestly that is definitely something we will be adapting into a resource. Utter legend.
As for Ideas to Implementation: It really isn't a formula-based topic, so having a glace through the dotpoint you actually don't need to know ANY formulas that aren't on your formula sheet!
Jake
Post by: sire123 on February 05, 2016, 12:34:33 am
Post by: sire123 on February 05, 2016, 01:38:20 am
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?
Post by: Happy Physics Land on February 05, 2016, 09:51:24 am
And btw, I'm not sure with some of the ones that aren't underlined.
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?
Hello Sire 123:
Honestly man, these problems that you have put forward are such good questions and this will not only help you, but also other students in NSW as well. Honestly heaps good questions. Ok, those formulae that I have not underlined are definitely not directly related to the syllabus, however they are beneficial for you to know because it will help you understand the theory better and you can definitely refer to them in some questions.
I understand that in projectile motion, air resistance is completely ignored for the sake of simplicity of calculation. However, when we draw the free body diagram for rocket launch, we know that there will be three forces acting upon the rocket: thrust force, weight force and air resistance. You will not be required to perform a calculation based on this formula but in questions where they ask you the forces experienced by the rocket, these three forces MUST be mentioned.
And yes the thrust force and rocket velocity are not directly related to the syllabus HOWEVER IT IS RELEVANT for answering questions in HSC EXAMS, they enable you to see these quantities are derived and you can still refer to them in HSC exam questions. Sorry I can give you an example of a HSC question right now and I will show you the derivation process for this formula when I get back home this afternoon (Im being a naughty boy going on atarnotes during my chemistry lesson hehe).
And the last question is my absolute favourite. The reason why the value of EMF is negative is because of Lenz's law, which states that the EMF (and hence the current) induced is such that it will oppose the change in magnetic flux. Hence, we need to use the negative sign to indicate that this EMF induce is a resistive force that acts against the direction of change in magnetic flux. And if this happens to be a question in your exam, remember to refer to Faraday's law for the induction of EMF and Lenz's law for the negative value of EMF.
So yeah this afternoon when I get the time I will reply your questions on:
- HSC question for thrust force and rocket velocity
- Derivation of rocket velocity from law of conservation of momentum
Best Regards
Happy Physics Land
Post by: brenden on February 05, 2016, 10:12:45 am
(Im being a naughty boy going on atarnotes during my chemistry lesson hehe)Hahahahaha you cheeky boy HPL
Post by: Happy Physics Land on February 05, 2016, 12:56:32 pm
Hahahahaha you cheeky boy HPL
Sorry Brenden I cant help with this ATARNOTES addiction
Post by: jamonwindeyer on February 05, 2016, 06:19:59 pm
And btw, I'm not sure with some of the ones that aren't underlined.
Like with the rocket launch formula T - (Fweight-Fair), isn't air resistance disregarded in the whole topic?
Also, I'm assuming the formulas of thrust force and rocket velocity are outside of the syllabus? Could you explain how you derive it, and which kind of Qs you'll use it in (or can you actually use it if we technically haven't "learnt" it?)
Lastly, why is EMF produced by rotation of conductor negative value?
HPL, you are a legend in the making. However, I can't resist a derivation, plus I'm dying to use LaTex in the forums now that it is back, so I'll jump in here, if yours is different post it, I want to see how you do it! Sire123, let's start with a derivation of rocket thrust force.
We learned in Prelim that force is equal to the rate of change of momentum, that is:
In this case, F is our thrust force. Now, lets blend that with a formula for momentum p = mv, also from Year 11:
However, this thrust force is referring to the exhaust from the rocket engine. The exhaust is expelled at a constant velocity (this is just a property of rocket engines, to do with the conservation of chemical energy, but totally irrelevant). So, if velocity is constant, we know that it must be the mass that is changing in the top part of the fraction. This leaves us with the formula:
The derivation of rocket velocity is more mathematical. We know from the law of conservation of momentum that the momentum in an isolated system must be constant. That is, the momentum of the rocket and its fuel must stay the same. If the rocket is initially at rest, it obviously has zero momentum, so interestingly, the momentum of the rocket and its fuel must remain zero throughout launch . How can this be?
It is because the momentum of the fuel (i.e. - exhaust) is opposite in direction to the momentum of the rocket. One goes up, one goes down, and they 'cancel' to give a total momentum of zero. With this in mind, the proof is below (using the formula for momentum from above):
And there is the second formula! Note that I used fuel, HPL used Gas, we're both referring to the speed/mass of the exhaust from the rocket engines ;)
Post by: Happy Physics Land on February 05, 2016, 07:01:25 pm
HPL, you are a legend in the making. However, I can't resist a derivation, plus I'm dying to use LaTex in the forums now that it is back, so I'll jump in here, if yours is different post it, I want to see how you do it! Sire123, let's start with a derivation of rocket thrust force.
We learned in Prelim that force is equal to the rate of change of momentum, that is:
In this case, F is our thrust force. Now, lets blend that with a formula for momentum p = mv, also from Year 11:
However, this thrust force is referring to the exhaust from the rocket engine. The exhaust is expelled at a constant velocity (this is just a property of rocket engines, to do with the conservation of chemical energy, but totally irrelevant). So, if velocity is constant, we know that it must be the mass that is changing in the top part of the fraction. This leaves us with the formula:
The derivation of rocket velocity is more mathematical. We know from the law of conservation of momentum that the momentum in an isolated system must be constant. That is, the momentum of the rocket and its fuel must stay the same. If the rocket is initially at rest, it obviously has zero momentum, so interestingly, the momentum of the rocket and its fuel must remain zero throughout launch . How can this be?
It is because the momentum of the fuel (i.e. - exhaust) is opposite in direction to the momentum of the rocket. One goes up, one goes down, and they 'cancel' to give a total momentum of zero. With this in mind, the proof is below (using the formula for momentum from above):
And there is the second formula! Note that I used fuel, HPL used Gas, we're both referring to the speed/mass of the exhaust from the rocket engines ;)
Oh crap I totally forgot that I still had the job to derive this for you sire 123!! Im so sorry and thank you so much Jamon for remembering to derive the two formulae for sire 123!!! You are such a legend!
But yeah I dont feel like I have done my job properly so l will take the rocket propulsion proof a step further, way back to the first principles so that you can see how rocket propulsion is related to the Law of Conversation of Momentum, keeping in mind that when a rocket launches, the force with which gas acts on the rocket (F_{gr} = Force of gas on rocket) is equal to the force with which the rocket acts onto the gas (F_{rg} = Force of rocket on the gas).
(http://i.imgur.com/f2leS94.jpg)
So yeah anyways sorry sire for such a late reply, hope this will help you to understand how the law of conservation of momentum is involved with rocket launch, great question! :)
Best Regards
Happy Physics Land
Post by: jakesilove on February 05, 2016, 09:06:43 pm
Oh crap I totally forgot that I still had the job to derive this for you sire 123!! Im so sorry and thank you so much Jamon for remembering to derive the two formulae for sire 123!!! You are such a legend!
But yeah I dont feel like I have done my job properly so l will take the rocket propulsion proof a step further, way back to the first principles so that you can see how rocket propulsion is related to the Law of Conversation of Momentum, keeping in mind that when a rocket launches, the force with which gas acts on the rocket (F_{gr} = Force of gas on rocket) is equal to the force with which the rocket acts onto the gas (F_{rg} = Force of rocket on the gas).
(http://i.imgur.com/f2leS94.jpg)
So yeah anyways sorry sire for such a late reply, hope this will help you to understand how the law of conservation of momentum is involved with rocket launch, great question! :)
Best Regards
Happy Physics Land
Absolute legend, also can't pass up a good derivation so I'm so sad I missed out on this/
Jake
Post by: Happy Physics Land on February 05, 2016, 09:27:58 pm
Absolute legend, also can't pass up a good derivation so I'm so sad I missed out on this/
Jake
Don't worry Jake, God has got bigger derivation plans for you ahead
Post by: AladinX on February 05, 2016, 11:49:08 pm
I'm currently a Year 12 student in Physics and am soon going to be given an assessment task requiring me to make a rocket launcher ( testing our knowledge of projectile motion ). Any ideas on where to start for such an assignment. What to use maybe to launch the rocket to the required distance?
Much appreciated, thanks.
Post by: jakesilove on February 06, 2016, 10:23:24 am
Hey Jake.
I'm currently a Year 12 student in Physics and am soon going to be given an assessment task requiring me to make a rocket launcher ( testing our knowledge of projectile motion ). Any ideas on where to start for such an assignment. What to use maybe to launch the rocket to the required distance?
Much appreciated, thanks.
Hey AladinX!
Firstly, that's a seriously cool assignment! Tricky, but cool. I'm sure heaps of people on the forum will have great ideas, so hopefully everyone can contribute!
Secondly, I think we need a little more information about the project. Is the rocket launcher supposed to be accurate, or just launch a projectile further than a required distance? Is there a target? What is the required distance? How heavy is the projectile?
If you just need to launch a projectile greater than a required distance, and accuracy doesn't matter, I'd build yourself a good ol' fashion Potato Gun. Google the method, but it's cheap to make and works like a charm. You could vary the construction depending on how far you want it to go etc.
If you need accuracy, typically people build trebuchets and other similar launchers. Again, I can go more in depth when I have the answers to the above questions, but I'm thinking that building something like that would be really difficult. What I'm imagining is sort of "cheating" the question: if all that is required is launching a projectile a set distance, accurately, I would buy something like Hot Wheels tracks, set up a ramp and fling is to the point you need to get to. You can easily do the calculations, doesn't required difficult construction etc. etc.
Let me know more information! Hopefully other people on the forum have more ideas, because whilst I love theoretical Physics I am certainly not an Engineer.
Jake
Post by: Happy Physics Land on February 06, 2016, 09:38:52 pm
Hey Jake.
I'm currently a Year 12 student in Physics and am soon going to be given an assessment task requiring me to make a rocket launcher ( testing our knowledge of projectile motion ). Any ideas on where to start for such an assignment. What to use maybe to launch the rocket to the required distance?
Much appreciated, thanks.
Hey AladinX:
I havent had any experience with constructing a rocket model to test projectile motions, however I have found two very useful sites that outlines two different types of rocket launchers that you may be able to use to demonstrate your knowledge of projectile motion. You can use the IDEA from these two websites (not the same methods or results or safety of course) and twist them and changed them to avoid being accused of plagiarism. So yeah you can have a read of these two sample experiments to inspire some ideas for yourself!
http://www.scienceworld.ca/resources/activities/pop-bottle-rocket-part-ii-projectile-motion
http://www.arborsci.com/cool/projectile-motion-for-everyone
Best Regards
Happy Physics Land
Post by: AladinX on February 06, 2016, 09:42:10 pm
Hey AladinX!
Firstly, that's a seriously cool assignment! Tricky, but cool. I'm sure heaps of people on the forum will have great ideas, so hopefully everyone can contribute!
Secondly, I think we need a little more information about the project. Is the rocket launcher supposed to be accurate, or just launch a projectile further than a required distance? Is there a target? What is the required distance? How heavy is the projectile?
If you just need to launch a projectile greater than a required distance, and accuracy doesn't matter, I'd build yourself a good ol' fashion Potato Gun. Google the method, but it's cheap to make and works like a charm. You could vary the construction depending on how far you want it to go etc.
If you need accuracy, typically people build trebuchets and other similar launchers. Again, I can go more in depth when I have the answers to the above questions, but I'm thinking that building something like that would be really difficult. What I'm imagining is sort of "cheating" the question: if all that is required is launching a projectile a set distance, accurately, I would buy something like Hot Wheels tracks, set up a ramp and fling is to the point you need to get to. You can easily do the calculations, doesn't required difficult construction etc. etc.
Let me know more information! Hopefully other people on the forum have more ideas, because whilst I love theoretical Physics I am certainly not an Engineer.
Jake
Hopefully once I receive the formal assessment sheet and requirements I will be able to give you more information regarding it. Thanks for your help :)
Post by: mijomo on February 07, 2016, 03:34:13 pm
My first assessment task in physics this year is an open ended investigation which requires me to analyse the relationship between the surface area of an object and the rate that it falls, and then write it up as a scientific report (aim, method etc.) Just got three quick questions for you:
1. Am i right in thinking that rate of fall or the rate that an object falls at is referring to velocity and not acceleration or just the time taken to fall?
2. What are the main things I should include in my discussion?
3. When we talk about validity, reliability and accuracy, what does accuracy mean in in a scientific report context?
Thanks again for doing this AMA, LEGENDS!
Post by: jakesilove on February 07, 2016, 03:51:52 pm
Hey guys, thanks so much for having this AMA, you guys are awesome!
My first assessment task in physics this year is an open ended investigation which requires me to analyse the relationship between the surface area of an object and the rate that it falls, and then write it up as a scientific report (aim, method etc.) Just got three quick questions for you:
1. Am i right in thinking that rate of fall or the rate that an object falls at is referring to velocity and not acceleration or just the time taken to fall?
2. What are the main things I should include in my discussion?
3. When we talk about validity, reliability and accuracy, what does accuracy mean in in a scientific report context?
Thanks again for doing this AMA, LEGENDS!
Hey Mijomo!
Firstly, I reckon this is an epic experiment that I would have spent WAY too much time on in my HSC year. That being said, love your teacher for giving it to you!
Let's go through your questions chronologically.
Is it asking about velocity, acceleration or time?
Well, in perfect Physics world the answer would be all of the above. That's because you SHOULD be able to derive any of the other components by gathering information about one: Calculating start and end velocity will give acceleration, acceleration will give time using projectile motion equations etc.
Therefore, if your question is "What should I even be measuring?" my answer would be "That depends on the accuracy of your equipment". We will discuss what "accuracy' means further down, but my overall answer re: what the dependent and independent variable should be is as followed.
Obviously, the thing that you should "change" is the surface area of the object. The thing that you should measure will either be
a) The time the object takes to fall a set distance.
b) The velocity of the object during the fall.
c) The acceleration of the object during the fall.
You should discuss these methods, deciding which will be MOST ACCURATE. I would guess that the TIME will be the most accurate: Use your phone to film the drop, then try and slow down the footage so you can see EXACTLY how long it took to fall. This will be very accurate, and you can do subsequent calculations from there.
What should I include in my discussion
You should include things that worked, things that didn't, reason for error, explanation of results etc. etc. Basically just explain what you did, why you did it, and how successful it was. There are heaps of guides online with what you should include in a discussion: Try find whatever your textbook says!
Accuracy
Accuracy refers to how close the measured value is to the real value. For instance, the more decimal places you measure a value to be, the more accurate that measurement is. So by filming the descent, and getting the point of landing to the millisecond, is quite accurate.
Accuracy refers to the success of the instruments, so just talk about limitations to accuracy depending on the equipment you use.
I hope that this helps! Sounds like an amazing task, so if you want to keep us updated with your progress/ideas/places you need help, I would be super interested. I would also recommend googling what the relationship between surface area and velocity SHOULD be, so you know what you're looking for.
Hope this helps! Fantastic question.
Jake :)
Post by: Neutron on February 07, 2016, 04:07:49 pm
Neutron
Post by: jakesilove on February 07, 2016, 04:15:44 pm
Hey! I was just doing some homework and I was wondering how eddy currents cause flux leakage? So I'm doing the transmission process for electrical energy and I read somewhere that hysteresis losses can be reduced by laminating the iron core, which reduces the size of eddy currents formed but I was wondering how smaller eddy currents would help the flux leakage? Sorry if this seems dumb!
Neutron
Hey Neutron!
Not a dumb question at all! Basically, the solid magnetic core is NOT supposed to be used for eddy currents, its just to produce a Magentic field etc. So its use is as a magnet, rather than a cylinder.
However, being a conductive cylinder, and because of Lenz's law, the flowing current around the core will cause Eddy currents to be formed in the core. Remember, the purpose of the core is NOT to produce Eddy currents; that is just a result of the type of object it is.
Because of conservation of energy, any eddy currents produce must cause a reduction of energy elsewhere. Basically, this means energy is "lost" in the conversion process in the production of Eddy currents in the core, and those Eddy currents aren't useful in any way.
By laminating the core, it reduces the "size" and therefore the amount of energy removed by the Eddy currents. Essentially you put a sheet of insulator in between each sheet of metal, so the Eddy currents can't become really really big. By reducing the size of eddy currents, less energy is loss.
I hope this helps! Just remember that we don't WANT eddy currents to be produced in this process; that's just an unfortunate byproduct. By reducing the size of eddy currents, we reduce the energy lost!
Jake
Post by: Neutron on February 07, 2016, 04:32:46 pm
Neutron
Post by: Neutron on February 07, 2016, 05:20:17 pm
Neutron
Post by: Happy Physics Land on February 07, 2016, 07:40:43 pm
Thanks Jake but how do these Eddy currents cause flux leakage? Like I understand that they dissipate heat and all but the topic I'm studying right now is transformers (if that helps!) :D
Neutron
Hey Neutron:
I think Jake did a really great job explaining the lamination of soft iron cores to reduce hysteresis losses and I will just reinforce some of his statements and just tell you a bit about flux leakage. It's a good question and if l dont do a good job explaining this please tell me because it is a hard physics concept.
Two concepts you definitely shouldnt confuse are heat loss and flux leakage. Flux by definition is just the total number of magnetic field lines that passes through a perpendicular area. When a magnetic device keeps operating for a period of time, there will be residual magnetic field lines that remains for a short amount of time. An analogy to this is that when you turn on the stove for a while and when you suddenly turn it off its not gonna to immediately cooldown. So since there is a residual magnetic field, there will be magnetic field lines produced over an area (flux), and without your soft iron core, the flux wouldnt be able to have a path to pass through to the other coil. Hence a non-conductive plastic core or wooden core would lead to an absence of pathway for magnetic field lines to pass through and hence there will be a flux leakage (i.e. not all the flux will be transferred to the second coil in the transformer).
In regards to eddy currents, they are produced as a result of change in magnetic flux and EMF and Lenz's law etc (Im assuming you would be familiarised with this process).So it is reasonable to think that when there is a high value of eddy current flowing through the transformer, there will be a stronger residual magnetic field after the device is turned off, and so there is more flux leakage.
Post by: Happy Physics Land on February 07, 2016, 08:28:36 pm
And also, another question (sorry!), what are the benefits of the glass disc insulators' shape (still talking about the transmission tower oops) and without it, would the current just travel through the tower and be earthed? Like if the live wires were not separated from their supporting pylons, would all the electricity just be earthed? Thank you!
Neutron
Now this glass disc insulator questions (just a reminder, disc insulators can be all types of ceramics, glass is just a type of ceramic that is made in epoxy resin). Glass disc insulators take on a slope-shape to prevent the accumulation of moisture and dust to ensure a long conductive pathway for any spark discharge across the insulator. In addition, the metal links in glass insulators are isolated from each other, and the fibreglass is a non-conductor, so there is no continuity of conduction, which is exactly the purpose of an insulator.
If you have any further questions dont hesitate to ask! :)
Best Regards
Happy Physics Land
Post by: mijomo on February 07, 2016, 09:43:04 pm
Post by: jakesilove on February 07, 2016, 09:47:29 pm
Thanks so much for your reply Jake, greatly appreciated. Great minds must think alike because i did use my ipod to film the drops then counted how many frames the falling body was in the air for and worked out the time in seconds. I'm able to work out average velocity since i know what vertical height I dropped the metal sheets from and this gives quite a nice graph showing a trend that indicates that as the surface area was decreased the average velocity of the falling body increased! I suppose one limitation that I could talk about in my discussion is that fact that I can only calculate average velocity as opposed to instantaneous velocity which would better show when the object approached terminal velocity etc. Maybe using a data logger would have done the trick, or a background erected behind for the fall which had 10cm increments lined on it, which I could use in conjunction with a frame by frame analysis to see when uniform motion was reached. My teacher is always going on about trying to find a straight line relationship when you have a graph so that's what i'm now attempting to do. I was able to solve a differential equation, (Sum of forces acting) = m (dv/dt) = mg-F(drag), (with surface area proportional to the drag force), but it doesn't really give a straight line relation, more logarithmic. Reckon it's worth putting in, or is that over analysing the problem? Anyway, very interesting stuff, thanks again for your help!
Glad that you had such success with your experiment!
I'd definitely say that you can NEVER over analyse a Physics question. As long as you're doing the correct maths (which it seems like you are), any further calculations are beneficial. If you want to knock the socks off your teacher, I would definitely try to cater to what he wants! Straight line relationships are standard in higher levels of Physics, and so if you can work something out you absolutely should. That being said, if you don't get a straight line (but seem to have some other nice function) I would still put that in, with maybe an attempted explanation as to why the graph looks as it does. Just make sure to explain each step, label everything, and increase a "discussion" of each point (ie. this graph proves that.....).
Loving your enthusiasm! Keep it up.
Jake :)
Post by: mijomo on February 07, 2016, 09:47:46 pm
Post by: jakesilove on February 07, 2016, 10:07:43 pm
Sorry one last question. My iPod shoots video in 30 fps. That means each frame is 1/30th of a second (0.03s roughly). Does that mean when converting back into time in seconds I can have times with 3 sig figs? Something like 1.11s or only 2 sig figs (1.1s), or only 1 sig fig (1s). I feel like 3 is the way to go but i'm not really sure on the 'rules' of sig figs haha. Thanks again!
Don't be sorry! Honestly these questions are all fantastic ones, and shows that you have a very in depth understanding of the concepts.
I would stick with two sig figs. Remember that you always select the LOWEST number of sig figs you used in the calculation. In this case, 30 fps could be 1 or 2 sig figs, but I think it's fair to assume that it's 2. So I would write something like "assuming 30 fps is correct to 2 sig figs, t = 1.1s".
Jake :)
Post by: Happy Physics Land on February 07, 2016, 10:19:36 pm
Post by: jamonwindeyer on February 07, 2016, 11:30:44 pm
And also, another question (sorry!), what are the benefits of the glass disc insulators' shape (still talking about the transmission tower oops) and without it, would the current just travel through the tower and be earthed? Like if the live wires were not separated from their supporting pylons, would all the electricity just be earthed? Thank you!
Neutron
Hey Neutron! Before I begin, once again I'll give props to HPL for an awesome answer to both your questions. His first answer was based on a definition of magnetic flux and, essentially, that the rate of change of magnetic flux cannot be infinite (instantaneous change is not possible). His second answer is also an awesome explanation of exactly why insulators are disc shaped. However, I do want to add my two cents here.
In terms of transformers, flux leakage refers to the fact that not all of the magnetic flux from the primary winding will pass through the secondary winding, inducing a new EMF. It "leaks" through the surrounding oil or winding insulation. How do Eddy Currents contribute to Flux Leakage (not energy loss, the more general notion focused on in the HSC syllabus).
Eddy Currents are produced by a changing magnetic flux in the iron core of a transformer. We know that Lenz's Law/Faraday's Law (Lenz's Law is just a common way of explaining/understanding the Conservation of Energy in Faraday's Law) says that any induced EMF (and thus, currents) will be induced in such a direction that it opposes that process which created it. But how do Eddy Currents oppose the changing flux from the primary winding?
Eddy Currents have their own magnetic field. The spinning currents induce new magnetic fields (magnetic flux) which act in the opposite direction to the flux from the primary winding. We know the new magnetic fields act in the opposite direction as an extension of Lenz's Law from above; opposing current, opposing magnetic field. So, to put it very colloquially, the flux from the Eddy Currents cancels some of the flux from the primary winding out!
Now, for your second question, HPL's answer is more than sufficient (and actually contains stuff I didn't think of, nice!). To answer your extra little question at the end, if the live power lines were not insulated, it is very likely they would be earthed. At voltages of 25000+, you need lots of insulation to prevent sparking, and since the power poles are earthed, I would say that yes, that is a very logical conclusion. ;D
Post by: dhungelsajal123 on February 08, 2016, 06:09:13 pm
Post by: Syndicate on February 08, 2016, 06:17:41 pm
Hey Jake! I attended the ATAR notes lectures, do you know where I can access the powerpoints?
Hey,
You will find them over here
Post by: sire123 on February 08, 2016, 11:18:24 pm
Post by: jakesilove on February 09, 2016, 09:38:49 am
Can anyone explain the slingshot effect? And how in depth we would needa know for a possible hsc q on it, cheers :)
Hey Sire!
You really don't need to know the slingshot effect to any great depth. You know that the slingshot effect is basically when a space shuttle uses a planet's gravitational field to "turn"; let's examine the benefits.
The first benefit is that usually, when accelerating (ie. turning), the shuttle would have to use thrusters to create a turning force. This would waste fuel, and quite a lot of fuel. Using a planet's gravitational field gives a "free" turning force: the object will be attracted towards the planet, and as it moves forward will "turn" towards to planet. Thrusters and still used to get into and out of the planet's gravitational field, but to a much lower extent.
Another benefit is that the spacecraft will not LOSE speed, and it can even GAIN speed. But how that that make sense? Conservation of momentum. Basically, the spacecraft will "steal" some of the planet's momentum to increase its own. Do you need to understand that in any more depth? Definitely not.
You would never get an in depth question about the slingshot effect. If you have a general understanding of the above answer, you will seriously be totally fine. It's cool stuff!
Go watch the Martian if you want some more details, and a great understanding from a great movie.
Hope this helps!
Jake
Post by: Happy Physics Land on February 09, 2016, 11:34:37 am
Can anyone explain the slingshot effect? And how in depth we would needa know for a possible hsc q on it, cheers :)
Hello Sire 123,
Seeing that Jake has provided so many valid points and the explanation is so great, I cannot help but to give in my 2 cents in terms of the conversation of momentum. As the space probe passes tangentially to the large planet (usually Jupiter or Saturn), it gains a significant increase in velocity whilst the large planet loses a tiny amount of rotation velocity. This is because the planet is thousands of times larger than the space probe, and since momentum of the system has to be conserved, the momentum of space probe increases and the momentum of the planet descreases in its momentum. Since the mass of the space probe and planet remains constant, then velocity must change for both objects to allow the momentum of each object to increase/decrease.
Let's have a look at this example: suppose the space probe is 10kg and the planet is 1x10^8 kg, the space probe is travelling at 100m/s and the planet is rotating with a velocity of 100m/s as well.
momentum of planet = 10^8 x 100 = 10^10 Ns
momentum of space probe = 10 x 100 = 1000 Ns
when the space probe passes tangentially near the planet, it "steals" 200 m/s off the planet
momentum of space probe = 10 x 300 = 3000 Ns
momentum of planet = 10^10 - (3000-1000) = 9.99998 x 10^9 Ns (since the momentum of the system has to be conserved, then it follows that if the space probe gains 2000 Ns momentum, then planet must lose 2000 Ns momentum, which is not a significant amount for the planet)
Velocity of planet = 9.999998 x 10^9 / 10^8 = 99.98 m/s, which is virtually the same as the 100m/s prior to slingshot effect (i.e. a trivial decrease in the velocity of planet)
Through this example (definitely not a real circumstance, but just to let you see whats going on), we can see that the planet only needs to lose a little bit of velocity for the space probe to gain a significant change in velocity, and this is because of the law of conservation of momentum.
I hope that was a good explanation on how the law of conservation of momentum relates to the slingshot effect. If you have any questions, please dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: Happy Physics Land on February 09, 2016, 09:33:26 pm
The question is as follows,
A car and driver of total mass 850 kg are travelling east along a straight road at a constant speed of 75.0 km/h. The car collides with a rubbish bin of mass 120 kg, which had been left on the road. The bin becomes wedged under the car within 0.350 s. The driver removed his foot from the accelerator and did not apply the brakes during collision.
1) Calculate initial momentum of the car
2) what is the final momentum of car-bin system
3) determine the speed of the car immediately after collision.
4)What is the change in momentum of the rubbish bin during the collision?
5) determine the magnitude of the average force exerted on the bin by the car.
6) What is the direction of the average force exerted by the bin on the car?
Here is the solution:
(http://i.imgur.com/LgXu1oZ.jpg)
Thank you Jelly Beanz, very valuable stuff for our community, the question is certainly going to do all preliminary students a big favour.
N.B. THANK YOU to adequace for pointing out my mistake in question 4. In question four I subtracted initial momentum from final momentum, however I divided by time which is an unnecessary step in question 4. All you would have to do is to subtract initial momentum from final momentum for question 4, and in question five, you use the value from question 4 and divide that by 0.35 seconds because the formula for impulse (or change in momentum) is I = F x t. Much apologies for this mistake and any inconveniences caused
Post by: Neutron on February 14, 2016, 10:49:02 am
Post by: jakesilove on February 14, 2016, 01:11:28 pm
Hey guys! I need help with this question :D "A square has a magnetic flux of intensity B through it. What is the new intensity of the area of the square is doubled?" The answer is 0.25 B and I have no idea why (I thought it was 0.5 :( ) And could you please also clarify magnetic flux and magnetic flux density, thank you 😅
Hey Neutron!
Firstly, to define Magnetic Flux and Magnetic Flux Density:
Magnetic Flux:
The total flux measured across a surface. This value is a Scalar.
Density per volume, essentially.
Magnetic Flux Density:
The flux measured at a specific point. This value is a Vector.
Density per area, essentially.
Same as a Magnetic Field at a point.
The total flux measured across a surface. This value is a Scalar.
Density per volume, essentially.
Magnetic Flux Density:
The flux measured at a specific point. This value is a Vector.
Density per area, essentially.
Same as a Magnetic Field at a point.
As for your actual question: unless I am reading the formulas wrong, I think that 0.5B is the correct answer (and maybe your textbook has got it wrong?).
Check with your teacher, but if you have x magnetic field lines across a square, causing a flux density of B, then doubling the area of the square (maintaining x magnetic field lines) should result in a flux density half that of the original.
Unless the question states that the side length doubles (in which case the area is multiplied by 4), I think your textbook may have just got this wrong.
Jake
Post by: Maz on February 16, 2016, 08:59:45 pm
Can someone please help me with this?
Alpha Centauri is a star 4.367light year from earth. A time traveller taking this journey at near light speed in a spaceship was confused by the fact that he made the journey in less than 4 years based on his clock. how would you explain this situation to the traveller? How would this dilemma be viewed by an observer on Earth.
As an answer i put: To the observer on Earth, time dilation has affected their view of time taken by the the ship to be more than time taken relative to the traveler. Time dilation is more noticeable as speed increases.
i was wandering if that would be enough as an answer- or if i was even right and how you would calculate that?
Thankyou so much in advance :)
Post by: jakesilove on February 16, 2016, 09:12:59 pm
hello
Can someone please help me with this?
Alpha Centauri is a star 4.367light year from earth. A time traveller taking this journey at near light speed in a spaceship was confused by the fact that he made the journey in less than 4 years based on his clock. how would you explain this situation to the traveller? How would this dilemma be viewed by an observer on Earth.
As an answer i put: To the observer on Earth, time dilation has affected their view of time taken by the the ship to be more than time taken relative to the traveler. Time dilation is more noticeable as speed increases.
i was wandering if that would be enough as an answer- or if i was even right and how you would calculate that?
Thankyou so much in advance :)
Hey!
I think that you would definitely need to be a little bit clearer with your answer, just to make sure you fully get across to the marker that you know what you're talking about.
How would you explain this situation to the traveler?
Firstly, I would start by explaining that his situation is impossible. If something is 4.367 light years from earth, then it takes light 4.367 years to get from Alpha Centauri to Earth. Therefore, it is impossible to get there in less than 4 years as this would be faster than the speed of light.
Maybe his clock reads more than 4.367 years (let's just say 5 years). In that case, the fact that he was traveling near the speed of light would explain why he got there so quickly. Length contraction occurs at high speeds, meaning that, to the traveler, the distance between himself and Alpha Centauri has decreased. This allows him to get there quicker. Remember that to the traveler, time operates exactly as normal. To the traveler, it is the outside world that appears to be moving more slowly. So time dilation should not actually make a difference to a traveler viewing a clock on the space ship: to him, the clock moves as normal.
Observer on Earth
To an observer on earth, time dilation would mean that the time experienced by a traveler appears to be slower. To the person on Earth, the clock on the spaceship will move more slowly. However, time will move forward as normal on Earth, and so it will appear that the traveler takes longer than 5 years to get there. The observer on earth will also notice the mass of the spaceship increase (due to mass dilation) and the length of the spaceship decrease (based on length contraction).
I hope this helps! Take a look at the formulas in order to know exactly what's going on when something moves very quickly. Remember that a person moving near to the speed of light will still experience everything (time, length etc.) to be the same; it is only the outside world that appears to be different.
Jake
Post by: Maz on February 16, 2016, 09:45:38 pm
Hey!
I think that you would definitely need to be a little bit clearer with your answer, just to make sure you fully get across to the marker that you know what you're talking about.
How would you explain this situation to the traveler?
Firstly, I would start by explaining that his situation is impossible. If something is 4.367 light years from earth, then it takes light 4.367 years to get from Alpha Centauri to Earth. Therefore, it is impossible to get there in less than 4 years as this would be faster than the speed of light.
Maybe his clock reads more than 4.367 years (let's just say 5 years). In that case, the fact that he was traveling near the speed of light would explain why he got there so quickly. Length contraction occurs at high speeds, meaning that, to the traveler, the distance between himself and Alpha Centauri has decreased. This allows him to get there quicker. Remember that to the traveler, time operates exactly as normal. To the traveler, it is the outside world that appears to be moving more slowly. So time dilation should not actually make a difference to a traveler viewing a clock on the space ship: to him, the clock moves as normal.
Observer on Earth
To an observer on earth, time dilation would mean that the time experienced by a traveler appears to be slower. To the person on Earth, the clock on the spaceship will move more slowly. However, time will move forward as normal on Earth, and so it will appear that the traveler takes longer than 5 years to get there. The observer on earth will also notice the mass of the spaceship increase (due to mass dilation) and the length of the spaceship decrease (based on length contraction).
I hope this helps! Take a look at the formulas in order to know exactly what's going on when something moves very quickly. Remember that a person moving near to the speed of light will still experience everything (time, length etc.) to be the same; it is only the outside world that appears to be different.
Jake
Thankyou so so much- especially for explaining it so well
Post by: Loki98 on February 16, 2016, 09:55:24 pm
Post by: jamonwindeyer on February 16, 2016, 10:13:39 pm
hello
Can someone please help me with this?
Alpha Centauri is a star 4.367light year from earth. A time traveller taking this journey at near light speed in a spaceship was confused by the fact that he made the journey in less than 4 years based on his clock. how would you explain this situation to the traveller? How would this dilemma be viewed by an observer on Earth.
As an answer i put: To the observer on Earth, time dilation has affected their view of time taken by the the ship to be more than time taken relative to the traveler. Time dilation is more noticeable as speed increases.
i was wandering if that would be enough as an answer- or if i was even right and how you would calculate that?
Thankyou so much in advance :)
Hey mq123! I want to add my interpretation to Jake's answer above. I think this situation would be possible, and here is my reasoning.
We remember that 4.367 light years is a measure of distance. As the time traveller moves at relativistic speeds, length contraction will occur. We normally consider the effects of length contraction on the moving object as viewed by an observer (EG - a moving spaceship compressing to external view). However, length contraction also occurs as viewed from the moving object. From the frame of reference of the time traveller, the 4.367 light years of distance will shrink by length contraction. It actually will travel a smaller distance, and thus, arrive sooner than you would expect without considering Special Relativity! This, in my opinion, causes the anomaly described in Part 1.
As for Part 2, I agree with Jake's reasoning. If the clock on the ship reads 4 years, this will actually represent much longer in earth years, since time passes more slowly for moving observers. This is by the normal effects of time dilation.
I hope this interpretation adds to Jake's and that between us we've given you a solid understanding of the question! The amount of detail you would respond with would depend on how many marks it is worth in the HSC Exam, but any response which covers Length Contraction of the distance travelled, and time dilation causing the clocks to fall out of sync, to a decent level of detail, would be right on track ;D
Post by: RuiAce on February 16, 2016, 10:14:06 pm
Sorry for such a broad question, could u please explain the differences and similarities between ac and dc motors and ac and dc generators. I'm mainly having trouble with the purpose of the commutator in each case.
There are tons of differences. But in terms of the commutator:
In an AC generator, the commutator ONLY exists to ensure current flow exists. But in a DC generator, there exists an actual need for the polarity of the current to be reversed so that the current produced is always the same in polarity (conversely, for a DC motor it's to ensure that the DC motor maintains torque in the same direction).
Some other differences include:
The slip ring commutator, due to not having the obvious split, will not be as susceptible to sparking as opposed to that of the split ring commutator making it more safer to use. The split also causes wear and tear to be more frequent. So these mean that AC generators will typically be safer and require less maintenance.
AC generators, of course, can have their voltages stepped up or down by transformers and thus can be:
a) Transmitted more effectively, as heat losses are minimised according to P_loss=RI^2
b) High voltage requiring appliances (TVs) and low voltage requiring ones (computers) can both be used simultaneously.
c) Power stations can be built together in one rural site, instead of several 100s of rural AND urban regions.
Yet, for only a very specific voltage range (that range I have forgotten), DC is more powerful than AC. Furthermore, AC requires synchronisation between power stations to ensure power transmission is effective.
As far as I recall, a lot more information on this I obtained from Physics in Focus during my study.
Post by: Happy Physics Land on February 16, 2016, 10:31:13 pm
Ok so essentially the most significant difference between an AC motor and a DC motor is indeed their commutators, and the mechanism of AC motor can vary depending on the type of motor you are talking about. In this case I will talk to you about the difference between an AC universal motor and a DC motor (If Jake would like to spot out the difference between AC induction motor or synchronous motor and DC motor, please feel free to do so too!).
A DC motor requires a current flow being supplied to the coil placed inside a magnetic field. As a result of the motor effect, which states that a current-carrying current will experience a force in magnetic field, the coil will experience a torque because of the two forces acting on both sides of the coil. As the coil is rotating, it encounters the problem of not being able to rotate consistently in one direction because torque direction will reverse every half a turn. This problem is effectively overcome through the implementation of split-ring commutators, which reverses the direction of current flow in each coil every half-revolution. This ensures that the torque on the coil is ALWAYS in the same direction. Do keep in mind that change of current direction is only due to the change in position of each half of the split ring (i.e. if one half of the split ring originally is in contact with the positive brush, then after half a turn it will come in contact with negative brush). This ensures torque in constant direction.
Similarly, universal motor operating on AC also employs a split ring commutator to keep the direction of current in the rotor coils the same. In contrast to a DC motor, an AC universal motor involves the reversing of the direction of the magnetic field produced by the stator coils (electromagnets) every half-turn, whereas in an DC motor, the magnetic field is kept permanent. The reason for doing so is that, because of the commutator, we have effectively kept the direction of AC current in the rotor coils the same, and to keep torque in a constant direction, it is necessary to change the polarity of the electromagnets every half turn.
All generators involves the input of external mechanical force and torque applied which causes the armature and coil to rotate in a magnetic field. As the rotor coil experiences a change in flux due to change in relative motion, alternating EMF and current are induced in the coil (Faraday's Law).
In an AC generator, the stator coils produces a magnetic field that is similar to that provided by a pair of permanent magnets. Its crucial difference from DC generators is the utilisation of SLIP rings. Each end of the coil is connected to one of the two slip rings on the axle. Brushes press against the slip rings providing a sliding electrical contact. The slip rings remain in contact with the same brush and the same end of the coil throughout the rotation of an AC generator.
Similarly, in a DC generator, the stator coils are also producing a magnetic field that is similar to that provided by a pair of permanent magnets. However, the difference between DC and AC generator is its use of SPLIT ring commutators. The commutator provides points of contact between thr rotor coils and the external electric circuit, it serves to reverse the direction of current flow in the external circuit once every half turn of the generator. This ensures that the current produced is always in the same direction (which is essentially DC which flows only in one direction).
I really hoped that my explanation would help you Loki and if you have any doubts or questions, please dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: jamonwindeyer on February 16, 2016, 10:42:32 pm
Sorry for such a broad question, could u please explain the differences and similarities between ac and dc motors and ac and dc generators. I'm mainly having trouble with the purpose of the commutator in each case.
Sorry for such a broad question, could u please explain the differences and similarities between ac and dc motors and ac and dc generators. I'm mainly having trouble with the purpose of the commutator in each case.
Not at all Loki98! Broad questions are awesome, I get more freedom to write what I want to say ;)
So, a big question, I could write pages, I'll try and keep it fairly concise by focusing only on Universal AC Motors/Generators.
Motors and generators should be viewed as similar in design, but opposite in function. Generators take kinetic energy and transform it into electrical energy, via the principle of electromagnetic induction. Motors take electrical energy and convert it into kinetic energy via the Motor Effect. Opposite in function, however, they have quite a few similarities. All generators and motors (as studied in this course) have:
- A stator, normally a permanent magnet. This creates a magnetic field, inside which is placed:
- A rotor, the spinning part of the apparatus, which consists of coils wrapped around an iron core.
In motors, the rotor spins due to the Motor Effect, which causes a rotational force due to the interaction between the magnetic field and the electric current in the armature. In generators, we spin the rotor and a current is induced in the coils, due to the changing magnetic field.
Now, the commutators. This is a little confusing. Commutators are little metal rings which allow the external circuit to maintain contact with the coils as they spin. There are two kinds, which I'll go into below, but to understand the explanations we first need to go through something else.
For both the Motor Effect and Electromagnetic Induction, the resultant force or current is dependent on the direction of the magnetic field. You should have learned in class that, as the motor/generator spins, this actually results in the direction of the force or current changing every half turn (180 degrees). This can be deduced by applying the "Right Hand Rule."
So, the two commutators:
Slip ring commutators simply maintain contact, nothing else. For AC Generators, this means that the direction of the generated current is unaffected, and so, continues to change every half turn. This creates an AC current with no further complications.
Split ring commutators are, as the name implies, split in half. This means the connection with the external circuit (either load or supply) is interrupted every half turn. In a DC motor, this is vital, as it actually allows the direction of current in the coil (say, clockwise/anticlockwise) to be reversed every half turn. It is quite hard to explain how this works with words or even pictures, I found some awesome videos on Youtube, like this one. This was literally just a bookmark I had leftover in my HSC folder, it helped me heaps back when I did the course.
And the principle in a DC Generator is the same. The induced current reverses every half turn normally, and the split ring commutator reverses it again, so the polarity remains constant!
This is just a quick run down of the concepts. Not everything is explained in full detail (feel free to ask if you need more), but hopefully it helps (especially in combination with that posted by those above) ;D
Post by: Loki98 on February 17, 2016, 01:34:21 pm
Post by: brenden on February 17, 2016, 01:53:21 pm
Thanks for all the help guys. Helped me out heaps. :Dare you checking ATAR Notes at school? if so you're my favourite type of person
Post by: Maz on February 17, 2016, 08:09:57 pm
Can someone please help me out with this?
Two trains are moving at different but constant velocities. Are the any conditions under which they are
in the same inertial reference frame?
Do 2 non-inertial reference frames imply that one frame is accelerating?
i'm not really understanding relativity...
Thankyou so much in advance
Post by: Happy Physics Land on February 17, 2016, 09:33:47 pm
Hey
Can someone please help me out with this?
Two trains are moving at different but constant velocities. Are the any conditions under which they are
in the same inertial reference frame?
Do 2 non-inertial reference frames imply that one frame is accelerating?
i'm not really understanding relativity...
Thankyou so much in advance
Hey mq123:
Ok so hmm let's start with defining inertial frames of reference and non-inertial frames of reference shall we?
An inertial frame of reference is one that is either at rest or travelling at constant velocity. Within the frame of reference, the occupant cannot perform any experiments to detect whether he is travelling at a constant velocity or at rest because he experiences no force. Imagine yourself being in a black box. You cannot determine whether the black box is travelling at a constant speed or at rest because you dont feel anything (IF the box is accelerating, you would feel a force according to Newton's second law F = ma). So in regards to your first question, if you are an occupant inside the train, then both trains would be the same inertial frame of reference because you cant perform any experiment to detect that the two trains are actually travelling at different velocities. If you are a person inside the train, you wouldnt feel a single difference between travelling at 20km/h and travelling at 2000km/h.
Non-inertial frames of references are basically the opposite. The one thing you'll have to know is that if the train is accelerating, then it cant be an inertial frame of reference (i.e. it will have to be a non-inertial frame of reference). In a non-inertial frame of reference, the occupant can actually detect if he is decelerating or accelerating because he feels a force. So umm in response to your second question I'm not too sure what you meant but I would tell you that both non-inertial frames of reference are accelerating, because the occupants in both frames would be experiencing a force that is created by acceleration.
If you have further doubts or questions, please dont hesitate to ask. If you would like to read more about special relativity, I found a site for you that covers everything you'd need to know http://physics.mq.edu.au/~jcresser/Phys378/LectureNotes/VectorsTensorsSR.pdf. You probably would only need to read chapters 1-4 because thats all what you are going to be assessed on.
Best Regards
Happy Physics Land
Post by: Maz on February 17, 2016, 10:59:40 pm
Hey mq123:thankyou so much
Ok so hmm let's start with defining inertial frames of reference and non-inertial frames of reference shall we?
An inertial frame of reference is one that is either at rest or travelling at constant velocity. Within the frame of reference, the occupant cannot perform any experiments to detect whether he is travelling at a constant velocity or at rest because he experiences no force. Imagine yourself being in a black box. You cannot determine whether the black box is travelling at a constant speed or at rest because you dont feel anything (IF the box is accelerating, you would feel a force according to Newton's second law F = ma). So in regards to your first question, if you are an occupant inside the train, then both trains would be the same inertial frame of reference because you cant perform any experiment to detect that the two trains are actually travelling at different velocities. If you are a person inside the train, you wouldnt feel a single difference between travelling at 20km/h and travelling at 2000km/h.
Non-inertial frames of references are basically the opposite. The one thing you'll have to know is that if the train is accelerating, then it cant be an inertial frame of reference (i.e. it will have to be a non-inertial frame of reference). In a non-inertial frame of reference, the occupant can actually detect if he is decelerating or accelerating because he feels a force. So umm in response to your second question I'm not too sure what you meant but I would tell you that both non-inertial frames of reference are accelerating, because the occupants in both frames would be experiencing a force that is created by acceleration.
If you have further doubts or questions, please dont hesitate to ask. If you would like to read more about special relativity, I found a site for you that covers everything you'd need to know http://physics.mq.edu.au/~jcresser/Phys378/LectureNotes/VectorsTensorsSR.pdf. You probably would only need to read chapters 1-4 because thats all what you are going to be assessed on.
Best Regards
Happy Physics Land
Post by: Maz on February 17, 2016, 11:04:58 pm
can i please get some help on another question? I've done part a...and the rest...i just don't know if it's right cos i
don't have any answers...
it's attached...my teachers are having a debate on what lo and l are- i think l0 is 600...
any help would be appreciated
Post by: jakesilove on February 18, 2016, 11:08:09 am
Hey humans...again
can i please get some help on another question? I've done part a...and the rest...i just don't know if it's right cos i
don't have any answers...
it's attached...my teachers are having a debate on what lo and l are- i think l0 is 600...
any help would be appreciated
Hey MQ!
THIS QUESTION IS NOT APPLICABLE TO HSC STUDENTS
Questions like this aren't found in the HSC, as they are slightly harder than standard length contraction/time dilation questions. The issue is that the distance the light needs to travel actually changes based on whether it is traveling to the nose of the aircraft, or the tail of the aircraft. Anyway, below are my answers! I only did them quickly, so feel free to correct me :)
(http://i.imgur.com/eMJUzcG.png?1)
(http://i.imgur.com/jHar3Tx.png?1)
(http://i.imgur.com/rbffQoP.png?1)
Jake
Post by: JellyBeanz on February 18, 2016, 07:27:30 pm
Hey MQ!
THIS QUESTION IS NOT APPLICABLE TO HSC STUDENTS
Questions like this aren't found in the HSC, as they are slightly harder than standard length contraction/time dilation questions. The issue is that the distance the light needs to travel actually changes based on whether it is traveling to the nose of the aircraft, or the tail of the aircraft. Anyway, below are my answers! I only did them quickly, so feel free to correct me :)
(http://i.imgur.com/eMJUzcG.png?1)
(http://i.imgur.com/jHar3Tx.png?1)
(http://i.imgur.com/rbffQoP.png?1)
Jake
Haha sorry Jake for us VCE kids posting you questions at times XD, thanks heaps for answering them.
You guys are awesome.
Post by: Maz on February 18, 2016, 07:28:05 pm
Hey MQ!
THIS QUESTION IS NOT APPLICABLE TO HSC STUDENTS
Questions like this aren't found in the HSC, as they are slightly harder than standard length contraction/time dilation questions. The issue is that the distance the light needs to travel actually changes based on whether it is traveling to the nose of the aircraft, or the tail of the aircraft. Anyway, below are my answers! I only did them quickly, so feel free to correct me :)
(http://i.imgur.com/eMJUzcG.png?1)
(http://i.imgur.com/jHar3Tx.png?1)
(http://i.imgur.com/rbffQoP.png?1)
Jake
thankyou so much- you guys are so lucky not having to do these types of questions :) we have sooo many like them
really appreciate the help :)
Post by: JellyBeanz on February 18, 2016, 07:35:42 pm
thankyou so much- you guys are so lucky not having to do these types of questions :) we have sooo many like them
really appreciate the help :)
We aren't doing special relativity, doing further electronics instead for a detailed study, how are you finding special relativity now? XD
Post by: Maz on February 18, 2016, 07:46:38 pm
We aren't doing special relativity, doing further electronics instead for a detailed study, how are you finding special relativity now? XDat first it was really really confusing- i mean getting your head around mass changing at speed and the length of something actually getting bigger or smaller- but it's really interesting :)
Post by: JellyBeanz on February 18, 2016, 07:48:11 pm
at first it was really really confusing- i mean getting your head around mass changing at speed and the length of something actually getting bigger or smaller- but it's really interesting :)
I find it weird, you guys have begun with special relativity, maybe doing motion before special relativity would have aided you?
Post by: Maz on February 18, 2016, 07:59:07 pm
I find it weird, you guys have begun with special relativity, maybe doing motion before special relativity would have aided you?thats what i thought- but my teachers found some logic in doing unit 4 before unit 3 :D
Post by: chloe9756 on February 19, 2016, 07:24:59 pm
answer: 34.0ms^1 and 10.3 s.
Not quite sure how to do this question. I created a triangle with a hypotenuse of a and opposite side of 9.8 due to gravity, but didn't get the right answer.
Post by: jamonwindeyer on February 19, 2016, 08:22:26 pm
A truck starts from rest at the top of a uniform slope and coasts down it a distance of 175 m, as a result its height above the ground is reduced by 59.1m. The plane is frictionless. What is the trucks velocity when it reaches the 175m mark, and how long does it take to reach this mark?
answer: 34.0ms^1 and 10.3 s.
Not quite sure how to do this question. I created a triangle with a hypotenuse of a and opposite side of 9.8 due to gravity, but didn't get the right answer.
Hey Chloe! No worries, I'll step through it for you!
So, rather than being a question on forces, acceleration and the like, this is actually a very deceitful question on conservation of energy. Say what?
EDIT: RuiAce has a solution below, which resolves forces to obtain the solution to Part B first, then Part A. Both are correct, use whichever seems simpler to you ;D
So, the truck starts from rest at the top of the hill. At this stage, the energy possessed by the truck is equal to the gravitational potential energy, which can be expressed as the product of its mass, its height, and the acceleration due to gravity. That is:
There is some things here we don't know, but bear with me.
Now, once the truck hits the 175 metre mark, the truck has two types of energy. It has potential energy like before, at a new height which is 59.1 metres less than the initial height. It also has kinetic energy. We can express this as:
Now, the conservation of energy says that the energy of the truck must be conserved. The question specifies that the plane is frictionless, so we know that no energy is lost to friction. In Tertiary Physics, we call this a conservative system. So, we can equate the two expressions for energy and solve for the velocity (we will find lots of cancellation):
So that is Part A! For Part B, we know the final velocity, and we can use this to calculate the acceleration of the truck as it rolls (it will be constant since gravity is the cause, the only reason we can't use gravity directly is because it is on an incline). We use the following formula from the formula sheet:
And thus we can find the time taken:
Note that there is a few ways to do Part B, but this is my preferred method due to its simplicity.
Post by: RuiAce on February 19, 2016, 08:26:54 pm
Post by: jamonwindeyer on February 19, 2016, 09:11:37 pm
I resolved forces instead.(http://uploads.tapatalk-cdn.com/20160219/deae719c7bc0ec5e581cbae0b3e9e1be.jpg)
Oh, I jumped to the energy method so quickly I didn't even consider that there would be enough information to do it this way ;D
Post by: sire123 on February 20, 2016, 08:21:07 am
Post by: jakesilove on February 20, 2016, 09:42:48 am
Yo guys, how much time shuld I spend on each mark for phys and chem? So 3hr exam, shuld 1 mark be equivalent to 2mins?
Hey!
For a 100 mark exam, sat in 180 minutes, then yes I would spend about 2 minutes per mark. However it won't be as black and white as that: some 3 markers will take you 3 minutes, and some 8 markers will take you 20. I think that the best way to judge how on track you are is to have a look through the whole paper in reading time, and assessing roughly where you should be at the end of each hour. If you think you're a little behind by the end of the first hour, pick up the pace. If you're halfway through the exam, maybe take a bit of extra time to think about the question.
There really is no 'rule' about this sort of thing; just do whatever works for you!
Jake
Post by: Syndicate on February 20, 2016, 04:40:51 pm
Yo guys, how much time shuld I spend on each mark for phys and chem? So 3hr exam, shuld 1 mark be equivalent to 2mins?
Multiple choice don't usually take 2 minutes to solve.
As Jake asserted, some of the questions attracting more marks tend to take longer time to finish.
What I would suggest is that you should start with short answer/ extended response questions, and then move your way down. Leave the harder questions for the end, and do the rest (this would allow you to get the maximised mark, as you won't be missing out on the other questions by spending most of your time trying to figure out the answer for the harder question(s)). Do the multiple choice in the end, as they are quite easy, and like I said, they won't take you 2 minutes to solve.
Good Luck
Post by: jamonwindeyer on February 20, 2016, 05:30:19 pm
Yo guys, how much time shuld I spend on each mark for phys and chem? So 3hr exam, shuld 1 mark be equivalent to 2mins?
To throw my opinion in with the others, I personally did my Elective first. I just preferred to get it done, then I gave myself a break from writing with the Multiple Choice (I preferred to do MC before the short answers to get myself rolling, generate some momentum I suppose you should say), then finishing with the largest section. Then I'd usually have time to go back and do questions I skipped/check answers.
In general, I used MC as a writing break/chance to get my brain switched on. It's a nice warm up. But it is also a nice one to do last, since you can guess. Each to their own ;D
Post by: RuiAce on February 20, 2016, 06:08:12 pm
Multiple choice don't usually take 2 minutes to solve.
As Jake asserted, some of the questions attracting more marks tend to take longer time to finish.
What I would suggest is that you should start with short answer/ extended response questions, and then move your way down. Leave the harder questions for the end, and do the rest (this would allow you to get the maximised mark, as you won't be missing out on the other questions by spending most of your time trying to figure out the answer for the harder question(s)). Do the multiple choice in the end, as they are quite easy, and like I said, they won't take you 2 minutes to solve.
Good Luck
No. I always take at least 15 out of the allowed 35 minutes to attempt the 20 marks worth of multiple choice. Some questions require serious breaking down and others are actually full on calculations in the multiple choice.
If you're taking 2 minutes to do 20 multiple choice questions you are absolutely rushing it in HSC physics.
Edit: Ok, I just realised you aren't in NSW. In that case, please be advised of the differences in the courses; HSC physics focuses on context but throws in more annoying calculations. Please refer to some papers such as this one: http://boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Post by: jamonwindeyer on February 20, 2016, 06:41:17 pm
No. I always take at least 15 out of the allowed 35 minutes to attempt the 20 marks worth of multiple choice. Some questions require serious breaking down and others are actually full on calculations in the multiple choice.
If you're taking 2 minutes to do 20 multiple choice questions you are absolutely rushing it in HSC physics.
Edit: Ok, I just realised you aren't in NSW. In that case, please be advised of the differences in the courses; HSC physics focuses on context but throws in more annoying calculations. Please refer to some papers such as this one: http://boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
I think Syndicate meant each individual question wouldn't take 2 minutes, meaning MC wouldn't take 40 minutes to complete. I agree that 15 minutes minimum is required to properly attempt the questions.
Post by: sire123 on February 20, 2016, 07:32:16 pm
Post by: RuiAce on February 20, 2016, 08:09:04 pm
I think Syndicate meant each individual question wouldn't take 2 minutes, meaning MC wouldn't take 40 minutes to complete. I agree that 15 minutes minimum is required to properly attempt the questions.That'd make more sense.
Post by: RuiAce on February 20, 2016, 08:11:43 pm
Aite thanks guys! And btw could anyone say a list of commonly chucked 6 markers? Ik the impacts of transformers come out a lot, and the MM experiment. Anything else?Those two are the main popular ones. Note that transformers can be asked as AC generators as well.
And then, you have a variety of occasionally appearing ones.
Applications of superconductors
Applications of semiconductors
Launching a rocket into space, slingshot and reentry
Hertz experiment all the way down to Einstein with regards to photoelectric effect
Edison VS Westinghouse appears rarely
And that's as far as my memory goes.
Post by: jakesilove on February 21, 2016, 12:47:28 pm
Aite thanks guys! And btw could anyone say a list of commonly chucked 6 markers? Ik the impacts of transformers come out a lot, and the MM experiment. Anything else?
My advice would be to look through past HSC papers from something like 2005-2015, and keep a list of any questions over 6 marks. Write out answers, get your teachers to mark them, write out answers again until they are perfect. Then, put them into your Physics notes, because without a doubt you will be asked the exact same questions in your HSC (Not necessarily all of them, but I promise you that you'll see at least a few). That way, when you go to answer questions you know precisely what information you need to include to get full marks!
Glad to see everyone so active on the forums and helping each other out!
Jake :)
Post by: Charli33 on February 24, 2016, 04:50:32 pm
I am pretty lost in regards to the Motors and Generators course and was wondering if someone could explain the concept of back EMF?
It would be much appreciated!
Post by: Happy Physics Land on February 24, 2016, 06:28:09 pm
Hi guys
I am pretty lost in regards to the Motors and Generators course and was wondering if someone could explain the concept of back EMF?
It would be much appreciated!
Hey Charlie:
Awesome! Back emf is actually a concept that my classmates all struggled with for two lessons before they eventually understood what it is, simply because you just feel like you have current and emf produced everywhere and you have your motor effect and all these other fancy physics principles and you would kinda feel lost. So don't worry, you are in the same boat as majority of the other students and I will definitely help you to understand this concept better.
So lets think back to out first principles - the motor effect. We know that when we supply a current to a coil inside a magnetic field, the coil experiences a force (motor effect) and it rotates. (N.B. if you would like to know why it experiences a force, it is because the current passing through the coil actually produces a magnetic field that interacts with the external magnetic field caused by a pair of magnets. This exerts a force onto the coil, making it spin). We call this current that is supplied to the rotor the "supply current". Since a current requires voltage(which pushes the electrons through the conductor) to flow through the coil, we call this voltage "supply voltage" or "supply emf".
We have also learnt however, that whenever there is a change in magnetic flux, an emf is induced and hence a current is induced (Faraday's law). Essentially there will be a change in magnetic flux whenever there is a relative motion which cuts magnetic field lines. The magnitude of this current/emf produced will be proportional to the rate of change of flux. This induced emf will be in such a direction that it opposes the change in magnetic flux (Lenz's law).
Ok now, we need to combine the two parts I talked about above to explain the back emf. According to the motor effect, the rotor coil would be rotating in the external magnetic field. As it rotates, there is relative motion between the rotor coil and the magnetic field. As a result, it constantly cuts different amount of magnetic field lines, causing change in magnetic flux and induces an emf (Faraday's law). But according to lenz's law, this induced emf will be in a direction to oppose the change in magnetic flux, hence this emf will be acting against the supply emf to try to maintain the previous magnetic flux condition. Because it opposes the supply emf, we call this INDUCED EMF "back emf".
When the armature (rotor coil) is parallel to the magnetic field, the back emf induced will be the greatest because torque is the greatest when armature is parallel to magnetic field lines (T = nBIAcostheta, theta=0, costheta = 1, hence T is maximum when theta=0 i.e. parallel). This maximum torque means that the armature will be cutting the magnetic field lines at a greater rate, and according to faraday's law, more emf will be produced because there is a faster change in magnetic flux. When the armature coil is perpendicular to the magnetic field, i.e. theta=90 degrees, the torque will be minimum, meaning that change in magnetic flux is minimum and hence back emf produced in minimum.
Ok so I hope my explanation made sense to you and if you didnt understand any of those, just remember two things at least:
1. back emf acts against the supply emf
2. back emf is induced due to the rotor's rotation which causes change in magnetic flux
Best Regards
Happy Physics Land
Post by: Charli33 on February 25, 2016, 05:03:44 pm
Post by: Maz on March 03, 2016, 07:26:07 pm
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?
one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?
thankyou soo much :)
Post by: RuiAce on March 03, 2016, 08:54:00 pm
Hey
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?
one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?
thankyou soo much :)
Admittedly, I'm beginning to forget this part of physics so don't rush to quote me.
When the two protons are moving towards each other, they will see each other approaching at the speed of light 'c'. This is because the speed of light is constant in all frames of references.
Note that the 'light' that the proton gives off to the other proton, is what the other proton, effectively, 'sees'. (Analogy: When you look at your friend, you're just seeing light reflected off their body)
If they were moving in the same direction however, at the same speed, then no experiment can be done to distinguish those two frames of references. They are the same frame of reference and they will appear stationary to each other.
On the initial question, the stationary observer himself is in a rest frame. He will see them at 0.75c.
Post by: jamonwindeyer on March 03, 2016, 09:12:09 pm
Admittedly, I'm beginning to forget this part of physics so don't rush to quote me.
When the two protons are moving towards each other, they will see each other approaching at the speed of light 'c'. This is because the speed of light is constant in all frames of references.
Note that the 'light' that the proton gives off to the other proton, is what the other proton, effectively, 'sees'. (Analogy: When you look at your friend, you're just seeing light reflected off their body)
If they were moving in the same direction however, at the same speed, then no experiment can be done to distinguish those two frames of references. They are the same frame of reference and they will appear stationary to each other.
On the initial question, the stationary observer himself is in a rest frame. He will see them at 0.75c.
I agree on all counts except the "approach each other at the speed of light" aspect. If I see a car approaching me, and it was travelling at 100 kilometres per hour, while I'm moving towards it at 100 kilometres an hour, then I would say it was travelling 200 kilometres an hour. Relative velocity. So I don't think we can automatically say that the answer is 'c' based on the fact that the reflected light travels at this speed.
That being said, the method of analysis above doesn't sit right with me either. It would mean that each proton sees the other approaching at 1.5c, which violates the Special Theory of Relativity. So, the answer could very well be c, I am not sure, but I don't think we can conclude it based on this reasoning.
So yeah, everything else I totally agree with, just that one bit I don't think the explanation is quite correct. The correct answer eludes me for the moment, give me a bit! Unless someone else wants to weigh in ;D
Post by: Maz on March 03, 2016, 09:27:08 pm
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight
so if this is the 'Pion' travelling at 0.95c in ---> direction <----- 'X' 'Y'---->
and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula: u+v/1+((uv)/c2)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c2)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance
Post by: jamonwindeyer on March 03, 2016, 09:36:21 pm
i think i still don't quite get the idea...
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight
so if this is the 'Pion' travelling at 0.95c in ---> direction <----- 'X' 'Y'---->
and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula: u+v/1+((uv)/c2)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c2)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance
Hmm, I think this is one of the areas that you guys do that HSC students don't cover. I've never seen that formula before. I just did some asking around and this is only covered at the highest level of First Year University Physics (UNSW). Aka, way beyond my capabilities. However, with regard to that question, I am quite certain the -0.95c comes from the motion of the pion. When it decays, the two photons are emitted at the speed of light with respect to the pion's reference frame. I know the speed of light is constant, but I think relative velocity still applies here, which is where the 0.95c would come from (the pion's speed).
Sorry I can't be of more help, but you have officially exceeded the limits of my knowledge! Let me do some more sleuthing and see if I can get you a better answer ;D
Post by: jakesilove on March 03, 2016, 09:46:29 pm
i think i still don't quite get the idea...
heres another question...
A pion is an unstable particle that decays into 2 photons. A particular pion travelling at 0.95c, with respect to an observer at rest, decays into 2 photons, X and Y, which then travel in opposite directions. photons travel at the speed of ight
so if this is the 'Pion' travelling at 0.95c in ---> direction <----- 'X' 'Y'---->
and there is a stationary observer
when calculating the velocity of photon Y with respect to the stationary observer...
wouldn't it be:
Formula: u+v/1+((uv)/c2)
so the answer is saying (c-0.95c)/((1-0.95c x c)/c2)=c (i know how the speed of light is constant...but idk how they got to it)
i don't really get how they got to that as they are both travelling in the same direction...so why is there a '-0.95'?
i don't think i really get how to do these...
thankyou in advance
Hey MQ!
Unfortunately, as Jamon said, this is an area that falls outside of the HSC curriculum. As such, as far as Relativity questions go, I'm afraid we won't be of much help. Whilst I've touched on this briefly, I don't feel like I could adequately explain relative velocity in terms of Relativistic effect. However, if you click on this link, hopefully you'll get a good idea of how the formulas work.
Sorry we couldn't be of more help!
Jake
Post by: Maz on March 03, 2016, 09:57:55 pm
Hey MQ!hey
Unfortunately, as Jamon said, this is an area that falls outside of the HSC curriculum. As such, as far as Relativity questions go, I'm afraid we won't be of much help. Whilst I've touched on this briefly, I don't feel like I could adequately explain relative velocity in terms of Relativistic effect. However, if you click on this link, hopefully you'll get a good idea of how the formulas work.
Sorry we couldn't be of more help!
Jake
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page :P
Post by: jakesilove on March 03, 2016, 10:14:08 pm
hey
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page :P
Try now!
Post by: jamonwindeyer on March 03, 2016, 10:17:00 pm
hey
no problem...but thank you for trying...and the link isn't working properly i think...not unless it was supposed to go to my Facebook page :P
It was Jake's way of saying the answer is inside you, you just have to find it...
Metaphors are strong ;)
Post by: Maz on March 03, 2016, 10:27:41 pm
It was Jake's way of saying the answer is inside you, you just have to find it...haha :D
Metaphors are strong ;)
and thank you both guys
Post by: RuiAce on March 04, 2016, 08:44:49 am
I agree on all counts except the "approach each other at the speed of light" aspect. If I see a car approaching me, and it was travelling at 100 kilometres per hour, while I'm moving towards it at 100 kilometres an hour, then I would say it was travelling 200 kilometres an hour. Relative velocity. So I don't think we can automatically say that the answer is 'c' based on the fact that the reflected light travels at this speed.
That being said, the method of analysis above doesn't sit right with me either. It would mean that each proton sees the other approaching at 1.5c, which violates the Special Theory of Relativity. So, the answer could very well be c, I am not sure, but I don't think we can conclude it based on this reasoning.
So yeah, everything else I totally agree with, just that one bit I don't think the explanation is quite correct. The correct answer eludes me for the moment, give me a bit! Unless someone else wants to weigh in ;D
I'll try to explain a bit of my thought :P
Model A uses classical physics
Model B uses relativistic physics
The relative velocity of 200km/hr seems reasonable enough at those speeds you gave, however I reckon this is actually why. At low velocities, the principles of classic physics can be regarded as true.
This occurs, because there is no need for the Lorentz Factor of √(1-v^2/c^2). Note that by plugging v=200/3.6 ms^-1 this value is negligible.
At relativistic speeds, we need to specifically consult relativistic physics in order to understand what is going on. Recall that Einstein postulated that speed was an absolute quantity. Length and time are what become relative - this is why at relativistic speeds we have length contraction and time dilation taking place.
(We do not see length contraction or time dilation so clearly at speeds that we can associate with. Yet it's there - It's just so negligible. But we will observe this happening quite clearly at the relativistic speeds. )
Post by: FallonXay on March 04, 2016, 08:52:49 pm
The answers say High AC voltage frequency and Low Electrical resistant of the pot base.
I don't understand why the base of the cooking base pot would need to have low resistance, as wouldn't you want the induced eddy currents to face resistance to convert the energy into heat?
Thanks.
Post by: Happy Physics Land on March 04, 2016, 11:27:46 pm
Question 19) in the 2012 HSC Physics exam asks: Which set of conditions would result in the most rapid heating of the base of the cooking pot in an induction cooking system?
The answers say High AC voltage frequency and Low Electrical resistant of the pot base.
I don't understand why the base of the cooking base pot would need to have low resistance, as wouldn't you want the induced eddy currents to face resistance to convert the energy into heat?
Thanks.
Hey Fallon Xay:
Interesting! This is actually one of the problems that I had an argument on with my teacher. So the way I understood the question is exactly the same as what you are thinking now, i.e. "the resistance of the metallic base of the pot required to heat it up as rapidly as possible". The question in itself is quite vague and one lesson l learn from this is to never assume anything.
So lets just start analysing this question. Of course we would need a high AC voltage frequency in order to cause constant change in magnetic flux in order to induce increasing eddy currents flowing through the base of the pot. Tick, thats the right answer. Now we are up to the electrical resistance of the base of the pot. The base of the pot would have to be metallic in order for eddy current to flow through and any conductive metal would have to have relatively low electrical resistance compared to things like plastic or rubber. This is why the answer would be low electrical resistance not high.
But yes I beyond all doubts agree with what you are saying. If we just assume that they are talking about a metallic pot base, then high electrical resistance would be the correct answer, because with more resistance in the metal, more heat is produced as the eddy current experiences more impedance. I.e. copper pot would be unsuitable because it has an extremely low electrical resistance and therefore a lot of eddy current may flow through but heat may not necessarily be produced. A steel pot on the other hand, because it has carbon atoms interstitially alloyed to iron, the lattice structure is more solid and its electrical resistance is higher than copper, hence able to produce a lot of heat but is still able to conduct electricity. In this question however, I think the key idea is not to assume that the question is only talking about metal, but just generally the electrical resistance of the material used for pot base. So, generally the material of pot base would need to have low electrical resistance to allow the flow of eddy current.
Best Regards
Happy Physics Land
Post by: jamonwindeyer on March 05, 2016, 01:01:14 am
Hey Fallon Xay:
Interesting! This is actually one of the problems that I had an argument on with my teacher. So the way I understood the question is exactly the same as what you are thinking now, i.e. "the resistance of the metallic base of the pot required to heat it up as rapidly as possible". The question in itself is quite vague and one lesson l learn from this is to never assume anything.
So lets just start analysing this question. Of course we would need a high AC voltage frequency in order to cause constant change in magnetic flux in order to induce increasing eddy currents flowing through the base of the pot. Tick, thats the right answer. Now we are up to the electrical resistance of the base of the pot. The base of the pot would have to be metallic in order for eddy current to flow through and any conductive metal would have to have relatively low electrical resistance compared to things like plastic or rubber. This is why the answer would be low electrical resistance not high.
But yes I beyond all doubts agree with what you are saying. If we just assume that they are talking about a metallic pot base, then high electrical resistance would be the correct answer, because with more resistance in the metal, more heat is produced as the eddy current experiences more impedance. I.e. copper pot would be unsuitable because it has an extremely low electrical resistance and therefore a lot of eddy current may flow through but heat may not necessarily be produced. A steel pot on the other hand, because it has carbon atoms interstitially alloyed to iron, the lattice structure is more solid and its electrical resistance is higher than copper, hence able to produce a lot of heat but is still able to conduct electricity. In this question however, I think the key idea is not to assume that the question is only talking about metal, but just generally the electrical resistance of the material used for pot base. So, generally the material of pot base would need to have low electrical resistance to allow the flow of eddy current.
Best Regards
Happy Physics Land
Spot on, a little bit interpretive there. Here is another thought.
The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:
This heating is, as suggested by the formula, caused primarily by the passage of an electrical current through resistance. A phenomenon known as hysteresis also plays a role, but most engineers argue that it isn't as important as Joule heating.
So, this formula suggests that, given a constant current, that increasing the resistance will increase the heating effect, and this is true. However, it is important to remember that our resistance impacts the size of the induced eddy currents.
Let me explain. The size of the induced EMF/voltage in the pot base is proportional to the rate of change of magnetic flux. This is Faraday's Law. Let's assume we have a constant voltage. We know that:
So, given our voltage is constant, if we want a higher current, we need a smaller resistance. Now, going back to the heating formula above, doubling the current will cause the heating effect to be quadrupled (since the current is squared), whereas doubling the resistance only results in the heating effect to be doubled.
Based on this interpretation, it is definitely more effective to have a lower resistance, and thus, larger eddy currents in the pot base, to achieve the maximum heating effect. This becomes more obvious if we use ohms law to re-write the heat dissipation law:
So, in fact, having a lower resistance means more current will flow, and thus, more heat will be generated than it would be by having a high resistance. This is why a low valued resistor, plugged into a huge voltage, blows up (for lack of a better term).
This is a very simplified model and by no means totally accurate, but it is an interesting way to look at the problem.
Now, this does not get rid of the idea that high resistance = more heat. It does. But we need a sizeable voltage drop across the said resistance for it to matter. This is quite a complex topic, definitely not worth investigating in depth. In actuality, there are many things at play here.
Just a different way of looking at the question. At its core, I am saying the same as Happy Physics Land. If the resistance goes too high, no current flows, and so heating does not occur. This is just a more detailed/mathematical explanation of that intuition ;D
Post by: RuiAce on March 07, 2016, 08:35:26 pm
Spot on, a little bit interpretive there. Here is another thought.Keep in mind that
The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:
is actually taught in the HSC when mentioning resistive heating. This can be easily combined with the definition of power:
Where W = work done, measured in joules
Post by: jamonwindeyer on March 08, 2016, 11:00:53 pm
Keep in mind that
is actually taught in the HSC when mentioning resistive heating. This can be easily combined with the definition of power:
Where W = work done, measured in joules
Good call, this is a much better way to explain it in terms of the HSC syllabus.
Post by: Neutron on March 10, 2016, 09:38:51 pm
1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?
2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?
For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!
Post by: jamonwindeyer on March 10, 2016, 10:17:41 pm
Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!
1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?
2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?
For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!
Hey Neutron! I'll tackle Question 2 first, then I'll work on a diagram for Question 1. For Question 2, we have a satellite placed into orbit. This means that the force acting on the satellite is caused by gravity. We equate Newton's Formula for Gravitational Force to Nerwtons 2nd Law to get the answer. Note that we use r+h, where r is the earth's radius and h is the altitude. Centripetal acceleration is taken from the centre of earth.
To derive formulae for centripetal acceleration,orbital radius, etc etc, you will be looking at combinations of Newton's 2nd Law, Newton's Law of Universal Gravitation, and the Formula for Centripetal Acceleration. Totally depends on what you want the value with respect to. Did you have any specific derivations in mind?
Post by: jamonwindeyer on March 10, 2016, 10:35:01 pm
Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!
1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?
2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?
For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!
Okay, so I put together a diagram in Word, and then it decided to crash. Lesson: Save often when running a Microsoft Program on an Apple Machine. ;)
I'll have another crack at a diagram tomorrow, but a quick explanation in the mean time. What is happening is best explained using vectors and forces.
We assume that the scale measures the weight force as directed downwards onto its surface. That is, it measures the magnitude of the equivalent force pushing directly down on it. When the plank and scale are flat, all the weight force is directed downwards and so 100% of the weight force is registered. As the plank is raised, the scale becomes inclined. However, the students weight force continues to be directed downwards. If we break this into two vector-components, one pointing perpendicular to the scales surface and one pointing parallel to it, we can see that the component of the weight force directed in the correct direction (such that the scale registers the weight) is lessened. In English, the weight force is redistributed so that less and less acts directly downwards on the scale. The rest is (and this is irrelevant to the question), balanced by frictional forces.
This explanation makes much more sense with a diagram, I'll get on it for you tomorrow as soon as I can! ;D
Post by: Happy Physics Land on March 10, 2016, 10:41:22 pm
Hey guys! Neutron here, I've got a few basic physics problems that I got kinda stuck with D: I was wondering whether you guys could help!
1. A student was standing on bathroom scales placed on a plank of wood and was recording the reading of weight from the scales. The plank was slowly raised at one end, so that the angle of the inline to the horizontal increased. What would the student observed about the weight reading?
2. The Earth's radius is 638 x 10^6 m and mass 5.98 x 10^24 kg. A 670 kg satellite is placed into circular orbit where teh acceleration due to gravity is 0.225 ms^-2. What is the altitude of this orbit?
For question 1, I was wondering how you can explain it through physics principles? Like I know that the weight would decrease but how would you describe it? Also, for question 2, I kinda forgot which equation to use. And also, how do you derive the equations for orbital radius and centripedal acceleration? Thank you so much!
Hey Neutron:
Wow there are a whole heap of questions right there haha! But thats ok, I thoroughly enjoy them all. I will start off with the easy works of derivation first then I will move on to answering your questions yeah?
For centripetal acceleration:
F = ma (Newton's 2nd law)
Hence F(c) = ma(c) (where F(c) means centripetal force and a(c) is centripetal acceleration)
But using the centripetal force formula we also know that F(c) = mv^2/r
Hence mv^2/r = ma(c)
Cancelling m from both sides, we get a(c) = v^2/r
So the formula for centripetal acceleration is v^2/r
For Orbital Velocity:
*Theres a lot of ways to work out orbital radius, and theres no set formula for it, the most common equation we use for working out orbital radius would be using Kepler's third law, i.e. r^3/T^2 = GM/4pi^2. So Im just assuming that you are asking for a derivation of orbital velocity instead of orbital radius here?
When an object is in uniform circular motion (orbit) around a planet, centripetal force = gravitational force of attraction
Hence applying the formula for both, we get:
mv^2/r = GMm/r^2
Cancelling m from the numerator of both sides, we get:
v^2/r = GM/r^2
Cancelling r from the denominator of both sides, we get:
v^2 = GM/r
Hence v(orb) = squareroot(GM/r)
Right, I hope you can understand my derivation (I havent really Learnt LaTex yet, sorry for not being able to present this in normal words and format)
Answer to question 1:
So when the plank of wood is elevated (i.e. angle of inclination to the horizontal ground rises), the weight force you exert onto the scale is actually broken into components. I have included a diagram below to aid my explanation. Essentially, when we just weigh ourselves normally, the scale shows our weight as a measure of our normal force, which is the reaction force to your weight force. However when you are elevated to an angle, (let the weight force you exert be W), your normal force no longer equals to your weight force, but are broken into the vertical component of Wsin(theta) and horizontal force of Wcos(theta). So now, the normal force that is measured by the scale becomes the vertical component Wsin(theta), and since sine of any acute angle would be less than 1, hence Wsin(theta) is less compared to W, and hence the reading on the scale also decreases.
(http://i.imgur.com/Mebe8a3.png)
Answer to question 2:
So here I would be using the formula for gravitational acceleration: g = GM/r^2
So what we WANT TO figure out here is r, because once we find r, we get to know what the altitude is by doing r - 638 x 10^6
We know what (http://G is (universal constant)), we know what M is (5.98 x 10^24), and we also know what g is (0.225ms^-2).
Now, you might ask, whats the purpose of the question giving us the mass of satellite then? Well there is absolutely no purpose, its just a red herring. The question wants to trick you into thinking that you need to use m somewhere, so some students might end up applying the formula GMm/r^2. But thats not the formula for gravitational acceleration! The formula for gravitational acceleration would be GM/r^2 and is independent from the mass of the orbiting object!
Now, substitute in every value:
0.225 = 5.98x10^24 x 6.67x10^-11 / r^2
Hence r^2 = 1.77 x 10^15
Hence r = 4.21 x 10^7
So now, we do r - 6.38x10^6, we get 3.572x10^7 m altitude
I might have made a mistake in a place or two, just double check with your answer. And double check also that the Earth's radius is 6.38 x 10^6m not 638 x 10^6 m?
But overall great questions! Thanks for posting and if you have any further questions, dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 10, 2016, 10:42:45 pm
Okay, so I put together a diagram in Word, and then it decided to crash. Lesson: Save often when running a Microsoft Program on an Apple Machine. ;)
I'll have another crack at a diagram tomorrow, but a quick explanation in the mean time. What is happening is best explained using vectors and forces.
We assume that the scale measures the weight force as directed downwards onto its surface. That is, it measures the magnitude of the equivalent force pushing directly down on it. When the plank and scale are flat, all the weight force is directed downwards and so 100% of the weight force is registered. As the plank is raised, the scale becomes inclined. However, the students weight force continues to be directed downwards. If we break this into two vector-components, one pointing perpendicular to the scales surface and one pointing parallel to it, we can see that the component of the weight force directed in the correct direction (such that the scale registers the weight) is lessened. In English, the weight force is redistributed so that less and less acts directly downwards on the scale. The rest is (and this is irrelevant to the question), balanced by frictional forces.
This explanation makes much more sense with a diagram, I'll get on it for you tomorrow as soon as I can! ;D
Jamon fam I got your back covered! :)
Post by: Neutron on March 10, 2016, 11:10:07 pm
A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases
The answer is b and I was wondering why? Thank you!
(Sorry for the amount of questions I have D:)
Neutron
Post by: jamonwindeyer on March 10, 2016, 11:11:52 pm
Jamon fam I got your back covered! :)
Thanks HPL, glad to know I've got backup ;) remind me I need to teach you LaTex! It is super easy once you know what is going on 8)
Post by: jamonwindeyer on March 10, 2016, 11:17:05 pm
Thank you so much to you both for the comprehensive responses! Although HPL, where did you get the equation you used for Q2 from? I don't seem to be able to find it on the data sheet D: And also, I was revising some past papers and I forgot the relationship between the voltage in DC motors and the resulting speed current. The question I"m having trouble with is:
A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases
The answer is b and I was wondering why? Thank you!
(Sorry for the amount of questions I have D:)
Neutron
The formula he used was from equating Newton's 2nd Law with Newton's Law for Universal Gravitation, the exact same as my response. You end up with:
As for your question, consider the motor as just a general resistor in a circuit. As we increase the voltage, the current will also increase by Ohm's Law (Back EMF will play a part, but the current will still increase). So the current increases for sure.
Now, the rotational force, or torque, is what causes a rotor to spin. The larger this force, the greater the rotational acceleration, and the rotor will end up spinning faster.
The formula for torque is:
So, since current is increasing, torque is increasing. Therefore, the motor speed will also increase. Hence the Answer of B ;D
Was that overview good or did you need some help with Ohm's Law and the ideas behind it? Just let me know :D
Post by: Happy Physics Land on March 10, 2016, 11:17:29 pm
Thank you so much to you both for the comprehensive responses! Although HPL, where did you get the equation you used for Q2 from? I don't seem to be able to find it on the data sheet D: And also, I was revising some past papers and I forgot the relationship between the voltage in DC motors and the resulting speed current. The question I"m having trouble with is:
A 12V DC motor has the input voltage increased from 0V to 12V. What happens to the speed and current?
a) The rotor speed is constant but the current in the motor increases
b) The rotor speed increases and the current in the motor increases
c) The rotor speed is constant but the current in the motor decreases
d) The rotor speed increases and the current in the motor decreases
The answer is b and I was wondering why? Thank you!
(Sorry for the amount of questions I have D:)
Neutron
Ok so jamon adopted the kepler's third law approach whereas I adopted the gravitational acceleration approach. Of course, both are viable, since we all got the same answer, and in both cases, the mass of the orbiting object is a red herring.
So to derive the equation I used:
We all know that gravitational force constitutes to weight
weight = mg
Newton's law of gravitational force: F(g) = GMm/r^2 (where F(g) means gravitational force of attraction)
so if we equate the two values, we get:
GMm/r^2 = mg
Cancelling m from both sides, we get g = GM/r^2 which is where I got my equation from
Post by: jamonwindeyer on March 10, 2016, 11:20:49 pm
Ok so jamon adopted the kepler's third law approach whereas I adopted the gravitational acceleration approach. Of course, both are viable, since we all got the same answer, and in both cases, the mass of the orbiting object is a red herring.
So to derive the equation I used:
We all know that gravitational force constitutes to weight
weight = mg
Newton's law of gravitational force: F(g) = GMm/r^2 (where F(g) means gravitational force of attraction)
so if we equate the two values, we get:
GMm/r^2 = mg
Cancelling m from both sides, we get g = GM/r^2 which is where I got my equation from
I think we actually did the same thing! I just equated straight to Newton's 2nd Law whereas you included a contextual formula for weight force, which is probably a little more correct. Tertiary Physics has me taking shortcuts ;)
Post by: smiley2101 on March 13, 2016, 08:48:53 am
Post by: Happy Physics Land on March 13, 2016, 10:35:29 am
Hi! If a projectile has a time of flight of 7.5 seconds and a range of 1200m then how would you work out the maximum height? Thank you!!
Hey Smiley!
This question sounds a little vague to me. If this is the full question, my solution will be shown below. If you have the full question, I would gladly want to have a look at it!
Ok so in projectile motion, the object is only acted upon by gravity, meaning that horizontal velocity stays constant (i.e. initial horizontal velocity = final horizontal velocity)
Let u(x) = initial horizontal velocity
Range = u(x) x time
1200 = u(x) x 7.5
u(x) = 160m/s
So Im assuming here that you are launching your projectile from ground-level and it is going to trace out a parabolic trajectory. In this case, I would really need the launching angle to be able to figure out the initial vertical velocity. So now we have figured out the horizontal velocity to be 160m/s, you can just use your trigonometry to figure out the initial vertical velocity. Because we know that at max height, vertical velocity = 0 (i.e. v = 0), apply the v = u + at formula to figure the time at which max height occurs. Then apply the formula y = ut + 1/2 at^2 to figure out the max height.
Hope my answer helps, if you have any further concerns please dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: Meckenza on March 14, 2016, 04:03:45 pm
I am confused about the answer to the following question:
A copper ring is suspended near to a fixed solenoid S as shown.
When the switch is closed a large current flows through S. As a result, the ring R
Answer: is repelled towards S as long as the switch is closed.
Any help would be much appreciated. Thanks!
Post by: FallonXay on March 14, 2016, 05:16:01 pm
Why does maximum torque on a current carrying loop occur when the angle between the loop's plane and magnetic field vector is 90 degrees if using the formula T=nBIAcosTheta, cos90=0 and cos 0=1?
Post by: Happy Physics Land on March 14, 2016, 05:21:42 pm
Hello!
I am confused about the answer to the following question:
A copper ring is suspended near to a fixed solenoid S as shown.
When the switch is closed a large current flows through S. As a result, the ring R
Answer: is repelled towards S as long as the switch is closed.
Any help would be much appreciated. Thanks!
Hey Meckenza:
Thanks for posting! Definitely an interesting question! (quite tricky actually) :)
Let's have a look at this together. Whenever I approach this type of question, I just pick out the most useful information and briefly state them. So in this case:
- Copper ring suspended
- Current flows through solenoid when circuit is closed
- In the circuit, current flows clockwise (electric current flows from positive to negative terminal, and you can see the direction of current in the circuit by looking at the battery/power source)
- Copper is a conductor, and we can deduce that the core of the solenoid is also a conductor (large current means that the core must have enabled a larger current to be produced through concentrating the magnetic field of the solenoid)
Ok so I have done a deep analysis of the question. Now its time to tackle the question! Look at the direction in which the current in the circuit travels, and looking at how the solenoid is wound onto the core, we can tell that the current direction is out of the page when below the core and into the page when above the core. Using other right hand coil rule, we can effective determine that the North pole is on the left hand side of the solenoid and South is on the right hand side of the solenoid. So because of this, since the magnetic field lines would be travelling from North to South, you would get more field lines out of the page when you observe through the ring. (I have attached a diagram below to help explaining what Im trying to say).
(http://i.imgur.com/Ad3f2uq.png)
Because there is a change in the copper ring's magnetic flux, then according to Lenz's law there will be a current flowing in the ring in such a direction that it opposes this change in flux. So in order to minimise the amount of out of page magnetic field lines, the current in the ring would want to flow anticlockwise and again by applying the right hand coil rule, we can ascertain that North is on the left hand side of the ring and South is on the right hand side. Therefore the ring would be attracted to the solenoid. Really hard question actually!
(http://i.imgur.com/2juTdJB.png)
I hope my answer helps you in some sort of way haha. If you have any questions please dont hesitate to ask! :)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 14, 2016, 05:27:49 pm
Heyy. I'm just looking over a practice exam and the answers say that there is maximum torque on a current currying loops occurs when the angle between the loop's plane and magnetic field vector is 90 degrees.
Why does maximum torque on a current carrying loop occur when the angle between the loop's plane and magnetic field vector is 90 degrees if using the formula T=nBIAcosTheta, cos90=0 and cos 0=1?
Hey FallonXay!
The loop's plane should be parallel to the magnetic field in order to produce the maximum torque, just like what you have correctly stated, cos 0 = 1. But the answer from your practise exam could perhaps be saying that the current through loop is at 90 degrees to the magnetic field? I highly doubt the correctness of the exam answer here. If this is a question with an image, would you mind to send the image as well? That may cause the answer to be different. But yeah the loop has to be parallel to magnetic field for maximum torque.
Best Regards
Happy Physics Land
Post by: Meckenza on March 14, 2016, 06:26:23 pm
Thanks!
Post by: Happy Physics Land on March 14, 2016, 07:55:44 pm
sorry, would you be able to clarify why the current flows anticlockwise in the ring? I don't quite understand that part. Also, If the ring were attracted to the solenoid wouldn't the answer be A - I have attached an image of the full question -? (The answer page says the correct answer is B).
Thanks!
Hello Meckenza:
What makes this question confusing is that it has the option "attracted towards" and "repelled towards". I dont quite understand how the ring can be repelled "towards" the solenoid, because usually what happens is the ring is either "attracted towards" or "repelled away". I do remain dubious of the answer, I will try to find out the answer for you to the best of my ability.
In regards to the anticlockwise flow of current inside the ring, it is because when you look through the ring (hypothetically) you can see magnetic field lines coming out of page. So originally there wasnt any magnetic field lines travelling through the ring, but now because we established a complete circuit by closing the switch, we have made magnetic field lines that will come out of the page. So Lenz's law states that the current in the ring will flow in such a direction that it will oppose the change in magnetic flux. Since in this case the change in flux is a result of more out of page magnetic field lines, then consequently the current in the ring will flow in such a direction that it will produce into the page magnetic field lines that oppose this increase in out of page magnetic field lines. So if we apply the right hand grip rule, with our thumb pointing away from us (i.e.into the page), you can see that your fingers will curl in an anti-clockwise direction.
In regards to this question I will go find out the answer for you and I will get back to you soon!
Best Regards
Happy Physics Land
Post by: Meckenza on March 14, 2016, 08:30:45 pm
Post by: Maz on March 14, 2016, 09:46:03 pm
we have a test on 'the standard model' coming up, and i was wandering if you could please help me with this question?
when a muon and an anti-muon collide they can annihilate each other and release their mass-energy as 2 photons. assuming that these two photons are identical,
a) what will each of their energies be
b) what wavelength will they have
c) why does there need to be 2 photons produced and not just one?
d) in what directions would they have to travel relative to each other and why?
e) in what part of the electromagnetic spectrum are they located?
i don't have any answers to these so i even the ones that i have attempted, i don't know if the answer is right or wrong
thankyou soooo much in advance :)
Post by: Happy Physics Land on March 14, 2016, 10:28:35 pm
ok thank you so much! :)
Hey Meckenza!
Sorry for letting you wait for so long. It just seemed like I got my directions wrong. So when I said using right hand grip rule we get North being on the left hand side of the solenoid and South being on the right hand side, I decided that magnetic field line is going to go towards the right. But I made a conceptual error here. This north and south is OUTSIDE the solenoid. Inside the solenoid its different. Inside the solenoid the South pole is on the left and North pole is on the right, meaning that flux is actually going left. So according to Lenz's law, the current inside the ring should be flowing in a direction that it opposes the flux change to the left (i.e. the current will flow in side the ring in such a direction that it produces a flux going to the right). So if you use your right hand grip rule the current is actually flowing anticlockwise inside the ring. Hence to the left of the ring would be South and to the right of the ring would be North. Hence it repels away from the solenoid (the wording in the answer is really bad, there is absolutely no such thing as "repel towards", only "repel away from").
Sorry for all my previous tedious explanations, I hope you understand this explanation of mine haha.
Best Regards
Happy Physics Land
Post by: Meckenza on March 15, 2016, 06:49:00 am
Hey Meckenza!
Sorry for letting you wait for so long. It just seemed like I got my directions wrong. So when I said using right hand grip rule we get North being on the left hand side of the solenoid and South being on the right hand side, I decided that magnetic field line is going to go towards the right. But I made a conceptual error here. This north and south is OUTSIDE the solenoid. Inside the solenoid its different. Inside the solenoid the South pole is on the left and North pole is on the right, meaning that flux is actually going left. So according to Lenz's law, the current inside the ring should be flowing in a direction that it opposes the flux change to the left (i.e. the current will flow in side the ring in such a direction that it produces a flux going to the right). So if you use your right hand grip rule the current is actually flowing anticlockwise inside the ring. Hence to the left of the ring would be South and to the right of the ring would be North. Hence it repels away from the solenoid (the wording in the answer is really bad, there is absolutely no such thing as "repel towards", only "repel away from").
Sorry for all my previous tedious explanations, I hope you understand this explanation of mine haha.
Best Regards
Happy Physics Land
Ahh, ok I see. Thanks for the explanation Happy Physics Land!
One last thing though: Why is the ring permanently repelled when the switch is closed. Wouldn't the ring only be momentary, similar to Faraday's experiment with the primary and secondary coil wrapped around the wooden block - because the changing magnetic flux is only changing when the power supply is turned on/off?
Post by: jamonwindeyer on March 15, 2016, 01:16:34 pm
Just a little clarification of HPL's answer.
He is correct in saying that flux flows to the left in the solenoid. However, don't think of it as there being an imaginary south pole on the inside of the solenoid.
Instead, consider it this way. A solenoid with a current flowing through acts as an electromagnet, and behaves very similarly to a bar magnet (especially for your purposes). For a bar magnet, the magnetic field lines form closed loops. That is, magnetic field lines (which represent the direction of magnetic flux) must always loop back on themselves. We also have to abide by the convention that field lines go out from North and go in at South. Logically, the only way for this to occur and the diagrams to work, is for the lines to loop outwards from North, around the magnet, into South, and back through to the beginning. There is no secret south pole involved, just the principle of closed magnetic field lines applied to a magnetic dipole.
As for your question on momentary vs ongoing repulsion. As the magnetic field turns on, the Eddy Currents flow to cause the ring to be repelled from the solenoid, as HPL discussed above. Now, the magnetic field stays the same, we have a DC power supply so once it is on, the solenoid acts as a bar magnet. This would lead me to believe that eddy currents stop flowing and the ring falls back into place after a momentary repulsion. That your answers suggest otherwise is interesting, it is definitely not what I would say would be occurring... I'm not sure!
Post by: Happy Physics Land on March 15, 2016, 01:25:29 pm
Ahh, ok I see. Thanks for the explanation Happy Physics Land!
One last thing though: Why is the ring permanently repelled when the switch is closed. Wouldn't the ring only be momentary, similar to Faraday's experiment with the primary and secondary coil wrapped around the wooden block - because the changing magnetic flux is only changing when the power supply is turned on/off?
Hey Meckenza:
Good question actually. My teacher suggested that this case is perhaps different to Faraday's experiment. In this case, we've found out that the ring being suspended in air is repelled away from the solenoid. Originally I would have picked D, but if you think about it, is there a force there to bring the ring back to its original position? Clearly we dont have any evidence of a restoring force acting on the ring to bring it back to its original position. So as long as there is flux the ring would just stay there because the current would keep on circulating in the ring and the same poles between ring and solenoid would cause them to keep being repelled. Its quite a tough question!
Best Regards
Happy Physics Land
Post by: jamonwindeyer on March 15, 2016, 01:26:50 pm
hi,Hey mq123! I think we have another case of the WA course not overlapping with ours, so I can't give full solutions. I don't have all the info. However, the help I can give:
we have a test on 'the standard model' coming up, and i was wandering if you could please help me with this question?
when a muon and an anti-muon collide they can annihiate each other and release their mass-energy as 2 photons. assuming that these two photons are identical,
a) what will each of their energies be
b) what wavelength will they have
c) why does there need to be 2 photons produced and not just one?
d) in what directions would they have to travel relative to each other and why?
e) in what part of the electromagnetic spectrum are they located?
i don't have any answers to these so i even the ones that i have attempted, i don't know if the answer is right or wrong
thankyou soooo much in advance :)
For Part A, you would need the total energy of the muon/anti-muon system. I wager this would be some combination of their masses, and perhaps their speed. Once you have this, the conservation of energy dictates that this energy must now be contained in the two photons. So,
For Part B, use the formula linking a photons energy to its frequency, and then convert this to a wavelength. In one step, it looks like this:
For Part C/D, I think your answer lies in the Conservation of Momentum. When two particles collide, they are moving (presumably) in opposite directions. There is momentum in one direction, and then an identical momentum in the opposite direction, and so the total momentum of the system is likely to be zero (same mass, velocities in opposite directions, and so the momentum of one cancels out the other when we add them as vectors). Photons, being a particle, must obey this principle, and so the total momentum of the photons must also be 0. Thus, there must be two photons, as they move in opposite directions to obey the Conservation of Momentum. We can't have just 1, because there would then be momentum in one direction, and the momentum in the opposite direction would have been "lost." Let me know if this is a little confusing and I will elaborate.
For Part E, take your wavelength and map it to the EM Spectrum, it will probably be in the visible spectrum.
I hope this helps! ;D
Post by: jamonwindeyer on March 15, 2016, 01:28:04 pm
Hey Meckenza:
Good question actually. My teacher suggested that this case is perhaps different to Faraday's experiment. In this case, we've found out that the ring being suspended in air is repelled away from the solenoid. Originally I would have picked D, but if you think about it, is there a force there to bring the ring back to its original position? Clearly we dont have any evidence of a restoring force acting on the ring to bring it back to its original position. So as long as there is flux the ring would just stay there because the current would keep on circulating in the ring and the same poles between ring and solenoid would cause them to keep being repelled. Its quite a tough question!
Best Regards
Happy Physics Land
Depends on if we naturally assume the presence of gravity or not. Gravity is our restoring force if it is there. I would also assume that the copper ring has resistance and thus, any induced currents will dissipate very quickly. Fair enough that it isn't specifically in the question, but I would naturally assume these things. It's a matter of theoretical versus practical Physics! :)
Post by: Happy Physics Land on March 15, 2016, 01:29:53 pm
Depends on if we naturally assume the presence of gravity or not. Gravity is our restoring force if it is there. Fair enough that it isn't specifically in the question, but I would naturally assume :)
The question is either too ambiguous or too hard l reckon, my friend is find your notes very helpful by the way.
Post by: jamonwindeyer on March 15, 2016, 01:32:44 pm
The question is either too ambiguous or too hard l reckon, my friend is find your notes very helpful by the way.
It is difficult and definitely ambiguous. I don't think this was in a HSC Paper, I don't remember seeing it in the last few years anyway :D awesome! I'm glad they are of help :D
Post by: monique.degiovanni on March 15, 2016, 04:33:59 pm
Post by: Happy Physics Land on March 15, 2016, 07:58:28 pm
anyone know bohr's postulates there seems to be multiple answers
Hey Monique!
Nice question! And I would certainly expect there to be several answers because bohr's postulate indeed encompasses quite a few concepts. (I haven't done in depth studies in regards to this but I did some research and here are some responses to your question!)
First postulate: Electrons in an atom exist in STATIONARY STATES. Bohr stated that electrons orbit the nucleus WITHOUT EM emission and any permanent change in their motion must be accompanied by a complete transition from one stationary state to another. Problem with this postulate is that Bohr cannot explain this very effectively.
Second Postulate: Transmission between stationary states produces/absorbs EM Radiation. When an electron moves between stationary states, it is accompanied by the emission or absorption of a photon whose equation can be given mathematically as ΔE=hf. Thus, EM radiation is produced by the movement of electrons between energy states which account for Planck's "atomic oscillators."
Third Postulate: The ANGULAR MOMENTUM of a stationary electron is quantised. If you recall from Rutherford's model of the atom the electrons orbited in circular orbits at ANY RADII. If this coincided with Bohr's 2nd postulate, it would mean that every element can emit a FULL SPECTRUM as any transition would be possible. As this is not the case, it would imply that electrons orbited at FIXED RADII.
Hope these informations can help you Monique! If anyone has done/is doing from quanta to quarks, you may feel welcome to add any extra information!
Best Regards
Happy Physics Land
Post by: Neutron on March 16, 2016, 05:05:54 pm
Okay this is actually sorta dumb but I'm having the biggest brain fart.. Where's the armature and what does it do again? Like I keep getting confused when I look at diagrams as to which part's the coil, the armature and the axle D: Thanks guys
Neutron
Post by: RuiAce on March 16, 2016, 05:51:13 pm
Yo fam,
Okay this is actually sorta dumb but I'm having the biggest brain fart.. Where's the armature and what does it do again? Like I keep getting confused when I look at diagrams as to which part's the coil, the armature and the axle D: Thanks guys
Neutron
The armature is literally the coil. I remember getting that confused for so long and realising oh wait they're the same thing.
An axle is not necessary for a motor. On a generator, you probably want an axle so that you have something to spin the actual generator. (This is of course, for an ordinary generator you built or used as a play toy; not for large scale electricity production)
Post by: Castform99 on March 18, 2016, 08:44:33 pm
1. A projectile has a time of flight of 8.0s and a range of 1120m.
a) What maximum height does it reach?
b) At what velocity is it projected?
2. The model rocket has a pre-launch mass of 94.2g, of which 6.24g is solid propellant. It is able to deliver a thrust of 4.15N for a period of 1.2s. Assuming that the rocket is fired directly up, determine:
a) The initial rate of acceleration and g force
b) The final rate of acceleration and g force just prior to exhaustion of the fuel
3. Bill is selected for the space trip to Alpha Centauri a distance of 4.3 light years from Earth. It is planned for them to travel at 0.9c. If his body can withstand a force of 3g, how long will it take to accelerate them to this speed(ignoring relativistic effect)?
Post by: Stylo on March 18, 2016, 08:58:55 pm
Having a bit of trouble understanding how a synchronus motor works. Anyone be able to just briefly explain it?
Cheers
Post by: Happy Physics Land on March 18, 2016, 09:38:24 pm
Hi, I need help with some questions:
1. A projectile has a time of flight of 8.0s and a range of 1120m.
a) What maximum height does it reach?
b) At what velocity is it projected?
2. The model rocket has a pre-launch mass of 94.2g, of which 6.24g is solid propellant. It is able to deliver a thrust of 4.15N for a period of 1.2s. Assuming that the rocket is fired directly up, determine:
a) The initial rate of acceleration and g force
b) The final rate of acceleration and g force just prior to exhaustion of the fuel
3. Bill is selected for the space trip to Alpha Centauri a distance of 4.3 light years from Earth. It is planned for them to travel at 0.9c. If his body can withstand a force of 3g, how long will it take to accelerate them to this speed(ignoring relativistic effect)?
Hey Castform!
The solution to question one is posted before, if there are any obscurities in my solutions please dont hesitate to ask! :D
(http://i.imgur.com/z3tUscf.jpg)
(http://i.imgur.com/jTBs2br.jpg)
Best Regards
Happy Physics Land
Post by: jamonwindeyer on March 18, 2016, 09:41:52 pm
Yo
Having a bit of trouble understanding how a synchronus motor works. Anyone be able to just briefly explain it?
Cheers
Hey Stylo! Sure thing, a quick explanation coming up ;D
So, as shown in the diagram below, a synchronous motor (very similar to an AC Induction Motor has a rotor surrounded by three pairs of electromagnets. For a synchronous motor, the rotor is a permanent magnet.
Now, the electromagnets which make up the stator are fed 3 Phase AC Current. Without going into technicalities, what this means is that the three pairs of electromagnets turn on in sequence: Pair 1, Pair 2, Pair 3, Pair 1, etc, one pair at a time. This happens at a speed proportional to the frequency of the AC current fed into the motor.
Now, as the electromagnets turn on, the magnet rotates to align itself with the resultant magnetic field. Since the electromagnet pairs turn on and off in sequence, the magnet rotates to follow. This is what causes the rotation.
An AC Induction Motor operates on very similar principles, except the rotor is not magnetised. Instead, eddy currents formed in the rotor generate the magnetic field required for the rotation of the rotor. This can be modelled by taking a conductive disk, suspending it from string, and moving a magnet around it in a circle. The disc will chase the magnet. AC Induction motors work off that same principle ;D
(http://www.cvel.clemson.edu/auto/actuators/images/shenoy-synchronous-motor.png)
Post by: Happy Physics Land on March 18, 2016, 10:03:55 pm
Hi, I need help with some questions:
1. A projectile has a time of flight of 8.0s and a range of 1120m.
a) What maximum height does it reach?
b) At what velocity is it projected?
2. The model rocket has a pre-launch mass of 94.2g, of which 6.24g is solid propellant. It is able to deliver a thrust of 4.15N for a period of 1.2s. Assuming that the rocket is fired directly up, determine:
a) The initial rate of acceleration and g force
b) The final rate of acceleration and g force just prior to exhaustion of the fuel
3. Bill is selected for the space trip to Alpha Centauri a distance of 4.3 light years from Earth. It is planned for them to travel at 0.9c. If his body can withstand a force of 3g, how long will it take to accelerate them to this speed(ignoring relativistic effect)?
Hey Castform:
Solution to question 2 is posted below. The main formulae I used were newton's 2nd law and the g-force formula, and also the weight formula. The quantity 1.2s can fundamentally be ignored here, it only indicates the amount of time the fuel is gonna last for but this is kind of irrelevant to solving the question. So essentially its a red herring.
(http://i.imgur.com/tSG2B7K.jpg)
(http://i.imgur.com/Hz8C76P.jpg)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 18, 2016, 10:15:44 pm
Hi, I need help with some questions:
1. A projectile has a time of flight of 8.0s and a range of 1120m.
a) What maximum height does it reach?
b) At what velocity is it projected?
2. The model rocket has a pre-launch mass of 94.2g, of which 6.24g is solid propellant. It is able to deliver a thrust of 4.15N for a period of 1.2s. Assuming that the rocket is fired directly up, determine:
a) The initial rate of acceleration and g force
b) The final rate of acceleration and g force just prior to exhaustion of the fuel
3. Bill is selected for the space trip to Alpha Centauri a distance of 4.3 light years from Earth. It is planned for them to travel at 0.9c. If his body can withstand a force of 3g, how long will it take to accelerate them to this speed(ignoring relativistic effect)?
Hey Castform:
Here is your solution to the third question. The trick with this part is remembering to use v = u +at. This formula is not limited only to projectile motions, it can be applied versatilely across all physics calculations. In this case, because initially the rocket doesnt have any velocity, hence u=0 m/s. IF this question wanted us to consider the relativistic effects as well, it would be harder because relativistic effects of length contraction and time dilation would be different at each stage of our voyage, since velocity is gradually increasing and this has an impact upon the lorentz factor.
(http://i.imgur.com/bsbPen3.jpg)
Best Regards
Happy Physics Land
UPDATE: thanks Lazydreamer for spotting my mistake. The value of "a" that I used in v = u +at was 9.8 when it was meant to be 19.6 (i.e. the acceleration that I calculated using g-force formula). This would change the result by twofold (i.e. just multiply the final result by 2 to get the correct answer)
Post by: lazydreamer on March 19, 2016, 04:01:10 pm
Hey Castform:
Here is your solution to the third question. The trick with this part is remembering to use v = u +at. This formula is not limited only to projectile motions, it can be applied versatilely across all physics calculations. In this case, because initially the rocket doesnt have any velocity, hence u=0 m/s. IF this question wanted us to consider the relativistic effects as well, it would be harder because relativistic effects of length contraction and time dilation would be different at each stage of our voyage, since velocity is gradually increasing and this has an impact upon the lorentz factor.
(http://i.imgur.com/bsbPen3.jpg)
Best Regards
Happy Physics Land
Hey HPL, was looking at your solution and in the section of 'Implementing v=u+at', i've noticed you used acceleration due to gravity as 'a' instead of the one calculated from the g-force.
Just making sure haha, your solutions are otherwise great, esp. that handwriting, so pretty :D
Post by: Happy Physics Land on March 19, 2016, 06:36:26 pm
Hey HPL, was looking at your solution and in the section of 'Implementing v=u+at', i've noticed you used acceleration due to gravity as 'a' instead of the one calculated from the g-force.
Just making sure haha, your solutions are otherwise great, esp. that handwriting, so pretty :D
Hey Lazydreamer!
Thanks for pointing out that fatally careless error! Yes it was indeed 19.6 not 9.8. This is an exemplar of what happens when you go on late night practise hahahaha. Thanks for your kind words regarding my handwriting as well, appreciate it! :)
Post by: Neutron on March 19, 2016, 10:27:44 pm
In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I2R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?
Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks :D
Neutron
Post by: Happy Physics Land on March 20, 2016, 01:31:46 am
Hey guys!
In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I2R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?
Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks :D
Neutron
Hey neutron!
Okay great questions, I especially like the second one because I was stuck on that question for a period of time as well. In regards to the induction cooktop, heat is generated when you have a current going through a high resistance conductor. Just like you stated, P = I2R, if we just have a look at the operation of induction cooktop quantatively, then we can see that Power (in the form of heat) increases as current increases and as resistance increases. If we think about this qualitatively, it makes sense as well. If we have higher current (i.e. more charges) travelling through a highly resistant conductor, then there is a higher possibility that the charges will collide with the conductor's crystal lattice, and this microscopic collision would produce heat as a result of friction experienced by these charges as they travel through the conductor. Remember however, when I say high resistance, I mean a conductor that has high internal resistance, relative to the insulators, they would still have low resistance. If a material has TOO high of an internal resistance, then eddy current wouldnt be able to pass through and we get no heat ... of course!
In regards to ohm's law and P=VI, they do seem contradictory. But if we think about it, in an ideal transformer, the POWER is the same in both the primary and secondary coil, not the RESISTANCE! Similarly in electricity transmission, we are dealing with POWER loss, and hence we use P=VI to determine that V and I are inversely proportional. If the situation doesnt deal with power, but with resistance, then ofc according to ohm's law I would be directly proportional to V. Tricky stuff there!
Thanks for asking that question buddy, if you have any further problems dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: jamonwindeyer on March 20, 2016, 03:28:23 am
Hey guys!
In induction cooktops, you know how the resistance in the base of the saucepan produces the heat required to heat the food? I was wondering which equation this was from? Is it the P=I2R? So in this case, would the induced eddy currents also assist in producing heat? If this was the case, wouldn't having low resistance be better since it produces higher eddy currents and since the heat produced is proportional to current squared, it would have more effect?
Also, in transformers, in the power equation P=IV, current and voltage are essentially inversely proportional however in ohm's law V=IR, they are directly proportional so when do we use which relationship? Thanks :D
Neutron
Hey Neutron! HPL has you covered well for both of those questions, we had a similar question on that high vs low resistance argument earlier in the thread. My thinking (copied from earlier):
The effect we are discussing here is called Joule Heating (or Resistive Heating), and essentially, the formula for the heating of an object is:
This heating is, as suggested by the formula, caused primarily by the passage of an electrical current through resistance. A phenomenon known as hysteresis also plays a role, but most engineers argue that it isn't as important as Joule heating.
So, this formula suggests that, given a constant current, that increasing the resistance will increase the heating effect, and this is true. However, it is important to remember that our resistance impacts the size of the induced eddy currents.
Let me explain. The size of the induced EMF/voltage in the pot base is proportional to the rate of change of magnetic flux. This is Faraday's Law. Let's assume we have a constant voltage. We know that:
So, given our voltage is constant, if we want a higher current, we need a smaller resistance. Now, going back to the heating formula above, doubling the current will cause the heating effect to be quadrupled (since the current is squared), whereas doubling the resistance only results in the heating effect to be doubled.
Based on this interpretation, it is definitely more effective to have a lower resistance, and thus, larger eddy currents in the pot base, to achieve the maximum heating effect. This becomes more obvious if we use ohms law to re-write the heat dissipation law:
So, in fact, having a lower resistance means more current will flow, and thus, more heat will be generated than it would be by having a high resistance. This is why a low valued resistor, plugged into a huge voltage, blows up (for lack of a better term).
This is a very simplified model and by no means totally accurate, but it is an interesting way to look at the problem.
Now, this does not get rid of the idea that high resistance = more heat. It does. But we need a sizeable voltage drop across the said resistance for it to matter. This is quite a complex topic, definitely not worth investigating in depth. In actuality, there are many things at play here.
Just a different way of looking at the question. At its core, I am saying the same as Happy Physics Land. If the resistance goes too high, no current flows, and so heating does not occur. This is just a more detailed/mathematical explanation of that intuition ;D
Post by: polpark on March 22, 2016, 02:01:43 pm
do i asssume all circuts in the hsc are ohmic??
Post by: polpark on March 22, 2016, 02:10:13 pm
for example compare the torque created between motor 1 and motor 2?
would you show a ratio?
Post by: Happy Physics Land on March 22, 2016, 04:27:16 pm
hello !!!
do i asssume all circuts in the hsc are ohmic??
Hey Polpark!
Yes indeed! In HSC, if they ask you questions about electric circuits, they must all obey ohm's law, or equations that are derived from Ohm's law (E.g. P=IV, but according to ohm's law we can also express it in the form of P = I2R). After all, ohm's law is all that we have been taught!
Best Regards
Happy Physics Land
Post by: RuiAce on March 22, 2016, 04:45:31 pm
Hey Polpark!
Yes indeed! In HSC, if they ask you questions about electric circuits, they must all obey ohm's law, or equations that are derived from Ohm's law (E.g. P=IV, but according to ohm's law we can also express it in the form of P = I2R). After all, ohm's law is all that we have been taught!
Best Regards
Happy Physics Land
Just be careful there. P=RI2 is specifically for power loss.
hello !!!
do i asssume all circuts in the hsc are ohmic??
Now. In terms of long distance power generation we assume that power is constant. In P=VI, we effectively assume that voltage and current are what vary. (Naturally, this is the whole fundamental basis of the AC current transistor)
But keep in mind that this MAY mean resistance does change as well. So I wouldn't say all conductors are, ohmic conductors. Ohmic conductors assumes that the resistance is constant whilst we change the voltage, whereas non-ohmic assumes that it is variable.
Typically, HSC questions that require V=IR are all related to power generation now. For these questions, generally you simply cannot be wrong when you assume POWER is the constant.
also how would you answer compare questions?
for example compare the torque created between motor 1 and motor 2?
would you show a ratio?
This is completely ambiguous. Present a past question for us to comment on it.
Post by: jamonwindeyer on March 22, 2016, 08:02:34 pm
This is completely ambiguous. Present a past question for us to comment on it.
I think what Rui meant to say is that we're not 100% sure what you mean, but if you could give us some more details we're super eager to help you out as quick as we can! ;D
Post by: Neutron on March 23, 2016, 03:11:42 pm
Okay okay this is a sort of long winded question but I feel like I haven't grasped transmission properly.. I was wondering whether you could explain what the answer is for each one and why it is? D;
Which generator (AC or DC) is:
a) More reliable
b) Suffers less wear on critical parts
c) Can be transmitted over long distances with less loss of energy
d) Is easier to connect to houses
e) Is more likely to have shorting
f) Can have cables with less insulation
g) Produces less electrical interference
h) Is likely to experience a higher impedance from coiled cables
Thank you so much omg you guys are lifesavers xx
Neutron
Post by: Happy Physics Land on March 23, 2016, 05:32:39 pm
Hey guys!
Okay okay this is a sort of long winded question but I feel like I haven't grasped transmission properly.. I was wondering whether you could explain what the answer is for each one and why it is? D;
Which generator (AC or DC) is:
a) More reliable
b) Suffers less wear on critical parts
c) Can be transmitted over long distances with less loss of energy
d) Is easier to connect to houses
e) Is more likely to have shorting
f) Can have cables with less insulation
g) Produces less electrical interference
h) Is likely to experience a higher impedance from coiled cable
Thank you so much omg you guys are lifesavers xx
Neutron
Hey Neutron!
a)More reliable: DC, because commutator is in contact with carbon brushes, ensuring a more consistent flow of electricity
b) Suffers less wear on critical parts: AC, because slip rings are not directly in contact with carbon brushes, meaning that they dont suffer as much wear as commutators which will eventually wear out due to physical contact with brushes
c) Can be transmitted over long distances with less loss of energy: AC, because AC generators produce AC which are able to be stepped-up or stepped-down. Through stepping up AC to a very high voltage (550000V transmission voltage in Australia), the amount of currents being delivered has been reduced (Voltage and current are inversely proportional), hence less power loss because less heat dissipation.
d) Is easier to connect to houses: AC, because many appliances at homes require the standard 240V AC to operate
e) Is more likely to have shorting: AC (Im not too sure of the reason, perhaps when others come around they can provide you with the explanation?)
f) Can have cables with less insulation: DC, AC is a lot of dangerous than DC because it has the ability to cause fibrillation and can fatally propagate through human body. Hence theres more necessity to insulate AC than DC.
g) Produces less electrical interference: DC, because AC travels in a sinusoidal wave, hence its rising and falling propagation pattern makes it more likely to interfere with other waves. In contrast, DC currents only travel in straight lines, hence it is unlikely for it to come in contact and interfere with other waves.
h) Is likely to experience a higher impedance from coiled cables: DC, this is because DC travels only in a straight line, so if the cable is in the shape of a coil, DC would encounter a lot more resistance. Imagine yourself trying to run around in a circular maze, if you run in straight lines, you would frequently collide into walls and then you would turn and run straight in another direction, but again you would collide into another wall. Similarly, this is why DC would experience a lot more impedance as it travels through the coil. AC travels in a sine curve that is easier to propagate through the coil and because it has higher frequencies, it would experience less impedance
I hope these explanations all make sense to you, the technicality in my reasoning may not be very strong but Im sure the answers are correct.
Best Regards
Happy Physics Land
Post by: Meckenza on March 23, 2016, 07:44:14 pm
I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?
Post by: Happy Physics Land on March 23, 2016, 09:50:43 pm
Hii,
I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?
Hey Meckenza!
Very nice question there! I completely agree with your reasoning and you should also be satisfied that you understand the main concern of this question, well done!
Like you have correctly stated, as speed increases, back emf increases and this opposes the supply current, hence causing the current to decrease. Now let's analyse what happens physically. Initially, at speed = 0, the DC motor is first started. At this time, the coil is stationary so the back emf is zero. Therefore the current flowing through the coil is high and equal to supply current. Once the armature rotates faster, the back EMF increases (This explains the relatively sharp drop in current at the start) and the difference between constant supply EMF and back EMF becomes smaller. However, because back EMF can never exceed the value of supply emf, there has to be a point where supply current pretty much stops decreasing because the impact of back emf upon it becomes smaller and smaller. This explains that the current is decreasing at a decreasing rate. Hence you have that curve that looks like as if its reaching a horizontal asymptote (you can think about it this way if you are maths-orientated).
Hope that helped! If you have further questions please dont hesitate to ask! :D
Best Regards
Happy Physics Land
Post by: jamonwindeyer on March 23, 2016, 09:52:07 pm
Hii,
I was looking at a multiple choice question that plotted Current (y-axis) vs speed (x-axis) of an electric motor, but the graph appeared to have a curved shape that approached zero as the speed increased. I understand that A) would be the most correct because as speed increases so does back emf which causes a decrease in current, but would someone be able to explain the shape of the curve please?
Hey Meckenza! Ah, Back EMF, this concept drove me nuts in the HSC aha ;)
So you've correctly said that A makes the most sense. I'll try to explain the curved shape as best I can, though it actually isn't something I know about for sure!
Back EMF is caused by the change in magnetic flux experienced by the coil in a motor as it spins. As it rotates, the coil moves through different parts of the field, and this changing flux induces a current. By Lenz's Law, this current acts against the supply. It sounds like you understand this pretty well, but let me know if you need a little more!
To answer your question on the shape of the curve, it probably makes sense to first consider the alternative. What if we had a straight line relationship? That would imply that we could push the speed of a motor high enough that no current flows. As speed increased, eventually the current would drop to zero. That can't be possible, since that would mean the motor would not be spinning! So a straight line relationship definitely doesn't work.
In actuality, Back EMF only increases until it reaches a point very similar to the supply EMF. This is the horizontal part of the graph you are seeing!
What we have here is an inversely proportional relationship. As speed increases, current decreases at a proportional rate, since Back EMF is proportional to speed. This creates a hyperbolic graph, roughly represented by:
If you graph a hyperbola, it has a shape similar to what you are showing in that question.
Now I am sure it is more complex than this, but this explanation gives you the idea of why the line is curved. I hope it helps! If you wanted a more concrete, detailed explanation I can definitely do some research for you, I am sure I have some more detailed explanations in my Electrical Engineering textbooks ;D
Post by: jamonwindeyer on March 23, 2016, 09:52:49 pm
Hey Meckenza!
Very nice question there! I completely agree with your reasoning and you should also be satisfied that you understand the main concern of this question, well done!
Like you have correctly stated, as speed increases, back emf increases and this opposes the supply current, hence causing the current to decrease. Now let's analyse what happens physically. Initially, at speed = 0, the DC motor is first started. At this time, the coil is stationary so the back emf is zero. Therefore the current flowing through the coil is high and equal to supply current. Once the armature rotates faster, the back EMF increases (This explains the relatively sharp drop in current at the start) and the difference between constant supply EMF and back EMF becomes smaller. However, because back EMF can never exceed the value of supply emf, there has to be a point where supply current pretty much stops decreasing because the impact of back emf upon it becomes smaller and smaller. This explains that the current is decreasing at a decreasing rate. Hence you have that curve that looks like as if its reaching a horizontal asymptote (you can think about it this way if you are maths-orientated).
Hope that helped! If you have further questions please dont hesitate to ask! :D
Best Regards
Happy Physics Land
Teamwork man ;)
Post by: Happy Physics Land on March 23, 2016, 10:05:46 pm
Teamwork man ;)
You and your engineering swag have illuminated my understanding of physics Jamon! :L
Never realised you can use hyperbola here, maybe this graph A is a conic section? 8)
Post by: Meckenza on March 23, 2016, 10:10:30 pm
Hey Meckenza! Ah, Back EMF, this concept drove me nuts in the HSC aha ;)
So you've correctly said that A makes the most sense. I'll try to explain the curved shape as best I can, though it actually isn't something I know about for sure!
Back EMF is caused by the change in magnetic flux experienced by the coil in a motor as it spins. As it rotates, the coil moves through different parts of the field, and this changing flux induces a current. By Lenz's Law, this current acts against the supply. It sounds like you understand this pretty well, but let me know if you need a little more!
To answer your question on the shape of the curve, it probably makes sense to first consider the alternative. What if we had a straight line relationship? That would imply that we could push the speed of a motor high enough that no current flows. As speed increased, eventually the current would drop to zero. That can't be possible, since that would mean the motor would not be spinning! So a straight line relationship definitely doesn't work.
In actuality, Back EMF only increases until it reaches a point very similar to the supply EMF. This is the horizontal part of the graph you are seeing!
What we have here is an inversely proportional relationship. As speed increases, current decreases at a proportional rate, since Back EMF is proportional to speed. This creates a hyperbolic graph, roughly represented by:
If you graph a hyperbola, it has a shape similar to what you are showing in that question.
Now I am sure it is more complex than this, but this explanation gives you the idea of why the line is curved. I hope it helps! If you wanted a more concrete, detailed explanation I can definitely do some research for you, I am sure I have some more detailed explanations in my Electrical Engineering textbooks ;D
Wait, so how did you derive that current is proportional to 1/speed?
Post by: Happy Physics Land on March 23, 2016, 10:48:46 pm
Wait, so how did you derive that current is proportional to 1/speed?
Hey Meckenze!
I will answer this question for jamon (If he doesnt mind). Essentially that expression "supply current is proportional to 1/speed of motor" is the same as "supply current is inversely proportional to the speed of motor". Like what you have correctly stated, supply current is inversely proportional to back emf. But because speed is proportional to back emf, then it follows that current is also inversely proportional to speed.
Post by: Meckenza on March 24, 2016, 06:49:29 am
Post by: jamonwindeyer on March 24, 2016, 08:01:22 am
Ahhhk, ok. Thanks Happy Physics Land and Jamonwindeyer!!! ;D
You are most welcome! Yeah HPL is on the answer, it is basically just the fact that as Speed goes up, current goes down, an inversely proportional relationship (if you take it to the simplest level). Sorry for being a little ambiguous! ;D
Hey Meckenze!
I will answer this question for jamon (If he doesnt mind). Essentially that expression "supply current is proportional to 1/speed of motor" is the same as "supply current is inversely proportional to the speed of motor". Like what you have correctly stated, supply current is inversely proportional to back emf. But because speed is proportional to back emf, then it follows that current is also inversely proportional to speed.
You can answer any question at any time HPL! ;)
And with regard to your comment, it most definitely could be, it is most definitely more complex than the relationship in the previous message. I might even do some reading and see if I can find a proper equation! I've tried a couple of derivations but can't come up with anything that would suggest the curved shape mathematically just yet, I'll ask around the brainiac electrical engineers at uni ;)
Post by: Happy Physics Land on March 24, 2016, 07:30:26 pm
You are most welcome! Yeah HPL is on the answer, it is basically just the fact that as Speed goes up, current goes down, an inversely proportional relationship (if you take it to the simplest level). Sorry for being a little ambiguous! ;D
You can answer any question at any time HPL! ;)
And with regard to your comment, it most definitely could be, it is most definitely more complex than the relationship in the previous message. I might even do some reading and see if I can find a proper equation! I've tried a couple of derivations but can't come up with anything that would suggest the curved shape mathematically just yet, I'll ask around the brainiac electrical engineers at uni ;)
Engineering god can I please have a hand with a mechanics question??? :D
The school bus with a total mass of 10 tonnes travels at a constant velocity of 40km/h down a long hill with a slope of 15 degrees. Determine the braking force required to stop the bus within 50 metres from when the driver applies the brakes.
Thank you so much in advance! :)))
Post by: Neutron on March 24, 2016, 10:09:49 pm
Hey Neutron!
a)More reliable: DC, because commutator is in contact with carbon brushes, ensuring a more consistent flow of electricity
b) Suffers less wear on critical parts: AC, because slip rings are not directly in contact with carbon brushes, meaning that they dont suffer as much wear as commutators which will eventually wear out due to physical contact with brushes
c) Can be transmitted over long distances with less loss of energy: AC, because AC generators produce AC which are able to be stepped-up or stepped-down. Through stepping up AC to a very high voltage (550000V transmission voltage in Australia), the amount of currents being delivered has been reduced (Voltage and current are inversely proportional), hence less power loss because less heat dissipation.
d) Is easier to connect to houses: AC, because many appliances at homes require the standard 240V AC to operate
e) Is more likely to have shorting: AC (Im not too sure of the reason, perhaps when others come around they can provide you with the explanation?)
f) Can have cables with less insulation: DC, AC is a lot of dangerous than DC because it has the ability to cause fibrillation and can fatally propagate through human body. Hence theres more necessity to insulate AC than DC.
g) Produces less electrical interference: DC, because AC travels in a sinusoidal wave, hence its rising and falling propagation pattern makes it more likely to interfere with other waves. In contrast, DC currents only travel in straight lines, hence it is unlikely for it to come in contact and interfere with other waves.
h) Is likely to experience a higher impedance from coiled cables: DC, this is because DC travels only in a straight line, so if the cable is in the shape of a coil, DC would encounter a lot more resistance. Imagine yourself trying to run around in a circular maze, if you run in straight lines, you would frequently collide into walls and then you would turn and run straight in another direction, but again you would collide into another wall. Similarly, this is why DC would experience a lot more impedance as it travels through the coil. AC travels in a sine curve that is easier to propagate through the coil and because it has higher frequencies, it would experience less impedance
I hope these explanations all make sense to you, the technicality in my reasoning may not be very strong but Im sure the answers are correct.
Best Regards
Happy Physics Land
legend <3
Post by: jamonwindeyer on March 25, 2016, 12:12:44 am
Engineering god can I please have a hand with a mechanics question??? :D
The school bus with a total mass of 10 tonnes travels at a constant velocity of 40km/h down a long hill with a slope of 15 degrees. Determine the braking force required to stop the bus within 50 metres from when the driver applies the brakes.
Thank you so much in advance! :)))
Definitely not an Engineering God, but totally man! Let's have a look.
Let's start by converting everything to SI units.
Okay, so there is a few things to consider here. I think we should start by determining the acceleration required to slow the bus in that distance!
We have a formula linking velocity, acceleration, and distance:
Now this is directed along the slope the bus is travelling on.
Now, I am actually a little confused as to how the slope comes in to this question. At first I thought we'd have to account for the weight force of the bus, but that force is acting when the bus is travelling at the constant speed as well. Therefore, we have already accounted for it in the acceleration calculation. So, I have a funny feeling it is there as a trick, do you think I'd be right there?
Let's proceed assuming the simple case, that we now just need to calculate the force required to generate the above acceleration. We need to create the acceleration above, we have a mass, this is a job for Newton:
So this is my response for now, I hope it helps either as a solution or to get you started with one! I'm curious what you think about the slope thing (sorry if I explained it poorly)? My brain is telling me that we can ignore it! ;D the other option is we need to counteract weight force of the bus, which would be a matter of calculating the weight force, finding the component of that force which is directed down the slope, and then adding that value to the answer. I can go into more detail if you need, but yeah, super keen to hear your thoughts!
Post by: jamonwindeyer on March 25, 2016, 12:22:23 am
Definitely not an Engineering God, but totally man! Let's have a look.
Let's start by converting everything to SI units.
Okay, so there is a few things to consider here. I think we should start by determining the acceleration required to slow the bus in that distance!
We have a formula linking velocity, acceleration, and distance:
Now this is directed along the slope the bus is travelling on.
Now, I am actually a little confused as to how the slope comes in to this question. At first I thought we'd have to account for the weight force of the bus, but that force is acting when the bus is travelling at the constant speed as well. Therefore, we have already accounted for it in the acceleration calculation. So, I have a funny feeling it is there as a trick, do you think I'd be right there?
Let's proceed assuming the simple case, that we now just need to calculate the force required to generate the above acceleration. We need to create the acceleration above, we have a mass, this is a job for Newton:
So this is my response for now, I hope it helps either as a solution or to get you started with one! I'm curious what you think about the slope thing (sorry if I explained it poorly)? My brain is telling me that we can ignore it! ;D the other option is we need to counteract weight force of the bus, which would be a matter of calculating the weight force, finding the component of that force which is directed down the slope, and then adding that value to the answer. I can go into more detail if you need, but yeah, super keen to hear your thoughts!
Actually, yep, changed my mind, I think we need to account for weight force here. Sorry! ;D
So the acceleration bit is the same as above! Now, not only do we need to produce this acceleration against the bus, we also need to produce a force to counteract the weight force of the bus, which is pulling it down the hill! How do we do this?
Well, the bus is on a slant. It's weight force acts directly downwards, but we only want the weight force directed parallel to the slope of the hill. We use vector components here and we get:
So instead, our braking force is:
I'm rolling with this now, sorry HPL! It has been a while since I have done mechanics, does this help you out at all?
Post by: Happy Physics Land on March 25, 2016, 01:31:40 am
Actually, yep, changed my mind, I think we need to account for weight force here. Sorry! ;D
So the acceleration bit is the same as above! Now, not only do we need to produce this acceleration against the bus, we also need to produce a force to counteract the weight force of the bus, which is pulling it down the hill! How do we do this?
Well, the bus is on a slant. It's weight force acts directly downwards, but we only want the weight force directed parallel to the slope of the hill. We use vector components here and we get:
So instead, our braking force is:
I'm rolling with this now, sorry HPL! It has been a while since I have done mechanics, does this help you out at all?
I understood everything you said Jamon, thank you so much!!!! It was one of those mechanics questions that was really confusing (came from a CSSA paper so no surprise there aye hehe). Initially l thought the angle didnt matter and I tried to do kinetic energy = work. Because the bus is coming down the slope with 40km/h, so we can calculate its kinetic energy Ek = 1/2 mv2. And I saw that 50m, so I thought this probably has something to do with the amount of work (energy) to stop this bus (because work = Fs). Im not sure if you would approve of my method but yes I definitely benefitted from your solution! Thank you again Jamon you are my saviour! :D
Post by: wils013 on March 25, 2016, 11:11:20 am
How does it work that End X is negative?
Post by: jamonwindeyer on March 25, 2016, 12:03:59 pm
I understood everything you said Jamon, thank you so much!!!! It was one of those mechanics questions that was really confusing (came from a CSSA paper so no surprise there aye hehe). Initially l thought the angle didnt matter and I tried to do kinetic energy = work. Because the bus is coming down the slope with 40km/h, so we can calculate its kinetic energy Ek = 1/2 mv2. And I saw that 50m, so I thought this probably has something to do with the amount of work (energy) to stop this bus (because work = Fs). Im not sure if you would approve of my method but yes I definitely benefitted from your solution! Thank you again Jamon you are my saviour! :D
I like that! Cool idea! Never even considered it that way. For anyone looking at this, HPL means:
Now this is a different solution than I got, but I realised I made a mistake calculating the component of weight force. Just the numbers were wrong. The actual answer I got was:
So, I would say we are in agreement! Both methods work very nicely! The difference is probably rounding errors, I rounded harshly when I calculated acceleration and such last night ;)
Great thinking HPL! I would never have thought to do it that way ;D
Post by: jamonwindeyer on March 25, 2016, 12:10:39 pm
http://imgur.com/YYMQBlW
How does it work that End X is negative?
Hey Wils013! This is actually a pretty obscure question concerning forces on charges! At least, that's the way I see it ;D
We know that moving charges inside a magnetic field experience a force, let's see what that force does.
So, let's consider individual charges inside that rod. The rod is moved up through the magnetic field. So the direction of motion is upwards. Now, the magnetic field lines move from North to South between the magnets. So they go to the right of the page, if you like.
Using these two directions, you should apply something like the Right Hand Slap Rule, but every school does something a little bit different. Whatever method you were taught, use it, and you'll find that the resultant force points towards End Y of the rod.
However! This represents the direction of the force on a positive particle. The force on a negative particle will be in the opposite direction! So, while the positive charges experience a force in the direction of End Y, the negative charges experience a force towards End X. Thus, all the negative charges accumulate at End X, and it becomes negatively charged!
I hope this helps! Let us know if you need a picture of that directional reasoning, it is a little hard to explain, but give it a go! ;D
Post by: wils013 on March 25, 2016, 12:51:13 pm
Hey Wils013! This is actually a pretty obscure question concerning forces on charges! At least, that's the way I see it ;D
We know that moving charges inside a magnetic field experience a force, let's see what that force does.
So, let's consider individual charges inside that rod. The rod is moved up through the magnetic field. So the direction of motion is upwards. Now, the magnetic field lines move from North to South between the magnets. So they go to the right of the page, if you like.
Using these two directions, you should apply something like the Right Hand Slap Rule, but every school does something a little bit different. Whatever method you were taught, use it, and you'll find that the resultant force points towards End Y of the rod.
However! This represents the direction of the force on a positive particle. The force on a negative particle will be in the opposite direction! So, while the positive charges experience a force in the direction of End Y, the negative charges experience a force towards End X. Thus, all the negative charges accumulate at End X, and it becomes negatively charged!
I hope this helps! Let us know if you need a picture of that directional reasoning, it is a little hard to explain, but give it a go! ;D
Yes! Thank you so much I didn't recognise to use the hand rule, that makes so much sense thanks!
Post by: Happy Physics Land on March 25, 2016, 01:11:25 pm
I like that! Cool idea! Never even considered it that way. For anyone looking at this, HPL means:
Now this is a different solution than I got, but I realised I made a mistake calculating the component of weight force. Just the numbers were wrong. The actual answer I got was:
So, I would say we are in agreement! Both methods work very nicely! The difference is probably rounding errors, I rounded harshly when I calculated acceleration and such last night ;)
Great thinking HPL! I would never have thought to do it that way ;D
Wait Jamon do you mean mgsin(15) + ma??? But yeah thanks for approving my solution!!! It's always good to receive approval from you! :D
Post by: jamonwindeyer on March 25, 2016, 01:59:25 pm
Wait Jamon do you mean mgsin(15) + ma??? But yeah thanks for approving my solution!!! It's always good to receive approval from you! :D
Actually no! If you consider the weight force versus the braking force, they are actually acting in opposite directions! Hence, subtraction. It is probably a little easier to see with a diagram, and in the following form ;D
Post by: Happy Physics Land on March 25, 2016, 03:59:24 pm
Actually no! If you consider the weight force versus the braking force, they are actually acting in opposite directions! Hence, subtraction. It is probably a little easier to see with a diagram, and in the following form ;D
AH YES! i SEE I SEE!!! Everything's clear now! Thanks Jamon! :) Honestly they just shove so much hard mechanics questions in past papers but even then engineering still has such a low scaling :3
Post by: jamonwindeyer on March 25, 2016, 11:54:32 pm
AH YES! i SEE I SEE!!! Everything's clear now! Thanks Jamon! :) Honestly they just shove so much hard mechanics questions in past papers but even then engineering still has such a low scaling :3
Really? I've always been of the thinking that Engineering scaled pretty well!
It's a foreign concept for me, my school has never run Engineering Studies, I didn't even know the subject existed until about halfway through Year 11! :o
Post by: Happy Physics Land on March 25, 2016, 11:55:12 pm
Really? I've always been of the thinking that Engineering scaled pretty well!
It's a foreign concept for me, my school has never run Engineering Studies, I didn't even know the subject existed until about halfway through Year 11! :o
You can always repeat year 11 you know?
Post by: jamonwindeyer on March 26, 2016, 12:02:17 am
You can always repeat year 11 you know?
Hmm, it is tempting, there are times sitting in some of my lectures that I do miss the preliminary content, that's for sure. But I'll have to pass for now ;)
Post by: Neutron on March 26, 2016, 10:30:10 am
I was wondering if you guys could explain how to determine which time is the dilated time and which time is the proper time (Einstein's special theory of relativity) when given questions TT^TT So for like this one:
"A spaceship travelling through space experiences a 90% time dilation. How fast is it travelling?"
I thought the dilated time was 0.9 and proper time was 2, but then you'd eventually have to square root a negative so that doesn't work out :/
And for this one
"The high-speed muons produced for an experiment by the Fermilab accelerator are measured to have a lifetime of 5.0 microseconds. When these muons are brought to rest, their lifetime is measured to be 2.2 microseconds"
I thought you could look from the perspective of the muon (since relative motion, doesn't matter whether you look from muon or lab) so you'd have proper time to 5 microseconds and dilated time to be 2.2 but apparently not :/ that too gives you a square root of a negative number. But the other way doesn't make sense to me, if proper time is 2.2 seconds and dilated time is 5 seconds, won't that mean in 2.2 seconds, 5 seconds passed for the muon? Meaning each second is faster idk
So basically I was wondering how you tell which one is proper and which one is dilated :'(
Love y'all,
Neutron
Post by: jamonwindeyer on March 26, 2016, 03:52:31 pm
Hello, it's me
I was wondering if after all these years you'd like to meet ;)
Quote
I was wondering if you guys could explain how to determine which time is the dilated time and which time is the proper time (Einstein's special theory of relativity) when given questions TT^TT So for like this one:
"A spaceship travelling through space experiences a 90% time dilation. How fast is it travelling?"
I thought the dilated time was 0.9 and proper time was 2, but then you'd eventually have to square root a negative so that doesn't work out :/
And for this one
"The high-speed muons produced for an experiment by the Fermilab accelerator are measured to have a lifetime of 5.0 microseconds. When these muons are brought to rest, their lifetime is measured to be 2.2 microseconds"
I thought you could look from the perspective of the muon (since relative motion, doesn't matter whether you look from muon or lab) so you'd have proper time to 5 microseconds and dilated time to be 2.2 but apparently not :/ that too gives you a square root of a negative number. But the other way doesn't make sense to me, if proper time is 2.2 seconds and dilated time is 5 seconds, won't that mean in 2.2 seconds, 5 seconds passed for the muon? Meaning each second is faster idk
So basically I was wondering how you tell which one is proper and which one is dilated :'(
Love y'all,
Neutron
Hey Neutron! This confused the heck out of me in Year 12 as well! I'll start by explaining the two examples you provided, then give you the trick to make sure you have everything in the right direction ;D
So, your first example. I would interpret "90% time dilation" as time passes by at 10% of the normal rate. That is, the dilated time is 10 times greater than the undiluted time. A spaceship travelling extremely quickly will have time pass by much slower, as measured from earth. I could be misinterpreting slightly, but in any case, we have some kind of effect on time passage.
Remembering our time dilation formula, I'll refer to the time term on the LHS as dilated time, and the RHS as unaffected time.
In this case, the dilated time is 10 times greater than the actual, unaffected time. Therefore:
This one is a little hard to understand, the next one is a bit easier.
In your second example, 5.0 microseconds is the dilated time. This is what we measure when the particle is at speed. It moving affects our time measurements. Then, the 2.2 microseconds is the unaffected time. Your statement about "if proper time is 2.2 seconds and dilated time is 5 seconds, won't that mean in 2.2 seconds, 5 seconds passed for the muon", is correct! What the muon would experience in 5 seconds only takes us 2.2 seconds to experience. The issue with your approach is that the time measurements are made from the earths frame of reference, not the muons. So you have to go from the earth as unaffected time. That is what had you a bit backwards.
If you are at all confused, your check is simple. Dilated time must ALWAYS be larger than unaffected time. Time dilation causes time to pass by slower, so, the dilated time difference is larger. This is an easy way to check your reasoning against the question.
I hope this helps! This whole thing is a little confusing, your dilated time represents the time as measured by a stationary observer. It will thus always be larger ;D
Post by: Loki98 on March 26, 2016, 10:50:01 pm
Astronauts on a long space journey are playing golf inside their spaceship, which is travelling away from the Earth with speed 0.6c. One of the astronauts hits a drive exactly along the length of the spaceship (in its direction of travel) at speed 0.1c in the frame of the spaceship. What is the speed of the golf ball as observed from Earth?
Post by: jakesilove on March 26, 2016, 11:19:11 pm
Need help with this question,
Astronauts on a long space journey are playing golf inside their spaceship, which is travelling away from the Earth with speed 0.6c. One of the astronauts hits a drive exactly along the length of the spaceship (in its direction of travel) at speed 0.1c in the frame of the spaceship. What is the speed of the golf ball as observed from Earth?
Hey Loki!
I'm not sure whether you can get a question like this in the HSC, as it uses the relativistic addition of velocities (Lorentz transformations) which falls outside of the course. Are you working through special relativity from another state? The required formula is
Subbing in u' as 0.1c and v as 0.6c, you get the required answer of 0.66c!
Hope this helps :)
Jake
Post by: Happy Physics Land on March 26, 2016, 11:25:14 pm
Hey Loki!
I'm not sure whether you can get a question like this in the HSC, as it uses the relativistic addition of velocities (Lorentz transformations) which falls outside of the course. Are you working through special relativity from another state? The required formula is
Subbing in u' as 0.1c and v as 0.6c, you get the required answer of 0.66c!
Hope this helps :)
Jake
That was actually what I was thinking too, I remember seeing the question from a james ruse paper a while ago. I agree that the only way to solve it is probably via relativistic addition of velocities. The only other way I can think of using hsc methods is probably to work out a contracted length and a dilated time and then use the contracted length to divide by the dilated time to obtain a resultant velocity? I havent exactly experimented with it and I think its better to stick with relativistic addition...
Post by: Maz on March 27, 2016, 12:54:19 pm
i was wandering if someone could please explain the difference between the strong force and the strong nuclear force?
thakyou so much :)
Post by: ATWalk on March 27, 2016, 03:03:27 pm
I'm having trouble with this question on forces during a rocket launch because I basically had to teach this part of the syllabus to myself. The question is: A 20 000 kg rocket exhausts gases at a constant 700kg per second at 300m/s. (a) Calculate the momentum of the exhaust gases.
Do I just substitute 700kg and 300m/s into p=mv? Or have I overlooked/misunderstood something?
Thank you in advance for the help. I really appreciate it.
Post by: Happy Physics Land on March 27, 2016, 04:11:30 pm
hey
i was wandering if someone could please explain the difference between the strong force and the strong nuclear force?
thakyou so much :)
Hey mq123!
Ok Im not too familiar with your syllabus, so I will just explain my perception of these two concepts. Strong force can be defined in two ways. One way to describe it is a push, pull or twist that exceeds or nearly exceeds the limit which a body/object can undertake. In quantum and quarks, strong force binds quarks together in clusters to make more-familiar subatomic particles, such as protons and neutrons. It also holds together the atomic nucleus and is essential to interactions between all particles containing quarks. Strong nuclear force is a strong interaction between the quarks that make up protons and neutrons.
Hope it helped! :)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on March 27, 2016, 04:17:41 pm
Hi,
I'm having trouble with this question on forces during a rocket launch because I basically had to teach this part of the syllabus to myself. The question is: A 20 000 kg rocket exhausts gases at a constant 700kg per second at 300m/s. (a) Calculate the momentum of the exhaust gases.
Do I just substitute 700kg and 300m/s into p=mv? Or have I overlooked/misunderstood something?
Thank you in advance for the help. I really appreciate it.
Hey ATWalk!
I think what makes this question hard is that they expressed the mass of the gas as a rate, not simply 700kg. Personally I would say you can just sub in 700kg and 300m/s into p = mv. What I found interesting is, if you times 700kg/s by 300m/s, you actually get a value in kgms-2, which is the unit for force! For now I agree with your solution, but if Jake or Jamon happens to come around later on, they can probably correct me if l happen to be wrong.
Best Regards
Happy Physics Land
Post by: jakesilove on March 27, 2016, 04:30:23 pm
Hi,
I'm having trouble with this question on forces during a rocket launch because I basically had to teach this part of the syllabus to myself. The question is: A 20 000 kg rocket exhausts gases at a constant 700kg per second at 300m/s. (a) Calculate the momentum of the exhaust gases.
Do I just substitute 700kg and 300m/s into p=mv? Or have I overlooked/misunderstood something?
Thank you in advance for the help. I really appreciate it.
Hey!
I agree with HPL's answer: you just need to sub in the values to the momentum formula, and get a momentum per second! In regards to your 'units' query, that's a really brilliant thing to bring up. Momentum is calculated in kgm/s, or Newton seconds. By adding in an extra 'per second', momentum becomes force! So the answer to your question itself would be "a momentum of x kgm/s, per second" or, more accurately, "x Newtons".
Again, this is definitely over complicating things but making sure to take into account the "700kg per second" thing is definitely a good idea.
Jake
Post by: Happy Physics Land on March 28, 2016, 01:29:48 pm
I always tend to be stuck with this kind of question, would you like to lend me a hand? Thank you very much!!! (The answer is D)
(http://i.imgur.com/dWvfBF0.png)
Best Regards
Happy Physics Land
Post by: jakesilove on March 28, 2016, 02:14:32 pm
Hey Jake!
I always tend to be stuck with this kind of question, would you like to lend me a hand? Thank you very much!!! (The answer is D)
(http://i.imgur.com/dWvfBF0.png)
Best Regards
Happy Physics Land
Hey HPL!
Glad to be able to help out! I didn't know that coefficients of static friction etc. were even in the HSC course! Are you sure you're looking at HSC past papers? If not, good on you for taking a look at some difficult Physics outside of the curriculum.
In terms of the actual answer, I've checked my working a few times and I can't seem to find an error! I get the answer to be D, but that is only logically (the Force required to keep the block STABLE is greater than C, therefore logically to push the block UP the slope the force required must also be greater than C, leaving D as the only possible answer). However, I would have expected the number D to just pop out of my maths somewhere, which it didn't.
Take a look at my answer, and maybe see if you can spot any errors or problems I missed!
(http://i.imgur.com/YS2YXzP.jpg?1)
Jake
Post by: Happy Physics Land on March 28, 2016, 03:00:20 pm
Hey HPL!
Glad to be able to help out! I didn't know that coefficients of static friction etc. were even in the HSC course! Are you sure you're looking at HSC past papers? If not, good on you for taking a look at some difficult Physics outside of the curriculum.
In terms of the actual answer, I've checked my working a few times and I can't seem to find an error! I get the answer to be D, but that is only logically (the Force required to keep the block STABLE is greater than C, therefore logically to push the block UP the slope the force required must also be greater than C, leaving D as the only possible answer). However, I would have expected the number D to just pop out of my maths somewhere, which it didn't.
Take a look at my answer, and maybe see if you can spot any errors or problems I missed!
(http://i.imgur.com/YS2YXzP.jpg?1)
Jake
Yeah hehe its not a physics question. Its actually from an engineering paper but I thought its physics related anyways. I eventually did 30 minutes of research and managed to figure it out through breaking up force P and break up force W=mg into horizontal and vertical components. But yeah I had the same diagram as you, what I initially did wrong was the fact that I mixed up the direction of my friction force and forgot to include the vertical component of P. Sorry for causing such a hussle and thanks for the help Jake!!! ^^
Post by: jakesilove on March 28, 2016, 03:26:36 pm
Yeah hehe its not a physics question. Its actually from an engineering paper but I thought its physics related anyways. I eventually did 30 minutes of research and managed to figure it out through breaking up force P and break up force W=mg into horizontal and vertical components. But yeah I had the same diagram as you, what I initially did wrong was the fact that I mixed up the direction of my friction force and forgot to include the vertical component of P. Sorry for causing such a hussle and thanks for the help Jake!!! ^^
That's very, very cool that engineering includes questions like this! I'm glad you got to the answer in the end: with questions like this, the trick is ALWAYS to draw a diagram with all forces and then resolve horizontal and vertical forces. Usually, the algebra just sort of pops out after that! Let me know if you have other questions like that one: to be honest, they're way more fun than the usual Physics-Maths questions.
Jake
Post by: Neutron on March 28, 2016, 04:20:30 pm
I kinda forgot how this works but why does the acceleration due to gravity decrease with altitude? I might've gotten mixed up with electrical potential energy increasing with altitude ah idk D: Thanks in advance, highly appreciated!
Neutron
Post by: jakesilove on March 28, 2016, 04:28:55 pm
Ahhh I have another dumb question D:
I kinda forgot how this works but why does the acceleration due to gravity decrease with altitude? I might've gotten mixed up with electrical potential energy increasing with altitude ah idk D: Thanks in advance, highly appreciated!
Neutron
Hey Neutron!!
You're absolutely right! As altitude increase, the acceleration due to gravity decreases! This is due to one very simple formula, which I would absolutely recommend you have up your sleeve:
So, as the radius (or altitude) from the centre of the planet increases, the gravitational acceleration will decrease in an inverse square relationship! Note also that M is the mass of the planet.
Hope this helps!
Jake
Post by: Loki98 on March 28, 2016, 07:55:25 pm
-Thank you
Post by: jakesilove on March 28, 2016, 08:44:58 pm
Could someone please tell me some of the questions that can be asked about Edison vs Westinghouse.
-Thank you
Hey!
Literally the only question I have ever seen asked was something along the lines of "Describe the differences of opinion between Edison and Westinghouse, with regards to AC and DC". If you have a brief history memorised, you'll be totally fine answering any question on that dot point!
Jake
Post by: ATWalk on March 28, 2016, 09:15:14 pm
Ahhh I have another dumb question D:
I kinda forgot how this works but why does the acceleration due to gravity decrease with altitude? I might've gotten mixed up with electrical potential energy increasing with altitude ah idk D: Thanks in advance, highly appreciated!
Neutron
Hey,
Just adding on to what Jake said, I thought I'd pitch in with the derivation for the equation, because I've been doing a lot of physics HSC questions lately in preparation for half yearlies and they often ask questions related to deriving equations that aren't on your data sheet based on equations that are on the sheet. You might already know this, but this is super useful so that you can remember exactly how we get:
So you know the formula
So there's the classic rule that all Fs are equal, and as they're both the exact same force of gravity acting on an object, we get:
And yeah, from there you can see exactly what Jake said about how g is inversely proportional to the square of the radius, so that's why the further away from the centre Earth you get, the less gravitational acceleration you experience. Hope this helps.
Post by: Happy Physics Land on March 28, 2016, 11:35:16 pm
Could someone please tell me some of the questions that can be asked about Edison vs Westinghouse.
-Thank you
Hey Loki:
Just suggesting some alternative possibilities in addition to Jake's. The syllabus dotpoint says "analyse secondary information on the competition between Westinghouse and Edison to supply electricity to cities". Now our directional word here is "analyse", so theoretically, HSC exams and school trial exams cannot give you questions that are more higher order (i.e.discuss/assess/evaluate) . However, unfortunately there are numerous occasions where your teacher in trials or half yearlies, or sometimes even in HSC exams, that they test you on more specific things that are beyond what the syllabus dotpoint requires you to do. Sometimes they can be even be questions up to 6 marks (I've never seen anything more than 6 marks though).
Some questions they can ask include:
- Assess the outcomes of the competition between Edison and Westinghouse.
- Evaluate how the competition between Edison and Westinghouse had changed the modern system of electricity transmission.
- Discuss the role of transformers in the competition between Edison and Westinghouse
- Evaluate the importance of transformers in helping Westinghouse's AC system to defeat Edison's DC system
That's all from me, hope they can guide you through the type of exam questions they can give you on the dotpoint!
Best Regards
Happy Physics Land
Post by: Neutron on March 30, 2016, 04:42:29 pm
Okay so my half yearlies are tomorrow (pray for me) and I still have a few questions D: I was wondering how you would know how to do this one:
16. A wire has an AC current running through it. What happens to the force acting on a stationary +1C charge placed near the wire?
A) The force will vary sinusoidally from +Fmax to -F max
B) The force will increase from zero to +Fmax
C) The force will vary sinusoidally from +Fmax to zero
D) The force will constantly be zero
And also, in laminated iron cores, what material is chosen as the insulation? One last thing, leading on from the Eddison and Westinghouse thing, what are some points that basically have to be mentioned in order to get the marks? Thanks a lot!
Neutron :)
Post by: Neutron on March 30, 2016, 07:10:38 pm
Neutron
Post by: lazydreamer on March 30, 2016, 08:21:41 pm
Also, one more thing (sorry!!), if you dropped a bar magnet into a coil that's connected to a globe, how come the globe will turn on, switch off and turn on momentarily (whilst the magnet falls through). Like even though the magnet is in the middle of the coil, isn't there still relative motion and won't it still induce emf? Thanks :Da short video : https://www.youtube.com/watch?v=ItFggklNX0o
Neutron
hopefully it helps, i've got half-yearlies tomorrow as well so i know the feels, rip me lol :P best of luck!!
Post by: jamonwindeyer on March 30, 2016, 09:45:02 pm
Hey guys!
Okay so my half yearlies are tomorrow (pray for me) and I still have a few questions D: I was wondering how you would know how to do this one:
16. A wire has an AC current running through it. What happens to the force acting on a stationary +1C charge placed near the wire?
A) The force will vary sinusoidally from +Fmax to -F max
B) The force will increase from zero to +Fmax
C) The force will vary sinusoidally from +Fmax to zero
D) The force will constantly be zero
And also, in laminated iron cores, what material is chosen as the insulation? One last thing, leading on from the Eddison and Westinghouse thing, what are some points that basically have to be mentioned in order to get the marks? Thanks a lot!
Neutron :)
Hey Neutron!! Good luck in advance for your half yearly exams ;D
Okay, for your multiple choice question. An AC current varies sinusoidally over time. The current produces a radial magnetic field around the wire, which will also vary sinusoidally. So, the charge is exposed to a changing magnetic field.
Now, we know by Faraday's law that charges exposed to a changing magnetic field will experience a force: this is why we have induced currents. Now, since the magnetic field varies sinusoidally, its rate of change will also vary sinusoidally (trigonometric calculus, not super important). Therefore, I believe the answer would be A.
For laminated iron cores, I'm not 100% sure. A quick read suggest layers of oxide or insulating varnish are used for laminations, or even plastics?
For Edison v Westinghouse, you are discussing advantages of AC over DC (pretty much DC has no advantages, besides the fact that it is, volt for volt, more powerful than AC, and it is more safe). The big things to mention are advantages for transmission (almost 100% efficient), and ability to be transformed. If you cover all of this stuff, you'll be on your way to solid marks!
Hope this helps, good luck!! ;D
Post by: ATWalk on March 31, 2016, 07:09:12 pm
I'm having some trouble with this question right here. I'm not really sure what to say about the validity or accuracy. For example, I found Vp/Vs to be greater than np/ns. That would mean that energy is lost, so would that affect the experiment's validity? Would the addition of a switch to limit heat be a good thing to mention? And also, is it enough just to say that the experiment should be repeated in order to improve its reliability? Basically, what exactly should I say for this question?
Thanks a lot in advance.
Post by: jamonwindeyer on March 31, 2016, 08:22:19 pm
Hi,
I'm having some trouble with this question right here. I'm not really sure what to say about the validity or accuracy. For example, I found Vp/Vs to be greater than np/ns. That would mean that energy is lost, so would that affect the experiment's validity? Would the addition of a switch to limit heat be a good thing to mention? And also, is it enough just to say that the experiment should be repeated in order to improve its reliability? Basically, what exactly should I say for this question?
Thanks a lot in advance.
Hey ATWalk!
I always hated the validity/reliability/accuracy questions, got them confused heaps!
Validity concerns how well the variables in an experiment are controlled. That is, if we want to investigate the voltage relationship in a transformer, the validity of that investigation would hinge on how well we can achieve flux linkage between the coils. In this situation, you have correctly identified through some calculations, that we have definitely lost energy to the environment. We haven't effectively controlled all variables. Therefore, your assessment would be that this experiment is not completely valid. It does not ask you to suggest an improvement for validity, so this is all you need.
To improve reliability, you are looking at larger sample sizes/more repetition. However, it wants a change to the apparatus, so perhaps "repetition" isn't the best answer. Perhaps replace the coils with solenoid-like apparatus where the amount of turns can be adjusted, then you could get results for a variety of different turn-to-turn relationships, thus improving reliability ;D
Post by: mijomo on April 01, 2016, 03:36:37 pm
Q) Just after launch the engines of a spaceship (mass 6.8x10^5 kg) produce a thrust of 10^7N. What is the force that a 100kg astronaut exerts on the launch lounge at this time?
(a) 100N
(b) 490N
(c) 980N
(d) 1470N
Post by: jakesilove on April 01, 2016, 03:56:41 pm
Hi everyone, just wondering if someone could explain how to do this question thanks.
Q) Just after launch the engines of a spaceship (mass 6.8x10^5 kg) produce a thrust of 10^7N. What is the force that a 100kg astronaut exerts on the launch lounge at this time?
(a) 100N
(b) 490N
(c) 980N
(d) 1470N
Hey Mijomo!
This question requires multiple applications of the formula
First, you have to identify that the Thrust force is the same as the force on the rocket, ie.
Using this equation, we can solve to find the total acceleration of the rocket, which is
Now, we can figure out the downward force that the Astronaut exerts based on this acceleration! Using F=ma again,
And so, the answer is D.
Hope that this helped!
Jake
Post by: RuiAce on April 01, 2016, 04:02:13 pm
Observe that the acceleration of the rocket = the acceleration on the astronaut. This occurs because the astronaut is essentially inside the rocket, so they are the same system. You could be fancy, and say that they are in fact the same non-inertial frame of reference as well.
This is why the acceleration calculated for the rocket can be applied immediately to that of the astronaut.
Post by: gstaah on April 03, 2016, 03:32:35 pm
Post by: RuiAce on April 03, 2016, 03:57:12 pm
Hey, this is from the 2005 HSC physics paper. From what I've learnt I would think it's B, but its C. Isn't there still minuscule eddy currents generated in R?
From memory this question has been disputed. The conclusion among many people I've consulted with have agreed that it should be B
Post by: gstaah on April 03, 2016, 06:06:44 pm
From memory this question has been disputed. The conclusion among many people I've consulted with have agreed that it should be B
Oh no wonder. Thanks!
Post by: Meckenza on April 05, 2016, 04:16:33 pm
Q: The mass spectrometer is used to determine the mass of particles. It operates by projecting particles, with known charge and velocity, through an evacuated chamber into a region of uniform magnetic field acting perpendicular to the velocity of the particles. Explain how the charged particles will behave as they enter the magnetic field and describe how this behaviour allows the mass to be determined.
Thanks!
Post by: jakesilove on April 05, 2016, 06:04:46 pm
Hello, I'm having trouble understanding the answer of circular motion to this question. Any clarification would be appreciated!
Q: The mass spectrometer is used to determine the mass of particles. It operates by projecting particles, with known charge and velocity, through an evacuated chamber into a region of uniform magnetic field acting perpendicular to the velocity of the particles. Explain how the charged particles will behave as they enter the magnetic field and describe how this behaviour allows the mass to be determined.
Thanks!
Hey!
Let's start by thinking about what is actually going on in your scenario. We have a charged particle, fired into an area with a constant magnetic field. This means that, by the right hand rule, there will be a force acting on the particle. An example of this is displayed in the image.
(http://www.physbot.co.uk/uploads/1/2/5/0/12507040/5221086_orig.png)
However, remember that the force acting on the particle is CONSTANT (as it is proportional to the magnetic field, and the charge of the particle, neither of which are changing). Recalling the formula
We can see that if the force is constant, and the mass is constant, there will also be a constant acceleration! This means that the particle is 'rotating' through the field at the same, constant rate. The curve, therefore, will be uniform, and therefore traces out a circular path! See the image below for clearer details.
(http://cnx.org/resources/b48abc4b98dab06d4db2cf621fac642d21dfaf7c/mm3a.gif)
You can use the right hand rule along the entire path of the particle (just move your hand as though it IS the particle) to show that the force will always be towards the centre of the circle (ie. it is always accelerating towards the centre). If the circle is entirely contained within the magnetic field, it will trace the following path.
(http://cnx.org/resources/6058801887e8688ecef75938dce3d3b11e7a5553/mm1.gif)
Okay great, so we know that the particle follows a curved path, accelerating towards the centre only. This sounds pretty familiar! Space shuttles, and orbits, work exactly the same way! So we can apply formulas used in the Space section here!
The next derivation is similar to the method used by JJ Thompson to figure out the mass:charge ratio of electrons. It's important to know the derivation for the HSC, although very few questions ask about it.
We know that the force acting on a moving charge in a uniform magnetic field is
However, we also know the force exerted on an object in orbit. The formula for centripetal force is
Since these two forces must be the same, we can equate them!
Since we know the charge, velocity and Magnetic field strength, and we can measure the radius of the spinning charge, we can figure out the mass!
Very rarely, HSC questions will make you derive things like this. You just need to think about what is actually HAPPENING, and try work things out from there. I hope that this explanation helped! Let me know if I can clarify anything.
Jake
Post by: Meckenza on April 05, 2016, 06:38:48 pm
Post by: jakesilove on April 05, 2016, 07:15:45 pm
Ahh, ok! That makes sense. Thanks Jake!
No problem! That was quite a difficult (and, as far as my experience goes, unique) question. Keep looking through difficult material, and I look forward to seeing you on the forums!
Jake
Post by: RuiAce on April 05, 2016, 08:39:45 pm
Some students even remember the equations in JJ Thomson's experiment and know the full derivation to q/m=E/(rB2)
To be fair, it is hard, but it something that you should know how to do after first time exposure to it at least.
Post by: jakesilove on April 05, 2016, 08:45:11 pm
Actually can I make a comment... questions that ask you to equate FB=FC for magnetic field deflection of an electron are actually pretty common. It's expected that students should be able to handle this.
Some students even remember the equations in JJ Thomson's experiment and know the full derivation to q/m=E/(rB2)
To be fair, it is hard, but it something that you should know how to do after first time exposure to it at least.
Just following on from this (and I totally agree with RuiAce; you should try to learn the JJ. Thompson equations derivation, if possible), I thought I would post the derivation for anyone interested. If you want a comprehensive formula sheet, including a whole bunch of important information, check out the link here!
(http://i.imgur.com/szHYn80.png?1)
Jake
Post by: Neutron on April 12, 2016, 10:27:12 pm
So I started learning about Cathode Ray Tubes and I was wondering how the control electrode actually controls the number of electrons coming off the cathode? And also, how does the accelerating anode work? Like I get how the electrons are attracted to the anode but how does the electric field produced by the accelerating anode actually speed up the electrons? Many thanks!
Neutron :D
Post by: jamonwindeyer on April 13, 2016, 10:48:10 am
Hey guys!
So I started learning about Cathode Ray Tubes and I was wondering how the control electrode actually controls the number of electrons coming off the cathode? And also, how does the accelerating anode work? Like I get how the electrons are attracted to the anode but how does the electric field produced by the accelerating anode actually speed up the electrons? Many thanks!
Neutron :D
Hey Neutron! First off, super impressed into the detail you are going into learning this stuff, this is definitely extra knowledge you are asking! Love it, I'll do my best to answer ;D
Your second question first, how does the accelerating anode speed up the electrons? Basically, the accelerating anode is usually cylindrical in shape, around the path of the electrons. The electric field produced by that anode will pull an electron towards it and accelerate it. But why? The answer is simple. Electrons are negatively charged, and the anode has an extremely large positive charge. Keep in mind, a positive voltage means an accrual of positive charge (i.e. - a complete absence of electrons), and a voltage generates an electric field. In any case, positive charge and negative charge attracts, so the electron is accelerated.
There is a formula governing the force experienced by the electron, linked to the net field strength of the accelerating anode. It is:
where F is the force, E is the field strength (in volts per metre), and q is the charge on, in this case, the electron.
It sounds like you understand most of this already, does this help at all? Any particular part of this you are confused about?
In terms of controlling the number of electrons coming off the cathode, it is completely dependent on the electric fields/voltages involved. Apply a larger potential difference between the cathode and anode, and you get more electrons. We also use a technique called thermionic emission, where heat energy excites electrons and emits them from the cathode. This can be used to increase electron emission at a set potential difference ;D
Hope this helps!! :)
Post by: Happy Physics Land on April 13, 2016, 11:04:34 am
Hey guys!
So I started learning about Cathode Ray Tubes and I was wondering how the control electrode actually controls the number of electrons coming off the cathode? And also, how does the accelerating anode work? Like I get how the electrons are attracted to the anode but how does the electric field produced by the accelerating anode actually speed up the electrons? Many thanks!
Neutron :D
Hey Neutron!
I've just started doing Ideas into Implementation as well. I did a bit of research on control electrode and how it works. It is situated at the cathode which contains a heating filament/small heater (depending on whether you are talking about CRO/TV). This heating filament would heat up and "boil" off electrons from the cathode, producing electron clouds. What the control electrode does is that it would vary the output of electrons according to the varying signals that are applied. So to put it into simplest terms, if you require low brightness on the screen you can just emit less electrons through the control electrode and if you want a high brightness you just emit more electrons using control electrode. The actual mechanism of control electrode is not examinable in the scope of the syllabus so I will just briefly explain it. Essentially this control electrode is connected to the external voltage circuit of the cathode and depending on the amount of input voltage (i.e. the amount of input signal), the amount of electrons released can be controlled.
In regards to the accelerating anodes, we will have to recall a concept we learnt in year 11: currents flow from higher potential to lower potential. The current here refers to positive charges (i.e. conventional current). Now, we have two anodes, one is the focusing anode and another is the accelerating anode. The accelerating anode has HIGHER VOLTAGE than the focus anode, meaning that for a positive charge it will travel from accelerating anode to focusing anode. However, because cathode rays are NEGATIVELY CHARGED ELECTRONS, therefore they will travel from the focusing anode to the accelerating anode.
Now, recall that higher voltage leads to higher electric field (E = V/d, hence E is directly proportional to V) and this will lead to higher force exerted upon the electrons (F=Eq, increase in E --> increase in F). Since the mass of electrons is constant, and force applied is constant (due to constant supply of voltage), then it will accelerate according to F=ma. Because accelerating anode has higher voltage than focusing anode, there is more force being applied to the electrons and hence it will accelerating as it travels through the accelerating anode.
Post by: Happy Physics Land on April 13, 2016, 11:06:47 am
Hey Neutron! First off, super impressed into the detail you are going into learning this stuff, this is definitely extra knowledge you are asking! Love it, I'll do my best to answer ;D
Your second question first, how does the accelerating anode speed up the electrons? Basically, the accelerating anode is usually cylindrical in shape, around the path of the electrons. The electric field produced by that anode will pull an electron towards it and accelerate it. But why? The answer is simple. Electrons are negatively charged, and the anode has an extremely large positive charge. Keep in mind, a positive voltage means an accrual of positive charge (i.e. - a complete absence of electrons), and a voltage generates an electric field. In any case, positive charge and negative charge attracts, so the electron is accelerated.
There is a formula governing the force experienced by the electron, linked to the net field strength of the accelerating anode. It is:
where F is the force, E is the field strength (in volts per metre), and q is the charge on, in this case, the electron.
It sounds like you understand most of this already, does this help at all? Any particular part of this you are confused about?
In terms of controlling the number of electrons coming off the cathode, it is completely dependent on the electric fields/voltages involved. Apply a larger potential difference between the cathode and anode, and you get more electrons. We also use a technique called thermionic emission, where heat energy excites electrons and emits them from the cathode. This can be used to increase electron emission at a set potential difference ;D
Hope this helps!! :)
Combine my answer with Jamon's you will basically understand to a professional level how an electron gun works...
Post by: jamonwindeyer on April 13, 2016, 04:46:27 pm
Combine my answer with Jamon's you will basically understand to a professional level how an electron gun works...
The teamwork is strong ;) again I'll stress, the HSC will never test to this level of detail. You guys are definitely pushing for a Band 6 with this kind of knowledge ;D
Post by: Neutron on April 13, 2016, 07:05:20 pm
Negative- It repels the electrons back into the cathode (so not as many electrons can move onto the anode and thus be focussed and accelerated)
Positive- Encourages electrons to come out of the cathode and thus, increases the amount of electrons
Yeah, so I guess the control electrode uses more than just thermionic emission! Just thought I might throw that out there heh
Although if you guys don't mind answering another question, I was wondering how Hertz's antenna actually generated radiowaves? I was reading up on how spark-gap transmitters work but they just say that it sparks across the gap and produces electromagnetic radiation but I'm having trouble understanding how D: And also, how are electromagnetic waves produced naturally? So what actually provides the initial changing magnetic or electric fields? Thanks you guys, you're legends <3
Neutron
Post by: RuiAce on April 13, 2016, 10:30:33 pm
Yesss thank you so much you two! I too, did some research online and I discovered that the control electrode can either be positive or negative (this helps in controlling the amount of electrons and thus brightness etc), meaning if the control electrode is:
Negative- It repels the electrons back into the cathode (so not as many electrons can move onto the anode and thus be focussed and accelerated)
Positive- Encourages electrons to come out of the cathode and thus, increases the amount of electrons
Yeah, so I guess the control electrode uses more than just thermionic emission! Just thought I might throw that out there heh
Although if you guys don't mind answering another question, I was wondering how Hertz's antenna actually generated radiowaves? I was reading up on how spark-gap transmitters work but they just say that it sparks across the gap and produces electromagnetic radiation but I'm having trouble understanding how D: And also, how are electromagnetic waves produced naturally? So what actually provides the initial changing magnetic or electric fields? Thanks you guys, you're legends <3
Neutron
Quoted off Physics in Focus:
"An accelerating or oscillating charge produces EMR."
"EMR can cause charges to accelerate or oscillate."
As for how this phenomena is true, that you may want to research. (For the most part I am no longer a physicist.)
The extremely compact version of the long story is that Hertz's spark gap basically had charges that oscillated. Basically, the current that was fed INTO the induction coil was oscillating.
Post by: Happy Physics Land on April 13, 2016, 10:48:51 pm
Yesss thank you so much you two! I too, did some research online and I discovered that the control electrode can either be positive or negative (this helps in controlling the amount of electrons and thus brightness etc), meaning if the control electrode is:
Negative- It repels the electrons back into the cathode (so not as many electrons can move onto the anode and thus be focussed and accelerated)
Positive- Encourages electrons to come out of the cathode and thus, increases the amount of electrons
Yeah, so I guess the control electrode uses more than just thermionic emission! Just thought I might throw that out there heh
Although if you guys don't mind answering another question, I was wondering how Hertz's antenna actually generated radiowaves? I was reading up on how spark-gap transmitters work but they just say that it sparks across the gap and produces electromagnetic radiation but I'm having trouble understanding how D: And also, how are electromagnetic waves produced naturally? So what actually provides the initial changing magnetic or electric fields? Thanks you guys, you're legends <3
Neutron
Hey Neutron!
Thats some really interesting information on control electrode! Turns out that it functions somewhat like a control grid in thermionic triodes!!! Good stuff there!
Ah yeah lesson 3 is confusing aye, apparently the most difficult lesson in the book. I will begin with answering the electromagnetic radiation part. In nature, radio waves can be produced through lightning. Why? Because when lightning strikes the air, the air molecules are ionised and charged particles are released. Because lightning hits with an extremely high voltage, it would put charged particles into motions. When the charged particles are travelling in a certain direction, a current is formed. Using our Right Hand Coil Rule, we ascertain the existence of a magnetic field around the current. What does this tell us? This tells us that when a charged particle is put into motion (i.e. a current is formed), there is a changing electric field and consequently a magnetic field is produced. James Maxwell further proposes that a changing magnetic field will then produce a electric field which will produce a magnetic field and so on. So this explains the fact that electromagnetic waves are self-sustaining mutual generation of electric field and magnetic fields.
However keep in mind, that only CHANGING magnetic field or CHANGING electric field may induce a changing electric field or changing magnetic field. This means that the charged particle will HAVE TO BE IN MOTION (i.e. oscillate) in order for a changing electric field to be produced. This is why electromagnetic waves propagate! Because the electric field is constantly changing due to motion!
Think about this: if an electron is travelling at constant speed in a consistent direction, will a magnetic field be produced? No. Because its motion is constant which DOES NOT lead to a CHANGING electric field. So in order for it to produce a changing electric field, it must be accelerated or decelerated. Therefore the fundamental mechanism responsible for the production of electromagnetic waves is the acceleration of charged particles which produce changing electric field. This theory can be proven through Heinrich Hertz's experiments. Going back to lightning producing EM waves, since lightning provide the charged particles with a load of force, the particles will accelerate (F=ma) and hence producing changing electric fields for producing changing magnetic fields and so on.
The spark gap transmitter itself isnt the thing that creates the radiowave, but sure it does contribute to the production of EM wave through providing a sufficient voltage in the inductor coil, hence maintaining an arc across the spark gap. The actual formation of EM wave is this: one electrode is connected to the positive terminal of the power source, another electrode is connected to the negative terminal of the power source. Now, this spark gap transmitter acts like a step-up transformer, providing a high voltage to both electrodes. So the positive electrode would have a very high positive potential whilst the negative electrode would have a very high negative potential. The significant difference between these two potentials allows air to actually become a conductor as current passes from negative to positive electrode, ionising all the air molecules along the way, causing discharge arcs. These ionised air particles (charged particles) keep on fluctuating inside the gap because there is a constant forward and backward exchange of energy into the capacitor. This exchange of energy resembles alternating current in nature and hence a radio wave is created (Alternating current provides constant acceleration of discharge ions, hence constantly changing electric field --> constantly changing magnetic field --> constantly changing electric field and so on).
Post by: Happy Physics Land on April 13, 2016, 10:49:38 pm
Quoted off Physics in Focus:
"An accelerating or oscillating charge produces EMR."
"EMR can cause charges to accelerate or oscillate."
As for how this phenomena is true, that you may want to research. (For the most part I am no longer a physicist.)
The extremely compact version of the long story is that Hertz's spark gap basically had charges that oscillated. Basically, the current that was fed INTO the induction coil was oscillating.
Downfall of all actuaries
Post by: jamonwindeyer on April 13, 2016, 11:08:45 pm
Hey Neutron!
Thats some really interesting information on control electrode! Turns out that it functions somewhat like a control grid in thermionic triodes!!! Good stuff there!
Ah yeah lesson 3 is confusing aye, apparently the most difficult lesson in the book. I will begin with answering the electromagnetic radiation part. In nature, radio waves can be produced through lightning. Why? Because when lightning strikes the air, the air molecules are ionised and charged particles are released. Because lightning hits with an extremely high voltage, it would put charged particles into motions. When the charged particles are travelling in a certain direction, a current is formed. Using our Right Hand Coil Rule, we ascertain the existence of a magnetic field around the current. What does this tell us? This tells us that when a charged particle is put into motion (i.e. a current is formed), there is a changing electric field and consequently a magnetic field is produced. James Maxwell further proposes that a changing magnetic field will then produce a electric field which will produce a magnetic field and so on. So this explains the fact that electromagnetic waves are self-sustaining mutual generation of electric field and magnetic fields.
However keep in mind, that only CHANGING magnetic field or CHANGING electric field may induce a changing electric field or changing magnetic field. This means that the charged particle will HAVE TO BE IN MOTION (i.e. oscillate) in order for a changing electric field to be produced. This is why electromagnetic waves propagate! Because the electric field is constantly changing due to motion!
Think about this: if an electron is travelling at constant speed in a consistent direction, will a magnetic field be produced? No. Because its motion is constant which DOES NOT lead to a CHANGING electric field. So in order for it to produce a changing electric field, it must be accelerated or decelerated. Therefore the fundamental mechanism responsible for the production of electromagnetic waves is the acceleration of charged particles which produce changing electric field. This theory can be proven through Heinrich Hertz's experiments. Going back to lightning producing EM waves, since lightning provide the charged particles with a load of force, the particles will accelerate (F=ma) and hence producing changing electric fields for producing changing magnetic fields and so on.
The spark gap transmitter itself isnt the thing that creates the radiowave, but sure it does contribute to the production of EM wave through providing a sufficient voltage in the inductor coil, hence maintaining an arc across the spark gap. The actual formation of EM wave is this: one electrode is connected to the positive terminal of the power source, another electrode is connected to the negative terminal of the power source. Now, this spark gap transmitter acts like a step-up transformer, providing a high voltage to both electrodes. So the positive electrode would have a very high positive potential whilst the negative electrode would have a very high negative potential. The significant difference between these two potentials allows air to actually become a conductor as current passes from negative to positive electrode, ionising all the air molecules along the way, causing discharge arcs. These ionised air particles (charged particles) keep on fluctuating inside the gap because there is a constant forward and backward exchange of energy into the capacitor. This exchange of energy resembles alternating current in nature and hence a radio wave is created (Alternating current provides constant acceleration of discharge ions, hence constantly changing electric field --> constantly changing magnetic field --> constantly changing electric field and so on).
Nice HPL! Only thing I'd add is just a stressed fact: The EM Waves are not exclusively emitted from the spark itself. Rather, the whole antennae emits the waves. As HPL correctly describes, the spark itself is caused by the voltage exceeding some value and causing part of the air itself to become conductive. This is called electrical breakdown. When this happens, the circuit goes from having zero current, to having a huge current, virtually instantaneously.
What this does is what I'm studying right now. Basically, such a huge change happening so quickly causes oscillations. Picture holding a pendulum in your hand, and quickly moving your hand forward, like throwing a punch. This virtually instantaneous change does not just let the pendulum hang, it swings. This is a very poor analogy for what happens in an electrical circuit which has a HUGE current applied instantaneously (for those interested, research the Fourier Transform and the Dirac Delta function). These oscillations occur in the whole circuit at a variety of frequencies, and as HPL correctly states, it is these oscillations which generate EM waves (with the different frequencies of oscillation being what causes the different wavelengths of EM radiation).
Remember, charges produces electric fields. Moving charges produce magnetic fields. Accelerating (oscillating) charges produce EM waves.
Hopefully between Rui, HPL and myself, we have given you a solid understanding of what is a very complex phenomenon. Again, you have gone way beyond what you need to know for HSC Physics! ;D
Post by: Happy Physics Land on April 14, 2016, 06:28:09 pm
Nice HPL! Only thing I'd add is just a stressed fact: The EM Waves are not exclusively emitted from the spark itself. Rather, the whole antennae emits the waves. As HPL correctly describes, the spark itself is caused by the voltage exceeding some value and causing part of the air itself to become conductive. This is called electrical breakdown. When this happens, the circuit goes from having zero current, to having a huge current, virtually instantaneously.
What this does is what I'm studying right now. Basically, such a huge change happening so quickly causes oscillations. Picture holding a pendulum in your hand, and quickly moving your hand forward, like throwing a punch. This virtually instantaneous change does not just let the pendulum hang, it swings. This is a very poor analogy for what happens in an electrical circuit which has a HUGE current applied instantaneously (for those interested, research the Fourier Transform and the Dirac Delta function). These oscillations occur in the whole circuit at a variety of frequencies, and as HPL correctly states, it is these oscillations which generate EM waves (with the different frequencies of oscillation being what causes the different wavelengths of EM radiation).
Remember, charges produces electric fields. Moving charges produce magnetic fields. Accelerating (oscillating) charges produce EM waves.
Hopefully between Rui, HPL and myself, we have given you a solid understanding of what is a very complex phenomenon. Again, you have gone way beyond what you need to know for HSC Physics! ;D
Fourier transform :D :D :D
We learnt about this thing called effective inductance and capacitance today, not too sure what they mean. Apparently they are specific to electrical engineering???? :D
Post by: jamonwindeyer on April 14, 2016, 07:58:16 pm
Fourier transform :D :D :D
We learnt about this thing called effective inductance and capacitance today, not too sure what they mean. Apparently they are specific to electrical engineering???? :D
Yep! My life for the next few years ;)
As a quick run down, inductance is an electrical property whereby the circuit element in question resists changes in current through it. Note that this is different to resistance. Capacitance is the ability to store charge.
Both inductors and capacitors store energy. Inductors are normally large coils of wire, where energy is stored in the form of magnetic fields. Capacitors are normally two plates, separated by some distance, where energy is stored in the form of an electric field.
Both are only really useful for AC circuits. At DC, an inductor essentially disappears (since the magnetic fields required to store energy require a changing current), and a capacitor essentially breaks the circuit (we reach a point where the capacitor is fully charged and current cannot flow anymore). They can be used in AC circuits to transfer what we call reactive power between circuit elements, they are in essence, an energy storage mechanism ;D
Post by: Happy Physics Land on April 14, 2016, 10:33:13 pm
Yep! My life for the next few years ;)
As a quick run down, inductance is an electrical property whereby the circuit element in question resists changes in current through it. Note that this is different to resistance. Capacitance is the ability to store charge.
Both inductors and capacitors store energy. Inductors are normally large coils of wire, where energy is stored in the form of magnetic fields. Capacitors are normally two plates, separated by some distance, where energy is stored in the form of an electric field.
Both are only really useful for AC circuits. At DC, an inductor essentially disappears (since the magnetic fields required to store energy require a changing current), and a capacitor essentially breaks the circuit (we reach a point where the capacitor is fully charged and current cannot flow anymore). They can be used in AC circuits to transfer what we call reactive power between circuit elements, they are in essence, an energy storage mechanism ;D
Well l didnt really get that last sentence but I guess Im not gonna go anywhere near electrical engineering. I would imagine now why so many people drop out after the first year --- its really damn hard.
Post by: jamonwindeyer on April 14, 2016, 10:40:51 pm
Well l didnt really get that last sentence but I guess Im not gonna go anywhere near electrical engineering. I would imagine now why so many people drop out after the first year --- its really damn hard.
Happy to explain it a little better, but it is totally irrelevant to anything HSC related ;) yes, Elec is quite intense, extremely mathematical! One of my subjects this year has something like a 20% failure rate, it's very difficult stuff, definitely not for anyone who isn't into their math ;D
Post by: Planck on April 16, 2016, 11:52:20 am
I was wondering how the equation w=qv was derived for work done in an electric field? Thanks :D
Post by: Happy Physics Land on April 16, 2016, 12:32:09 pm
Hey guys!
I was wondering how the equation w=qv was derived for work done in an electric field? Thanks :D
Hey Professor Max Planck!
I thought you are the professor in this area, not sure why you would be asking us for help but I guess you are too great to remember minute details such as this. After all I'm sure you are busier with your investigation into the blackbody radiation and E=hf. But it is certainly my honour to help a person like you who have receive nobel prizes for your significant contribution to our physics world. :D :D (Dont worry, I know you are not Max Planck hehehehe)
Ok so the simpler version of this derivation is simply substituting into the formula for work, where the force is the electric force experienced by a charged particle in an electric field and displacement is distance the charged particle has moved as a result of this electric force.
Work done = Force x Displacement
Work = qE x d
Work = qV
The more complicated version is actually more intuitive for me, so if you think you can follow my logic here, you can have a read. I feel like for me this is a better way of truly understanding how W=qV really came about.
Recall that 1 Voltage = 1 Joule of energy available to move 1 Coulomb of charge. When voltage is applied to a pair of parallel plates, an electric field is produced ( E = V/d ). Now, in this electric field, a charged particle will experience a force ( F = qE ). This force will cause the charged particle to accelerate ( F = ma ) and hence the velocity of this charged particle will increase. Because the velocity increases, the charged particle's kinetic energy also increase ( KE = 1/2 mv2 ). Since kinetic energy is a form of work done on an object, it has the unit Joules. If we now compare the unit for voltage, which is Joules/Coulomb with the unit for kinetic energy (work), which is Joules, then we can see that all we need to do to convert for voltage to work is to multiply voltage by q (i.e. J/C x C = J). Hence formula for work is W = qV.
Best Regards
Happy Physics Land
Post by: RuiAce on April 16, 2016, 05:37:27 pm
Hey guys!
I was wondering how the equation w=qv was derived for work done in an electric field? Thanks :D
Actually, whilst HPL makes a valid point...
...the whole definition of voltage is that V=W/q, i.e. the potential difference across a circuit is the work required to move each unit charge across it.
Post by: Neutron on April 19, 2016, 09:18:33 pm
Post by: jamonwindeyer on April 19, 2016, 11:44:47 pm
Hey guys! On the topic of p-type semiconductors, why is the acceptor level above the valence band if electrons are moving DOWN from the outer valence electron shells to fill up the holes of the covalent bonds within the doped silicon lattice?
Hey Neutron! Answering this in detail is tough, and is covered properly at the tertiary level (I did Solid State Physics at the back end of my first year) ;) Let's have a look!
To answer, first consider an N-Type semiconductor. N-Type semiconductors are formed by doping with a Group 5 Impurity atom, which introduces additional electrons into the lattice. The energy level of these donor electrons is higher than the energy level of the others, and they are easily excited into the conduction band. Thus, electrons act as majority charge carriers.
Right, now lets consider p-type semiconductors. These are formed by doping with a Group 3 impurity atom, which introduces additional electron vacancies or holes into the lattice. It is very easy for electrons to be excited from their place in the valence band to fill this vacancy, leaving holes in the valence band. These holes are then the majority charge carriers.
We note that the energy levels involved here can be referred to as activation energies. For N-Type, this is the energy required to excite an electron donation from the impurity. For P-Type, this is the energy required to excite an electron acceptance from the impurity.
So, my understanding would be that electrons do not jump DOWN from the P-type donor. The holes do. The electrons are actually excited UP to that energy level you are referring to :D
Post by: Happy Physics Land on April 20, 2016, 10:04:45 pm
Hey guys! On the topic of p-type semiconductors, why is the acceptor level above the valence band if electrons are moving DOWN from the outer valence electron shells to fill up the holes of the covalent bonds within the doped silicon lattice?
Yeah l think jamon's explanation was quite sufficient. I will just drop in my 2 cents here.
The acceptor level is really determined by the band structure of the acceptor atom. So for example, in the case of gallium, the acceptor level would appear above the valence band because the acceptor level would be the same as the highest level in the valence band of gallium. The reason why this acceptor level is helpful is that it significantly reduces the forbidden energy gap for electrons from valence band to become mobile (this would only take about 0.05eV). This in turn leaves excess holes in the valence band, allowing for current to be conducted through the principal charge carriers of positively charged holes.
Valence band is essentially the lowest band in the valence shell, and if the acceptor level is below the valence band, that kinda means that the electrons have now moved into a lower energy shell. But we know that only the valence electrons in semiconductors can act as charge carriers. So if we think about it logically, whats the point of doping if all we are doing is moving an electron into a lower energy shell where it cant act as a charge carrier?
Another way to think about it: in an intrinsic semiconductor, the electron move up into conduction band in order to become mobile charge carriers and leave holes behind in the valence band. So if this is the case, then shouldn't the same be applied to extrinsic semiconductors too? Where the electrons would also move up in band to become mobile charge carriers?
Basically my first point about the acceptor level being the valence band of the acceptor atom is the more scientific explanation of why acceptor level is above the valence band for host semiconductor. The next two points I think are good ways to logically think about the existence of acceptor level in such manner.
Post by: FallonXay on April 20, 2016, 10:54:18 pm
Post by: Happy Physics Land on April 20, 2016, 11:58:57 pm
Heyy, I'm having trouble understanding the relation between the blackbody radiation curve and Planck's formula E=hf (as well as light being quantized). How does this concept work? Thanks in advance! :)
Hey FallonXay!
E=hf is Planck's second postulate, which means that quantitatively, energy possessed by each quantum is directly proportional to the frequency of incident electromagnetic radiation. A quantum is essentially a packet of energy, and this energy is quantised, meaning that they only exist in discrete levels of well-defined energy. So for example, if say I need 10 J of energy to jump onto a stair, I cant gain 9J otherwise I would fall when I almost reach the stairs, nor can I gain 11 J because I would also fall because I have taken a step higher than the staircase.
One aspect of blackbody radiation curve is the ultraviolet catastrophe, which means that intensity at short wavelength, high frequency UV, gamma and X-rays are displayed to be 0. This issue cannot be understood by classic theory of physics, because according to classical theory, higher frequency would cause atoms to oscillate with greater energy and hence according to classical theory the intensity of UV radiation should reach infinity because wavelength is approaching 0, meaning that frequency is very high. Experimental data however oppose this theory. In fact, if infinity amount of UV radiations are emitted by blackbodies, then humans cant really exist because we all know that UV rays are harmful to human bodies.
What Planck is suggesting through E=hf is that, for high frequency UV, X-ray and gamma rays, there will be high amounts of energy emitted in discrete packets called quanta (plural for quantum). These quanta are only released when the atom undergoes a change in discrete energy level (imagine yourself stepping down from the second stair on the staircase to the first stair). However, because UV has extremely high frequency, using E=hf, we know that an extremely high amount of energy are emitted. This means, that there must be a significant change in energy level in an atom of the blackbody in order to emit quanta that contain such high energy values. But because we dont have such a wide range of energy levels in an atom, this would be impossible to achieve. Therefore there are no UV, X-ray and gamma rays being emitted.
Another feature you would have noticed in a blackbody radiation curve is the peak wavelength radiation. So according to Planck's postulate, the intensity of radiation emitted (or you can understand it as the number of quanta emitted) is proportional to the number of atoms undergoing a change in energy level. That means, for certain wavelength/frequency radiations, there is a higher probability for certain changes in energy levels to take place. Therefore these more probable changes of energy would result in more intense emission of the corresponding wavelength of radiation.
These are the main features you would need to know. The peak wavelength radiation part seems irrelevant to E=hf, but if you recall what I said before, E=hf implies that quanta are emitted when there is a change in energy level. So thats how E=hf relates to peak wavelength radiation. Hope it made sense! :D
Best Regards
Happy Physics Land
Post by: jamonwindeyer on April 21, 2016, 12:53:51 am
Heyy, I'm having trouble understanding the relation between the blackbody radiation curve and Planck's formula E=hf (as well as light being quantized). How does this concept work? Thanks in advance! :)
Love your work HPL! ;D To add just a couple of points to what is a fantastic explanation already:
- The characteristic frequency is what we call the peak frequency (and wavelength), and it is dependent on the temperature of the object, which relates indirectly to Planck's formula E=hf, but not in any way that you need to know for this course.
- Since this frequency depends on temperature, black body curves vary drastically between objects. The rule that governs their shape is called Planck's Law, which you don't need to know specifically. The only important thing is that you recognise that the curve shape will change depending on the object.
- As HPL stated, it is found that the intensity of high frequency radiation is quite low for black bodies, but this is only true for objects like our sun or similar. There do exist black body radiation curves that have a characteristic frequency in the gamma range, perhaps from a gamma burst from a supernova. Again, not totally important, but again the focus is that the curves will change shape drastically depending on the object!
Pretty much the theme of these additions is this: The shape of the curve has to do with the object itself, and thus changes rapidly. E=hf is just our mechanism for deriving why that curve looks the way it does, instead of the Ultraviolet Catastrophe Curve (which instead stems from Rayleigh-Jean's Law).
Between this post and HPL's, you have way more detail than you need for this part of the course, hope it helps ;D
Post by: FallonXay on April 21, 2016, 09:37:02 am
But because we dont have such a wide range of energy levels in an atom, this would be impossible to achieve.However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?
(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers! ;D)
Post by: jamonwindeyer on April 21, 2016, 01:36:47 pm
However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?
(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers! ;D)
You don't need to for HSC Physics! All good, electron energy levels in atoms are not overly important in this course.
All you need to understand is that the energy in an emitted quanta is determined based on some change in the atomic structure of the black body. An electron is demoted, a reaction takes place, something like that. This releases a photon with an energy exactly the same as the energy change in the atom, whatever that may be.
Gamma photons (for example) have a massive amount of energy. Massive energy changes are less common, and thus, we usually don't get many gamma photons emitted from a black body. Sometimes we do! But it is rare ;D Does that help at all? Anything still a little confusing?
(HPL or another chemist may wish to answer your questions specifically, but again, not necessary to understand in this course!) ;D
Post by: Happy Physics Land on April 21, 2016, 06:48:20 pm
Ahh ok.
However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?
(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers! ;D)
Oh yeah hehehe no worries! What Im really saying here is that, for example you have sodium and sodium has the electron configuration of 1s22s22p63s1. Now for UV radiation to be emitted, according to E=hf, there would need to be an infinitely high amount of energy being released in the form of quanta. Each quanta is released by a change in energy level. This means that we need a very significant change in energy level in the atom to emit that much energy, and this change in energy level might be, say, from 2s shell into 6f shell. But since 6f shell isnt anywhere to be found in sodium, this change in energy level cannot exist! So what Im implying here is, if energy emitted becomes infinity, then there will be infinite change in energy shell levels, and hence this is not gonna happen with any atoms simply because they cant supply this many shells.
Post by: Happy Physics Land on April 21, 2016, 06:50:02 pm
Post by: FallonXay on April 22, 2016, 05:37:20 pm
I will just make an extra post here to hit 200 comments, YAY IM A RESPECTED MEMBER NOW :D :D :D !!!Also Congratulations!!! and thanks for your contribution ;)
Post by: Happy Physics Land on April 22, 2016, 07:44:15 pm
oh ok, that makes sense :)
Also Congratulations!!! and thanks for your contribution ;)
Ahahahaha how kind of you! Thanks mate! :D :D :D
Post by: RuiAce on April 22, 2016, 09:41:48 pm
I will just make an extra post here to hit 200 comments, YAY IM A RESPECTED MEMBER NOW :D :D :D !!!
M8.
Post by: jamonwindeyer on April 22, 2016, 11:55:46 pm
M8.
M(9-1)
Post by: odonataa on April 27, 2016, 09:28:46 pm
Post by: jamonwindeyer on April 27, 2016, 09:31:58 pm
What causes the striation bands to appear? THANK YOU
Hey odonataa!!
Basically, it is the electrons in the Cathode Ray Tube colliding with air particles still in the tube. It is not a perfect vacuum, so electrons hit the air still in there and this causes the striation patterns. Changing the air pressure in the tube will change the pattern.
Hope this helps!! ;D
Post by: RuiAce on April 27, 2016, 09:54:46 pm
Speaking of which can someone enlighten me on this topic :P
Post by: jamonwindeyer on April 27, 2016, 09:59:25 pm
Better question would be why do the different gas pressures exhibit those specific results and not something else
Speaking of which can someone enlighten me on this topic :P
Do you mean like the stripes? I was always funny on this, but my best explanation would be as follows (may be slightly off, apologies if I offend any quantum physicists ;)).
Electrons moving from the cathode strike atoms in the tube. If the electron has enough energy, it will be 'absorbed' by the atom and dislodge another, with the excess energy being emitted as a photon of light, similar to the photoelectric effect in reverse I suppose? I had it likened to a relay race of sorts, one electron tags in the next.
The new electron then has to speed up before it strikes another atom and the cycle continues. The time it takes to accelerate to the required speed corresponds to a gap in the stripes. The stripes change distance apart based on how far apart the atoms are on average, which is dependent on the air pressure.
Post by: RuiAce on April 27, 2016, 10:09:17 pm
Do you mean like the stripes? I was always funny on this, but my best explanation would be as follows (may be slightly off, apologies if I offend any quantum physicists ;)).
Electrons moving from the cathode strike atoms in the tube. If the electron has enough energy, it will be 'absorbed' by the atom and dislodge another, with the excess energy being emitted as a photon of light, similar to the photoelectric effect in reverse I suppose? I had it likened to a relay race of sorts, one electron tags in the next.
The new electron then has to speed up before it strikes another atom and the cycle continues. The time it takes to accelerate to the required speed corresponds to a gap in the stripes. The stripes change distance apart based on how far apart the atoms are on average, which is dependent on the air pressure.
I suppose so yea but also why different gas pressures cause different patterns. Striations appear for a certain range and then you have Faraday's Dark space and etc., but then if you go further down Crooke's dark space dominates the entire thing and you get this green light and nothing else.
If you go a bit more up then you just get these horizontal purple lines appearing rather than those stripes so to speak! So how does pressure affect that?
Post by: jamonwindeyer on April 27, 2016, 10:26:02 pm
I suppose so yea but also why different gas pressures cause different patterns. Striations appear for a certain range and then you have Faraday's Dark space and etc., but then if you go further down Crooke's dark space dominates the entire thing and you get this green light and nothing else.
If you go a bit more up then you just get these horizontal purple lines appearing rather than those stripes so to speak! So how does pressure affect that?
Hmm, I suppose it would just be different manifestations of the same idea! I know that Crooke's Dark Space dominates once the air pressure becomes low enough that the electrons stop colliding with particles along the way. Most of them get through and hit the end of the tube, causing the green glow you describe.
I think I remember Faraday's dark space to be the region between the Cathode and Anode (after the negative glow), where the electrons are travelling further and further before they hit an atom (on average) with enough speed to cause fluorescence. Decrease the pressure, increase this gap.
The colours are indicative of the gas inside the tube.
And that's about all I've got, which explains a few of those phenomenon, but not all of them ???
Post by: Happy Physics Land on April 27, 2016, 11:10:45 pm
I suppose so yea but also why different gas pressures cause different patterns. Striations appear for a certain range and then you have Faraday's Dark space and etc., but then if you go further down Crooke's dark space dominates the entire thing and you get this green light and nothing else.
If you go a bit more up then you just get these horizontal purple lines appearing rather than those stripes so to speak! So how does pressure affect that?
I think Rui you are talking about violet streamers appearing in the cathode ray tube. This indicates a high pressure (20mm of Hg) inside the discharge tube. This high pressure indicates a high number of gas particles inside the tube. The reason why you see violet streamers is because collision is occurring everywhere inside the tube. We talking about how when energetic collision between cathode ray and air molecules occur, fluorescence occurs. So when collision occurs everywhere, there will be fluorescence occurring everywhere, forming violet streamers (or purple lines as you call it). These might not be "lines", but our eyes perceive it to be purple lines because collisions are happening at spots very close to one another inside the tube.
Post by: Happy Physics Land on April 27, 2016, 11:12:54 pm
Do you mean like the stripes? I was always funny on this, but my best explanation would be as follows (may be slightly off, apologies if I offend any quantum physicists ;)).
Electrons moving from the cathode strike atoms in the tube. If the electron has enough energy, it will be 'absorbed' by the atom and dislodge another, with the excess energy being emitted as a photon of light, similar to the photoelectric effect in reverse I suppose? I had it likened to a relay race of sorts, one electron tags in the next.
The new electron then has to speed up before it strikes another atom and the cycle continues. The time it takes to accelerate to the required speed corresponds to a gap in the stripes. The stripes change distance apart based on how far apart the atoms are on average, which is dependent on the air pressure.
I think Jamon's second paragraph really effectively summarises about the striation patterns. Basically when you see gaps between stripes, it is a brief region of dark space where no collisions occur because the cathode ray is accelerating without being interfered by any air molecules along the day.
Post by: RuiAce on April 27, 2016, 11:21:06 pm
I think Rui you are talking about violet streamers appearing in the cathode ray tube. This indicates a high pressure (20mm of Hg) inside the discharge tube. This high pressure indicates a high number of gas particles inside the tube. The reason why you see violet streamers is because collision is occurring everywhere inside the tube. We talking about how when energetic collision between cathode ray and air molecules occur, fluorescence occurs. So when collision occurs everywhere, there will be fluorescence occurring everywhere, forming violet streamers (or purple lines as you call it). These might not be "lines", but our eyes perceive it to be purple lines because collisions are happening at spots very close to one another inside the tube.
You're right I forgot that the correct term was "streamers". But he did provide quite a fair bit of info on what I needed - I just wanted to deduce what pressure exactly meant. Collisions as a word by itself doesn't explain the change in what's observed to me
Post by: jamonwindeyer on April 27, 2016, 11:32:59 pm
You're right I forgot that the correct term was "streamers". But he did provide quite a fair bit of info on what I needed - I just wanted to deduce what pressure exactly meant. Collisions as a word by itself doesn't explain the change in what's observed to me
"Collisions" is probably better described as electrons 'interacting' with positive ions with missing electrons lost to the flow of charge in the tube. The atoms regaining their electron is what causes fluorescence. 'Pressure' corresponds to the spacing between the atoms for which this occurs, thus changing the patterns. At least that's how I would explain it ;D
Post by: jamonwindeyer on April 27, 2016, 11:33:58 pm
Post by: smiley2101 on May 01, 2016, 09:14:31 pm
Post by: smiley2101 on May 01, 2016, 09:17:20 pm
thank you!
Post by: smiley2101 on May 01, 2016, 09:18:59 pm
I just don't get it haha
why is the answer D and not c?
Post by: smiley2101 on May 01, 2016, 09:34:05 pm
Post by: smiley2101 on May 01, 2016, 09:37:32 pm
Post by: jakesilove on May 01, 2016, 09:45:42 pm
last one!! haha :'(
Hey Smiley!
The answer to this multiple choice is going to be C. We can figure this out by process of elimination: gravity is actually the MAIN force acting on any spacecraft orbiting our planet, as it causes the orbit to occur in the first place (the shuttle is dragged down to earth, but is traveling so fast that it 'misses' the ground due to the curvature of the planet!), so the answer can't be A. The 'orbiting around the Sun' etc. thing doesn't even make any sense, and the whole 'inversely proportional' relationship thing would require some sort of mathematics that told you WHEN the astronaut would stay with the spacecraft, and when it would not. As the forces on the two objects are the same (namely, gravity), there is no reason that their ACCELERATION would be different!.
Remember, gravity has units that are meters per second, per second. There is no 'kilograms' or anything like that in the units, meaning that it operates INDEPENDENTLY of the mass of the object!
Jake :)
Post by: jakesilove on May 01, 2016, 09:48:38 pm
why is the answer C not D? thank you so much
For this question, you need to look at the formula for Force on a current carrying wire: namely
By creating a formula of I versus B, it is pretty clear that there will be an inverse (hyperbolic) relationship.
Therefore, the relevant graph will be the hyperbolic one!
Jake
Post by: jakesilove on May 01, 2016, 09:51:36 pm
I always stuff up the frame of reference like in this question. Do you have any tips or tricks that you used to know what tv or to should be substituted by?
The main trick I use is to first think about whether your value should be getting smaller or bigger. If you are dividing a number by the whole square root thing (which will equal less than one), then you are making the number bigger. If you are multiplying by the whole square root thing, you are making the number smaller. Therefore, if you are given a time and you are expecting it to get bigger, sub it into whichever time gets divided by the square root factor. If you get a length and you are expecting it to get smaller, sub it into whichever length gets multiplied by the square root factor!
Jake
Post by: jakesilove on May 01, 2016, 10:02:53 pm
why is the answer B?
thank you!
We know that the force between Q and P is F newtons. The question wants us to find the total force on R.
The relevant formula we need to use is
When the distance double, the force will clearly halve. Therefore, the force on R due to P will be half that of Q due to P, and in the opposite direction (as the currents have switched directions). Therefore, the force on R due to P will be
The force on R due to Q will be the same as the force on Q due to P. Therefore, the force on R due to Q will be
So, the total force will be
Jake
Post by: jakesilove on May 01, 2016, 10:05:31 pm
sorry I have so many questions!!!
I just don't get it haha
why is the answer D and not c?
Remember that the output voltage is going to be proportionate to the speed of the rotation (ie. the change in flux), as per Faraday's law. Therefore, the faster the thing is moving, the greater the output voltage will be! If it is slowed down, therefore, the output voltage has to decrease, making the answer D (note also that the amount it is slowed down is proportionate to the amount that the voltage decreases, ie. a factor of half).
Jake
Post by: jakesilove on May 01, 2016, 10:06:15 pm
Jake
Post by: jamonwindeyer on May 01, 2016, 11:07:43 pm
Also, as a final note, welcome to the forum! Hope we are able to help out as much as possible :)
Jake
For future reference to all forum users, this man is a machine ;D
Post by: jakesilove on May 02, 2016, 12:22:24 am
For future reference to all forum users, this man is a machine ;D
I have to procrastinate writing up my Physics lab somehow! But thank you Jamon, one of the few times I could have a crack at some Physics questions before you've snapped them all up with beautifully LaTexed responses.
Post by: jamonwindeyer on May 02, 2016, 08:36:19 am
I have to procrastinate writing up my Physics lab somehow! But thank you Jamon, one of the few times I could have a crack at some Physics questions before you've snapped them all up with beautifully LaTexed responses.
I see you are trying your hand at LaTex yourself! It will change your life!
Post by: brontem on May 03, 2016, 10:03:12 pm
Post by: FallonXay on May 03, 2016, 10:32:01 pm
I was wondering (In regards to the photoelectric effect), If a metal surface is hit with an incident EMR above the threshold frequency, electrons are emitted. What happens to the missing electrons in the metal? Are they replaced? How so?
thanks.
Post by: jamonwindeyer on May 03, 2016, 10:39:16 pm
Hey :) I'm slightly confused about this question, I kinda get it, kinda don't.. and the answer in the book is really vague. Can anyone help me?? Thanks!! :)
Hey Brontem!! Don't worry, I hated these questions in the HSC, and I think a lot of people do, because a lot of books give vague answers for these sorts of things ;)
I actually think the question is vague in this case as well, since the direction of the magnetic field is not given. The plates are identical except for their orientation, so the direction of the field makes a pretty big difference.
Essentially though, the principle is this. As you may have guessed, the aluminium plates being dropped in the magnetic field will generate eddy currents. These eddy currents will decelerate the plates, because, by Lenz's Law, the induced currents will oppose the change that created them. Therefore, dropping the plates will cause an induced current which pushes them back up (opposes the changes). This isn't enough, and the plate still falls, but they will both fall at a slower rate.
The question is, what role do those slots play? The closest comparison is probably laminations in the core of a transformer. Essentially, those gaps limit the size of the induced eddy currents in the plate, if they are at the appropriate orientation.
The way to tell how much slots/gaps in the conductor will impede eddy currents is to draw the magnetic field lines through it. The diagram below shows the way the eddy currents are affected by slots. Smaller circles = Less Eddy Currents = Less Deceleration = Plate Falling Faster.
Without seeing the field it is being dropped into I can't give a specific answer, but does this help elaborate on your answer at all? Anything specific in their answer that you wanted elaborated? I really hope this helps a little!! ;D
(https://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Laminated_core_eddy_currents_2.svg/2000px-Laminated_core_eddy_currents_2.svg.png)
Post by: jamonwindeyer on May 03, 2016, 10:42:59 pm
Hiya!
I was wondering (In regards to the photoelectric effect), If a metal surface is hit with an incident EMR above the threshold frequency, electrons are emitted. What happens to the missing electrons in the metal? Are they replaced? How so?
thanks.
Hey FallonXay!!
It depends on the context a little bit. If we are just talking about a metal, totally isolated from anything else, then no! The electrons are not replaced, the metal will gradually build up a positive charge.
If the metal is earthed in any way, either as part of a circuit or just in the circumstances, then electrons from earth will be attracted to the positive charge in the metal (caused by things called holes, which is covered in the next section) and replace the missing electrons. Remember, earth just defines an infinite reservoir of electrons, so if the metal is attached to anything like this, the electrons are replaced ;D I hope this helps!!
Post by: brontem on May 03, 2016, 10:51:58 pm
Post by: jamonwindeyer on May 03, 2016, 11:46:24 pm
Hey!! Thanks so much that definitely helps, way better than the 2 line answer in the book :P
Aha totally! Used to hate that when I was doing my HSC :o Sorry I couldn't be a little more specific, either X or Y will definitely fall faster, but I'm not sure which one with the info in front of me ;D
Post by: FallonXay on May 04, 2016, 06:42:55 am
Hey FallonXay!!
It depends on the context a little bit. If we are just talking about a metal, totally isolated from anything else, then no! The electrons are not replaced, the metal will gradually build up a positive charge.
If the metal is earthed in any way, either as part of a circuit or just in the circumstances, then electrons from earth will be attracted to the positive charge in the metal (caused by things called holes, which is covered in the next section) and replace the missing electrons. Remember, earth just defines an infinite reservoir of electrons, so if the metal is attached to anything like this, the electrons are replaced ;D I hope this helps!!
oh ok, thanks!
Post by: FallonXay on May 05, 2016, 09:11:30 pm
Post by: jamonwindeyer on May 05, 2016, 09:13:25 pm
Hiiii, I've attached a question from the 2014 Physics HSC exam (Q26b) and I don't quite understand how to solve it. I'm assuming it has something to do with the stopping voltage equation? But i'm not sure how to come to this conclusion.
I sat this paper, it was nasty let me tell you! I have a nicely typed solution to this, bear with me ;D
Post by: jamonwindeyer on May 05, 2016, 09:21:14 pm
Hiiii, I've attached a question from the 2014 Physics HSC exam (Q26b) and I don't quite understand how to solve it. I'm assuming it has something to do with the stopping voltage equation? But i'm not sure how to come to this conclusion.
Okay, this question is actually quite tricky, and requires a bit of thought. Consider the initial experiment; when the frequency of the light is 0 (no light), we have no current flow. For any other frequency, we do have current flow. What this means is that the potential difference of 4.1V exactly takes care of the work function of the metal; any frequency of light (and thus, any amount of energy) will release electrons. The work function is, therefore, 4.1eV.
When the external voltage is removed, this work function comes back into play, and here lies our work. We need to calculate the x-intercept of the new line, by calculating the frequency required for photoelectrons to have precisely 4.1eV of energy. This is the new threshold frequency, the minimum frequency to now overcome this work function. The gradient of the new line is the same (Planck’s Constant), and so doing the math, the new line is as follows (remember to convert between electron-volts and Joules, the conversion is in your data sheet):
So, you would draw a new line with the same gradient (the gradient is always Planck's Constant), with an x-intercept at this frequency. Without the voltage, this is the frequency you need to overcome the work function.
For the second part of the question, you just use Planck's formula and the work function:
I hope this helps a little bit!! ;D It's a little confusing actually, is there anything here you needed clarified? Happy to help if so ;D
Post by: FallonXay on May 05, 2016, 09:50:40 pm
I sat this paper, it was nasty let me tell you! I have a nicely typed solution to this, bear with me ;Doh haha. yeah, a lot of tricky questions in the exam :P
Okay, this question is actually quite tricky, and requires a bit of thought. Consider the initial experiment; when the frequency of the light is 0 (no light), we have no current flow. For any other frequency, we do have current flow. What this means is that the potential difference of 4.1V exactly takes care of the work function of the metal; any frequency of light (and thus, any amount of energy) will release electrons. The work function is, therefore, 4.1eV.Ooooh, I see, thanks! Interesting question with an interesting solution ;D
When the external voltage is removed, this work function comes back into play, and here lies our work. We need to calculate the x-intercept of the new line, by calculating the frequency required for photoelectrons to have precisely 4.1eV of energy. This is the new threshold frequency, the minimum frequency to now overcome this work function. The gradient of the new line is the same (Planck’s Constant), and so doing the math, the new line is as follows (remember to convert between electron-volts and Joules, the conversion is in your data sheet):
So, you would draw a new line with the same gradient (the gradient is always Planck's Constant), with an x-intercept at this frequency. Without the voltage, this is the frequency you need to overcome the work function.
For the second part of the question, you just use Planck's formula and the work function:
I hope this helps a little bit!! ;D It's a little confusing actually, is there anything here you needed clarified? Happy to help if so ;D
Only part I don't understand is how you concluded that the work function is 4.1eV. Is there some direct relation between potential difference/volts and electronvolts?
Also, is the SI unit for work function always joules?
Post by: jamonwindeyer on May 05, 2016, 09:57:31 pm
oh haha. yeah, a lot of tricky questions in the exam :P
Ooooh, I see, thanks! Interesting question with an interesting solution ;D
Only part I don't understand is how you concluded that the work function is 4.1eV. Is there some direct relation between potential difference/volts and electronvolts?
Also, is the SI unit for work function always joules?
Yep! So the definition of an electron volt is: "a unit of energy equal to the work done on an electron accelerating it through a potential difference of one volt."
I suppose a way to think of it is this. Since the voltage source of 4.1 volts exactly cancels the work function, we can conclude that the work function is deaccelerating the electrons (keeping them on/within the metal) with an energy equal to that provided by the 4.1 volts. Accelerating an electron out of a 4.1 volt potential difference, takes 4.1eV of energy, by the definition. So, accelerating the electron OUT of the metal, must take 4.1eV of energy also. And there is the answer.
And yep, SI unit for energy is Joules. If you try to use eV with Planck's Constant, for example, the answer won't be correct. Always convert ;D
Post by: FallonXay on May 05, 2016, 10:00:20 pm
Yep! So the definition of an electron volt is: "a unit of energy equal to the work done on an electron accelerating it through a potential difference of one volt."ahh ok, thanks a ton!!! ;D ;D ;D
I suppose a way to think of it is this. Since the voltage source of 4.1 volts exactly cancels the work function, we can conclude that the work function is deaccelerating the electrons (keeping them on/within the metal) with an energy equal to that provided by the 4.1 volts. Accelerating an electron out of a 4.1 volt potential difference, takes 4.1eV of energy, by the definition. So, accelerating the electron OUT of the metal, must take 4.1eV of energy also. And there is the answer.
And yep, SI unit for energy is Joules. If you try to use eV with Planck's Constant, for example, the answer won't be correct. Always convert ;D
Post by: jamonwindeyer on May 05, 2016, 10:21:35 pm
ahh ok, thanks a ton!!! ;D ;D ;D
Not a problem at all, happy to help!! ;D
Post by: Happy Physics Land on May 06, 2016, 04:03:49 pm
Hiya!
I was wondering (In regards to the photoelectric effect), If a metal surface is hit with an incident EMR above the threshold frequency, electrons are emitted. What happens to the missing electrons in the metal? Are they replaced? How so?
thanks.
Hey FallonXay!
Just reinforcing Jamon's point. When a high frequency light ray hits the surface of the metal, the electron is ejected when the frequency is greater than the threshold frequency (i.e. f > f0) and this means that the energy given off by each photon is able to overcome the work function of the metal (hf > hf0).
Ok but this doesn't really answer your question. So when the electrons is ejected (i.e. photoelectron), it can end up in different places depending on the scenario, just like what jamon said. I will just explain some of the following that are related to the hsc course:
1. If a circuit is provided, the photoelectron will travel through the circuit in the form of a photocurrent
2. If no circuits are provides and the metal is just exposed to air, then the photoelectron is ejected into the air and will perhaps ionise air molecules (This is seen in Hertz's experiment)
3. If a voltage is applied to the metal, then the electron ejected from the metal will accelerate through the air (This is seen in einstein's photoelectric experiment, when the photoelectron emitted from the cathode metal accelerates towards the anode metal)
4. If the light ray is shone upon the depletion layer of a p-n junction (or a solar cell), then the ejected electron can travel to the n-type semiconductor layer and this creates a potential difference between n-type and p-type layers. Consequently, when a circuit is connected to the solar cell, electricity can flow.
Post by: FallonXay on May 06, 2016, 04:36:26 pm
Hey FallonXay!
Just reinforcing Jamon's point. When a high frequency light ray hits the surface of the metal, the electron is ejected when the frequency is greater than the threshold frequency (i.e. f > f0) and this means that the energy given off by each photon is able to overcome the work function of the metal (hf > hf0).
Ok but this doesn't really answer your question. So when the electrons is ejected (i.e. photoelectron), it can end up in different places depending on the scenario, just like what jamon said. I will just explain some of the following that are related to the hsc course:
1. If a circuit is provided, the photoelectron will travel through the circuit in the form of a photocurrent
2. If no circuits are provides and the metal is just exposed to air, then the photoelectron is ejected into the air and will perhaps ionise air molecules (This is seen in Hertz's experiment)
3. If a voltage is applied to the metal, then the electron ejected from the metal will accelerate through the air (This is seen in einstein's photoelectric experiment, when the photoelectron emitted from the cathode metal accelerates towards the anode metal)
4. If the light ray is shone upon the depletion layer of a p-n junction (or a solar cell), then the ejected electron can travel to the n-type semiconductor layer and this creates a potential difference between n-type and p-type layers. Consequently, when a circuit is connected to the solar cell, electricity can flow.
ok, makes sense, thanks for the additional info regarding possible scenarios!
However one thing I wanted to clear up (another question :) ) is in regards to the photon if the frequency is below the threshold frequency. If the photon is below the threshold frequency, is the energy still transferred from the photon into the electron - consequently making the electron 'jump' up a 'shell' however since there is insufficient energy, the electron in the new shell is unstable and returns to its original shell and in the process, emits the energy back off? Or what happens to the photon, is it just reflected off the surface of the metal?
Post by: jamonwindeyer on May 06, 2016, 05:53:19 pm
ok, makes sense, thanks for the additional info regarding possible scenarios!
However one thing I wanted to clear up (another question :) ) is in regards to the photon if the frequency is below the threshold frequency. If the photon is below the threshold frequency, is the energy still transferred from the photon into the electron - consequently making the electron 'jump' up a 'shell' however since there is insufficient energy, the electron in the new shell is unstable and returns to its original shell and in the process, emits the energy back off? Or what happens to the photon, is it just reflected off the surface of the metal?
Hey FallonXay! That's a little trickier, I'll give you the simple explanation and the more complex one (which is way beyond the scope of the course):
Simple: The photon will be re-emitted, and it happens pretty much exactly as you say it would. The electron is excited to a new energy level, which is unstable, and when it falls back the photon is then re-emitted. The photon may also just simply not give its energy and, it has to give everything or nothing. This is called the All or Nothing Principle. Either way, the outcome is the same, photons are then re-emitted from the metal as reflected light.
Complex: There is an equation in quantum mechanics called the Schrodinger Equation which dictates the probability certain events occurring in a system, in this case, an electron/photon system. This equation actually says that it is possible for an electron to be excited into a higher energy level, but it can only realistically occur when the energy of a photon almost exactly matches the difference in energy between the electron shells. At all other energies, the probability of it occurring is very low (in Quantum Physics, every outcome has a probability of occurring, even those which seemingly defy classical physical logic). When the frequency is too low, the highest probability lies with the option above, the photon is re-emitted and the electron is unaffected. This is the manifestation of the All or Nothing Principle.
It is important to mention that, if a light has a very high intensity, and there are HEAPS of photons striking the metal, it is actually very slightly possible for a photon to strike an electron, excite it to a higher energy level, and then another photon to induce emission. Teamwork! However, for this to happen, the next photon has to strike the electron AFTER it has been excited, but BEFORE it jumps back down. This is highly unlikely, and so we definitely don't get any substantial amounts of electrons being emitted from the metal ;D
Reminder: The latter two paragraphs are definitely NOT required knowledge for this course! However, I'm happy to expand on some of it, if you feel like getting a head start on university physics ;)
Post by: FallonXay on May 06, 2016, 07:05:49 pm
Hey FallonXay! That's a little trickier, I'll give you the simple explanation and the more complex one (which is way beyond the scope of the course):Wouldn't mind hearing some more, if you don't mind. Seems like an interesting phenomenon ^.^
Simple: The photon will be re-emitted, and it happens pretty much exactly as you say it would. The electron is excited to a new energy level, which is unstable, and when it falls back the photon is then re-emitted. The photon may also just simply not give its energy and, it has to give everything or nothing. This is called the All or Nothing Principle. Either way, the outcome is the same, photons are then re-emitted from the metal as reflected light.
Complex: There is an equation in quantum mechanics called the Schrodinger Equation which dictates the probability certain events occurring in a system, in this case, an electron/photon system. This equation actually says that it is possible for an electron to be excited into a higher energy level, but it can only realistically occur when the energy of a photon almost exactly matches the difference in energy between the electron shells. At all other energies, the probability of it occurring is very low (in Quantum Physics, every outcome has a probability of occurring, even those which seemingly defy classical physical logic). When the frequency is too low, the highest probability lies with the option above, the photon is re-emitted and the electron is unaffected. This is the manifestation of the All or Nothing Principle.
It is important to mention that, if a light has a very high intensity, and there are HEAPS of photons striking the metal, it is actually very slightly possible for a photon to strike an electron, excite it to a higher energy level, and then another photon to induce emission. Teamwork! However, for this to happen, the next photon has to strike the electron AFTER it has been excited, but BEFORE it jumps back down. This is highly unlikely, and so we definitely don't get any substantial amounts of electrons being emitted from the metal ;D
Reminder: The latter two paragraphs are definitely NOT required knowledge for this course! However, I'm happy to expand on some of it, if you feel like getting a head start on university physics ;)
Post by: jamonwindeyer on May 07, 2016, 12:13:24 am
Wouldn't mind hearing some more, if you don't mind. Seems like an interesting phenomenon ^.^
You have opened Pandora's Box 8)
Okay, so in classical physics we learn about Newton's 2nd Law, which can be used (in various forms) to predict the behaviour of a system over time. Classically, this is absolutely accurate. In quantum mechanics, we cannot predict the outcome of an event until it happens, we can only calculate the probabilities of certain events occurring. This links to the whole idea of Schrodingers Cat.
Picture a cat inside a box (we can't see inside), and next to it is a contraption with a radioactive isotope, a detector, and a flask of poison. Two things in this 'system' can occur. The particle can decay, and if this happens, the resultant gamma radiation will trigger the detector, release the poison, and kill the cat. If the particle does not decay, nothing happens and the cat stays alive.
Certain interpretations of quantum mechanics (Copenhagen Interpretations) would suggest that, since we cannot know whether the cat is alive or dead until we check, the cat is simultaneously alive and dead. This is known as a quantum superposition. Only when we open the box, does this superposition collapse into a reality where the cat is alive (yay!) or dead (wahh...). Really, the cat here is just enabling us to extend an atomic state (whether a particle decays or not) to a macroscopic, real living thing.
So, how do we predict the probability of the cat dying or living, or more generally, how do we predict the outcomes of a system? This is where the Schrodinger Equation comes in. This is a partial differential equation which describes how the quantum state of a system evolves over time. The equation looks like this, the subject of the equation being the weird symbol Psi appearing on both sides of the equation (the rest are an essay in itself)
The subject is the wave equation. Now, the probability of certain quantum states is proportional to the square of this wave function. To solve for the wave equation, therefore, is to have complete understanding of a system and the probability of any event occurring. In some interpretations of quantum physics, the wave function can be used to predict the evolution of HUGE systems, perhaps even the entire universe. Solving the Schrodinger Equation for a particle is hard enough though, for a universe is simply impossible ;D
The Schrodinger Equation has many implications:
- Depending on measurements, the states of a system are quantised, and thus, energy is quantised in all forms (not just for electromagnetic quanta). This is verified already, for example, electron energies in atoms are proven to be quantised.
- Under the Copenhagen interpretation of quantum mechanics, particles do not have set positions, and thus, the result when we measure is drawn from a probability distribution (wave function gives us this). Flowing on from this, we cannot know the precise position of a particle, unless we completely abandon any attempt to measure its momentum. The product of error in momentum and error in position of a particle must be larger than Planck's constant divided by 2 pi, this is Heisenberg's Uncertainty Principle.
- There are, in certain systems, small probabilities of classical physics being completely broken. This is called Quantum Tunnelling. For example, there is always a slight probability that a particle will pass through a classically insurmountable barrier (the microscopic and basic equivalent of me appearing next to you, from where I am, with no lead up, right as you finish this sentence).
This is a very quick run through of a SUPER fascinating topic, you should definitely do some extra research if you can! And study Physics at uni, it is awesome ;D
Post by: FallonXay on May 07, 2016, 07:19:35 pm
You have opened Pandora's Box 8)
Okay, so in classical physics we learn about Newton's 2nd Law, which can be used (in various forms) to predict the behaviour of a system over time. Classically, this is absolutely accurate. In quantum mechanics, we cannot predict the outcome of an event until it happens, we can only calculate the probabilities of certain events occurring. This links to the whole idea of Schrodingers Cat.
Picture a cat inside a box (we can't see inside), and next to it is a contraption with a radioactive isotope, a detector, and a flask of poison. Two things in this 'system' can occur. The particle can decay, and if this happens, the resultant gamma radiation will trigger the detector, release the poison, and kill the cat. If the particle does not decay, nothing happens and the cat stays alive.
Certain interpretations of quantum mechanics (Copenhagen Interpretations) would suggest that, since we cannot know whether the cat is alive or dead until we check, the cat is simultaneously alive and dead. This is known as a quantum superposition. Only when we open the box, does this superposition collapse into a reality where the cat is alive (yay!) or dead (wahh...). Really, the cat here is just enabling us to extend an atomic state (whether a particle decays or not) to a macroscopic, real living thing.
So, how do we predict the probability of the cat dying or living, or more generally, how do we predict the outcomes of a system? This is where the Schrodinger Equation comes in. This is a partial differential equation which describes how the quantum state of a system evolves over time. The equation looks like this, the subject of the equation being the weird symbol Psi appearing on both sides of the equation (the rest are an essay in itself)
The subject is the wave equation. Now, the probability of certain quantum states is proportional to the square of this wave function. To solve for the wave equation, therefore, is to have complete understanding of a system and the probability of any event occurring. In some interpretations of quantum physics, the wave function can be used to predict the evolution of HUGE systems, perhaps even the entire universe. Solving the Schrodinger Equation for a particle is hard enough though, for a universe is simply impossible ;D
The Schrodinger Equation has many implications:
- Depending on measurements, the states of a system are quantised, and thus, energy is quantised in all forms (not just for electromagnetic quanta). This is verified already, for example, electron energies in atoms are proven to be quantised.
- Under the Copenhagen interpretation of quantum mechanics, particles do not have set positions, and thus, the result when we measure is drawn from a probability distribution (wave function gives us this). Flowing on from this, we cannot know the precise position of a particle, unless we completely abandon any attempt to measure its momentum. The product of error in momentum and error in position of a particle must be larger than Planck's constant divided by 2 pi, this is Heisenberg's Uncertainty Principle.
- There are, in certain systems, small probabilities of classical physics being completely broken. This is called Quantum Tunnelling. For example, there is always a slight probability that a particle will pass through a classically insurmountable barrier (the microscopic and basic equivalent of me appearing next to you, from where I am, with no lead up, right as you finish this sentence).
This is a very quick run through of a SUPER fascinating topic, you should definitely do some extra research if you can! And study Physics at uni, it is awesome ;D
:o I think my brain just imploded. In a good way :P
Post by: Happy Physics Land on May 07, 2016, 10:09:45 pm
You have opened Pandora's Box 8)
Okay, so in classical physics we learn about Newton's 2nd Law, which can be used (in various forms) to predict the behaviour of a system over time. Classically, this is absolutely accurate. In quantum mechanics, we cannot predict the outcome of an event until it happens, we can only calculate the probabilities of certain events occurring. This links to the whole idea of Schrodingers Cat.
Picture a cat inside a box (we can't see inside), and next to it is a contraption with a radioactive isotope, a detector, and a flask of poison. Two things in this 'system' can occur. The particle can decay, and if this happens, the resultant gamma radiation will trigger the detector, release the poison, and kill the cat. If the particle does not decay, nothing happens and the cat stays alive.
Certain interpretations of quantum mechanics (Copenhagen Interpretations) would suggest that, since we cannot know whether the cat is alive or dead until we check, the cat is simultaneously alive and dead. This is known as a quantum superposition. Only when we open the box, does this superposition collapse into a reality where the cat is alive (yay!) or dead (wahh...). Really, the cat here is just enabling us to extend an atomic state (whether a particle decays or not) to a macroscopic, real living thing.
So, how do we predict the probability of the cat dying or living, or more generally, how do we predict the outcomes of a system? This is where the Schrodinger Equation comes in. This is a partial differential equation which describes how the quantum state of a system evolves over time. The equation looks like this, the subject of the equation being the weird symbol Psi appearing on both sides of the equation (the rest are an essay in itself)
The subject is the wave equation. Now, the probability of certain quantum states is proportional to the square of this wave function. To solve for the wave equation, therefore, is to have complete understanding of a system and the probability of any event occurring. In some interpretations of quantum physics, the wave function can be used to predict the evolution of HUGE systems, perhaps even the entire universe. Solving the Schrodinger Equation for a particle is hard enough though, for a universe is simply impossible ;D
The Schrodinger Equation has many implications:
- Depending on measurements, the states of a system are quantised, and thus, energy is quantised in all forms (not just for electromagnetic quanta). This is verified already, for example, electron energies in atoms are proven to be quantised.
- Under the Copenhagen interpretation of quantum mechanics, particles do not have set positions, and thus, the result when we measure is drawn from a probability distribution (wave function gives us this). Flowing on from this, we cannot know the precise position of a particle, unless we completely abandon any attempt to measure its momentum. The product of error in momentum and error in position of a particle must be larger than Planck's constant divided by 2 pi, this is Heisenberg's Uncertainty Principle.
- There are, in certain systems, small probabilities of classical physics being completely broken. This is called Quantum Tunnelling. For example, there is always a slight probability that a particle will pass through a classically insurmountable barrier (the microscopic and basic equivalent of me appearing next to you, from where I am, with no lead up, right as you finish this sentence).
This is a very quick run through of a SUPER fascinating topic, you should definitely do some extra research if you can! And study Physics at uni, it is awesome ;D
Your pandora's box just caused a supernova in my brain ... every sentence I read felt like a quantum plucking out my hair. I think I will just rewrite the greek mythology here.
Board of Studies sent Pandora down to earth and gave her as a present to Epimetheus. BOSTES told Epimetheus that he should marry Pandora. Also, BOSTES sent Pandora with a little box, with a big lock on it and said not to ever open the box, and he gave the key to Epimetheus. But Pandora was very curious about what was in the box. One day Pandora stole the key and opened the box.
Oh! Under Newton's 2nd law, F-ma, every kind of trouble in the box that people had never known about before experienced a force and jumped out at 300N West! Special theory of relativity, Schrodinger's equation and the quantum theory all began to fly away like little bugs by overcoming the gravitational field and spread all over the place. Pandora was very sorry now that she had opened the box! She tried to catch them and put them back in the box but its too late! They all accelerated in another direction under the slingshot effect and landed in the brains of all NSW students!
But the very last thing to fly out of the box, as Pandora sat there crying, was not as ugly as the others. In fact it was beautiful. It was MATHS EXTENSION II, which Board of Studies sent to keep people going when all the nasty things got them down.
Post by: jamonwindeyer on May 07, 2016, 11:41:57 pm
Your pandora's box just caused a supernova in my brain ... every sentence I read felt like a quantum plucking out my hair. I think I will just rewrite the greek mythology here.
Board of Studies sent Pandora down to earth and gave her as a present to Epimetheus. BOSTES told Epimetheus that he should marry Pandora. Also, BOSTES sent Pandora with a little box, with a big lock on it and said not to ever open the box, and he gave the key to Epimetheus. But Pandora was very curious about what was in the box. One day Pandora stole the key and opened the box.
Oh! Under Newton's 2nd law, F-ma, every kind of trouble in the box that people had never known about before experienced a force and jumped out at 300N West! Special theory of relativity, Schrodinger's equation and the quantum theory all began to fly away like little bugs by overcoming the gravitational field and spread all over the place. Pandora was very sorry now that she had opened the box! She tried to catch them and put them back in the box but its too late! They all accelerated in another direction under the slingshot effect and landed in the brains of all NSW students!
But the very last thing to fly out of the box, as Pandora sat there crying, was not as ugly as the others. In fact it was beautiful. It was MATHS EXTENSION II, which Board of Studies sent to keep people going when all the nasty things got them down.
Made my night ;) also realise the irony of saying that MX2 is your escape from 'nasty' things ;)
Post by: Happy Physics Land on May 08, 2016, 01:09:59 pm
Made my night ;) also realise the irony of saying that MX2 is your escape from 'nasty' things ;)
That situational irony was totally intentional
Post by: jamonwindeyer on May 08, 2016, 09:15:30 pm
That situational irony was totally intentional
You should use that post above as a Related Text for English ;)
Post by: smiley2101 on May 09, 2016, 11:25:32 pm
Hey Smiley!
The answer to this multiple choice is going to be C. We can figure this out by process of elimination: gravity is actually the MAIN force acting on any spacecraft orbiting our planet, as it causes the orbit to occur in the first place (the shuttle is dragged down to earth, but is traveling so fast that it 'misses' the ground due to the curvature of the planet!), so the answer can't be A. The 'orbiting around the Sun' etc. thing doesn't even make any sense, and the whole 'inversely proportional' relationship thing would require some sort of mathematics that told you WHEN the astronaut would stay with the spacecraft, and when it would not. As the forces on the two objects are the same (namely, gravity), there is no reason that their ACCELERATION would be different!.
Remember, gravity has units that are meters per second, per second. There is no 'kilograms' or anything like that in the units, meaning that it operates INDEPENDENTLY of the mass of the object!
Jake :)
but the answer says it is D?
Post by: Happy Physics Land on May 10, 2016, 10:00:49 am
but the answer says it is D?
I cant really see why the answer would be D but according to Newton's 2nd law F=ma, it is true that mass and acceleration would be inversely proportional. Since the gravitational force acting on both the spaceship and the person are constant (since both objects are in the same orbit), the person should be accelerating towards the centre at a higher value than the spaceship which has a much larger mass. I think l agree with with Jake's explanation except for that point.
Post by: jakesilove on May 10, 2016, 10:59:04 am
I cant really see why the answer would be D but according to Newton's 2nd law F=ma, it is true that mass and acceleration would be inversely proportional. Since the gravitational force acting on both the spaceship and the person are constant (since both objects are in the same orbit), the person should be accelerating towards the centre at a higher value than the spaceship which has a much larger mass. I think l agree with with Jake's explanation except for that point.
https://www.youtube.com/watch?v=E43-CfukEgs
An object with a greater mass does not accelerate towards the earth more quickly. The terminal velocity may be different, but as you can see, a bowling ball and a feather falls at the same rate (and by extension, a person and a space ship!). Pretty cool, and pretty counter intuitive.
Jake
Post by: RuiAce on May 10, 2016, 12:23:47 pm
Your pandora's box just caused a supernova in my brain ... every sentence I read felt like a quantum plucking out my hair. I think I will just rewrite the greek mythology here.
Board of Studies sent Pandora down to earth and gave her as a present to Epimetheus. BOSTES told Epimetheus that he should marry Pandora. Also, BOSTES sent Pandora with a little box, with a big lock on it and said not to ever open the box, and he gave the key to Epimetheus. But Pandora was very curious about what was in the box. One day Pandora stole the key and opened the box.
Oh! Under Newton's 2nd law, F-ma, every kind of trouble in the box that people had never known about before experienced a force and jumped out at 300N West! Special theory of relativity, Schrodinger's equation and the quantum theory all began to fly away like little bugs by overcoming the gravitational field and spread all over the place. Pandora was very sorry now that she had opened the box! She tried to catch them and put them back in the box but its too late! They all accelerated in another direction under the slingshot effect and landed in the brains of all NSW students!
But the very last thing to fly out of the box, as Pandora sat there crying, was not as ugly as the others. In fact it was beautiful. It was MATHS EXTENSION II, which Board of Studies sent to keep people going when all the nasty things got them down.
Yeah nah mate Jamon's stuff made more sense
I cant really see why the answer would be D but according to Newton's 2nd law F=ma, it is true that mass and acceleration would be inversely proportional. Since the gravitational force acting on both the spaceship and the person are constant (since both objects are in the same orbit), the person should be accelerating towards the centre at a higher value than the spaceship which has a much larger mass. I think l agree with with Jake's explanation except for that point.
To back up Jake's comment.
F=ma implies that force varies according to mass. But not acceleration.
(Note that the concept of a terminal velocity comes straight out of elementary particle kinematics.)
Post by: jamonwindeyer on May 10, 2016, 01:02:59 pm
but the answer says it is D?
Just to qualify, the answer would be D.
Examine C. It says that the force due to gravity on each the astronaut and the spacecraft is equal. Now we know that gravitational force is equal to the centripetal force in an orbit, so we can equate the force due to gravity to centripetal force for both the astronaut (the top equation) and the spaceship (the bottom):
What we notice is that, for the astronaut and the spaceship to not drift apart, we require their velocities to be equal, and their radius of orbit to be equal. The gravitational force on both must be equal IF C is to be correct.
However, that means EVERYTHING except the masses of the objects has to be equal. This then, for the equations above to be true, means the astronaut must have the same mass as the spaceship. Unless we have a ship made of balsa wood, or the astronaut has eaten too much pie in toothpaste form, this just doesn't make sense. Thus, C has to be incorrect.
D is therefore correct by the process of elimination. And indeed, if we examine the formula for acceleration, and then substitute in centripetal force formula:
This does show that the acceleration of the astronaut/spaceship is inversely proportional to its mass, and then when we substitute, we find that the acceleration of each body is dependent only on velocity and radius of orbit, which are equal by definition.
This last bit is a little roundabout in nature, I personally think that it is easier to answer this one by eliminating options A-C, which is usually the best way to avoid silly mistakes anyway ;) hope this helps!!
Post by: Happy Physics Land on May 10, 2016, 05:07:45 pm
https://www.youtube.com/watch?v=E43-CfukEgs
An object with a greater mass does not accelerate towards the earth more quickly. The terminal velocity may be different, but as you can see, a bowling ball and a feather falls at the same rate (and by extension, a person and a space ship!). Pretty cool, and pretty counter intuitive.
Jake
Right, so Galileo's experiment can equally be applied to space things as well. Yeah sorry I was ignorant that acceleration due to gravity should be of the same value because spaceship and the astronaut are at the same position away from earth
Just to qualify, the answer would be D.
Examine C. It says that the force due to gravity on each the astronaut and the spacecraft is equal. Now we know that gravitational force is equal to the centripetal force in an orbit, so we can equate the force due to gravity to centripetal force for both the astronaut (the top equation) and the spaceship (the bottom):
What we notice is that, for the astronaut and the spaceship to not drift apart, we require their velocities to be equal, and their radius of orbit to be equal. The gravitational force on both must be equal IF C is to be correct.
However, that means EVERYTHING except the masses of the objects has to be equal. This then, for the equations above to be true, means the astronaut must have the same mass as the spaceship. Unless we have a ship made of balsa wood, or the astronaut has eaten too much pie in toothpaste form, this just doesn't make sense. Thus, C has to be incorrect.
D is therefore correct by the process of elimination. And indeed, if we examine the formula for acceleration, and then substitute in centripetal force formula:
This does show that the acceleration of the astronaut/spaceship is inversely proportional to its mass, and then when we substitute, we find that the acceleration of each body is dependent only on velocity and radius of orbit, which are equal by definition.
This last bit is a little roundabout in nature, I personally think that it is easier to answer this one by eliminating options A-C, which is usually the best way to avoid silly mistakes anyway ;) hope this helps!!
I didnt think to jump straight into Fg = mav2/r . I've always been trying to use the formula for orbital velocity which made me think its C
Post by: Maz on May 10, 2016, 06:44:49 pm
hey
could u please help me with this question?
i attached it :)
thankyou so much :)
Post by: jakesilove on May 10, 2016, 08:35:30 pm
hey
could u please help me with this question?
i attached it :)
thankyou so much :)
Hey! To be completely honest, I have never done this kind of physics in depth. After a touch of research, I think I have got an answer out for a) Essentially, you need to sum up the three torques (the first two are obvious, but apparently the third is just the perpendicular distance from the chord to the pivot point, which I guess makes sense, times to tension force) and set it equal to zero. My answer is below.
(http://i.imgur.com/KyZCrIc.jpg)
Apologies for the sideways-ness.
As for the other two parts, I'll keep thinking. My intuition told me that, since the pivot isn't moving, the sum of all forces on it must equal to zero. I can't imagine that's the answer you're looking for, though, but I can't seem to move beyond that mentality.
I'll keep thinking. For HSC students who are skimming through the forum, note that this is not able to be assessed in the HSC!
Jake
Post by: jakesilove on May 10, 2016, 08:46:14 pm
hey
could u please help me with this question?
i attached it :)
thankyou so much :)
Okay, I've found something that appears to answer your question. Now, these are formulas that I have found, and for a more detailed explanation (although, to be honest, not a very satisfactory one), take a look here. However, this appears to be what you need to do.
(http://i.imgur.com/wl8LJdi.jpg)
To be clear, I don't actually know much about 'classical' physics (that's next year for me!). Very likely, I'm just totally wrong.
Jake
Post by: jamonwindeyer on May 10, 2016, 09:49:06 pm
Right, so Galileo's experiment can equally be applied to space things as well. Yeah sorry I was ignorant that acceleration due to gravity should be of the same value because spaceship and the astronaut are at the same position away from earth
I didnt think to jump straight into Fg = mav2/r . I've always been trying to use the formula for orbital velocity which made me think its C
I always had trouble with when to use what fact, many of my practice papers are scribbled with a few different attempts at getting the result I want aha! ;D
Post by: JellyBeanz on May 10, 2016, 09:55:34 pm
I always had trouble with when to use what fact, many of my practice papers are scribbled with a few different attempts at getting the result I want aha! ;D
Jamon, i'm interested in the solution of the question that mq123 posted, can you please have a look at Jake's solution and comment on it?
Thanks :P
Post by: jamonwindeyer on May 10, 2016, 11:45:38 pm
Jamon, i'm interested in the solution of the question that mq123 posted, can you please have a look at Jake's solution and comment on it?
Thanks :P
Sure thing!! Jake and I's Physics background is virtually identical, so I don't think I'll add much ;D
The approach to the first question is (I'm quite sure) definitely correct, I saw something similar in Physics last year. I do think there would be other methods, perhaps examining the components of force on the beam, but I like Jake's approach!
Less sure about the next bit, but I like Jake's approach and the reasoning is sound. However, I think the fact that the sign is not at the centre of the beam will change the answer, because the distance from the pivot point makes a massive difference to the force on the beam.
In general, that sign is what makes this question tricky. I'm not quite sure how to compensate for it. One possibility would be to move the sign to the end of the beam, so that it can be compensated for more easily, then find the lesser, equivalent mass that makes this valid. For example, if we find that moving it to the end of the beam doubles the torque, we could then adjust the mass to return it to its initial value. This would allow us to compensate for the vertical force of the sign in our vector diagrams on the right of the beam, which not changing the overall effect.
Sorry if that last thing was not explained quite nicely, but on the whole, I like Jake's approach! Just that one finer point I'd disagree with, but this is a bit beyond my capabilities ;D
Post by: jakesilove on May 11, 2016, 08:22:40 am
Sure thing!! Jake and I's Physics background is virtually identical, so I don't think I'll add much ;D
The approach to the first question is (I'm quite sure) definitely correct, I saw something similar in Physics last year. I do think there would be other methods, perhaps examining the components of force on the beam, but I like Jake's approach!
Less sure about the next bit, but I like Jake's approach and the reasoning is sound. However, I think the fact that the sign is not at the centre of the beam will change the answer, because the distance from the pivot point makes a massive difference to the force on the beam.
In general, that sign is what makes this question tricky. I'm not quite sure how to compensate for it. One possibility would be to move the sign to the end of the beam, so that it can be compensated for more easily, then find the lesser, equivalent mass that makes this valid. For example, if we find that moving it to the end of the beam doubles the torque, we could then adjust the mass to return it to its initial value. This would allow us to compensate for the vertical force of the sign in our vector diagrams on the right of the beam, which not changing the overall effect.
Sorry if that last thing was not explained quite nicely, but on the whole, I like Jake's approach! Just that one finer point I'd disagree with, but this is a bit beyond my capabilities ;D
Having known really very little about classical mechanics, my reasoning for the final parts goes something like this. The example questions I found (linked above) bears a striking resemblance to our current predicament, except it doesn't include a sign. I think that the horizontal force approach is likely correct (it does not require taking into account any mass, just the Tension force and an angle). However, like Jamon said, the vertical force takes into account the mass of the beam, therefore perhaps the sign changes the formula (ie. it won't be just mass one + mass two * g).
However, that formula also doesn't require use of the length of the bar. Length appears to be independent of the force: all that is required is a Tension force, an angle and a 'mass'.
I'm not sure whether this is true, but this was my line of reasoning. If you add the mass of the sign at a point, this will change the center of mass. The new center (which will likely be somewhere *on* the point of contact between the sign and the rod, a % way along the infinitesimal connection) *should* have half of the mass on its right, and half of its mass on the left. Therefore, you could create a new rod (with a new length) that is the same mass as the original rod (including the sign), with a center of mass at the same point.
This new rod will have a mass of (mass one + mass two), probably be longer, and probably be denser. (Note: If you're worried about the length change affecting tension, just increase the density of the metal. No harm done) However, the formula doesn't care about the length of the rod (or density); only its mass. In all respects, this new rod is the same as the old, so I figure that the formula will still apply.
This could be completely off, I'm just explaining my reasoning from a currently-purely-quantum-mechanical-and-definitely-not-classical perspective.
I don't know if I explained myself well: hopefully it made sense!
Jake
Post by: jamonwindeyer on May 11, 2016, 09:56:54 am
Having known really very little about classical mechanics, my reasoning for the final parts goes something like this. The example questions I found (linked above) bears a striking resemblance to our current predicament, except it doesn't include a sign. I think that the horizontal force approach is likely correct (it does not require taking into account any mass, just the Tension force and an angle). However, like Jamon said, the vertical force takes into account the mass of the beam, therefore perhaps the sign changes the formula (ie. it won't be just mass one + mass two * g).
However, that formula also doesn't require use of the length of the bar. Length appears to be independent of the force: all that is required is a Tension force, an angle and a 'mass'.
I'm not sure whether this is true, but this was my line of reasoning. If you add the mass of the sign at a point, this will change the center of mass. The new center (which will likely be somewhere *on* the point of contact between the sign and the rod, a % way along the infinitesimal connection) *should* have half of the mass on its right, and half of its mass on the left. Therefore, you could create a new rod (with a new length) that is the same mass as the original rod (including the sign), with a center of mass at the same point.
This new rod will have a mass of (mass one + mass two), probably be longer, and probably be denser. (Note: If you're worried about the length change affecting tension, just increase the density of the metal. No harm done) However, the formula doesn't care about the length of the rod (or density); only its mass. In all respects, this new rod is the same as the old, so I figure that the formula will still apply.
This could be completely off, I'm just explaining my reasoning from a currently-purely-quantum-mechanical-and-definitely-not-classical perspective.
I don't know if I explained myself well: hopefully it made sense!
Jake
Ohhh okay, I'm with you, that sounds reasonable!! If it worked in other examples then yep I reckon you are right!! I'm drastically out of my element regardless ;)
Post by: jakesilove on May 11, 2016, 10:25:13 am
Ohhh okay, I'm with you, that sounds reasonable!! If it worked in other examples then yep I reckon you are right!! I'm drastically out of my element regardless ;)
I know exactly what you mean, I'll get back to you with a proper answer in about a year.
Jake
Post by: jamonwindeyer on May 11, 2016, 10:42:08 am
I know exactly what you mean, I'll get back to you with a proper answer in about a year.
Jake
I will hold you to this 8)
Post by: Maz on May 11, 2016, 06:24:20 pm
firstly, thank you for having a go at the question :)
i had another go at it and i did it a different way and then i showed my teacher and he said it's right...i've put what i did below:
the sum of the clockwise torques has to equal the sun of the anticlockwise torques: C.W.T=A.C.W.T
doing that you get
(12kg x 9.8ms-1 x 0.8 (half the beam's length)) + (45kg x 9.8 x 1.3 (1.6-0.3=1.3)) = Tsin20 x 1.6
evaluating that you get; 667.38= Tsin20 x 1.6
that gives the Tension (T) to equal 1.22x 103 N
so thats what u do for the first bit
b) The vertical component is
45x9.8= 441 (down) 12x 9.8= 117.6
so then to get vertical component: 117.6+ 441 = 1.22x 103 N + up
solving that gives a vertical component of 141.34N
Horizontal component
.22x 103 x cos20 = 1146.42
then for part c you just need to find the resultant which can be done with pythagorus. i don't know how to get a triangle on here, but if u visualise a right angled triangle the 141.34N will be the vertical side and the 1146.42 will be the horizontal (adjacent to theta) side. this will leave the hypotenuse:
a2 + b2 =c2
(1.15x103)2 + 141.342 = c2
solving this gives c as 1.16 x 103 which is the magnitude of the force
now as it is a vector we need an angle which is given simply by tan-1(141.5/(1.15x103)
which equals 7.015 degrees
so final answer for part c is 1.16 x 103N at 7.015 degrees
hope this helps :)
Post by: jamonwindeyer on May 11, 2016, 09:46:41 pm
Hey guys
firstly, thank you for having a go at the question :)
i had another go at it and i did it a different way and then i showed my teacher and he said it's right...i've put what i did below:
the sum of the clockwise torques has to equal the sun of the anticlockwise torques: C.W.T=A.C.W.T
doing that you get
(12kg x 9.8ms-1 x 0.8 (half the beam's length)) + (45kg x 9.8 x 1.3 (1.6-0.3=1.3)) = Tsin20 x 1.6
evaluating that you get; 667.38= Tsin20 x 1.6
that gives the Tension (T) to equal 1.22x 103 N
so thats what u do for the first bit
b) The vertical component is
45x9.8= 441 (down) 12x 9.8= 117.6
so then to get vertical component: 117.6+ 441 = 1.22x 103 N + up
solving that gives a vertical component of 141.34N
Horizontal component
.22x 103 x cos20 = 1146.42
then for part c you just need to find the resultant which can be done with pythagorus. i don't know how to get a triangle on here, but if u visualise a right angled triangle the 141.34N will be the vertical side and the 1146.42 will be the horizontal (adjacent to theta) side. this will leave the hypotenuse:
a2 + b2 =c2
(1.15x103)2 + 141.342 = c2
solving this gives c as 1.16 x 103 which is the magnitude of the force
now as it is a vector we need an angle which is given simply by tan-1(141.5/(1.15x103)
which equals 7.015 degrees
so final answer for part c is 1.16 x 103N at 7.015 degrees
hope this helps :)
Ahhhh I see, nicely done! Jake and I were close but no cigar it seems ;) great job!!
Post by: cajama on May 13, 2016, 05:26:27 pm
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.
Post by: RuiAce on May 13, 2016, 05:40:22 pm
Hi I've attached a multiple choice question from the 2008 paper (Q4.)
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.
No. You should be able to quote the relationship between time and vertical displacement in projectile motion.
The equation is given
Δy=uyt + 1/2 ay t2
It is clear that the relationship between height and time is a quadratic. This means that graphically:
If height was on the y-axis, and time on the x-axis, then you should have a parabola
Whereas if you INVERT IT:
If height is on the x-axis, and time is on the y-axis, then you should have a curve that looks like y=sqrt(x)
If you can't recognise the shape of y=sqrt(x) then that's a problem in itself. That's more of a maths problem.
Post by: Alalamc on May 14, 2016, 10:26:41 am
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou :)
Post by: RuiAce on May 14, 2016, 10:28:51 am
Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou :)
There are heaps of ways to do this. One way is just to move a magnet back and forth in a solenoid.
Post by: jamonwindeyer on May 14, 2016, 11:08:40 am
Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou :)
Rui is totally correct! Another (more interesting) way you can do it is get a really long bit of wire (a few metres) and get two people to swing it around like a skipping rope!! The wire is then moving through the earths magnetic field, and it is long enough that you actually do get a (very small) alternating current! You can even determine the direction of the earth's magnetic field by rotating the wire, the most current will be generated when the wire is perpendicular to the magnetic field, so when you have a maximum current, you know the wire is pointing East/West! At least with respect to magnetic north, which isn't perfectly north, but pretty close ;)
Post by: jamonwindeyer on May 14, 2016, 11:20:18 am
Hi I've attached a multiple choice question from the 2008 paper (Q4.)
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.
To flow on from Rui's answer, you definitely don't need to memorise the graph shapes. But don't stress, this is actually a really common thing people struggle with grasping!
There are two ways to think of it. If you are math-oriented, use Rui's approach above which requires the formula on your formula sheet!
Or, if you prefer, consider the graphs with some clever thinking. Note that the question says that the independent variable is graphed on the horizontal axis. The independent variable is what we are changing, that is, the height we are dropping from. Thus, graphs C and D are automatically excluded!! The variables aren't in the right spot. Additionally, graphs C and D are just rotated versions of Graphs A and B, they are perfectly valid graphs! Just not for this question ;D
With this done, we just need to choose between a straight line and a parabola. You can just know it is a parabola if you want. Alternatively, think of it this way. The ball will cover the first 10 metres of drop at some speed, then the next 10 metres faster, since it has accelerated. Then the next 10 metres will be faster again. So on. Every time we add 10 metres to the drop, it increases the time taken, but it increases it by a smaller and smaller increment each time. This way of thinking will give you the answer, Graph A ;D
Hope this helps! I had a bit of trouble with the reason the graphs were parabolic in questions like this when I first studied it, so I hope this helps!! ;D
Post by: Happy Physics Land on May 14, 2016, 08:45:13 pm
Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou :)
Another way to produce alternating current is just to simply turn the switch on and off quickly.
Post by: RuiAce on May 14, 2016, 09:26:02 pm
Another way to produce alternating current is just to simply turn the switch on and off quickly.
This is true if and only if we're speaking in the context of electromagnetic induction and basically a second coil is beside the first. Otherwise we just have a DC circuit that's being damaged because flicking the switch on and off constantly isn't even a good idea to begin with.
Post by: jamonwindeyer on May 14, 2016, 11:42:16 pm
This is true if and only if we're speaking in the context of electromagnetic induction and basically a second coil is beside the first. Otherwise we just have a DC circuit that's being damaged because flicking the switch on and off constantly isn't even a good idea to begin with.
Actually, it does work outside of induction too. Additionally, a switch turned on and off quickly (an input called a variety of names, most appropriately the Rectangular Function) is identical to producing an AC signal at a variety of frequencies, with the amplitude of those signals at some frequency related to the sinc function. So, a DC signal on for a short period is actually like a whole bunch of AC inputs at once ;D This is actually fine, depending on the circuit. For lightbulbs for example, since there is a massive inrush current every time it is switched on, it's definitely not a good idea ;)
Disclaimer: None of that is assessable. Rui, HPL and I tend to go off on tangents a lot.
Post by: smiley2101 on May 15, 2016, 09:37:55 am
Post by: RuiAce on May 15, 2016, 09:50:36 am
hi guys, I don't understand this graph ie why it starts from zero and why the amplitude is different for the first but the next two are the same?? thank you legends!
The reason why it starts at 0 can be analysed in multiple ways depending on which you prefer
The generator starts at the point where there is no change in flux. (Recall that induced EMF is proportional to the CHANGE in flux, not the flux itself.)
The generator starts where the flux passing through it is maximum, hence the change in flux is 0.
The generator is allowing the maximum amount of field lines through the coil. (If instead, it were cutting the field lines, then you would start at the maximum.)
You should be well aware of the typical setup of a generator and when the EMF is max. The analogies are similar to that of motors, and I know that physics in focus has good diagrams for the motor and when torque is maximised.
If the motor undergoes TWO revolutions instead of ONE in the SAME time, then we have effectively increased the SPEED that the motor is being turned at. This creates two implications
- Because the motor is being turned faster (two-fold) now, the period of the graph should decrease (inversely, so 1/2 here). This is simply a direct consequence of the definition of the period of a wave - the time taken for one full oscillation through a point.
- As the motor is being turned faster, the change in flux it experiences increases. The change in flux has to be directly proportional to the speed of which it is being changed at. If the change in flux increases, so must the magnitude of the EMF, thus the amplitude also gets doubled here.
Actually, it does work outside of induction too. Additionally, a switch turned on and off quickly (an input called a variety of names, most appropriately the Rectangular Function) is identical to producing an AC signal at a variety of frequencies, with the amplitude of those signals at some frequency related to the sinc function. So, a DC signal on for a short period is actually like a whole bunch of AC inputs at once ;D This is actually fine, depending on the circuit. For lightbulbs for example, since there is a massive inrush current every time it is switched on, it's definitely not a good idea ;)
Disclaimer: None of that is assessable. Rui, HPL and I tend to go off on tangents a lot.
Ah, fair enough lols
Post by: smiley2101 on May 16, 2016, 07:38:04 pm
Post by: RuiAce on May 16, 2016, 08:55:10 pm
What is the difference between gpe =mgh and The other formula and since a change gpe = work done, why is work done BY gravitational field not counted as work done when there is an obvious change in GPE
For starters, the formula Ep=mgh assumes that we are always on the Earth's surface. The formula E=-GmM/d works for essentially every single planet and wherever you are on the surface.
Whilst the error isn't massive, the formula Ep=mgh is doomed to be inaccurate to some noticeable extent. This formula already treats h with respect to the Earth's "surface" instead of its centre, but the reality is that the Earth simply is not a perfectly spherical body. The 'spheroid' shape of the Earth means that depending on which part of the surface you are on the Earth, there is slight deviation in what mgh apparently computes.
The formula Ep=-GmM/d caters for all of these problems perfectly
Post by: RuiAce on May 16, 2016, 08:56:01 pm
What is the difference between gpe =mgh and The other formula and since a change gpe = work done, why is work done BY gravitational field not counted as work done when there is an obvious change in GPEAlso, this second part of the question made no sense to me. What exactly are you getting at?
Post by: emarmma on May 21, 2016, 05:40:25 pm
how the repulsion force between two magnets can be affected if it was calculated on the North pole and on the equato
Post by: RuiAce on May 21, 2016, 06:16:55 pm
Hey guys,
how the repulsion force between two magnets can be affected if it was calculated on the North pole and on the equato
This is simply because the Earth has its own magnetic field, and at the poles the effect of the Earth's magnetic field is much greater there.
So when you try to calculate a force resulting from a magnetic interaction, the Earth's magnetic field impedes it a bit more at the poles than at the Equator
Post by: smiley2101 on May 21, 2016, 07:52:56 pm
Post by: smiley2101 on May 21, 2016, 08:12:18 pm
Post by: jakesilove on May 21, 2016, 10:18:58 pm
why was an induction coil used to generate voltage in the hertz experiment
Hey!
I'm not really going to answer your question; mainly because I don't know. However, you definitely don't need to know either. Hertz's experiments don't need to be understood in great depth; just get the importance of his 'discoveries', and how that impacted on future scientific exploration. Maybe also try to be a bit clearer with your questioning, just so that we can answer your specific question! Physics is full of difficult concepts, so if you could identify specifically what you're looking for help with we can best assist you.
Jake
Post by: emm22 on May 21, 2016, 10:21:49 pm
Post by: Happy Physics Land on May 21, 2016, 10:56:42 pm
Hi! Could you please explain how Planck's Law solved the Ultraviolet Catastrophe issue?
Hey emm!
Sure thing! Firstly let's just recall Max Planck's first postulate about the quantisation of energy levels in an atom. He stated that energies exist in discrete amounts, not just any random amount, and that in order for an electron to move from one energy level to another, a well-defined amount of energy has to be absorbed or emitted by the electron. Now let's think about the blackbody radiation curve. We know that ultraviolet radiations, X-rays and gamma rays all have low wavelength and hence high frequency, as described by the blackbody curve. According to Planck's equation E=hf, when wavelength becomes increasingly small, frequency would become increasingly large and hence energy would also becoming infinitely large. Such phenomenal energy change would involve a significantly large leap from one energy level to another. However, because such change in energy level is impossible in any atom, high frequency radiations such as UV cannot be emitted and thus for X-rays, UV rays and gamma rays, the intensity is 0. This hence provides an explanation for ultraviolet catastrophe.
Post by: Happy Physics Land on May 21, 2016, 11:03:24 pm
why was an induction coil used to generate voltage in the hertz experiment
Hey Smiley!
Recall the purpose of a transformer in Motors and Generators. The purpose of induction coils is approximately the same as that of a transformer. Basically we want to create a spark (an accelerator of charge) across the air. But air is an insulator and we would need a very high voltage in order to achieve a spark. Our home supply voltage is around 240 V which isn't high enough to cause a discharge to occur. Thus we need an induction coil to provide short high voltage surges to the electrodes at the gap of the transmitting antenna, causing electrons to have enough energy to accelerate through the air, ionising air molecules along the way, facilitating more electrons to travel through the air. With the induction coil, such high voltage cannot be achieved to cause the acceleration of electrons, which is the fundamental mechanism for the production of an electromagnetic wave.
Post by: emarmma on May 22, 2016, 12:12:38 am
This is simply because the Earth has its own magnetic field, and at the poles the effect of the Earth's magnetic field is much greater there.Thank you :) :)
So when you try to calculate a force resulting from a magnetic interaction, the Earth's magnetic field impedes it a bit more at the poles than at the Equator
Post by: RuiAce on May 22, 2016, 09:06:29 am
Hey emm!This last bit sounds a bit out of place. Aren't we looking for how the UV catastrophe is resolved?
Sure thing! Firstly let's just recall Max Planck's first postulate about the quantisation of energy levels in an atom. He stated that energies exist in discrete amounts, not just any random amount, and that in order for an electron to move from one energy level to another, a well-defined amount of energy has to be absorbed or emitted by the electron. Now let's think about the blackbody radiation curve. We know that ultraviolet radiations, X-rays and gamma rays all have low wavelength and hence high frequency, as described by the blackbody curve. According to Planck's equation E=hf, when wavelength becomes increasingly small, frequency would become increasingly large and hence energy would also becoming infinitely large. Such phenomenal energy change would involve a significantly large leap from one energy level to another. However, because such change in energy level is impossible in any atom, high frequency radiations such as UV cannot be emitted and thus for X-rays, UV rays and gamma rays, the intensity is 0. This hence provides an explanation for ultraviolet catastrophe.
Physics in focus definition here.
When a black body is heated to some temperature
in a vacuum, for example, by electric heating, it starts to emit radiation perfectly, known as black body radiation. This radiation can cover the entire range of the EMR spectrum with the intensity varying with the wavelength. If the individual wavelengths of this radiation are detected and the corresponding intensities measured experimentally, the data can then be plotted as ‘intensity’ versus
‘wavelength’, to produce a black body radiation curve.
in a vacuum, for example, by electric heating, it starts to emit radiation perfectly, known as black body radiation. This radiation can cover the entire range of the EMR spectrum with the intensity varying with the wavelength. If the individual wavelengths of this radiation are detected and the corresponding intensities measured experimentally, the data can then be plotted as ‘intensity’ versus
‘wavelength’, to produce a black body radiation curve.
The UV catastrophe arises when we apply classical physics calculations. If we use this, we deduce that the intensity of the EMR emitted should approach infinity as we reach the UV end of the EMS. But experimentally this was not true; for a certain temperature given, there would be a corresponding EMR wavelength that was emitted the most (graphically, this means that the blackbody radiation curve had a maximum point).
E=hf, a fundamental equation of quantum physics, was introduced to resolve this dilemma, not to explain it.
Hi! Could you please explain how Planck's Law solved the Ultraviolet Catastrophe issue?Now, I am not sure as to what exactly you mean by Planck's Law. Planck's law is described using much more complicated equations in Wikipedia and is far out of the scope of the syllabus. https://en.wikipedia.org/wiki/Planck%27s_law
I do not recall E=hf being given a name in the course, but Wikipedia also states that E=hf is called the "Planck-Einstein Relation"
Post by: Happy Physics Land on May 22, 2016, 12:33:23 pm
This last bit sounds a bit out of place. Aren't we looking for how the UV catastrophe is resolved?Physics in focus definition here.When a black body is heated to some temperature
in a vacuum, for example, by electric heating, it starts to emit radiation perfectly, known as black body radiation. This radiation can cover the entire range of the EMR spectrum with the intensity varying with the wavelength. If the individual wavelengths of this radiation are detected and the corresponding intensities measured experimentally, the data can then be plotted as ‘intensity’ versus
‘wavelength’, to produce a black body radiation curve.
The UV catastrophe arises when we apply classical physics calculations. If we use this, we deduce that the intensity of the EMR emitted should approach infinity as we reach the UV end of the EMS. But experimentally this was not true; for a certain temperature given, there would be a corresponding EMR wavelength that was emitted the most (graphically, this means that the blackbody radiation curve had a maximum point).
But classical physics couldnt explain the ultraviolet catastrophe and it was only explained through Planck's equation E=hf, implying that there is 0 intensity for high frequency radiations because none of these radiations can be released due to a significant change in energy levels being impossible in any atom.
Post by: RuiAce on May 22, 2016, 12:37:01 pm
But classical physics couldnt explain the ultraviolet catastrophe and it was only explained through Planck's equation E=hf, implying that there is 0 intensity for high frequency radiations because none of these radiations can be released due to a significant change in energy levels being impossible in any atom.
UV catastrophe was never real. It was a problem resulted from classical physics as you stated.
It's not really "explained" per se. More like corrected or justified.
Post by: smiley2101 on May 22, 2016, 03:32:01 pm
Post by: RuiAce on May 22, 2016, 03:58:49 pm
Thank you so much! You guys are lifesavers! Just another question- I don't really understand the black body radiation curve. How was the UV catastrophe line calculated in the first place (meaning why did scientist think this was what happened) and why is the actual graph a bell curve :) this topic is so hard
The actual graph is out of genuine experimentation. It just happened to be verified by Planck's idea that energy was quantised.
Classical calculations aren't really important but they basically predicted that smaller wave length would always mean intensity of EMR released is greater.
Post by: smiley2101 on May 22, 2016, 04:53:43 pm
Post by: RuiAce on May 22, 2016, 06:18:55 pm
Thanks rui! Also I'm finding it hard to see the connection with why the syllabus is asking us to look at cathode Ray's, black body and the photoelectric effect? Is there a connection?
Not really.
The topic is "From Ideas to Implementation". But in reality whilst they are all ideas it's really just four cleverly chosen ideas for the course in my opinion.
Post by: soldmychildforyeezys on May 22, 2016, 10:34:55 pm
I'm having trouble determining the change in flux just by analysing a simple scenario by eye.
For example, a bar magnet that passes through a coil moving right. When it exits the coil, I know the magnetic flux is decreasing. But is it decreasing to the left or to the right? Does this tell you the direction of the change in flux as well?
Even though the bar magnet is moving to the right, it doesn't mean the flux is decreasing in the same direction correct?
This problem makes it hard for me to answer worded problems. I also know that the magnetic field direction would be decreasing, but not sure about the direction. Hence the magnetic flux has to decrease, again not sure about the direction. Hence can't find the direction of the induced magnetic field, there can't find direction of induced current.
Sorry, this is a lot longer than it had to be. Probably because of how confused I am rip.
Post by: jamonwindeyer on May 22, 2016, 10:46:35 pm
Not really.
The topic is "From Ideas to Implementation". But in reality whilst they are all ideas it's really just four cleverly chosen ideas for the course in my opinion.
BOSTES perception, they are all advances in physics that resulted in useful applications for society ;D
Post by: jamonwindeyer on May 22, 2016, 10:51:03 pm
The actual graph is out of genuine experimentation. It just happened to be verified by Planck's idea that energy was quantised.
Classical calculations aren't really important but they basically predicted that smaller wave length would always mean intensity of EMR released is greater.
It isn't exactly necessary knowledge specifically, but the classical law which predicted the incorrect 'Catastrophe Curve' was Rayleigh Jeans Law, which was based on classical understanding of the propagation of waves in space ;D
Post by: jamonwindeyer on May 22, 2016, 11:05:36 pm
Hi,
I'm having trouble determining the change in flux just by analysing a simple scenario by eye.
For example, a bar magnet that passes through a coil moving right. When it exits the coil, I know the magnetic flux is decreasing. But is it decreasing to the left or to the right? Does this tell you the direction of the change in flux as well?
Even though the bar magnet is moving to the right, it doesn't mean the flux is decreasing in the same direction correct?
This problem makes it hard for me to answer worded problems. I also know that the magnetic field direction would be decreasing, but not sure about the direction. Hence the magnetic flux has to decrease, again not sure about the direction. Hence can't find the direction of the induced magnetic field, there can't find direction of induced current.
Sorry, this is a lot longer than it had to be. Probably because of how confused I am rip.
Hey there! It sounds like you've got a lot running through your head, but I think the focus of your question is the end sentence:
...direction of induced current.
That's the big thing that these sort of questions require. Direction of change of magnetic flux isn't really a thing, and the change in magnitude of the flux is something it sounds like you have a grasp of. Let me try and offer a solution to finding the direction of induced current, based on a more practical application based around Lenz's Law.
Consider the scenario you proposed, a bar magnet moving away from a coil. Let's assume the north pole is pointed towards the coil, and that it is moving away on the right side (that is, moving away to the right).
What we have here is a north pole moving away from the coil. Lenz's Law states that induced currents will oppose the change that created them. So, since the north pole is moving away, an induced current will act to bring the north pole back towards the coil.
What this means is that the induced current will set up a South pole on the right hand side of the coil, nearest the magnet, to pull the north pole back. Therefore, the north pole is on the left. We can now use the Right Hand Grip Rule to find the direction of current. The way it works is simple; take your right hand and give a thumbs up to yourself (yay Physics). The thumb points in the direction of the north pole. The fingers wrap around in the direction of current.
Using this analysis, we can find the direction of induced current in, essentially, two quick steps.
I hope this helps with the confusion. Again, the direction of change of magnetic flux isn't really a question that gets asked in the HSC. A little too ambiguous. Change in magnitude is asked, but this is okay, since if the magnet is moving away from the coil then the magnitude is decreasing. I mean, you could interpret it is something similar to: a magnetic flux to the left, decreasing, but yeah. Not really a thing to my knowledge ;D
Post by: soldmychildforyeezys on May 22, 2016, 11:16:54 pm
To clarify one last thing, is there a change in magnetic field when the magnet leaving further away from the coil in your example? I think I was confusing myself with thinking that the magnetic field will decrease when it moves further away, thus the flux would decrease. Therefore the induced current would create an induced flux to reinforce this decreasing flux.
Post by: jakesilove on May 22, 2016, 11:17:53 pm
Thanks man!
To clarify one last thing, is there a change in magnetic field when the magnet leaving further away from the coil in your example? I think I was confusing myself with thinking that the magnetic field will decrease when it moves further away, thus the flux would decrease. Therefore the induced current would create an induced flux to reinforce this decreasing flux.
Whilst I'll let Jamon answer your question, just wanted to jump in and say that I love you username. Bit too real though aha.
Jake
Post by: jamonwindeyer on May 23, 2016, 03:24:06 pm
Thanks man!
To clarify one last thing, is there a change in magnetic field when the magnet leaving further away from the coil in your example? I think I was confusing myself with thinking that the magnetic field will decrease when it moves further away, thus the flux would decrease. Therefore the induced current would create an induced flux to reinforce this decreasing flux.
No problem! Sorry if I've misinterpreted, is the question whether a change in magnetic flux occurs for a magnet moving away from the coil? If so, then yes absolutely! The magnitude of magnetic flux through the coil (and thus, magnitude of magnetic field strength) will decrease. Sorry for confusion, is that what you mean? ;D
Post by: soldmychildforyeezys on May 23, 2016, 05:23:07 pm
No problem! Sorry if I've misinterpreted, is the question whether a change in magnetic flux occurs for a magnet moving away from the coil? If so, then yes absolutely! The magnitude of magnetic flux through the coil (and thus, magnitude of magnetic field strength) will decrease. Sorry for confusion, is that what you mean? ;DNo worries, it was probably me being a bit confusing.
Last thing, so for a bar magnet moving through the coil. For your example, if the poles were reversed in the bar magnet. This would cause the induced magnetic field to be in the opposite direction, correct?
I think I was confusing myself by taking the direction of the moving bar magnet as the direction it was decreasing, opposed to looking at the actual magnetic field line directions.
Post by: Happy Physics Land on May 23, 2016, 07:39:38 pm
I have a physics research assessment coming up and there is a section that asks me to assess the validity, reliability and accuracy of the websites that I collected my information from. What do you guys reckon I should definitely have to evaluate these features?
Thank you lads :D :D
Best Regards
Happy Physics Land
Post by: RuiAce on May 23, 2016, 07:44:43 pm
Hey Guys!
I have a physics research assessment coming up and there is a section that asks me to assess the validity, reliability and accuracy of the websites that I collected my information from. What do you guys reckon I should definitely have to evaluate these features?
Thank you lads :D :D
Best Regards
Happy Physics Land
For a second hand investigation:
Validity - Does the information you obtain actually address the criteria of the question
Reliability - Is there consistency in your sources (cross referencing would be a good idea here...)
Accuracy - This is a bit hard for a second hand investigation as the accuracy of what you obtain is really subject to question - how can you really be sure of if you're obtaining the most accurate data. Perhaps reputable sources is what you want here
Post by: Happy Physics Land on May 23, 2016, 08:10:36 pm
For a second hand investigation:
Validity - Does the information you obtain actually address the criteria of the question
Reliability - Is there consistency in your sources (cross referencing would be a good idea here...)
Accuracy - This is a bit hard for a second hand investigation as the accuracy of what you obtain is really subject to question - how can you really be sure of if you're obtaining the most accurate data. Perhaps reputable sources is what you want here
Thanks heaps Rui! But do you reckon l should be doing cross referencing for accuracy and reputable sources for reliability??
Post by: RuiAce on May 23, 2016, 08:11:41 pm
Thanks heaps Rui! But do you reckon l should be doing cross referencing for accuracy and reputable sources for reliability??
Yep. They do mix between each other for a second hand investigation
Post by: jamonwindeyer on May 23, 2016, 08:39:50 pm
No worries, it was probably me being a bit confusing.
Last thing, so for a bar magnet moving through the coil. For your example, if the poles were reversed in the bar magnet. This would cause the induced magnetic field to be in the opposite direction, correct?
I think I was confusing myself by taking the direction of the moving bar magnet as the direction it was decreasing, opposed to looking at the actual magnetic field line directions.
You are spot on! The magnitude would be identical, but the field lines would be in the opposite direction, and so everything would swap!! ;D
Post by: jamonwindeyer on May 23, 2016, 08:42:20 pm
For a second hand investigation:
Validity - Does the information you obtain actually address the criteria of the question
Reliability - Is there consistency in your sources (cross referencing would be a good idea here...)
Accuracy - This is a bit hard for a second hand investigation as the accuracy of what you obtain is really subject to question - how can you really be sure of if you're obtaining the most accurate data. Perhaps reputable sources is what you want here
Yeah I concur, I did a massive research journal task for my HSC, and I took the approach of assuming that reputable sources were accurate, and through cross referencing with these sources, I could evaluate the accuracy of other sources. Another potential idea for you HPL ;D
Post by: Happy Physics Land on May 23, 2016, 09:04:20 pm
Yeah I concur, I did a massive research journal task for my HSC, and I took the approach of assuming that reputable sources were accurate, and through cross referencing with these sources, I could evaluate the accuracy of other sources. Another potential idea for you HPL ;D
That's what I am trying to do right now hehehehe
But thanks guys for your assistances, they were quite helpful!
Post by: jakesilove on May 23, 2016, 09:22:18 pm
That's what I am trying to do right now hehehehe
But thanks guys for your assistances, they were quite helpful!
From my understanding, cross-referencing is reliability and credentials of author etc. is accuracy. Like everyone else has said, these readily cross over, but I know at least that I was marked correctly when I used the above methodology. Regardless, it's all about waffling on for a paragraph or two, so I'm sure you'll manage to get the marks!
Jake
Post by: Happy Physics Land on May 23, 2016, 09:32:30 pm
From my understanding, cross-referencing is reliability and credentials of author etc. is accuracy. Like everyone else has said, these readily cross over, but I know at least that I was marked correctly when I used the above methodology. Regardless, it's all about waffling on for a paragraph or two, so I'm sure you'll manage to get the marks!
Jake
Maybe I will just do cross-referencing and credentials for both reliability and accuracy to cover my bases
Post by: jamonwindeyer on May 23, 2016, 09:36:14 pm
waffling on for a paragraph or two
The only time this works in HSC Physics 8)
Post by: jakesilove on May 23, 2016, 09:36:57 pm
Maybe I will just do cross-referencing and credentials for both reliability and accuracy to cover my bases
Alternatively, claim that you tried to do research for this area of the curriculum. Say you received multiple solutions to the same problem when cross-checking your information, but all of the people offering ideas had fantastic credentials.
Accuracy = Good? Bad? Reliability = Good? Bad?
????
Profit.
Post by: Neutron on May 25, 2016, 02:39:09 pm
I have a prac assessment on Friday (it's a written prac, so like data analysis as opposed to some hands on stuff :/) and we basically have to know all the pracs we've done in Motors and Generators as well as Ideas to Implementation.. For the prac we did for Hertz' experiment (basically we had an induction coil and a nearby radio and we just listened to the static when the induction coil was turned on), I read somewhere that the lower frequency EM wave produced but the induction coil, the larger the static noise is? I was wondering why that is.. And also, does the induction coil have to be connected to an AC or DC supply? Or does it not matter? Thank you guys so much! Ahh I'm having trouble preparing for this assessment cause normally I just do past papers but this time the teacher's only provided one so yeah.. How do you guys normally prepare for written pracs?
Thank youuu! :')
Neutron
Post by: jamonwindeyer on May 25, 2016, 06:33:22 pm
Yo yo yo, it's been a while! :D
I have a prac assessment on Friday (it's a written prac, so like data analysis as opposed to some hands on stuff :/) and we basically have to know all the pracs we've done in Motors and Generators as well as Ideas to Implementation.. For the prac we did for Hertz' experiment (basically we had an induction coil and a nearby radio and we just listened to the static when the induction coil was turned on), I read somewhere that the lower frequency EM wave produced but the induction coil, the larger the static noise is? I was wondering why that is.. And also, does the induction coil have to be connected to an AC or DC supply? Or does it not matter? Thank you guys so much! Ahh I'm having trouble preparing for this assessment cause normally I just do past papers but this time the teacher's only provided one so yeah.. How do you guys normally prepare for written pracs?
Thank youuu! :')
Neutron
Sup Neutron!! Long time no see ;)
Okay, so about the low frequency EM producing larger radio static, that is simply because radio waves ARE extremely low frequency waves. I'd imagine that the higher you go, you start to slide out of the range of the receiver in the radio. I have no idea what this range would be, but that would be my response. The radio receiver would be tuned to listen to those low frequency waves :D
Usually induction coils are connected to a DC power supply. A special type of transformer apparatus is used to 'step up' a reasonably low DC voltage to one high enough to induce sparking. The specifics of construction and how it works are a little complex, but yep, DC ;D
Edit: It might still actually be connected to AC mains and then rectified by the apparatus, I think it depends on the specific coil.
Practical tests are tricky. I'd brush up on your terms like accuracy, validity, reliability, and practice suggesting improvements to experiment design. Beyond that, remember all the practical tasks you've done in class; including the dependent and independent variables in each case, and be ready to analyse the data from those prices in some way. Not much else you can do for these, I never did a fully written practical task, so maybe someone else might have some better advice. Hope this helps Neutron!! ;D
Post by: Neutron on May 26, 2016, 01:59:53 am
Neutron
Post by: jamonwindeyer on May 26, 2016, 07:52:36 am
Ohh thank you so much! That was super helpful :D I have another question though.. Is there a relationship between the spark gap and the maximum distance the static can be received? If so, what is the relationship? Is it the larger the spark gap, the greater the distance? I have no idea, I'm making it up :') Does widening the spark gap increase the wavelength of the radio wave produced or something? Thank you so much guys!
Neutron
Just as a warning, you are straying into non-assessable content here.
Well, a larger spark gap would mean a higher breakdown voltage, which would mean a greater acceleration of electrons when the spark occurs. Thus, yes, the magnitude of the resultant EM waves would increase with the size of the spark gap, at least quantitatively, I'm not sure of the nature of the relationship there, but that would be my best guess ;D
As an edit, in terms of your question, this wouldn't likely translate to any measurable difference in the distance you can detect the static. Especially radio waves.
It's important to note that Spark Gap Transmitters (which is what you are looking at here) radiate EM energy across the EM spectrum, it just so happens that it is focused at the lower (radio) end. That's why you can hear the static interference on any radio channel (PS - the frequencies generated more intensely by these sort of gaps are now internationally banned for radio communication, because of this interference issue.[/b] Basically, the frequency spectrum of the waves is dependent on the construction of the coil. The EM wave production is caused by the oscillation of electrical charge, assisted by circuit elements called inductors and capacitors, and it is the value of these elements which affects the frequency. I don't think the size of the gap would have a measurable effect :)
Post by: FallonXay on June 07, 2016, 08:07:09 pm
Post by: jamonwindeyer on June 07, 2016, 08:11:20 pm
Hihi! In regards to the Braggs' Diffraction Experiment, why did they specifically use X-rays as opposed to other forms of EMR?
Hey FallonXay!! The reason is simply because the wavelength of an X-ray is very similar to the average spacing between atoms in a metallic crystal lattice, thus maximising the effects of scattering and creating a more accentuated diffraction pattern ;D I know this principle has a name, but it eludes me, and the specifics of it are beyond the syllabus anyway ;D
Post by: jakesilove on June 07, 2016, 08:37:28 pm
Hey FallonXay!! The reason is simply because the wavelength of an X-ray is very similar to the average spacing between atoms in a metallic crystal lattice, thus maximising the effects of scattering and creating a more accentuated diffraction pattern ;D I know this principle has a name, but it eludes me, and the specifics of it are beyond the syllabus anyway ;D
Yep, can confirm Jamon's answer having just recently done exactly this experiment! We fired x-rays into Sodium and Potassium, and used the fact that the wavelength was similar to the spacing to do crazy calculations. Principle-wise, we just used basic x-ray diffraction and various iterations of Braggs law :) Maybe some Compton shift stuff in there as well, but again this is well beyond the syllabus! For that matter, you hardly need to know anything about these experiments for the HSC; hardly ever assessed, barely even taught.
Jake
Post by: jamonwindeyer on June 07, 2016, 08:39:11 pm
Yep, can confirm Jamon's answer having just recently done exactly this experiment! We fired x-rays into Sodium and Potassium, and used the fact that the wavelength was similar to the spacing to do crazy calculations. Principle-wise, we just used basic x-ray diffraction and various iterations of Braggs law :) Maybe some Compton shift stuff in there as well, but again this is well beyond the syllabus! For that matter, you hardly need to know anything about these experiments for the HSC; hardly ever assessed, barely even taught.
Jake
Oh that is so cool. My experiments in Electrical are restricted to dodgy circuits on breadboards 8)
Post by: jakesilove on June 07, 2016, 08:41:55 pm
Oh that is so cool. My experiments in Electrical are restricted to dodgy circuits on breadboards 8)
Oh don't worry; next year I get to use a pendulum to find an approximation for Gravity again, so I have plenty of shit pracs to look forward to!
Post by: Neutron on June 13, 2016, 02:53:07 am
Been a while again, but I have a research assignment on Q2Q and I'm having some trouble with one of the questions and i was wondering whether you amazing genuises could shed some light? :o
"Explain how the rate of fission is controlled in a nuclear reactor and how adjustments are made over the lifetime of the fuel rods."
It's only the underlined part that I'm having trouble with and if you guys have could please help a gal out that would be great! Thank you!
Neutron
Post by: jakesilove on June 13, 2016, 10:13:49 am
Hey hey!
Been a while again, but I have a research assignment on Q2Q and I'm having some trouble with one of the questions and i was wondering whether you amazing genuises could shed some light? :o
"Explain how the rate of fission is controlled in a nuclear reactor and how adjustments are made over the lifetime of the fuel rods."
It's only the underlined part that I'm having trouble with and if you guys have could please help a gal out that would be great! Thank you!
Neutron
Hey Neutron!
I didn't do Q2Q myself, and haven't studied this area of physics at university, so nothing special to offer you. I did some research, and the link here has a paragraph about turning on and off the rods in order to adjust for changes in the rate of fission (to ensure a critical yield etc.). All other research (eg here) seems to support this idea; that Fuel rods are basically turned 'on and off' in order to stabalise the rate of reaction.
Hope that helps! Hopefully someone on here did Q2Q and can help out a little more!
Jake
Post by: znaser on June 21, 2016, 07:10:40 pm
Post by: jamonwindeyer on June 21, 2016, 10:15:31 pm
Hi. I'm just asking in regards to the formation of electron-hole pairs in extrinsic semi-conductors. So I understand that it is the movement of the positive holes in the valence band that constitutes current but I am confused as to whether there are movement of electrons in the conduction band.
Hey there!! So that's a pretty tricky question to answer, but essentially, no. However, remember that the holes moving in the valence band means that there will also be electrons moving in the valence band, in opposite directions! You might understand why this is already, but if not let me know!!
Now, this doesn't mean that there won't be some electrons moving in the conduction band. However, we call holes the majority charge carriers, because the majority of the charge movement is due to the hole movement in the valence band.
And just confirming, this covers the p-type extrinsic case. Is this what you were after? ;D
I'm also confused about the production of electron-hole pairs in an intrinsic semi-conductor. Is it that the electrons move from the valence band to the conduction band (when exposed to a high energy source) and leave behind a hole in the valence band and the process just continues or does it mean something else? Thanks in advance :)
Okay, so basically you've got the idea. Electrons graduate from the valence band to the conduction band via some external energy source (like a voltage), and yes. This leaves a hole in the valence band. Then these carry charge in the valence band. So in intrinsic semiconductors, we have a combination of conduction electrons and valence holes carrying charge for us. It is only when we dope the semiconductor that one starts to dominate over the other ;D
I hope this helps!! I'm happy to clarify if necessary ;D
Post by: znaser on June 22, 2016, 07:16:15 am
Post by: jamonwindeyer on June 22, 2016, 02:08:26 pm
Yep that's exactly what I'm after. Thank you that rlly helps! ;D
Awesome! No worries at all ;D
Post by: conic curve on June 22, 2016, 04:15:15 pm
Post by: Ahsun on June 22, 2016, 07:11:43 pm
I just need a guide on what i need to do i was thinking something to do with newtons thought expirements but not sure about the calculation part.
Post by: Swagadaktal on June 22, 2016, 09:24:22 pm
Physics here seems so interesting in comparison to vce physics.
if i didnt have an amazing teacher i dont think id be enjoying physics as much as i do atm but based off the questions in the hsc physics you guys seem like you're having a blast. some really fascinating ideas
Post by: jamonwindeyer on June 22, 2016, 10:51:38 pm
Is this for prelim physics or HSC physics?
Both! We even have some inter-state students asking questions (which we answer with some success), so ask away! ;D
Post by: jamonwindeyer on June 22, 2016, 11:06:10 pm
At one time it was proposed to launch satellites directly into a circular orbit at a constant distance from the Earth using a giant WW2 cannon. Assuming that the cannon would give the satellite a largely sufficient velocity, explain and use calculation to show why this proposal would not work ?
I just need a guide on what i need to do i was thinking something to do with newtons thought expirements but not sure about the calculation part.
Hey there Ahsun, welcome to the forums! ;D
Hmm, this question could be heading a few ways, but I think what it is looking at is the accelerations involved! Basically, your response to this question will revolve around the fact that the acceleration involved will create G-forces that cannot be tolerated by a satellite (or, well, anything). Let's do some calculations. The orbital velocity of the satellite:
Let's say we want the satellite in a low earth orbit, so put the radius equal to 300 000 metres. Thus, the velocity could approximated to be:
Now, a WW2 cannon would accelerate a satellite from rest, to this velocity, instantly. Let's be generous and say it takes 0.1 seconds, and let's say we have a 1000kg satellite. So we can say:
So, this is close to the gravitational force between the earth and the moon, experienced by something the size of a car. No chance in hell, torn to pieces.
The calculation might be a tad confusing (if so, let me know), and it's just guesswork for the most part ;D Importantly though, we recognise that using a cannon creates too much acceleration, and thus too much G-force for any satellite to withstand. Current launch techniques are designed to spread the acceleration over as much time as possible.
Just as an addition, the sort of forces above would turn a human into a smoothie.
This would be my interpretation!! Feel free to play around with the values in the calculation, I hope this helps! ;D
Post by: jamonwindeyer on June 22, 2016, 11:07:57 pm
hey can someone link me the hsc physics study design (or equivalent thing?)
Physics here seems so interesting in comparison to vce physics.
if i didnt have an amazing teacher i dont think id be enjoying physics as much as i do atm but based off the questions in the hsc physics you guys seem like you're having a blast. some really fascinating ideas
Hey!! The Prelim/HSC syllabus is here. Scroll down a bit through all the fluff ;) it's a super interesting course!! It definitely doesn't prepare you for tertiary physics very well, but in terms of generating interest and getting a broad understanding of how our world works, it's fantastic ;D
Post by: Ahsun on June 23, 2016, 07:07:10 pm
Hey there Ahsun, welcome to the forums! ;D
Hmm, this question could be heading a few ways, but I think what it is looking at is the accelerations involved! Basically, your response to this question will revolve around the fact that the acceleration involved will create G-forces that cannot be tolerated by a satellite (or, well, anything). Let's do some calculations. The orbital velocity of the satellite:
Let's say we want the satellite in a low earth orbit, so put the radius equal to 300 000 metres. Thus, the velocity could approximated to be:
Now, a WW2 cannon would accelerate a satellite from rest, to this velocity, instantly. Let's be generous and say it takes 0.1 seconds, and let's say we have a 1000kg satellite. So we can say:
So, this is close to the gravitational force between the earth and the moon, experienced by something the size of a car. No chance in hell, torn to pieces.
The calculation might be a tad confusing (if so, let me know), and it's just guesswork for the most part ;D Importantly though, we recognise that using a cannon creates too much acceleration, and thus too much G-force for any satellite to withstand. Current launch techniques are designed to spread the acceleration over as much time as possible.
Just as an addition, the sort of forces above would turn a human into a smoothie.
This would be my interpretation!! Feel free to play around with the values in the calculation, I hope this helps! ;D
Thanks so much this really helped me :)
Post by: jamonwindeyer on June 23, 2016, 07:08:36 pm
Thanks so much this really helped me :)
Anytime! I hope it all made sense, feel free to let me know if you needed any help finding anything around the forums ;D
Post by: soldmychildforyeezys on June 23, 2016, 08:03:27 pm
For these questions http://m.imgur.com/a/5WrVp
For Q55) I just want to check if my answer is correct, since the book states that Lenz's law determined the direction of the induced current which is given by the induced magnetic field. Would my answe be fine.
For Q56&57) I seem to have the opposites of the book's easier for both. How would the the induced magnetic field be to the right in 56, and anticlockwise for 57?
Cheers
Post by: Ahsun on June 23, 2016, 08:07:19 pm
Not sure exactly what I need to do any help would be appreciated
Thanks
Post by: jamonwindeyer on June 23, 2016, 08:15:02 pm
Hi
For these questions http://m.imgur.com/a/5WrVp
For Q55) I just want to check if my answer is correct, since the book states that Lenz's law determined the direction of the induced current which is given by the induced magnetic field. Would my answe be fine.
For Q56&57) I seem to have the opposites of the book's easier for both. How would the the induced magnetic field be to the right in 56, and anticlockwise for 57?
Cheers
Hey!!
For Q55, It's spot on in principle, though I think you might want to explain what Lenz' Law is a little bit better. I'd be doing something like:
Lenz's Law is the principle that any change must oppose the change that created it (due to the conservation of energy). In this experiment...
Just to make sure I got the 2 marks on offer ;D
For both 56 and 57, I agree with the book. Let me quickly explain them both using the Right Hand Grip Rule, have you heard of it?
If not, basically, hold your right hand in a thumbs up position. Looking at your hand, if your thumb points in the direction of the north pole of the induced field, then your fingers wrap in the direction of induced current.
Let's start with 57 because it's a tad easier. The south pole is moving away from the loop, so, the loop will want to set up a North pole facing towards the magnet to bring it back in (Lenz's Law). So, your thumb is pointing to the left of the page in this case. Thus, your current is wrapping anti-clockwise as viewed by the magnet (sort of hard to explain, but the opposite direction to what you drew).
Step 56 is similar. South Pole is coming in, so the loop will set up a South pole to push it back away. Magnetic field lines go towards the South pole. Thus, the magnetic field lines will be from left to right (assuming we only want the field lines of the induced field, if we considered the field of the magnet too we'd have some interactions to draw, which is complicated) ;D
I hope this helps!! If it doesn't seriously let me know and I'll break it down a little bit further :)
Post by: soldmychildforyeezys on June 23, 2016, 08:34:42 pm
Would my method by wrong entirely? Since I considered the direction of the bar magnet's magnetic field lines when looping back in to its South Pole, hence just said the induced magnetic field would oppose this..
Hm, I might've confused myself since when I learnt how to solve these questions initially, I might've just considered the direction of the magnetic field lines coming straight out of the North Pole. I think this is pretty much the way you did it.
Thanks again!
Post by: jamonwindeyer on June 23, 2016, 08:42:19 pm
Mathematically analyse, using diagrams, a rocket burning in terms of momentum conservation and the rate of fuel consumption R and derive the acceleration for a rocket during its launch stage.
Not sure exactly what I need to do any help would be appreciated
Thanks
Hey Ahsun! Ouch, that's a nasty question. I really like these though, they force you to be really analytical. Okay, well, heres a few ideas I have and I'll leave you to tie them together in a way that makes sense to you! ;D
For the diagram, I'd draw a simple picture of a rocket moving in one direction and the fuel exhaust moving in the other direction. Label it with any of the information I mention here that you think suits.
The big principle here is the Conservation of Momentum. The rocket and its fuel start from rest (that is, with zero momentum, since momentum is a product of velocity and mass. Therefore, the total momentum of the rocket and its fuel must remain zero, by the conservation of momentum. Of course, the rocket does move, so the solution is this: The momentum of the rocket must be equal and opposite to the momentum of the fuel being ejected in the other direction.
We can get a few expressions here in terms of R. The mass of the rocket will decrease by R every second, since it is ejecting R kilograms of fuel per second. So, if we let M be the total mass at the start, and t be time:
By the same logic, the mass of the fuel:
The momentum formula would now be:
Note that Rt < M since Rt was included in M.
So theres a few ideas. Let's derive that acceleration. The force upwards can be given by:
To be honest, I was a little unsure on linking the thrust force to the rate of fuel consumption, which is the last bit of the puzzle. But remember a formula from Prelim Physics linking momentum to the product of force and time, and the answer comes out:
Since R is just the mass of fuel ejected per unit time, or, m/t.
This is subject to a little inaccuracy, I've never done this before! ;D
I hope this gives you some ideas. I've done some pretty wacky math, so go away, try and follow what I've done, and if after some thought it doesn't quite make sense please let me know and I'll slow down a little! I kind of got away from myself ahaha ;D
Post by: jamonwindeyer on June 23, 2016, 08:44:45 pm
Thanks man!
Would my method by wrong entirely? Since I considered the direction of the bar magnet's magnetic field lines when looping back in to its South Pole, hence just said the induced magnetic field would oppose this..
Hm, I might've confused myself since when I learnt how to solve these questions initially, I might've just considered the direction of the magnetic field lines coming straight out of the North Pole. I think this is pretty much the way you did it.
Thanks again!
I think it is okay in principle, but for your initial answer, consider the magnetic field lines of the magnet and the loop right next to the South pole. If the induced field is going to the left, and the magnetic field at the South pole goes into the magnet, it is also going to the left!! They match, so you are actually assisting it! ;D
Post by: soldmychildforyeezys on June 23, 2016, 09:00:46 pm
I think it is okay in principle, but for your initial answer, consider the magnetic field lines of the magnet and the loop right next to the South pole. If the induced field is going to the left, and the magnetic field at the South pole goes into the magnet, it is also going to the left!! They match, so you are actually assisting it! ;DThanks, I actually meant the direction of the magnetic field lines when they are parallel to the bar magnet. So when it's in the middle of the process of coming back in to the South Pole.
I'll just shut up now and use your method since its obviously much clearer :P
Post by: Swagadaktal on June 23, 2016, 09:05:20 pm
I think it is okay in principle, but for your initial answer, consider the magnetic field lines of the magnet and the loop right next to the South pole. If the induced field is going to the left, and the magnetic field at the South pole goes into the magnet, it is also going to the left!! They match, so you are actually assisting it! ;DHey - do you reckon you could explain Lenz law a bit more for me?
As the south pole is approaching the loop - the magnetic field goes from N to S so the magnetic field is going right (-->) . Can you tell me the flaw in my logic. Soz i havent really read this part of the book yet
OH WAIT right next to the south pole? So at the S part the lines are going into the S so it's going from right to left?
ohhhh you're a legend
Post by: jamonwindeyer on June 23, 2016, 09:06:34 pm
Thanks, I actually meant the direction of the magnetic field lines when they are parallel to the bar magnet. So when it's in the middle of the process of coming back in to the South Pole.
I'll just shut up now and use your method since its obviously much clearer :P
Ohhhhh gotcha! Hm, I'm not sure why that doesn't work TBH, probably because the south pole is closest to the loop and so the effects there are the most pronounced? Not sure ;D
Definitely don't shut up ahaha, my methods are just what works for me, by all means not perfect ;D I do like the right hand grip rule though, it just made sense for me :) :)
Post by: jamonwindeyer on June 23, 2016, 09:07:06 pm
...
OH WAIT right next to the south pole? So at the S part the lines are going into the S so it's going from right to left?
ohhhh you're a legend
Yep that's right! ;D
Post by: zsteve on June 23, 2016, 09:13:14 pm
Hey - do you reckon you could explain Lenz law a bit more for me?
As the south pole is approaching the loop - the magnetic field goes from N to S so the magnetic field is going right (-->) . Can you tell me the flaw in my logic. Soz i havent really read this part of the book yet
OH WAIT right next to the south pole? So at the S part the lines are going into the S so it's going from right to left?
ohhhh you're a legend
Yo Swagdaktal :) you might find this interesting. Assume the opposite of Lenz's law holds, then imagine what would happen if there was a flux change - you would get infinite current ~ infinite energy.
So clearly, you have proved that the opposite of Lenz's law is impossible. Hence Lenz's law must hold.
Post by: jamonwindeyer on June 23, 2016, 09:15:41 pm
Yo Swagdaktal :) you might find this interesting. Assume the opposite of Lenz's law holds, then imagine what would happen if there was a flux change - you would get infinite current ~ infinite energy.
So clearly, you have proved that the opposite of Lenz's law is impossible. Hence Lenz's law must hold.
This is actually the best way of explaining Lenz's Law!! The formal definition for you Swagdaktal would be something like:
Lenz's Law: The direction of an induced current is always such that it opposes whatever change created it.
Post by: jakesilove on June 23, 2016, 09:34:24 pm
This is actually the best way of explaining Lenz's Law!! The formal definition for you Swagdaktal would be something like:
Lenz's Law: The direction of an induced current is always such that it opposes whatever change created it.
That's exactly how I always recommend describing and, in fact, 'proving', Lenz's law. The easiest analogy is looking at the Magnet falling down the copper tube. The induced current will obvious cause the magnet to experience a force; but will it be up or down? If it is down, then the increased speed will caused increased Eddy currents, which will cause increased speed, which will cause increased eddy currents, and so on and so on until it breaks the speed of light. Obviously, that's not OK (Physicists aren't sure of much, but breaking the universal speed limit? That's a not go). So, the current must OPPOSE the change created! It's a nice, logical argument, and the HSC markers love it if you ever need to explain back emf and lenz's law.
Jake
Post by: Swagadaktal on June 23, 2016, 09:43:33 pm
Yo Swagdaktal :) you might find this interesting. Assume the opposite of Lenz's law holds, then imagine what would happen if there was a flux change - you would get infinite current ~ infinite energy.OOh yeah - that makes sense.
So clearly, you have proved that the opposite of Lenz's law is impossible. Hence Lenz's law must hold.
That's exactly how I always recommend describing and, in fact, 'proving', Lenz's law. The easiest analogy is looking at the Magnet falling down the copper tube. The induced current will obvious cause the magnet to experience a force; but will it be up or down? If it is down, then the increased speed will caused increased Eddy currents, which will cause increased speed, which will cause increased eddy currents, and so on and so on until it breaks the speed of light. Obviously, that's not OK (Physicists aren't sure of much, but breaking the universal speed limit? That's a not go). So, the current must OPPOSE the change created! It's a nice, logical argument, and the HSC markers love it if you ever need to explain back emf and lenz's law.We actually did this in class.
Jake
Thanks guys
Just to clarify, if in that question there was a north pole facing the loop, then the magnetic field will be going left to right and therefore the field in the loop is going from right to left? And the current would go anticlockwise?
And if it were a north pole facing thing, and it was going away from the loop would the field of the loop be going from left to right (in the same direction of the magnet)?
Or would it still oppose the magnet but just have a weaker magnetic field?
And is my wording correct here?
Thanks heaps guys
Post by: Ahsun on June 23, 2016, 09:52:04 pm
Show Mathematically that he would have been able to calculate this q/m ratio from the formula: q^m = 2V^B2r2
Post by: jamonwindeyer on June 23, 2016, 11:11:03 pm
OOh yeah - that makes sense. We actually did this in class.
Thanks guys
Just to clarify, if in that question there was a north pole facing the loop, then the magnetic field will be going left to right and therefore the field in the loop is going from right to left? And the current would go anticlockwise?
Do you mean the question with the bar magnet and the loop? ;D If so, then yes. The magnetic field of the bar magnet (near the North pole) is going from left to right, and so the induced field must be from right to left. I think "induced by the loop" works a bit better language wise than "in the loop" (though I think it's still okay). And yes, the current would then be anti-clockwise! ;D
And if it were a north pole facing thing, and it was going away from the loop would the field of the loop be going from left to right (in the same direction of the magnet)?
Or would it still oppose the magnet but just have a weaker magnetic field?
And is my wording correct here?
Thanks heaps guys
Your first analysis is correct. The magnet is moving away, meaning the induced field will act in the same direction as the magnet's magnetic field to try and attract it back to the loop.
Note that if the induced field opposed the magnet, then the magnet would be pushed away faster, thus inducing a stronger field, then the magnet would be pushed away faster, thus inducing a stronger field... You get the idea. Lenz's Law broken! ;)
I hope that helps! ;D
Post by: Swagadaktal on June 23, 2016, 11:28:23 pm
Do you mean the question with the bar magnet and the loop? ;D If so, then yes. The magnetic field of the bar magnet (near the North pole) is going from left to right, and so the induced field must be from right to left. I think "induced by the loop" works a bit better language wise than "in the loop" (though I think it's still okay). And yes, the current would then be anti-clockwise! ;DYes this does help :D
Your first analysis is correct. The magnet is moving away, meaning the induced field will act in the same direction as the magnet's magnetic field to try and attract it back to the loop.
Note that if the induced field opposed the magnet, then the magnet would be pushed away faster, thus inducing a stronger field, then the magnet would be pushed away faster, thus inducing a stronger field... You get the idea. Lenz's Law broken! ;)
I hope that helps! ;D
Thanks a lot everyone <3
Post by: jamonwindeyer on June 24, 2016, 12:11:08 am
In one of the first attempts made by JJ Thomson to measure the q/m for cathode ray particles, he accelerated them through a voltage V, using this to calculate the kinetic energy, and hence the speed v, of the particles. He then allowed them to be bent into a circular path of radius r by a uniform magnetic field, B, at 90O.
Show Mathematically that he would have been able to calculate this q/m ratio from the formula: q^m = 2V^B2r2
Maaaan, these are brutal. Not sure about this one! Let me have a go.
Well the definition of a volt is that it is the potential difference required to give 1 coulomb of charge 1J of kinetic energy. Therefore, V volts will give V joules to 1 coulomb of charge. How many electrons are in 1 coulomb of charge? Well we obtain that by dividing by the charge per electron, so: 1/q. That means, putting that all together, that the kinetic energy per electron is K=Vq:
Doing some algebra:
Okay, so we have a formula for velocity. Now in the second part, the magnetic field provides a centripetal force. Identically to the analysis of the final version of this experiment, we can now equate centripetal and magnetic force (note that I will ignore the sine term in the formula for magnetic force, since the angle is 90 degrees, and sin90 = 1):
Equate the expression from earlier:
Whew, got it!! Aha I hope that makes sense, the top bit might be a little confusing, let me know if that needs any clarification! ;D
Post by: Ahsun on June 24, 2016, 07:36:16 pm
Maaaan, these are brutal. Not sure about this one! Let me have a go.Thanks so much made it so easy to understand thanks so much ;D
Well the definition of a volt is that it is the potential difference required to give 1 coulomb of charge 1J of kinetic energy. Therefore, V volts will give V joules to 1 coulomb of charge. How many electrons are in 1 coulomb of charge? Well we obtain that by dividing by the charge per electron, so: 1/q. That means, putting that all together, that the kinetic energy per electron is K=Vq:
Doing some algebra:
Okay, so we have a formula for velocity. Now in the second part, the magnetic field provides a centripetal force. Identically to the analysis of the final version of this experiment, we can now equate centripetal and magnetic force (note that I will ignore the sine term in the formula for magnetic force, since the angle is 90 degrees, and sin90 = 1):
Equate the expression from earlier:
Whew, got it!! Aha I hope that makes sense, the top bit might be a little confusing, let me know if that needs any clarification! ;D
Post by: RuiAce on June 24, 2016, 07:55:14 pm
Yo Swagdaktal :) you might find this interesting. Assume the opposite of Lenz's law holds, then imagine what would happen if there was a flux change - you would get infinite current ~ infinite energy.
So clearly, you have proved that the opposite of Lenz's law is impossible. Hence Lenz's law must hold.
This is actually important. It is a part of the syllabus. An explanation needs to be known as to how Lenz's law reflects the conservation of energy.
Post by: RuiAce on June 24, 2016, 07:57:27 pm
Thanks so much made it so easy to understand thanks so much ;D
A note on this as well
Any band 6 student will know how to derive
(Disclaimer: I have not been informed that this is a NECESSARY part of the syllabus though)
I think Physics in Focus did give a derivation of it. The short:
Post by: jamonwindeyer on June 24, 2016, 08:15:33 pm
A note on this
This is actually important. It is a part of the syllabus. An explanation needs to be known as to how Lenz's law reflects the conservation of energy.
Swag is from Victoria, he's just curious (but absolutely) ;D
A note on this as well
Any band 6 student will know how to derive
(Disclaimer: I have not been informed that this is a NECESSARY part of the syllabus though)
...
Love this. I was asked it as part of an in-school assessment, and I included it as part of a response in either my Trial or my HSC (I forget which off the top of my head). I teach it to everyone I tutor, and understanding it means you understand a solid chunk of the Ideas to Implementation topic, definitely worth knowing this proof properly ;D
Post by: Ahsun on June 24, 2016, 10:23:32 pm
Post by: jamonwindeyer on June 24, 2016, 11:30:50 pm
What are some applications of some type 1 superconductors? I know which ones are type 1 superconductors but I cant find any applications for any specific one of the type 1 superconductors
Hey Ahsun! Hmm, that's tricky, the difference between the two is really just their response to applied magnetic fields (and usually their critical temperatures too, I think). I know that Type 1 superconductors are usually pure metals, and to be honest, I think most applications of superconductors use Type 2 superconductors. This is purely because the alloys and ceramic oxides have higher critical temperatures and so are more practical ;D
So, in terms of specific applications of Type 1, I'm not sure! It would have to be something that needs a complete Meisner Effect, but I'm not sure about this one ;) anyone want to tag in here?
Post by: RuiAce on June 25, 2016, 01:38:44 pm
But then I did a Google search and this was the first link that popped up.
https://www.physicsforums.com/threads/uses-of-type-i-superconductors.237238/
Post by: jamonwindeyer on June 25, 2016, 01:42:44 pm
I thought it really doesn't matter about the type.
But then I did a Google search and this was the first link that popped up.
https://www.physicsforums.com/threads/uses-of-type-i-superconductors.237238/
I saw this as well, it looks like it's a tad beyond HSC scope, but for the original poster, it does (can't vouch for accuracy) say that Type 1's are used in SQUID Magnetometers ::)
Post by: RuiAce on June 25, 2016, 01:43:43 pm
I saw this as well, it looks like it's a tad beyond HSC scope, but for the original poster, it does (can't vouch for accuracy) say that Type 1's are used in SQUID Magnetometers ::)
At a physics day at USyd (compulsory school excursion) when they were talking about superconductors they did mention SQUID and what it stood for, but nothing about it.
Post by: jamonwindeyer on June 25, 2016, 01:50:51 pm
At a physics day at USyd (compulsory school excursion) when they were talking about superconductors they did mention SQUID and what it stood for, but nothing about it.
Yep I have no idea how they work, only that they are extremely sensitive magnetometers used in quantum research. That's all I knew back in the HSC too, so you definitely don't need any more than that ;D
I'll link to Wikipedia for SQUID though, in case anyone is curious, it seems to go through some pretty complex theory that I'm not overly familiar with ;D
Post by: Ahsun on June 26, 2016, 05:35:11 pm
Plan, choose equipment or resources for a hypothetical first hand investigation to
propose a design. Include free-hand sketch or a draw app for illustrating your design and briefly describe how you would test the module landing.
I have a design of the landing module but not sure how to test the module landing exactly
Post by: jamonwindeyer on June 26, 2016, 05:40:55 pm
Imagine you are the structural engineer at NASA for the design of a new moon landing module (not the rocket). You need to find the best unmanned module design to use for safe landing on the surface of the moon.
Plan, choose equipment or resources for a hypothetical first hand investigation to
propose a design. Include free-hand sketch or a draw app for illustrating your design and briefly describe how you would test the module landing.
I have a design of the landing module but not sure how to test the module landing exactly
Hey Ahsun! I'm not sure about this one, there would be a lot to test! I suppose the most important thing would be durability, perhaps you'd be looking at some sort of test where you drop the prototype module from different heights (in moon-like conditions) and determine the best design based on which can be dropped from the greatest height?
Hmm... That's about all I can think of right now!! I'll let you know if I come up with something else though ;D
Post by: Ahsun on June 26, 2016, 06:02:40 pm
Hey Ahsun! I'm not sure about this one, there would be a lot to test! I suppose the most important thing would be durability, perhaps you'd be looking at some sort of test where you drop the prototype module from different heights (in moon-like conditions) and determine the best design based on which can be dropped from the greatest height?
Hmm... That's about all I can think of right now!! I'll let you know if I come up with something else though ;D
Thanks for the ideas
Post by: jamonwindeyer on June 26, 2016, 06:28:46 pm
Thanks for the ideas
No problem! Someone else might have better ones than me if anyone wants to chip in ;)
Post by: Happy Physics Land on June 26, 2016, 07:12:00 pm
Imagine you are the structural engineer at NASA for the design of a new moon landing module (not the rocket). You need to find the best unmanned module design to use for safe landing on the surface of the moon.
Plan, choose equipment or resources for a hypothetical first hand investigation to
propose a design. Include free-hand sketch or a draw app for illustrating your design and briefly describe how you would test the module landing.
I have a design of the landing module but not sure how to test the module landing exactly
Hey ahsun!
An important thing about the module would be to make it out of a material that would be resistant against the effect of solar wind, and ensure that the internal parts are more robust against discharge arcs. Perhaps you can add protective coatings that can make electronic parts inside to be less vulnerable to solar wind radiations. Another aspect I would mention is more so the material science side of things. Firstly you would want your landing module to have 3 "legs" (tripod design) or 4 "legs" stand design. These stands can be made out of high strength to weight ratio aluminium alloys (after all, landing modules are quite massive and to minimise fuel cost on rockets you would wanna minimise weight). In fact majority of the landing module would be made out of titanium alloy or aluminium alloy for the same reasons. To ensure the strength of the entire module you would want to have a warren truss structure for the structural components such as the stand. When you are considering landing you must consider about the gravity of the moon, projectile motion principles (You can perhaps make up a situation and perform some calculations around it).
Hope the information helps!
Best Regards
Happy Physics Land
Post by: wyzard on June 26, 2016, 10:04:17 pm
Imagine you are the structural engineer at NASA for the design of a new moon landing module (not the rocket). You need to find the best unmanned module design to use for safe landing on the surface of the moon.
Plan, choose equipment or resources for a hypothetical first hand investigation to
propose a design. Include free-hand sketch or a draw app for illustrating your design and briefly describe how you would test the module landing.
I have a design of the landing module but not sure how to test the module landing exactly
Sounds pretty interesting. You can't really test the actual landing module on the moon of course. What you can do instead is to simulate the landing conditions. You can to consider at what speed will the landing module touch down and maybe the 'hardness' of the moon surface (think landing on hard concrete surface vs. grass).
With this simulated environment, you can then test out several different designs.
Post by: jamonwindeyer on June 27, 2016, 03:15:52 pm
Post by: conic curve on June 27, 2016, 10:21:31 pm
Thanks
Post by: Swagadaktal on June 27, 2016, 10:34:13 pm
So I really suck at maths physics and currently I'm doing the intensity square law and am finding it difficult (especially the maths to it). Could someone here please give me advice on how I should approach the inverse square law?Not sure how much you're meant to know within the HSC syllabus, but essentially the ratio of sound intensity is related in this fashion:
Thanks
I = 1/r^2 (this is about sound right? if this is a diff topic awks lmao)
So basically for a distance travelled of x m, the intensity decreases by a factor of x^2.
So if you travel 2 metres away from the source, the sound intensity decreases by a factor of 4 ( so divide by four)
Post by: jamonwindeyer on June 27, 2016, 10:44:32 pm
Not sure how much you're meant to know within the HSC syllabus, but essentially the ratio of sound intensity is related in this fashion:
I = 1/r^2 (this is about sound right? if this is a diff topic awks lmao)
So basically for a distance travelled of x m, the intensity decreases by a factor of x^2.
So if you travel 2 metres away from the source, the sound intensity decreases by a factor of 4 ( so divide by four)
This is a great example!
In terms of the mathematics behind it, this diagram might help! The best way to consider it is in terms of the surface area of a sphere with varying radius (formula below). If you double the radius, you quadruple the surface area. In terms of the inverse square law, this corresponds to some amount of energy being 'spread' over four times a larger area, and thus, being a quarter of the intensity. Intensity decreases in proportion to the square of the radius, and so:
(http://i.imgur.com/Gh9QmsF.png)
Post by: Swagadaktal on June 27, 2016, 10:55:02 pm
This is a great example!yeah that works too
In terms of the mathematics behind it, this diagram might help! The best way to consider it is in terms of the surface area of a sphere with varying radius (formula below). If you double the radius, you quadruple the surface area. In terms of the inverse square law, this corresponds to some amount of energy being 'spread' over four times a larger area, and thus, being a quarter of the intensity. Intensity decreases in proportion to the square of the radius, and so:
(http://i.imgur.com/Gh9QmsF.png)
hehe you guys are honestly amazing with your explanations :D
Post by: conic curve on June 28, 2016, 09:11:23 pm
1. A slide projector produces a rectangular image which has the dimensions 0.300m wide and 0.200m high when the projector is 0.500 from the screen. How far from the screen would the projector have to be placed in order to fill a screen 1.500 wide?
2. How will the intensity of the light falling on a surface be affected if the source of the light is moved to twice its original distance?
3. How will the intensity of the light falling on a surface be affected if the source of the light is moved to twice its original distance?
If someone could provide me with an indepth explanation, that'd be great ;D
Thanks :)
Post by: wyzard on June 28, 2016, 10:00:48 pm
So I'm having trouble with the following Inverse square law questions (there are actually much more than this)
1. A slide projector produces a rectangular image which has the dimensions 0.300m wide and 0.200m high when the projector is 0.500 from the screen. How far from the screen would the projector have to be placed in order to fill a screen 1.500 wide?
2. How will the intensity of the light falling on a surface be affected if the source of the light is moved to twice its original distance?
3. How will the intensity of the light falling on a surface be affected if the source of the light is moved to twice its original distance?
If someone could provide me with an indepth explanation, that'd be great ;D
Thanks :)
1) For the projected rectangular image, the length and height will increase proportional to the distance. So to have increase the width from 0.300m to 1.500m, you increase increase the distance five times.
2) As the length and height both increase proportional to the distance, this means that the area will increase proportional to the distance squared, as area is the product of length and height. Hence the light will spread out in the area, resulting in the intensity falling at a greater distance, being inversely proportional to the distance squared. So by doubling the distance, you decrease the intensity 4 times.
3) Just a repeat of Q2.
Hope this helps! :)
Post by: jakesilove on June 28, 2016, 11:15:51 pm
1) For the projected rectangular image, the length and height will increase proportional to the distance. So to have increase the width from 0.300m to 1.500m, you increase increase the distance five times.
2) As the length and height both increase proportional to the distance, this means that the area will increase proportional to the distance squared, as area is the product of length and height. Hence the light will spread out in the area, resulting in the intensity falling at a greater distance, being inversely proportional to the distance squared. So by doubling the distance, you decrease the intensity 4 times.
3) Just a repeat of Q2.
Hope this helps! :)
Thanks for the response mate, really succinct stuff! Definitely living up to your username
Post by: conic curve on July 05, 2016, 11:29:20 am
Post by: RuiAce on July 05, 2016, 11:37:24 am
How do you study the cosmic engine? I feel that this is like a content heavy rote learn topic and I dislike it very much. It is quite boringThat's cause it is.
Our teacher just gave us an assignment on it which was to essentially address all the points by yourself except for H-R diagrams.
Post by: conic curve on July 05, 2016, 11:57:18 am
Post by: RuiAce on July 05, 2016, 12:00:32 pm
As for open/closed book past papers that's always a matter of personal preference and how much preparation you already have. Just like however you want to study
Post by: conic curve on July 05, 2016, 12:09:12 pm
My teacher gave me a good mark for my assignment so I basically just kept studying the crap out of that.
As for open/closed book past papers that's always a matter of personal preference and how much preparation you already have. Just like however you want to study
lol okay thanks
Didn't they teach you some sort of maths in the cosmic engine though?
Post by: RuiAce on July 05, 2016, 12:20:42 pm
lol okay thanksNo
Didn't they teach you some sort of maths in the cosmic engine though?
Post by: conic curve on July 05, 2016, 12:24:41 pm
No
Oh well, okay
Post by: RuiAce on July 05, 2016, 12:28:20 pm
Oh well, okay:P
But yeah seriously none
Post by: conic curve on July 05, 2016, 12:43:00 pm
:P
But yeah seriously none
I guess that's great for me then because I suck at the maths side of physics
Also how much harder does the maths side of physics get in the HSC?
Post by: brontem on July 05, 2016, 01:25:56 pm
Post by: jakesilove on July 05, 2016, 01:42:47 pm
Can someone help with this question?? Probably simple but I've forgotten :)
Hey! This is a tiny area in the Semiconductor section, and it's just something you need to memorise. When you dope a semi-conductor with a Group 3 metal (ie. P-type), an extra band called the acceptor band is formed slightly above the valence band. When you dope a semi-conductor with a Group 5 metal (ie. N-type), an extra band called the donor band is formed slightly below the conduction band. These are totally arbitrary notions, and only make sense when you physically draw the levels. Both of these things makes it 'easier' for the electrons to jump between the valence and conduction bands, making it easier to conduct a charge!
(http://hyperphysics.phy-astr.gsu.edu/hbase/solids/imgsol/dban.gif)
Above courtesy of Hyperphysics, a great online resource!
So, to your actual question. Based on the location of the additional band, we're expecting a P-type semiconductor. This means it has been doped with a Group 5 element. The answer, therefore, is Boron.
Quick note; you could have quickly eliminated half of the potential solutions (B and D) noting that we can't dope semi-conductors with those elements. Then, you had a 50-50 chance of getting the question right, even if you had no idea what it was even asking!
Hope this is clear. It's a totally arbitrary section of the curriculum (and of Physics), so don't worry if you forgot how it works!
Jake
Post by: jakesilove on July 05, 2016, 01:46:19 pm
Can someone help with this question?? Probably simple but I've forgotten :)
Just adding another, clearer image.
(http://www.physics.udel.edu/~watson/scen103/semi2.gif)
Post by: RuiAce on July 05, 2016, 02:21:14 pm
Just adding another, clearer image.
(http://www.physics.udel.edu/~watson/scen103/semi2.gif)
Post by: brontem on July 05, 2016, 03:08:49 pm
Post by: jakesilove on July 05, 2016, 03:11:38 pm
Thankyou!! I can't ever recall learning about donor/acceptance :) will definitely add it to my notes :)
Yep, that's not uncommon; most teachers/tutors/notes leave it out, because its rarely assessed and generally doesn't make much sense (at least at the level of physics taught in the HSC!). Definitely worth knowing though, so great job finding a question like that whilst doing past papers! It's the weird, auxiliary sections that catch students off guard :)
Post by: Adriaclya on July 05, 2016, 05:59:40 pm
Is there an explanation as to why if an electric field exists, so too a magnetic field? Is there a reason why electricity and magnetism is strongly linked?
Thanks!
Post by: jakesilove on July 05, 2016, 06:23:27 pm
Hi!
Is there an explanation as to why if an electric field exists, so too a magnetic field? Is there a reason why electricity and magnetism is strongly linked?
Thanks!
Hey!
Look, to answer your question properly would take a semester of University, a hell of a lot of maths and some history lessons at a bare minimum. Essentially, though, the answer to your question is simple; Electricity and Magnetism are the same force. Up until 1873, Electricity and Magnetism were studied completely separately, and thought of as two of the fundamental universal forces. In steps Maxwell, a brilliant mathematician and Physicist, who essentially united the two forces into electromagnetism, formalised in Maxwell's equations. Electricity and Magnetism are two sides to the same coin; one necessarily implies the other. It just took us a while to realise it!
This isn't the most satisfying answer out there (if you want a better answer, try an electromag course at University!), but it is an answer. If an electric field exists, a magnetic fields exists, because it does, and that's how the universe works. It's almost like asking why the speed of light is 2.998*10^8; I can explain why it's a constant, I can explain what it implies, I can explain how it impacts us and I can use it to my advantage. I CAN'T explain why it is that number in particular (at least, not with the physics I know right now!). The universe is a funky place, with some fundamental laws. You've stumbled across one of them, and if the year you graduate in is correct, you've stumbled upon it really really early in your Physics career. So, congratulations, and good luck coming to terms with some of Physics' fundamental truths.
Jake
Post by: Meckenza on July 05, 2016, 06:26:59 pm
I was just wondering - why does a graph plotting g-forces against time in a spacecraft have a non-linear relationship?
Thanks.
Post by: Adriaclya on July 05, 2016, 06:47:46 pm
Whoops. Grad year 2017....
Also, is this correct:
Wind turns wind turbines = turns shaft= turns magnetic coil=generates current?
And for solar panels:
Light photons=knock electrons of silicon layer which has been doped in phosphorus=negative terminal=electrons carried through the silicon conductor to wires for use=comes back to silicon layer (that has been doped with boron)=electrons reset for use again??
Thanks once again!
Post by: jakesilove on July 05, 2016, 06:53:31 pm
Hi,
I was just wondering - why does a graph plotting g-forces against time in a spacecraft have a non-linear relationship?
Thanks.
Hey!
So, when plotting g-force against time, we have to think about what is actually happening to the shuttle. So, first let's look at the formula for g-force.
Where g is the force due to gravity, and a is the acceleration of the rocket.
So, since g is approximately constant in the initial stages of the launch (it will get weaker as the rocket moves away from the earth, but let's ignore this as it will hardly contribute to the graph, at least for small t), what we really want to know is the values for a. If a increases linearly, so will the g-force. If a decreases linearly, so will the g-force. And, if a increases non-linearly (as we will soon find that it does), so will the g-force.
Now, we know from Newton's laws that
This is how we figure out what the acceleration of the rocket is. The burning of fuel creates a fairly constant force downwards and, by principles of conservation of momentum, the rocket is therefore propelled upwards. However, let's think about what's actually happening. As the fuel is being burnt, resulting in a constant force (ie. the left hand side of the Newton's equation is constant), the mass of the rocket is decreasing by quite a lot. This is because most of the mass of most rockets is made up of fuel! So, over time, the mass is decreasing. However, if the left hand side of the equation is constant over time, the right hand side must be as well! So, if the mass is going down, the acceleration must be going up. In other words, acceleration increases over time.
Huh. Except, when you reach this point, you expect acceleration to increase linearly over time. Which means, as I've described above, that the g-force should increase linearly over time.
Damn.
Okay, so turns out I was wrong, but I'm going to leave my answer above for completeness, and to show the logical steps you should be taking (and, presumably, you did take, in order to reach the linear conclusion).
In that case, it is likely because g decreases as t increases (the rocket gets further from the earth). This will result in a non-linear relationship. Also turns out to be a way easier solution! Sorry for the rambliness but hopefully this all made sense; really great question!
Jake
Post by: jakesilove on July 05, 2016, 06:56:41 pm
Thank jake!
Whoops. Grad year 2017....
Also, is this correct:
Wind turns wind turbines = turns shaft= turns magnetic coil=generates current?
And for solar panels:
Light photons=knock electrons of silicon layer which has been doped in phosphorus=negative terminal=electrons carried through the silicon conductor to wires for use=comes back to silicon layer (that has been doped with boron)=electrons reset for use again??
Thanks once again!
Totally correct! Are you doing accelerated Physics or something? That's definitely not in the Prelim curriculum...
Post by: Adriaclya on July 05, 2016, 08:00:53 pm
Nope, just further reading haha!
Post by: jakesilove on July 05, 2016, 09:23:37 pm
Thanks jake.
Nope, just further reading haha!
That's seriously cool (I'm a nerd, and Physics is cool. Hopefully you agree!). Feel free to ask me anything, anytime, whether it's part of your/a curriculum or not :) Glad to see your enthusiasm!
Post by: Swagadaktal on July 05, 2016, 09:25:57 pm
Thanks jake.Yo when you're free hmu with some of that motivation/enthusiasm. Need some in yr 12 tbh :P
Nope, just further reading haha!
Post by: jamonwindeyer on July 05, 2016, 10:16:07 pm
So, to your actual question. Based on the location of the additional band, we're expecting a P-type semiconductor. This means it has been doped with a Group 5 element. The answer, therefore, is Boron.
Jake
Hey guys! Answer doesn't change for above, but in case it causes confusion later, this should read "Group 3 element" ;D
Post by: RuiAce on July 05, 2016, 10:18:53 pm
Hey guys! Answer doesn't change for above, but in case it causes confusion later, this should read "Group 3 element" ;DI thought you were the chemist Jake :P
_______________________
I remember when I did further reading for maths... but now it's like stick to what you've learnt
Post by: Spencerr on July 05, 2016, 11:27:24 pm
Could someone please explain to me the concept of mass defect and binding energy in the context of nuclear fission. E.g. If a Uranium nuclei splits into two daughter nuclei and releases a large amount of energy where does that energy come from? and does nuclear fusion also release energy?
Post by: jakesilove on July 06, 2016, 08:07:56 am
Hi there!
Could someone please explain to me the concept of mass defect and binding energy in the context of nuclear fission. E.g. If a Uranium nuclei splits into two daughter nuclei and releases a large amount of energy where does that energy come from? and does nuclear fusion also release energy?
Hey!
I can't answer much of your question, because I didn't do this option. I can briefly explain the whole 'mass defect' idea, I think, based on my knowledge of Physics.
When a nuclei splits, and releases energy, the energy itself actually comes from the mass loss. So, the 'mass defect' and energy release is the same! For instance, say the original nuclei weighed 1g (which is ridiculous, just using it for ease but obviously they weigh a lot less) and the two daughter nuclei weigh 0.4g each. Clearly, if you add the daughter nuclei up, you've lost 0.2g somewhere! In comes Einstein, with the most equation in Physics
We can sub the 0.2g of lost matter into this formula, and out pops a huge amount of energy! And that's where the energy release comes from :)
Hopefully someone else on the forum can help you with the rest!
Post by: conic curve on July 06, 2016, 09:26:08 am
Hi there!
Could someone please explain to me the concept of mass defect and binding energy in the context of nuclear fission. E.g. If a Uranium nuclei splits into two daughter nuclei and releases a large amount of energy where does that energy come from? and does nuclear fusion also release energy?
Is this Quanta to Quarks?
Post by: jakesilove on July 06, 2016, 04:13:03 pm
Is this Quanta to Quarks?
Yup it is! I did medical physics, so I can't be of much use for that option unfortunately... except for the Quantum physics I've done at Uni!
Post by: Syndicate on July 06, 2016, 04:14:13 pm
does nuclear fusion also release energy?
Decided to jump in the conversation :P
For two smaller atoms to fuse, they require a lot of energy and high temperatures. So, when they are fused together(for instance two hyrdogen atoms fuse together to produce deutrium), there is excess energy left over, which is emitted by the nuclei in the form of gamma radiation.
Post by: conic curve on July 06, 2016, 04:48:23 pm
Yup it is! I did medical physics, so I can't be of much use for that option unfortunately... except for the Quantum physics I've done at Uni!
Did jamon study quanta to quarks?
Also did you study industrial chemistry?
Post by: jamonwindeyer on July 07, 2016, 12:11:28 am
Did jamon study quanta to quarks?
Also did you study industrial chemistry?
I studied Medical Physics just like Jake did, but I bet there are lots of people hanging around who did Quanta ;D
Post by: conic curve on July 07, 2016, 11:02:29 am
I studied Medical Physics just like Jake did, but I bet there are lots of people hanging around who did Quanta ;D
Damm.....What if we ask things in Quanta to Quarks which require a high level of understanding? (I don't think I will be but I think others might)
Post by: jamonwindeyer on July 07, 2016, 11:21:57 am
Damm.....What if we ask things in Quanta to Quarks which require a high level of understanding? (I don't think I will be but I think others might)
We hope that someone else can answer!! Jake and I have some background knowledge with regard to the content, Jake more than me, so we'll handle what we can, otherwise we hope others can help out :)
Post by: RuiAce on July 07, 2016, 11:30:06 am
Post by: conic curve on July 07, 2016, 11:31:51 am
We hope that someone else can answer!! Jake and I have some background knowledge with regard to the content, Jake more than me, so we'll handle what we can, otherwise we hope others can help out :)
I hope that as well
Post by: jakesilove on July 07, 2016, 11:43:31 am
I hope that as well
I've done a fair amount of the Quanta curriculum, so I'm sure we'll be okay :)
Post by: Loki98 on July 07, 2016, 12:36:46 pm
The distance to the star alpha centuari is 4.367 light years as measured from Earth. Using a relevant equation, explain how a rocket could complete the journey from earth to alpha centuari in 3.28 years.
Post by: jamonwindeyer on July 07, 2016, 12:52:44 pm
Need help with this question,
The distance to the star alpha centuari is 4.367 light years as measured from Earth. Using a relevant equation, explain how a rocket could complete the journey from earth to alpha centuari in 3.28 years.
Hey Loki! The equation that this question mentions is the length contraction formula:
Essentially what is happening here is that if the rocket travels at a relativistic speed, the distance between the earth and Alpha Centauri (with respect to earth) contracts with respect to the frame of reference of the rocket. That is, the astronauts actually would believe the distance to be shorter due to length contraction. Thus, with respect to the earth, the astronauts would actually arrive earlier than they would have if we didn't take length contraction into account, because the distance travelled with respect to the rocket is shorter.
The answer to your question would explain that principle, preferably being a little more succinct ;) Does that make sense?
Post by: Loki98 on July 07, 2016, 01:11:31 pm
Hey Loki! The equation that this question mentions is the length contraction formula:
Essentially what is happening here is that if the rocket travels at a relativistic speed, the distance between the earth and Alpha Centauri (with respect to earth) contracts with respect to the frame of reference of the rocket. That is, the astronauts actually would believe the distance to be shorter due to length contraction. Thus, with respect to the earth, the astronauts would actually arrive earlier than they would have if we didn't take length contraction into account, because the distance travelled with respect to the rocket is shorter.
The answer to your question would explain that principle, preferably being a little more succinct ;) Does that make sense?
Yea it makes sense now. But how would i prove that the time taken would be 3.28 years?
Post by: conic curve on July 07, 2016, 01:13:57 pm
I've done a fair amount of the Quanta curriculum, so I'm sure we'll be okay :)
Fantastic :D
Post by: jamonwindeyer on July 07, 2016, 01:28:10 pm
Yea it makes sense now. But how would i prove that the time taken would be 3.28 years?
Oh, good point! This is a little beyond what would be asked in the HSC, but it would look like this. First, you'd use the speed equation:
This allows you to link the length quantities in the equation above to the times given in the question. You get:
We substitute those quantities into the length contraction formula and that lets us solve for the velocity:
I skipped a few steps in that algebraic working, let me know if you need me to clarify it, but this should give you the direction! ;D
This working is subject to inaccuracy, I've never actually done this calculation before! However, the answer seems sensible to me, and note that this is pretty bloody quick ;) again, it is unlikely you'd be asked to specifically provide the calculation in the HSC, and I think the question was looking for something a little more general ;D
Post by: RuiAce on July 07, 2016, 01:36:53 pm
Oh, good point! This is a little beyond what would be asked in the HSC, but it would look like this. First, you'd use the speed equation:Using S=D/t does pop around in the HSC I think. It's pretty rare though because it targets high band 5 in my opinion - have to know when to apply what formula, and in what order to apply them.
This allows you to link the length quantities in the equation above to the times given in the question. You get:
We substitute those quantities into the length contraction formula and that lets us solve for the velocity:
I skipped a few steps in that algebraic working, let me know if you need me to clarify it, but this should give you the direction! ;D
This working is subject to inaccuracy, I've never actually done this calculation before! However, the answer seems sensible to me, and note that this is pretty bloody quick ;) again, it is unlikely you'd be asked to specifically provide the calculation in the HSC, and I think the question was looking for something a little more general ;D
Post by: jamonwindeyer on July 07, 2016, 01:38:49 pm
Using S=D/t does pop around in the HSC I think. It's pretty rare though because it targets high band 5 in my opinion - have to know when to apply what formula, and in what order to apply them.
Definitely agree with that for standard projectile questions, know of a question like this (using it with relativity) appearing in any Trial exams? It's definitely never popped up in the HSC ;D
Post by: RuiAce on July 07, 2016, 01:41:21 pm
Definitely agree with that for standard projectile questions, know of a question like this (using it with relativity) appearing in any Trial exams? It's definitely never popped up in the HSC ;DAha ok maybe it wasn't there in the long run :P (pun intended but for wrong subject but you should know what i mean lol!)
I do remember seeing them out of somewhere though. My physics teacher had to go over several of them. Can't remember which paper though.
Post by: jamonwindeyer on July 07, 2016, 01:44:53 pm
Aha ok maybe it wasn't there in the long run :P (pun intended but for wrong subject but you should know what i mean lol!)
I do remember seeing them out of somewhere though. My physics teacher had to go over several of them. Can't remember which paper though.
Be careful mate, you are close to the long run shut down condition ;)
Ahh okay fair enough!! There you go, great question Loki, definitely one worth understanding if you are aiming high! ;D
Post by: Loki98 on July 07, 2016, 02:12:08 pm
Post by: brontem on July 07, 2016, 05:22:19 pm
Post by: RuiAce on July 07, 2016, 06:24:29 pm
What is photocurrent and why does it depend on intensity?? Or is it another deliberately brushed over/unimportant thing??
Recall from the definition of current that it is a measure of flow of electric charge. They are related by the formula q=It or equivalently I=q/t.
Clearly, when more electrons are present, the charge will be greater. If the amount of charge is greater, then the rate that charge flows will also increase.
But what is intensity? The intensity of the photocurrent IS by definition of electrons that flow out. Since a greater intensity is equivalent to saying greater charge, it also means a greater current.
(We just call it photocurrent because we are interested in the photoelectric effect. This means we are interested in photoelectrons - those that are emitted off the surface of the metal.)
Post by: jamonwindeyer on July 07, 2016, 10:34:57 pm
What is photocurrent and why does it depend on intensity?? Or is it another deliberately brushed over/unimportant thing??
I think I interpret your question differently than Rui, do you mean dependent on the intensity of incoming light? Saying "intensity of current" is a little bit incorrect, current doesn't have an intensity per say, but if you did use that term for current it would be in reference to the size of said current, measured in amperes ;D
As to why the size of a photocurrent is dependent on the intensity of light, we turn to the photon model of light. A greater intensity of light means that more photons are striking the surface of our metal! This results in a larger amount of photo-electrons emitted from the surface, as each photon (with a high enough energy) will release a single photoelectron. Let me know if you need this explained a little more! ;D
From here, Rui has you covered! More emitted electrons means a larger photocurrent, by definition ;)
Post by: smiley2101 on July 08, 2016, 08:48:25 am
Post by: RuiAce on July 08, 2016, 08:56:18 am
Hi! If v=IR why does increasing voltage decrease power loss if power loss =I^2R and according to v=IR voltage is proportional to current? Thank you! :)
This mistakenly assumes that resistance is constant. In fact, power is what is constant!
Because P=VI, we have V=P/I. So there is actually an inverse proportionality between voltage and current.
Post by: jamonwindeyer on July 08, 2016, 10:18:02 am
This mistakenly assumes that resistance is constant. In fact, power is what is constant!
Because P=VI, we have V=P/I. So there is actually an inverse proportionality between voltage and current.
Exactly, when we talk about increasing the voltage to reduce power loss, we are assuming that the same amount of power is being transmitted.
As an example, let's consider 1 Megawatt of power transmitted through a line resistance of 1 kilo-ohm at 100V and 1 MV respectively, and look at the power loss:
We can see that for 100V (top line), transmission is impossible. We can't transfer that much power through that resistance at that voltage, because too much power is dissipated in the line. The second line is much nicer, only 1000 Watts of power loss (this means the system is 99.999% efficient). This sort of calculation epitomises why we use high voltages for transmission ;D
Post by: jakesilove on July 08, 2016, 10:55:13 am
Fucking lol
Post by: conic curve on July 08, 2016, 12:04:58 pm
Fucking lol
how did he do that?
Post by: jamonwindeyer on July 08, 2016, 12:07:20 pm
how did he do that?
Do what? ;D
Post by: RuiAce on July 08, 2016, 12:16:06 pm
Fucking lol
Maybe we need a "rule" on scientific-notation
Post by: jamonwindeyer on July 08, 2016, 12:20:01 pm
Maybe we need a "rule" on scientific-notation
If I want to emphasise the size of a number, I don't think the index does it justice. Like, 100 Billion FREAKING WATTS MATE! I like the "oomph" ;)
Post by: RuiAce on July 08, 2016, 12:21:09 pm
If I want to emphasise the size of a number, I don't think the index does it justice. Like, 100 Billion FREAKING WATTS MATE! I like the "oomph" ;)Wooooow calm down :P of course i knew that
Post by: jamonwindeyer on July 08, 2016, 12:22:35 pm
Wooooow calm down :P of course i knew that
Ahaha see? The oomph is nice ;)
Post by: smiley2101 on July 09, 2016, 12:40:00 pm
Post by: jamonwindeyer on July 09, 2016, 01:34:46 pm
Ahh I see, thanks heaps! Also, I'm writing notes at the moment and was just wondering what sources you guys used to collate notes because reading your guides you have included information on Rayleigh jeans law etc and I can't find any textbook that mentions this information but I really like having an in depth understanding so just wanted to know what sources you guys used? :)
For me, Rayleigh Jeans came from uni. I've supplemented my HSC knowledge with my study at uni in a few places, you definitely don't need that for the HSC unless you want it! ;D
For reference though, in the HSC I primarily used:
- -Textbooks
- -Notes from my Teacher (they were amazing)
- -Hyperphysics
- -Youtube Videos on occasion if I happened to find something good
I'm a big believer in efficient study, only knowing what is necessary to get a Band 6 for the subject. I wrote these Physics notes with that in mind, condensed everything I know into exactly what you need to do well in the course, no fluff :) might be of interest to you! (forgive me shameless self advertising)
EDIT: These notes are freaking amazing, Jamon's just too modest. I literally tutor from them, and they have everything you need (plus great donut-related diagrams!). xoxo Jake
Post by: Swagadaktal on July 09, 2016, 02:59:32 pm
EDIT: These notes are freaking amazing, Jamon's just too modest. I literally tutor from them, and they have everything you need (plus great donut-related diagrams!). xoxo JakeLMAO AHAHAHA OMG
I love you guys omg
Post by: RuiAce on July 09, 2016, 03:01:23 pm
LMAO AHAHAHA OMGYea I cracked up at that as well.
I love you guys omg
Post by: Neutron on July 09, 2016, 03:37:46 pm
Trying to do some revision but I'm just dying instead.. Anyway, with Planck's quantum theory, he hypothesized that the black body emission curve was independent of the material of the blackbody. So basically the radiations emitted from copper and aluminium would produce the same curve, right? I'm just kinda confused because copper is a better thermal and electrical conductor than aluminium, and if Planck proposed that the source of radiation is FROM the oscillations of the atoms making up the material how can the type of atom (i.e. material) not have an effect on the subsequent radiation? Sorry if that sounded confusing ah..
Thanks guys!
Neutron
Post by: jamonwindeyer on July 09, 2016, 04:16:48 pm
Hey guys!
Trying to do some revision but I'm just dying instead.. Anyway, with Planck's quantum theory, he hypothesized that the black body emission curve was independent of the material of the blackbody. So basically the radiations emitted from copper and aluminium would produce the same curve, right? I'm just kinda confused because copper is a better thermal and electrical conductor than aluminium, and if Planck proposed that the source of radiation is FROM the oscillations of the atoms making up the material how can the type of atom (i.e. material) not have an effect on the subsequent radiation? Sorry if that sounded confusing ah..
Thanks guys!
Neutron
Hey there Neutron! Okay, I'm happy to be corrected on this answer because I'm not 100% sure, but here we go regardless ;D
Yes, copper is a better thermal and electrical conductor than aluminium. However, the black body radiation curves come from the thermal vibrations of the black body. So, even though copper is a better thermal conductor, all this means is that it is "easier" for copper to reach higher temperatures, so the black body radiation curve can shift to higher frequencies (say, visible light).
Although the atomic structure of the materials are different, black body radiation is thermal radiation. All atoms respond in the same way to thermal energy, by vibrating, and thus the radiation does not depend on the type of atom.
I hope that answer makes a little bit of sense, definitely happy for someone to give a better one ;D
Post by: RuiAce on July 09, 2016, 05:15:14 pm
Hey there Neutron! Okay, I'm happy to be corrected on this answer because I'm not 100% sure, but here we go regardless ;DSeems legit from a chemist's point of view as well.
Yes, copper is a better thermal and electrical conductor than aluminium. However, the black body radiation curves come from the thermal vibrations of the black body. So, even though copper is a better thermal conductor, all this means is that it is "easier" for copper to reach higher temperatures, so the black body radiation curve can shift to higher frequencies (say, visible light).
Although the atomic structure of the materials are different, black body radiation is thermal radiation. All atoms respond in the same way to thermal energy, by vibrating, and thus the radiation does not depend on the type of atom.
I hope that answer makes a little bit of sense, definitely happy for someone to give a better one ;D
Exactly. @Neutron - Just being a better thermal conductor only means it can absorb heat better.
Temperature can be seen as a measure of how much kinetic energy the particles are carrying overall. If the particles are placed in rooms with similar temperature, the kinetic energy of the particles will be around the same, making it equally easy/difficult for photons to hit electrons.
Conductance is also a measure of how easily electrons can flow THROUGH the substance in a circuit. It has nothing to do with the photoelectric effect, where they are ejected off.
Post by: smiley2101 on July 09, 2016, 06:27:14 pm
State how Hertz was able to determine the frequency of his e-m radiation
I thought that he just used an oscillator of known frequency yet the marking guideline states this : -"Correct statement involving resonance; Hertz used the condition giving the frequency when a circuit is in resonance and sparking optimally. "
I have never come across resonance in the hsc course.....? help :( thank you so much
Post by: jamonwindeyer on July 09, 2016, 11:07:48 pm
Hi again! Okay so I was doing a trial physics exam and one of the questions was stated as follows;
State how Hertz was able to determine the frequency of his e-m radiation
I thought that he just used an oscillator of known frequency yet the marking guideline states this : -"Correct statement involving resonance; Hertz used the condition giving the frequency when a circuit is in resonance and sparking optimally. "
I have never come across resonance in the hsc course.....? help :( thank you so much
Hmm, although I definitely know that resonance is applicable in this situation, I don't think it is necessary knowledge for the HSC either! I've definitely never seen it in any HSC textbook. Which Trial was this? Very curious indeed :o
Anyway, resonance in general is the idea that an oscillatory system can be made to oscillate at greater amplitudes when driven by external forces at a specific frequency (called the resonant frequencies). This is a mouthful, but you'll most definitely be familiar with the idea. Know the myth about opera singers breaking glasses with their voices? Based on this idea, the idea that singing at the resonant frequency of the glass will cause it to vibrate and break. Resonant frequencies are everywhere, anything that you can get "into a rhythm" is really just going at the resonant frequency.
Electrical resonance is similar, it is the idea that circuits will exhibit a greater response at specific input frequencies. In this situation, Hertz would absolutely have noticed the most sparking at the resonant frequency of the spark gaps. Why? Purely because the circuits, for reasons to complex to discuss here, just like that frequency. Hertz would use that frequency to make the sparking most obvious.
I hope this helps! Very interesting that that showed up in the criteria! :D
Post by: Sahar8642 on July 12, 2016, 09:35:30 pm
Does anyone have detailed notes on the practicals as we never covered them in school properly. As in, we watched while the teacher did the prac and didn't do proper write-ups
Thanks!
Post by: Sahar8642 on July 12, 2016, 09:39:31 pm
does anyone have concise notes on this dotpoint:
process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces
Thanks!
Post by: jamgoesbam on July 12, 2016, 10:02:08 pm
Post by: conic curve on July 12, 2016, 10:21:38 pm
A motor vehicle has a kerb weight of 1500kg and is moving with a velocity of 50 kilometres per hour. The mass of the driver is 80kg and there are no other passengers in the motor vehicle. This motor vehicle collides head on with a concrete wall (2 metres high and 1 metre thick) and comes to rest in 0.15 seconds.
Critically evaluate the change of energy which occurs during this collision.
Post by: Adriaclya on July 13, 2016, 02:43:13 pm
Didn't know how to do these questions :-\ ... could someone please explain them? Would be much appreciated! Thanks in advanced :)Hi! For the second question ie the second picture question??
An induced emf is produced because there is varying magnetic flux or a changing magnetic field relative to the wires - which are conductors. (This is true beacuse of faraday's law of electromagnetic induction.) The vary in magnetic field can be seen as the loop is moved away.
For the second part of that question, i think the current travels clockwise since the magnetic field is becoming weaker as the conductor moves further away from it. According to lenz's law, the direction of the current should produce a magnetic field that opposes the magnetic field which was induced by the straight current carrying conductor. Yeahhhhhh????
Hope it helps !
Post by: Happy Physics Land on July 13, 2016, 06:27:18 pm
Didn't know how to do these questions :-\ ... could someone please explain them? Would be much appreciated! Thanks in advanced :)
Hey Jamgoesbam!!!
The solution to your first question has been attached, the answers to your next question would come later. Basically for the first part, I clearly dont have a grid paper for me to draw on. But essentially you should draw just about everything it asks for and perhaps at the landing point you can further identify that the final vertical velocity is greater than the initial vertical velocity under the influence of gravitational field of attraction. The next part is basically a combination of your two formulae: v = u+at and distance = ut + 1/2 at2. You can perhaps try the other formula v2 = u2 + 2a(distance). Thats what I initially intended to do but in the end I decided the first method was the best in this situation. So, there you have it! :)
(http://i.imgur.com/VQ7xPD5.jpg)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on July 13, 2016, 06:40:13 pm
Didn't know how to do these questions :-\ ... could someone please explain them? Would be much appreciated! Thanks in advanced :)
For your next question. I will type out my answers here.
a) An emf will be induced in the loop because as the loop moves away from the current-carrying wire, there is relative motion between the loop and the magnetic field into the page. Consequently there will be a change in magnetic flux and according to Faraday's law, an emf will be induced. Using right hand palm rule, we can also ascertain that there will be a magnetic force pushing the positive charges downwards and negative charges upwards. As a result, a potential difference (hence emf) is created due to opposite charge accumulations at different ends of the loop.
b) The current direction is clockwise using right hand coil rule. Now this question doesnt require any explanation but I will explain here anyways. As the loop moves away from the wire, the magnetic field density decreases and hence there is less into the page magnetic flux penetrating through the loop. Therefore Lenz's law states that the current flow induced in the loop will be in a clockwise direction to compensate for the loss in magnetic flux into the page.
Post by: Happy Physics Land on July 13, 2016, 07:01:28 pm
Didn't know how to do these questions :-\ ... could someone please explain them? Would be much appreciated! Thanks in advanced :)
This last question is relatively difficult. Many people struggle to understand the relationship between stopping potential, work function and kinetic energy because they are so closely connected. But at the same time it is absolutely essential to distinguish between them because they are, after all, different concepts. So before I start explaining, I would just like to point out some essential things about this experiment and about these concepts.
Maximum Ek = hfincident light - Work function
Stopping voltage = minimum energy required to stop photocurrent from reaching the receiver (E=qV)
Just a sidenote, the circuit is connected in such a way that the receiver is attached to a negative potential and hence repels photoelectrons that try to come towards it. Essentially a current can only emerge if maximum kinetic energy can overcome the energy produced by stopping potential (E=qV). With these information in mind, I have attached my solution below and you can have a look for yourself. This is a challenging topic and in case if l get the answer wrong or if you dont understand anything please ask and I will be happy to help! :)
(http://i.imgur.com/risi5lg.jpg)
Best Regards
Happy Physics Land
Post by: Happy Physics Land on July 13, 2016, 07:14:54 pm
Hey,
does anyone have concise notes on this dotpoint:
process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces
Thanks!
Hey Sahar!
Yep yep this is one of those really tedious physics dotpoints that ask for the relationship between science, society and politics. Before I start, I must emphasise that despite how bs this can be, the syllabus wants us to recognise the disastrous impact science can have on the society even though it was never intended to cause these outcomes. Anyways,
Max Planck and Albert Einstein are similar in that:
- Both were leaders in the quantum physics field
- Both were born in Germany
- Both were involved in developing atomic bombs
- Both campaigned for the use of science in wars and politics
- Both were devoted to science however fail to recognise the adverse impacts of these scientific developments
Differences:
- Einstein was a pacifist, Planck was a nationalist
- Planck signed a document supporting the justification of war for Germany, Einstein refuses to sign the same document and instead signed an anti-war manifesto
- Planck was incredibly patriotic whereas Einstein called nationalism an infantile disease
- Planck was well-renowned scientist in Germany but Einstein was more famous worldwide and received little prestige within Germany because he opposed Germany's military actions
- Einstein served for Americans but Planck served for Germany
- Planck pursued science for his country but Einstein recognises science as a human endeavour
Conclusively, even though Einstein and Planck in their normal lives are very close friends, their political aspirations and scientific views differ. Evidently, science is closely related to politics and war in a variety of ways.
Post by: Swagadaktal on July 13, 2016, 07:31:50 pm
Hey Sahar!WAIT WTF DO YOU HAVE TO RELATE POLITICS IN PHYSICS?
Yep yep this is one of those really tedious physics dotpoints that ask for the relationship between science, society and politics. Before I start, I must emphasise that despite how bs this can be, the syllabus wants us to recognise the disastrous impact science can have on the society even though it was never intended to cause these outcomes. Anyways,
Max Planck and Albert Einstein are similar in that:
- Both were leaders in the quantum physics field
- Both were born in Germany
- Both were involved in developing atomic bombs
- Both campaigned for the use of science in wars and politics
- Both were devoted to science however fail to recognise the adverse impacts of these scientific developments
Differences:
- Einstein was a pacifist, Planck was a nationalist
- Planck signed a document supporting the justification of war for Germany, Einstein refuses to sign the same document and instead signed an anti-war manifesto
- Planck was incredibly patriotic whereas Einstein called nationalism an infantile disease
- Planck was well-renowned scientist in Germany but Einstein was more famous worldwide and received little prestige within Germany because he opposed Germany's military actions
- Einstein served for Americans but Planck served for Germany
- Planck pursued science for his country but Einstein recognises science as a human endeavour
Conclusively, even though Einstein and Planck in their normal lives are very close friends, their political aspirations and scientific views differ. Evidently, science is closely related to politics and war in a variety of ways.
soz from vce here... do you have to study the life of scientists and view their achievements through their position in society? Is this examinable or just in sacs (in school tests)
Pwoah just really awestruck here lol.
=
Post by: Sahar8642 on July 13, 2016, 08:10:47 pm
Hey Sahar!
Yep yep this is one of those really tedious physics dotpoints that ask for the relationship between science, society and politics. Before I start, I must emphasise that despite how bs this can be, the syllabus wants us to recognise the disastrous impact science can have on the society even though it was never intended to cause these outcomes. Anyways,
Max Planck and Albert Einstein are similar in that:
- Both were leaders in the quantum physics field
- Both were born in Germany
- Both were involved in developing atomic bombs
- Both campaigned for the use of science in wars and politics
- Both were devoted to science however fail to recognise the adverse impacts of these scientific developments
Differences:
- Einstein was a pacifist, Planck was a nationalist
- Planck signed a document supporting the justification of war for Germany, Einstein refuses to sign the same document and instead signed an anti-war manifesto
- Planck was incredibly patriotic whereas Einstein called nationalism an infantile disease
- Planck was well-renowned scientist in Germany but Einstein was more famous worldwide and received little prestige within Germany because he opposed Germany's military actions
- Einstein served for Americans but Planck served for Germany
- Planck pursued science for his country but Einstein recognises science as a human endeavour
Conclusively, even though Einstein and Planck in their normal lives are very close friends, their political aspirations and scientific views differ. Evidently, science is closely related to politics and war in a variety of ways.
Thank You so much!
Post by: RuiAce on July 13, 2016, 08:20:28 pm
WAIT WTF DO YOU HAVE TO RELATE POLITICS IN PHYSICS?Do you reckon I'm an advocate for the current physics course?
soz from vce here... do you have to study the life of scientists and view their achievements through their position in society? Is this examinable or just in sacs (in school tests)
Pwoah just really awestruck here lol.
=
They had to reduce its difficulty so that general maths students would not be deterred from enrolling. An immediate drawback is in how there is much more emphasis on the social context of physics and chemistry than actual physics/chemistry.
Post by: jamgoesbam on July 13, 2016, 08:30:48 pm
This last question is relatively difficult. Many people struggle to understand the relationship between stopping potential, work function and kinetic energy because they are so closely connected. But at the same time it is absolutely essential to distinguish between them because they are, after all, different concepts. So before I start explaining, I would just like to point out some essential things about this experiment and about these concepts.
Maximum Ek = hfincident light - Work function
Stopping voltage = minimum energy required to stop photocurrent from reaching the receiver (E=qV)
Just a sidenote, the circuit is connected in such a way that the receiver is attached to a negative potential and hence repels photoelectrons that try to come towards it. Essentially a current can only emerge if maximum kinetic energy can overcome the energy produced by stopping potential (E=qV). With these information in mind, I have attached my solution below and you can have a look for yourself. This is a challenging topic and in case if l get the answer wrong or if you dont understand anything please ask and I will be happy to help! :)
(http://i.imgur.com/risi5lg.jpg)
Best Regards
Happy Physics Land
Hey Happy Physics Land,
Thanks heaps for your answers - explained things much more clearly to me than the answers! Just for the last question though, about the photoelectric effect, they changes it back into eV (just fyi)! Other than that, awesome effort :)
Post by: jamonwindeyer on July 14, 2016, 12:07:51 am
Do you reckon I'm an advocate for the current physics course?
They had to reduce its difficulty so that general maths students would not be deterred from enrolling. An immediate drawback is in how there is much more emphasis on the social context of physics and chemistry than actual physics/chemistry.
Will vouch for fact that (and I chatted about this in my lecture today) HSC Physics does not get you prepared for university physics very well at all. I'm all for the current course because it is accessible to everyone and I think getting everyone interested in science is awesome!! ;D
That said, I've heard rumours of a Physics Extension course being developed that is more mathematical, and I would be a big fan of such a course ! ;D
Edit: Do still dislike the social aspect, not in principle, but in just how often it is integrated. I think it is important for scientists to consider the social impacts of their actions, but yeah, too much for my taste.
Post by: RuiAce on July 14, 2016, 12:11:23 am
Will vouch for fact that (and I chatted about this in my lecture today) HSC Physics does not get you prepared for university physics very well at all. I'm all for the current course because it is accessible to everyone and I think getting everyone interested in science is awesome!! ;DReally curious as to what physics extension will be like
That said, I've heard rumours of a Physics Extension course being developed that is more mathematical, and I would be a big fan of such a course ! ;D
Edit: Do still dislike the social aspect, not in principle, but in just how often it is integrated. I think it is important for scientists to consider the social impacts of their actions, but yeah, too much for my taste.
Post by: jamonwindeyer on July 14, 2016, 12:12:02 am
Hey,
Does anyone have detailed notes on the practicals as we never covered them in school properly. As in, we watched while the teacher did the prac and didn't do proper write-ups
Thanks!
Hey! Totally don't mean to do a sell of the notes I wrote, but all the practicals you need for the exam are covered in the ATAR Notes Physics HSC Notes (they cost $25.00). To be honest, I haven't seen much on the practicals outside of those notes, but keep in mind you won't need too much on them for the exam! Just an understanding of the physics principles involved at the basic level (check the syllabus for the details as to which practicals you need) ;D
Post by: Happy Physics Land on July 14, 2016, 12:14:05 am
Will vouch for fact that (and I chatted about this in my lecture today) HSC Physics does not get you prepared for university physics very well at all. I'm all for the current course because it is accessible to everyone and I think getting everyone interested in science is awesome!! ;D
That said, I've heard rumours of a Physics Extension course being developed that is more mathematical, and I would be a big fan of such a course ! ;D
Edit: Do still dislike the social aspect, not in principle, but in just how often it is integrated. I think it is important for scientists to consider the social impacts of their actions, but yeah, too much for my taste.
I also heard rumours about BOSTES changing the physics syllabus in 2018
Post by: RuiAce on July 14, 2016, 12:15:35 am
I also heard rumours about BOSTES changing the physics syllabus in 2018Well it's about time they got rid of stuff like TVs and old definition of Meissner effect and so on
Post by: brontem on July 14, 2016, 12:28:47 am
Do you reckon I'm an advocate for the current physics course?I'm going to blow in here and just give my two cents lol :P :P Of course HSC physics is not anywhere near uni level physics, they can't expect people to do uni level stuff in year 12 and excel at it.. how much actual physics and chemistry could possibly be squeezed into the curriculum? not to mention that a lot of the stuff everyone gets taught, regardless of the maths involved, is wrong
They had to reduce its difficulty so that general maths students would not be deterred from enrolling. An immediate drawback is in how there is much more emphasis on the social context of physics and chemistry than actual physics/chemistry.
Obviously not doing high level maths could be a drawback, if you commit to doing HSC physics then you can commit to putting in the extra effort for the maths involved, but don't go blaming us general kids for all the wordy crap in exams, we don't like it either ;) ;)
please don't hate me
Post by: RuiAce on July 14, 2016, 12:31:13 am
I'm going to blow in here and just give my two cents lol :P :P Of course HSC physics is not anywhere near uni level physics, they can't expect people to do uni level stuff in year 12 and excel at it.. how much actual physics and chemistry could possibly be squeezed into the curriculium? not to mention that a lot of the stuff everyone gets taught, regardless of the maths involved, is wrongLol I never said it had to be like "uni" physics.
Obviously not doing high level maths could be a drawback, if you commit to doing HSC physics then you can commit to putting in the extra effort for the maths involved, but don't go blaming us general kids for all the wordy crap in exams, we don't like it either ;) ;)
please don't hate me
Look at the exams before the year 2001. That's some real physics and chemistry.
I wouldn't say it's the general maths' students "fault" that the exam got dragged down in difficulty. That was a decision by BOS - not petitioned. But I doubt at the time they realised how many cons they'd introduce and that they potentially outweigh the pros.
Post by: brontem on July 14, 2016, 12:38:07 am
Look at the exams before the year 2001. That's some real physics and chemistry.
I'm good mate, I'll just stick to the 2016 fantasy physics ba dum tsh ;) ;)
Post by: jamonwindeyer on July 14, 2016, 12:38:59 am
I'm going to blow in here and just give my two cents lol :P :P Of course HSC physics is not anywhere near uni level physics, they can't expect people to do uni level stuff in year 12 and excel at it.. how much actual physics and chemistry could possibly be squeezed into the curriculium? not to mention that a lot of the stuff everyone gets taught, regardless of the maths involved, is wrong
Obviously not doing high level maths could be a drawback, if you commit to doing HSC physics then you can commit to putting in the extra effort for the maths involved, but don't go blaming us general kids for all the wordy crap in exams, we don't like it either ;) ;)
please don't hate me
Love the way you put this. You are absolutely right, and tbh, if it were up to me I'd maintain the current level of mathematics in Physics and introduce a new subject for those who want to move on and do tertiary physics, that is a bit more mathematical :)
PS - No one will ever hate you for your opinion!
Lol I never said it had to be like "uni" physics.
Look at the exams before the year 2001. That's some real physics and chemistry.
I wouldn't say it's the general maths' students "fault" that the exam got dragged down in difficulty. That was a decision by BOS - not petitioned. But I doubt at the time they realised how many cons they'd introduce and that they potentially outweigh the pros.
I don't think the difficulty was changed too much: It is just the difficulty shifted from the mathematical end of the spectrum to the analytic side of the spectrum that is more similar to the HSIE subjects ;D of course many students find the math stuff harder than the HSIE stuff,
Very interesting discussion guys, but probably let's keep it back to Physics Q+A now, I might do some research regarding the Physics Extension subject and start a discussion thread if I find more! Or feel free to start one now, don't let me interrupt the conversation ;)
I'm good mate, I'll just stick to the 2016 fantasy physics ba dum tsh
;)
Post by: brontem on July 14, 2016, 12:51:10 am
Why is it D? It's probably really obvious but idk? I'm pretty good with M&G but again I don't recall going through emf in any depth besides defining it basically
Thanks :) :)
Post by: Swagadaktal on July 14, 2016, 01:00:09 am
Since everyone seems to be upYo I think it's faraday's law (or some other physics hoe)( yo jamon or something will probs do this 100x better but if you need an answer in the mean time hehe :P )
Why is it D? It's probably really obvious but idk? I'm pretty good with M&G but again I don't recall going through emf in any depth besides defining it basically
Thanks :) :)
The equation goes: Induced EMF = -dflux/dt
So as you can see, there's a sin wave there for fluf. The dy/dx function of a sin wave is a cos wave f(x) = sin(x), f'(x) = cos(x)
Now, with induced current, the sign is reversed (positive becomes negative) - so you'd have a -cos(x) wave for induced emf. Which matches up with D -- are you expected to know diff of functions in HSC physics? I'm a vce nomad :P
Post by: brontem on July 14, 2016, 01:05:21 am
Yo I think it's faraday's law (or some other physics hoe)
100% going to be starting every answer to any Faraday question with this
But thanks :) What we're meant to know I'm not sure (hence me asking the question hahaha), graphs like this I just force into my memory ;D
Post by: jamonwindeyer on July 14, 2016, 01:27:33 am
100% going to be starting every answer to any Faraday question with this
But thanks :) What we're meant to know I'm not sure (hence me asking the question hahaha), graphs like this I just force into my memory ;D
Swag's got you covered, it's definitely Faraday, my favourite Physics hoe ;)
Faraday's Law goes like this:
In words, the induced EMF/voltage is given by the rate of change of magnetic flux with respect to time (there are more complex versions of this formula out there, but this is the principle).
Note that electromotive force (EMF) is interchangeable with the term "voltage" at this level, they are really one in the same. So, the voltage we generate is given by how quickly we are changing the magnetic field. The fact that the negative sign is there reflects Lenz's Law, the induced voltage should OPPOSE the changing field that created it, so essentially:
The two cancel each other out (as we'd expect), if they didn't that would violate Lenz's Law.
Anyway, in terms of the question, the formula itself is enough to answer mathematically. A sine curve is shown, so the negative derivative is a negative cosine curve, answer D. However, a Physics student isn't expected to know how to do this, so let's do it the longer way.
The rate of change of the magnetic field can be considered properly by taking the derivative of the sine curve, or just simply notice that the EMF should be at a peak when the magnetic field is changing rapidly, and when the magnetic field is at a peak (not changing), the induced EMF should be zero. Remember we only have induced EMF when magnetic flux is changing. That eliminates options A and C, because they don't have peaks/troughs in the correct places. We want either B or D.
Now, the next one is where we bring in Faraday's Law for ease of use. The negative sign makes selecting D easier to see. However, Faraday's Law isn't explicitly in the syllabus in the mathematical form, so instead, you could use a Lenz's Law explanation. The EMF must be opposite in sign to the rate of change of magnetic field, because the induced EMF must oppose the change that created it. So, as the rate of change of magnetic flux is positive (EG - at the start of the top graph), we expect a negative peak (trough) for the EMF. D matches this response ;D
Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? ;D
Post by: brontem on July 14, 2016, 01:35:19 am
Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? ;DAhh yep that definitley makes sense - got it now, thank you!! ;D its from the 2012 HSC exam ;D
Post by: jamonwindeyer on July 14, 2016, 01:39:37 am
Ahh yep that definitley makes sense - got it now, thank you!! ;D its from the 2012 HSC exam ;D
Oh lol, okay never mind BOSTES wanted to push with that question then!! ;)
To clarify, you definitely need to know Faraday's, but the mathematical version that makes that question easier is not compulsory as to my knowledge. Great to know about though if you like it!
Post by: MysteryMarker on July 14, 2016, 09:42:35 am
Need help with this question, thanks guys.
Post by: jakesilove on July 14, 2016, 10:23:34 am
An external observer on a planet witnesses a space probe to undergo the slingshot effect (on the same planet). Would an observer on the planet agree that the space probe's speed has increased? discuss your reasons.
Need help with this question, thanks guys.
I've blatantly stolen this answer from another source, but here ya go! Hope it makes sense :)
No, the observer will argue that the probe's speed has not increased.
Relative to the observer, the probe's speed will appear to be unchanged. As the probe approaches the assisting planet, its GPE will decrease and its kinetic energy will increase but as the probe moves away from the assisting planet, its GPE will increase and kinetic energy will decrease relative to the observer. This means that the speed at which the probe approaches will be the same as the speed at which the probe moves away from the assisting planet.
In actual fact, the velocity of the probe relative to the sun has changed in both magnitude and direction because the probe's output velocity is equal to the input velocity added to the velocity component of the assisting planet. This makes an observer from the sun argue that the probe's velocity (speed and magnitude) has increased.
Jake
Post by: MysteryMarker on July 14, 2016, 10:36:14 am
I've blatantly stolen this answer from another source, but here ya go! Hope it makes sense :)
No, the observer will argue that the probe's speed has not increased.
Relative to the observer, the probe's speed will appear to be unchanged. As the probe approaches the assisting planet, its GPE will decrease and its kinetic energy will increase but as the probe moves away from the assisting planet, its GPE will increase and kinetic energy will decrease relative to the observer. This means that the speed at which the probe approaches will be the same as the speed at which the probe moves away from the assisting planet.
In actual fact, the velocity of the probe relative to the sun has changed in both magnitude and direction because the probe's output velocity is equal to the input velocity added to the velocity component of the assisting planet. This makes an observer from the sun argue that the probe's velocity (speed and magnitude) has increased.
Jake
Oh, because what i was thinking is that for the maximum effect of the slingshot effect, the probe would gain a significant amount of energy, almost double the planets orbital speed. When this occurs there is a compromise. The planet itself will lose a bit of KE to compensate for the probe's gain, but due to the planets large mass, its velocity change is minimal. SO therefore, an observer on the planet will see an increase in speed.
Is this wrong or is the answer for the above question subjective?
Post by: Happy Physics Land on July 14, 2016, 10:44:22 am
Oh, because what i was thinking is that for the maximum effect of the slingshot effect, the probe would gain a significant amount of energy, almost double the planets orbital speed. When this occurs there is a compromise. The planet itself will lose a bit of KE to compensate for the probe's gain, but due to the planets large mass, its velocity change is minimal. SO therefore, an observer on the planet will see an increase in speed.
Is this wrong or is the answer for the above question subjective?
Hey Mysterymarker!
To be honest l dont like the way the question was written. You can think about Jake's answer in another way: suppose you are an observer standing on the assisting planet and you are rotating along with the planet's rotation about its central axis. The space probe undergoing slingshot effect will "steal" the planet's angular momentum, meaning that it will then be accelerating in the same direction of rotation as the assisting planet. This means that for you, there will be little change in relative velocity of the space probe since you are both rotating in the same direction with approximately the same angular velocity. So I can only say here that it is POSSIBLE that the observer wouldnt see a change in relative velocity. But for me, the best answer would be there would be a decrease in relative velocity observed compared to when you are an observer from another planet and therefore the increase in relative velocity would be too insignificant to be noticed.
Hope it helps!
Best Regards
Happy Physics Land
Post by: RuiAce on July 14, 2016, 11:03:23 am
Can I have some help pleaseWhat is he improving?
A weight lifter Can lift 140kg on earth he has been sent to Venus to improve his. Given the acceleration due to gravity on Venus is 8.93m/s/s, what should he be able to lift on Venus with the same effort?
Anyway, this question treats force as constant.
On Earth:
F = mEgE = 140kg * 9.8 m s-2 = 1372 N
On Venus:
mV = F/aV = 1372 N / 8.93 m s-2 = 153.63941...kg = 150 kg (2 s. f.)
Post by: conic curve on July 14, 2016, 11:20:36 am
What is he improving?
Anyway, this question treats force as constant.
On Earth:
F = mEgE = 140kg * 9.8 m s-2 = 1372 N
On Venus:
mV = F/aV = 1372 N / 8.93 m s-2 = 153.63941...N = 15 N (2 s. f.)
Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?
Thanks
Post by: RuiAce on July 14, 2016, 11:21:40 am
Will just clarify that that last line should read:Couldn't you just have edited my mistake (please) :P
Note the change in units and the extra 0 ;D
Post by: jamonwindeyer on July 14, 2016, 11:22:33 am
Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?
Thanks
That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth ;D
Post by: RuiAce on July 14, 2016, 11:23:16 am
Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?It's 9.8 m s-2 on EARTH.
Thanks
(In actuality it isn't, because the Earth is not a sphere - rather a spheroid. We just assume the Earth is a sphere for simplicity.)
Post by: jamonwindeyer on July 14, 2016, 11:23:54 am
Couldn't you just have edited my mistake (please) :P
With your permission I will definitely do that from now on (don't like to touch other peoples posts without permission - unless I have to of course - arbitrary use of moderator power ;) )
Post by: RuiAce on July 14, 2016, 11:24:39 am
With your permission I will definitely do that from now on (don't like to touch other peoples posts without permission - unless I have to of course - arbitrary use of moderator power ;) )Aha if it's just a correction then I permit ;) just make a note of what you fixed please :D
Post by: EEEEEEP on July 14, 2016, 11:31:47 am
Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?
Thanks
BEcause that's how strong gravity is on earth.
Post by: conic curve on July 14, 2016, 11:32:15 am
That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth ;D
So this is "by definition"
Post by: EEEEEEP on July 14, 2016, 11:34:29 am
So this is "by definition"NOt by definition. But it is the standard figure used for problems.
Post by: conic curve on July 14, 2016, 11:36:25 am
NOt by definition. But it is the standard figure used for problems.
Where's it derived from? It must be "by convention" then
It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity
Post by: RuiAce on July 14, 2016, 11:37:21 am
Where's it derived from? It must be "by convention" thenJamon already answered this.
It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity
That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth ;D
Post by: EEEEEEP on July 14, 2016, 11:40:42 am
Where's it derived from? It must be "by convention" thenIt's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.
It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity
Newtons law states that the sum of the forces equals the mass* acceleration
m*a = F
gravitational force is calculated as
F = G*m*M/r^2
where G is the gravitational constant (measured, look it up on wikipedia)
m is the mass of the falling body
M is the mass of the earth (or other planet)
r is the distance between the falling body and the center of the earth (or other planet)
m*a = G*m*M/r^2
cancel m
a = G*M/r^2
if you put in the correct values
G = 6.67428*10^(-11) m^3/(kgs^2)
M = 5.9736*10^(24) kg
r = 6.371*10^(6) m - average radius of earth
so if the mass is approximatly on the surface of the earth (say only a few feet above it) then the acceleration is
a = 9.8226 m/s^2
pretty close to the measured value of 9.80667---m/s^2. You'd get closer to that value if you used more significant digits in G, M and r.
Post by: RuiAce on July 14, 2016, 11:41:39 am
It's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.
Post by: jakesilove on July 14, 2016, 11:50:54 am
It's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.
Newtons law states that the sum of the forces equals the mass* acceleration
m*a = F
gravitational force is calculated as
F = G*m*M/r^2
where G is the gravitational constant (measured, look it up on wikipedia)
m is the mass of the falling body
M is the mass of the earth (or other planet)
r is the distance between the falling body and the center of the earth (or other planet)
m*a = G*m*M/r^2
cancel m
a = G*M/r^2
if you put in the correct values
G = 6.67428*10^(-11) m^3/(kgs^2)
M = 5.9736*10^(24) kg
r = 6.371*10^(6) m - average radius of earth
so if the mass is approximatly on the surface of the earth (say only a few feet above it) then the acceleration is
a = 9.8226 m/s^2
pretty close to the measured value of 9.80667---m/s^2. You'd get closer to that value if you used more significant digits in G, M and r.
Was literally about to post the derivation, thanks for sending that through! It's not too difficult a proof, even if it isn't directly examinable. A bit of cool Physics :)
Post by: MysteryMarker on July 14, 2016, 02:33:13 pm
Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.
Thanks guys.
Post by: RuiAce on July 14, 2016, 02:36:56 pm
An object that is travelling at a speed of 0.9c relative to an observer has a mass of 100kg measured by the observer. Calculate the mass of the object measured by the same observer when it is at rest relative to the observer?Can't explain it carefully anymore but the idea is to keep in mind that all frames of references are relative. Mass dilation says that moving objects become heavier. If the object is moving at 0.9c, it is going to appear way heavier than if it were at rest (relatively speaking).
Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.
Thanks guys.
Post by: jamonwindeyer on July 14, 2016, 02:37:38 pm
An object that is travelling at a speed of 0.9c relative to an observer has a mass of 100kg measured by the observer. Calculate the mass of the object measured by the same observer when it is at rest relative to the observer?
Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.
Thanks guys.
Hey! Okay, so it sounds like you know the idea here is mass dilation. The mass of an object (as measured by a stationary observer) changes based on its velocity due to this principle (the idea being that mass and energy are equivalent, and increasing kinetic energy also results in slight increase in mass). Since the measurement of 100kg is made when travelling close to the speed of light, the actual mass of the object must be much less. It is only be speeding the object to 0.9c that the mass dilates and increases to the value of 100kg. Hence, we use the formula to find the initial mass. Does that help? ;D
Edit: Combine Rui's comment about frames of reference with this for a pretty good explanation. We pretty much said the same thing ;)
Post by: MysteryMarker on July 14, 2016, 02:45:35 pm
Can't explain it carefully anymore but the idea is to keep in mind that all frames of references are relative. Mass dilation says that moving objects become heavier. If the object is moving at 0.9c, it is going to appear way heavier than if it were at rest (relatively speaking).
Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?
Thanks.
Post by: jamonwindeyer on July 14, 2016, 02:51:51 pm
Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?
Thanks.
It would mean that the two are travelling with the same velocity. Possibly 0.9c, possibly stationary, but that makes no difference either way (relativity) ;D
Post by: RuiAce on July 14, 2016, 02:52:42 pm
Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?Yeah they either travel at 0.9c relative to each other, or they're at rest relative to each other. Always remember no absolute.
Thanks.
Post by: MysteryMarker on July 14, 2016, 03:01:20 pm
Thanks guys.
Another question --> A plane flies high up in the atmosphere and then plunges towards the earth to produce weightlessness for the occupants inside. Comment on the use of the term weightlessness in this context.
The answer says that the would not be weightless as they are still under the influence of gravity, but they would 'feel' weightless because there is no contact force. Could someone expand on this and explain how they are not weightlessness but are at the same time?
Cheers.
Post by: Swagadaktal on July 14, 2016, 03:42:38 pm
Swag's got you covered, it's definitely Faraday, my favourite Physics hoe ;)Yeah see what I mean their explanations are on fleek
Faraday's Law goes like this:
In words, the induced EMF/voltage is given by the rate of change of magnetic flux with respect to time (there are more complex versions of this formula out there, but this is the principle).
Note that electromotive force (EMF) is interchangeable with the term "voltage" at this level, they are really one in the same. So, the voltage we generate is given by how quickly we are changing the magnetic field. The fact that the negative sign is there reflects Lenz's Law, the induced voltage should OPPOSE the changing field that created it, so essentially:
The two cancel each other out (as we'd expect), if they didn't that would violate Lenz's Law.
Anyway, in terms of the question, the formula itself is enough to answer mathematically. A sine curve is shown, so the negative derivative is a negative cosine curve, answer D. However, a Physics student isn't expected to know how to do this, so let's do it the longer way.
The rate of change of the magnetic field can be considered properly by taking the derivative of the sine curve, or just simply notice that the EMF should be at a peak when the magnetic field is changing rapidly, and when the magnetic field is at a peak (not changing), the induced EMF should be zero. Remember we only have induced EMF when magnetic flux is changing. That eliminates options A and C, because they don't have peaks/troughs in the correct places. We want either B or D.
Now, the next one is where we bring in Faraday's Law for ease of use. The negative sign makes selecting D easier to see. However, Faraday's Law isn't explicitly in the syllabus in the mathematical form, so instead, you could use a Lenz's Law explanation. The EMF must be opposite in sign to the rate of change of magnetic field, because the induced EMF must oppose the change that created it. So, as the rate of change of magnetic flux is positive (EG - at the start of the top graph), we expect a negative peak (trough) for the EMF. D matches this response ;D
Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? ;D
Btw can you teach me how to input formulas? - How you get that fancy writing :3
Post by: jakesilove on July 14, 2016, 03:45:21 pm
Yeah see what I mean their explanations are on fleek
Btw can you teach me how to input formulas? - How you get that fancy writing :3
Hey! Give this epic guide a read to learn how LaTeX works!
Post by: Skidous on July 14, 2016, 03:54:50 pm
Damn that just changed the whole game. *Mindblows*
Thanks guys.
Another question --> A plane flies high up in the atmosphere and then plunges towards the earth to produce weightlessness for the occupants inside. Comment on the use of the term weightlessness in this context.
The answer says that the would not be weightless as they are still under the influence of gravity, but they would 'feel' weightless because there is no contact force. Could someone expand on this and explain how they are not weightlessness but are at the same time?
Cheers.
Hi Mystery Marker
The reason as to why they are not weightless is due to the fact they are still within the gravitational field of earth and thus are experiencing their own weight force as described by W=mg
The reason why they experience this feeling of weightlessness is because they are experiencing no g-forces or 0g. This means that the force of gravity they feel exerted on them is not their regular weight (or g-forces<1), similar to that of going down in an elevator and feeling slightly less heavier (this is an actual thing that happens and an experiment you may want to try yourself). This gives them to sense that there is no gravity by they are just free falling in the craft.
There is also the contrast to this weightlessness as the ascent back up to that height also produces more g-forces (g-forces>1) and makes people feel heavier than they are.
I hope this answers your question
Skidous
Post by: Happy Physics Land on July 14, 2016, 06:26:14 pm
Hi Mystery Marker
The reason as to why they are not weightless is due to the fact they are still within the gravitational field of earth and thus are experiencing their own weight force as described by W=mg
The reason why they experience this feeling of weightlessness is because they are experiencing no g-forces or 0g. This means that the force of gravity they feel exerted on them is not their regular weight (or g-forces<1), similar to that of going down in an elevator and feeling slightly less heavier (this is an actual thing that happens and an experiment you may want to try yourself). This gives them to sense that there is no gravity by they are just free falling in the craft.
There is also the contrast to this weightlessness as the ascent back up to that height also produces more g-forces (g-forces>1) and makes people feel heavier than they are.
I hope this answers your question
Skidous
Josh made a really good point explaining weightlessness in terms of g-force. Good job on his part, definitely would include it in my answer.
I would just like to drop in my 2 cents as well. Weight always exists, unless you arrive at a point in the universe where there is 0 gravitational field (hypothetically) you can't not have weight. Now we experience the idea of "weight" in terms of normal force - something that we feel. We feel heavy because the ground pushes against us with greater magnitude if we exert too much weight force onto the ground (Newton's 3rd Law of Motion). The scale doesnt measure our "weight" but instead measures the amount of reaction force it has to exert on us when we stand onto the scale. Keeping that in mind, when the answer mentions contact force, they are really talking about NORMAL FORCE. When the plane plunges down into Earth, our inertia makes us want to remain in the same prior position (Newton's 1st law) and hence momentarily we would not be able to "feel" our weight because we are not exerting a force on anything (i.e. we arent really have physical contact with anything". So this is why they say "We still have weight, but we just FEEL like we are weightless".
Just to go a step further (you dont necessarily need this for you answer). You can perhaps use the word "free fall", but lm not going to risk that because free fall usually refers to when you and your frame of reference are accelerating in the same magnitude. E.g. You feel weightless in space because your spaceship and you are both experiencing 9.8 m/s2 towards Earth. Technically speaking, weight is a numerical concept but the FEELING of weight is really a relative concept. You can mention this in your response as well but what l wrote in the first paragraph combined with what Josh wrote was be sufficient enough.
Post by: MysteryMarker on July 14, 2016, 09:44:51 pm
Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?
i guess what I'm trying to ask is, how does a grid in a cathode ray tube work? :P
Cheers.
Post by: wyzard on July 14, 2016, 10:37:48 pm
For ideas to implementation,
Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?
i guess what I'm trying to ask is, how does a grid in a cathode ray tube work? :P
Cheers.
I've attached a simple schematic of a CRT from wikipedia. I'm guessing the grid you're talking about refers to the deflecting coils to control the electrons.
A CRT basically works by having an electron gun, the cathode and the heater, firing electrons to a fluorescent screen which lights up when an electron lands on it. The focus and deflecting coil will then control the direction of the fired electron. The coils are connected to a voltage source to produce an electric field, and they're usually oppositely charged. If the above is positively charged, the bottom part will be negatively charged which will deflect the electron upwards. Their polarity depends on where you want the electron to land on the screen.
The use of electric field though is the old design; more modern designs uses a varying magnetic field to deflect the electrons, which also requires voltage to drive current through the coils.
Last but not least, the deflecting coil usually does not affect the velocity, it just changes their direction; the velocity of electrons depends on the speed they are fired from the cathode.
Post by: RuiAce on July 14, 2016, 10:41:05 pm
This is copied out of Physics in Focus
Second, there exists another electrode in between the cathode and anode, which is named the grid. Making the grid more positive or negative with respect to the cathode controls the number of electrons reaching the anodes and hence striking the display screen per unit time, which consequently controls the intensity of the cathode ray and thus the brightness of the display.
Post by: jamonwindeyer on July 14, 2016, 10:47:50 pm
For ideas to implementation,
Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?
i guess what I'm trying to ask is, how does a grid in a cathode ray tube work? :P
Cheers.
Hey!! Okay, so this question has the potential to be very simple or extremely complicated! You'll never need the level of detail I'm about to give you in the HSC ;D
Before I start, the grid you are referring to is just another electrode like the cathode/anode. It is therefore, definitely connected to a voltage source, which will cause it to be positively or negatively charged ;D
Basically, Cathode Ray Tubes can contain a number of grids between the cathode and anode, which fulfil different functions. When you refer to the velocity/number of electrons, we can actually control those with separate grids. By placing a negatively charged grid immediately in front of the cathode, we can deflect a certain number of electrons based on how negatively charged the grid is. This way we can control the number of electrons. Placing the grid halfway between the cathode and anode won't change the number of electrons too much. Instead, it will just alter their velocity, because by this point they are already moving fast enough that they are almost definitely just going to continue to the anode :)
Pretty much, we can fulfil different functions based on where we place the grid and how it is placed. That can get quite complex (I simplified it a tad above, and it could be slightly off). Of course the simple version of this answer is just that the grid (or grids) is placed halfway between the cathode and anode, charged positively or negatively, to change the number of electrons striking the screen. That is the version Rui grabbed from Physics in Focus, and all that you'll need in any exam ;D
Post by: Happy Physics Land on July 14, 2016, 11:52:51 pm
100% agree. Who is josh, by the way? Is he an ATARNotes lecturer?
We all love the work you have done for economics and business!!! Extremely happy to see you involved and your work has been fabulous mate!!! (Is that what you want me to say???)
Post by: MysteryMarker on July 15, 2016, 03:21:04 pm
How would i go about this question to get the maximum? The marking criteria says 'an explanation of at least three observations', what does this mean?
Thanks guys.
Post by: conic curve on July 15, 2016, 04:59:23 pm
Outline Einsteins explanation for the photoelectric effect. (6marks)
How would i go about this question to get the maximum? The marking criteria says 'an explanation of at least three observations', what does this mean?
Thanks guys.
I think it means each observation is worth two marks
Post by: conic curve on July 15, 2016, 05:08:42 pm
Definitely not - there were way less than 6 observations.
I no longer remember the practical off by heart though - will wait for Jamon or Jacky or Jake someone. (Tbh I just wanted to use the 3 J's)
Sorry I meant 2. I'll change my comment
Post by: RuiAce on July 15, 2016, 05:13:13 pm
Sorry I meant 2. I'll change my commentLol all good.
Hard to say. Whilst you can keep that frame of mind (and I reckon you could be right), 6 markers become ambiguous because the examiner probably subtly wants more.
Post by: MysteryMarker on July 15, 2016, 05:29:51 pm
Also, what are some experimental evidence that the wave model of light could not explain for the photoelectric effect?
Thanks Guys.
Post by: Happy Physics Land on July 15, 2016, 06:18:35 pm
Could you please explain what the observations are, and how many Einstein accounted for with his 'photoelectric effect'?
Also, what are some experimental evidence that the wave model of light could not explain for the photoelectric effect?
Thanks Guys.
Hey MysteryMarker!
The question is a bit poorly worded here but I think I know approximately what you wanna ask, I will break my answer down into three sections, keeping in mind that Einstein wasn't the first to observe photoelectric effect, nor did he perform the experiment to ascertain the photoelectric effect (He just took the credit). And with "how many" I think you perhaps mean how many electrons did he see getting emitted? Well this emission of electrons is something thats theorised, even nowadays with the really advanced technologies we still can't see the movement of electrons, we can only propose likely reasons to explain the occurrence of certain events. So no it wouldnt be possible for him to see the amount of electrons being emitted. If you are talking about how many times he observed the photoelectric effect taking place, I wouldnt be sure and this is definitely something you dont need for physics.
First observation of photoelectric effect - By Heinrich Hertz:
- He conducted an experiment to ascertain the existence of electromagnetic radiation
- Because the spark at the receiver was perhaps a little hard to see, he decided to perform the experiment in a dark room
- He saw that the intensity of spark in a sun-lit bright room increased comparing to the intensity in a dark room
Photoelectric effect postulate - By Einstein:
- He theorised that when an incident light hits a metal surface, as long as the light possess frequency that is above the threshold frequency (i.e. if E=hf of incident light can overcome the work function), then a valence electron would be ejected with maximum kinetic energy (maximum because it takes minimum energy to eject a valence electron which is least tightly held to nucleus)
- He theorised that electrons and photons interact 1:1
- From this he theorised that light is a stream of photons and hence possess particle nature
Evidence for photoelectric effect - credited by Einstein and Planck:
- Blackbody radiation was a solid proof to photoelectric effect
- Einstein made his photoelectric effect postulates based upon the shape of the experimental black body radiation curve and Planck's explanation of the blackbody curve
- Experiment involving stopping voltage: This is the most important piece of evidence. So of course we can't seen current macroscopically, nor is there any microscopes to help Einstein to see the electrons. Therefore his assistant performed this experiment which aimed to show that yes there is a current (even though we couldnt see it) by shining a light onto the cathode and observe the reading on the ammeter. There was a current reading and therefore it was ascertained that a photocurrent was formed. To further detect the magnitude of the frequency of incident light ray, the anode was made negative (by supplying it with a negative potential) to repel the electron current. The moment ammeter reading --> 0, Einstein could apply the formula Ek = hf - work function to figure out frequency.
Post by: Goodwil on July 15, 2016, 06:46:25 pm
Post by: RuiAce on July 15, 2016, 07:00:36 pm
Hi, I'm having trouble understanding how to get the answer for this question (correct answer is A)I'll let someone else do the more comprehensive answer since I'm just on my phone but in short:
1. Derive the formula v=sqrt(GM/r) or just quote it cause it's multiple choice.
2. GM is fixed. r varies. Note that v/sqrt(r)=sqrt(GM)
Therefore first v/sqrt(r) equals second V/sqrt(R)
Post by: jamonwindeyer on July 15, 2016, 07:28:10 pm
I'll let someone else do the more comprehensive answer since I'm just on my phone but in short:
1. Derive the formula v=sqrt(GM/r) or just quote it cause it's multiple choice.
2. GM is fixed. r varies. Note that v/sqrt(r)=sqrt(GM)
Therefore first v/sqrt(r) equals second V/sqrt(R)
Can do ;)
First we derive the formula for orbital velocity by equating gravitational and centripetal forces:
So let's let that expression be the orbital velocity of P, since it suits. Now, if instead we have a distance of 2R:
Thus yielding the answer of A ;D
Post by: EEEEEEP on July 15, 2016, 07:30:27 pm
I'll let someone else do the more comprehensive answer since I'm just on my phone but in short:
1. Derive the formula v=sqrt(GM/r) or just quote it cause it's multiple choice.
2. GM is fixed. r varies. Note that v/sqrt(r)=sqrt(GM)
Therefore first v/sqrt(r) equals second V/sqrt(R)
((SLIGHTLY DIFF METHOD THan the other two.. but never the less ))
v=sqrt(GM/r)
>>> * Let the two different radius be 1 and 2 (planet 1 and 2 respectively ) >>>>
Planet one .
First we square the formula.
V^2 = GM/1
GM = V^2 (important for planet 2)
Planet two
Orbital velocity 2 = sqrt (GM/2)
THUS
Orbital velocity 2 = sqrt (V^2 / 2)
= V / sqrt (2)
= A
Post by: MysteryMarker on July 15, 2016, 07:48:34 pm
Hey MysteryMarker!
The question is a bit poorly worded here but I think I know approximately what you wanna ask, I will break my answer down into three sections, keeping in mind that Einstein wasn't the first to observe photoelectric effect, nor did he perform the experiment to ascertain the photoelectric effect (He just took the credit). And with "how many" I think you perhaps mean how many electrons did he see getting emitted? Well this emission of electrons is something thats theorised, even nowadays with the really advanced technologies we still can't see the movement of electrons, we can only propose likely reasons to explain the occurrence of certain events. So no it wouldnt be possible for him to see the amount of electrons being emitted. If you are talking about how many times he observed the photoelectric effect taking place, I wouldnt be sure and this is definitely something you dont need for physics.
First observation of photoelectric effect - By Heinrich Hertz:
- He conducted an experiment to ascertain the existence of electromagnetic radiation
- Because the spark at the receiver was perhaps a little hard to see, he decided to perform the experiment in a dark room
- He saw that the intensity of spark in a sun-lit bright room increased comparing to the intensity in a dark room
Photoelectric effect postulate - By Einstein:
- He theorised that when an incident light hits a metal surface, as long as the light possess frequency that is above the threshold frequency (i.e. if E=hf of incident light can overcome the work function), then a valence electron would be ejected with maximum kinetic energy (maximum because it takes minimum energy to eject a valence electron which is least tightly held to nucleus)
- He theorised that electrons and photons interact 1:1
- From this he theorised that light is a stream of photons and hence possess particle nature
Evidence for photoelectric effect - credited by Einstein and Planck:
- Blackbody radiation was a solid proof to photoelectric effect
- Einstein made his photoelectric effect postulates based upon the shape of the experimental black body radiation curve and Planck's explanation of the blackbody curve
- Experiment involving stopping voltage: This is the most important piece of evidence. So of course we can't seen current macroscopically, nor is there any microscopes to help Einstein to see the electrons. Therefore his assistant performed this experiment which aimed to show that yes there is a current (even though we couldnt see it) by shining a light onto the cathode and observe the reading on the ammeter. There was a current reading and therefore it was ascertained that a photocurrent was formed. To further detect the magnitude of the frequency of incident light ray, the anode was made negative (by supplying it with a negative potential) to repel the electron current. The moment ammeter reading --> 0, Einstein could apply the formula Ek = hf - work function to figure out frequency.
Thanks man I understand it a lot better now, just got a few questions though,
1. Would the electrons still liberated from the metals surface if the incident light has a frequency equal to the threshold frequency, and would there be a photocurrent?
2. Trial question - Discuss the ability of the wave model of light to explain the photoelectric effect? In your answer refer to observations that it could/could not explain.
Cheers guys.
Post by: jamonwindeyer on July 15, 2016, 07:55:45 pm
Thanks man I understand it a lot better now, just got a few questions though,
1. Would the electrons still liberated from the metals surface if the incident light has a frequency equal to the threshold frequency, and would there be a photocurrent?
2. Trial question - Discuss the ability of the wave model of light to explain the photoelectric effect? In your answer refer to observations that it could/could not explain.
Cheers guys.
For Question 1, yes, but they would have no kinetic energy!! So the photocurrent would be zero at this precise point; any higher and it would increase ;D note that in reality having an exact match between the two is pretty rare ;D
Question 2 is actually something I've not seen before, so happy for my answer to be expanded upon/corrected. The wave model of light could, I believe, explain why electrons were emitted from metals exposed to electromagnetic energy. Energy transfer caused excitation, this made sense. However, it could not explain why this effect was not dependent on the amount of incoming light (that is, if it wasn't happening, why more light wouldn't make it happen). It also couldn't explain why the effect only started happening at approximately the ultraviolet range. According to classical theory, an intense radio wave should have induced a similar effect to a weak ultraviolet wave ;D hope this helps!
Post by: conic curve on July 15, 2016, 08:03:39 pm
Post by: jamonwindeyer on July 15, 2016, 08:09:19 pm
What does it mean by one dimension, two dimension and three dimension in terms of waves?
Hmm, not 100% sure what you mean by the question, but it would perhaps mean how many dimensions the wave oscillates/propagates in?
EG - Consider sending a pulse along a slinky. That propagates along the slinky in a single dimension.
Next, consider tying one end of a string to a wall, then moving the other end of the string up and down so it looks like a sine wave (called a transverse wave). That oscillates up and down, but propagates to the left/right, so two dimensions.
Finally, light waves consist of magnetic and electric fields which oscillate in two dimensions, but the wave propagates in a third.
Does this sort of make sense? The question seems a little strange to me that's all ;D
Post by: conic curve on July 15, 2016, 08:12:19 pm
Hmm, not 100% sure what you mean by the question, but it would perhaps mean how many dimensions the wave oscillates/propagates in?
EG - Consider sending a pulse along a slinky. That propagates along the slinky in a single dimension.
Next, consider tying one end of a string to a wall, then moving the other end of the string up and down so it looks like a sine wave (called a transverse wave). That oscillates up and down, but propagates to the left/right, so two dimensions.
Finally, light waves consist of magnetic and electric fields which oscillate in two dimensions, but the wave propagates in a third.
Does this sort of make sense? The question seems a little strange to me that's all ;D
Yeah I don't think I was explicit enough although I remember being asked a question like that in a booklet given to me at school
Post by: Swagadaktal on July 15, 2016, 08:22:57 pm
Quote
2. Trial question - Discuss the ability of the wave model of light to explain the photoelectric effect? In your answer refer to observations that it could/could not explain.An observation it could not explain is the photoelectric effect : In theory if it were a wave it would reflect off a surface even at lower frequencies. However, there is a stopping voltage which can overcome the emission of light which suggests that light isn't a wave but rather a particle. - Not sure if I worded this correctly so any help would be appreciated :)
Post by: conic curve on July 15, 2016, 08:29:13 pm
Statement: waves are a transfer of energy disturbance that may occur in one, two or three dimensions. Different waves in different media can travel in different numbers of dimensions. A longitudinal wave or slinky is 1d. A transverse or water wave is 2d. Sound waves or EM are 3d
What other examples are ther?
Post by: jamonwindeyer on July 15, 2016, 08:31:14 pm
Jamon I think I'm clear on my question now.
Statement: waves are a transfer of energy disturbance that may occur in one, two or three dimensions. Different waves in different media can travel in different numbers of dimensions. A longitudinal wave or slinky is 1d. A transverse or water wave is 2d. Sound waves or EM are 3d
What other examples are ther?
Yep you've got it!! I can't think of any other examples off the top of my head though :P
Post by: conic curve on July 15, 2016, 08:34:43 pm
Yep you've got it!! I can't think of any other examples off the top of my head though :P
A transverse wave is a wave which oscillates perpendicular to the direction of propagation. Is this something by definition and something that we need to accept without questioning why?
Post by: zsteve on July 15, 2016, 08:39:06 pm
A transverse wave is a wave which oscillates perpendicular to the direction of propagation. Is this something by definition and something that we need to accept without questioning why?
That's a definition so yeah, you'll need to accept it without questioning :3
Basically people noted that waves could occur in two distinct types and named them. It's like why we call a cat a cat and a dog a dog.
However, note that there exist waves which combine both transverse and longitudinal motion (very convoluted waves, they are). So that's your cogs and dats :P
Post by: conic curve on July 15, 2016, 09:14:32 pm
That's a definition so yeah, you'll need to accept it without questioning :3
Basically people noted that waves could occur in two distinct types and named them. It's like why we call a cat a cat and a dog a dog.
However, note that there exist waves which combine both transverse and longitudinal motion (very convoluted waves, they are). So that's your cogs and dats :P
This is the reason why cats are clawed cats and why dogs are called dogs: e English word cat dates from before 900 A.D. - comes from the Old English words: catt (male) and catte (female). The word comes from Old High-German / Old Norse words. The English word dog comes from before 1050 A.D. - again from the Old English word docga
Anyways getting back to physics why do CROs represent sound waves as transverse waves?
What is the difference between pitch and noise?
What are some other applications of echoes other than SONAR and ultrasound imaging?
Post by: RuiAce on July 15, 2016, 09:17:20 pm
This is the reason why cats are clawed cats and why dogs are called dogs: e English word cat dates from before 900 A.D. - comes from the Old English words: catt (male) and catte (female). The word comes from Old High-German / Old Norse words. The English word dog comes from before 1050 A.D. - again from the Old English word docga"Noise" isn't even a term in physics.
Anyways getting back to physics why do CROs represent sound waves as transverse waves?
What is the difference between pitch and noise?
What are some other applications of echoes other than SONAR and ultrasound imaging?
Post by: jamonwindeyer on July 15, 2016, 09:20:50 pm
This is the reason why cats are clawed cats and why dogs are called dogs: e English word cat dates from before 900 A.D. - comes from the Old English words: catt (male) and catte (female). The word comes from Old High-German / Old Norse words. The English word dog comes from before 1050 A.D. - again from the Old English word dock
LOL
Anyways getting back to physics why do CROs represent sound waves as transverse waves?
What is the difference between pitch and noise?
What are some other applications of echoes other than SONAR and ultrasound imaging?
That first question is a loaded one, but basically, it is related to the operation of the CRO itself. The sound waves act as inputs to the CRO through a microphone, and CRO's are designed to represent the magnitude of the input R(in this case, the level of compression of the microphone) on the vertical axis of the screen. Thus, the longitudinal compressions are represented as transverse peaks, and the longitudinal rarefactions are represented as transverse troughs ;D
Pitch relates to how high or low a sound is, which is different to its volume. Noise is a generic term (it does have use in Physics though Rui, in electrical terms it refers to irregular fluctuations that accompany electrical signals), either way not really in the same category as pitch :)
RADAR also uses the idea of echo location imaging, just with radio waves instead of sound waves!! ;D
Post by: conic curve on July 15, 2016, 09:21:17 pm
"Noise" isn't even a term in physics.
Huh? It said pitch is related to frequency and volume to amplitude of sound waves. The higher the frequency, the higher the pitch the noise is. The higher the amplitude, the louder it is
Post by: jamonwindeyer on July 15, 2016, 09:24:31 pm
Huh? It said pitch is related to frequency and volume to amplitude of sound waves. The higher the frequency, the higher the pitch the noise is. The higher the amplitude, the louder it is
Yeah! So noise is just used in the general sense, like, it's a noise. Like a tree falling in the woods, it doesn't have any specific meaning beyond the normal ;D so we refer to the pitch of that noise, and the volume of that noise :)
Post by: RuiAce on July 15, 2016, 09:25:15 pm
Huh? It said pitch is related to frequency and volume to amplitude of sound waves. The higher the frequency, the higher the pitch the noise is. The higher the amplitude, the louder it isYes. Higher frequency = Higher pitch.
Noise is just a synonym for sound in that context.
Post by: MysteryMarker on July 16, 2016, 04:20:22 pm
Describe what is meant by the term drift velocity, in terms of electrons moving in a solid conductor. Compare its magnitude to the thermal motion of electrons in a solid.
Cheers.
Post by: conic curve on July 16, 2016, 04:24:07 pm
1. What does the gradient and the area represent in a displacement and velocity vs Time Graph?
2. Why do we let one side (either left or right) be positive whenever solving with acceleration equations?
3. How is a force defined (other than a push, pull or a twist)?
4. What does it mean by "magnitude and direction"?
Post by: RuiAce on July 16, 2016, 04:28:34 pm
Got a few questions to ask:1. If you had done calculus this would be more intuitive for you but in short
1. What does the gradient and the area represent in a displacement and velocity vs Time Graph?
2. Why do we let one side (either left or right) be positive whenever solving with acceleration equations?
3. How is a force defined (other than a push, pull or a twist)?
4. What does it mean by "magnitude and direction"?
Gradient of displacement -> velocity
Gradient of velocity -> acceleration
Area under velocity graph -> distance travelled
Area under acceleration graph -> how much your velocity increased/decreased
Gradient of acceleration and area under distance are useless for HSC physics. The latter is in fact just useless altogether.
2. Because if left is positive and right is positive at the same time you go nuts. If left is positive but right is negative (or vice versa) you can compare: positive numbers mean left, negative numbers mean right.
3. You can use Newton's second law of motion here. The sum of all forces acting on an object is the scalar product of its mass with the acceleration. (F=ma)
Otherwise, just use the junior science definition
4. Magnitude = Has a value. E.g. 1 , 4, 61, 100
Direction = Has a direction. E.g. east, 235 degrees True bearing, North 1 degrees West, into the book
If it just has magnitude (e.g. amount of energy you use a day) then it's a scalar
If it has magnitude and direction (e.g. precisely how far north you travelled) it is a vector
Post by: conic curve on July 16, 2016, 05:05:57 pm
1. Why is it difficult to see newton's first law?
2. What is the definition of a normal force (I tried to research this but don't understand it). Are normal forces the reaction force of gravity?
3. What is the relative velocity formula and what does it tell us?
4. For centripetal force, why do we say that objects travelling in a circular motion are still accelerating?
5. Why are speed limits a good idea
Post by: jakesilove on July 16, 2016, 05:08:54 pm
Thanks
1. Why is it difficult to see newton's first law?
2. What is the definition of a normal force (I tried to research this but don't understand it). Are normal forces the reaction force of gravity?
3. What is the relative velocity formula and what does it tell us?
4. For centripetal force, why do we say that objects travelling in a circular motion are still accelerating?
5. Why are speed limits a good idea
Where are you getting these questions from? Sounds like something out of a textbook, homework you're supposed to be doing or an Assignment. Don't mind if it's any but the last one, but just wondering!
Post by: conic curve on July 16, 2016, 05:10:48 pm
Where are you getting these questions from? Sounds like something out of a textbook, homework you're supposed to be doing or an Assignment. Don't mind if it's any but the last one, but just wondering!
A worksheet I was given. It's basically a worksheet for all the weak physics kids and require revision on the core basics
Post by: jamonwindeyer on July 16, 2016, 06:06:53 pm
Hey guys, need help with this question:
Describe what is meant by the term drift velocity, in terms of electrons moving in a solid conductor. Compare its magnitude to the thermal motion of electrons in a solid.
Cheers.
Hey there!! Drift velocity refers to the velocity attained by an electron due an applied electric field. A quick bit of research reveals that there is actually a formula for drift velocity!
Drift velocity is the product of the magnitude of the applied electric field and the electron mobility of the material (this is new for me too, I'm learning with you!)
Now, electrons normally vibrate randomly in solids due to thermal energy. Remember, heat causes atoms in solids to vibrate, that's what thermal energy is. This causes small random movements of election, but this doesn't cause any net drift velocity. Applying the electric field will cause larger movement, and they will be aligned in a single direction ;D i hope this helps!!
Post by: wyzard on July 16, 2016, 09:07:50 pm
Hey there!! Drift velocity refers to the velocity attained by an electron due an applied electric field. A quick bit of research reveals that there is actually a formula for drift velocity!
Drift velocity is the product of the magnitude of the applied electric field and the electron mobility of the material (this is new for me too, I'm learning with you!)
Now, electrons normally vibrate randomly in solids due to thermal energy. Remember, heat causes atoms in solids to vibrate, that's what thermal energy is. This causes small random movements of election, but this doesn't cause any net drift velocity. Applying the electric field will cause larger movement, and they will be aligned in a single direction ;D i hope this helps!!
Pretty much as explained 8)
As for the magnitude, if you run through the calculations you'll find that drift velocity have magnitudes around
The resulting motion is the electron jiggling around at really high speed, while slowly drifting in the opposite direction of the applied electric field as shown in the attached image.
Post by: MysteryMarker on July 16, 2016, 09:22:56 pm
Another Question, ;D for the dot point 'Identify Einstein contribution to quantum theory and its relation to black body radiation. What is his relation to black body radiation? Didn't he just extend upon Planck's model of energy being quantised and use it to explain the photoelectric effect? With threshold frequency, work function, E = hf and all that stuff? Just confused as to what Einstein had to do with 'Black body radiation'.
Cheers.
Post by: Swagadaktal on July 16, 2016, 09:24:30 pm
An observation it could not explain is the photoelectric effect : In theory if it were a wave it would reflect off a surface even at lower frequencies. However, there is a stopping voltage which can overcome the emission of light which suggests that light isn't a wave but rather a particle. - Not sure if I worded this correctly so any help would be appreciated :)I think this was lost in the sea of posts - can someone validate whether this is correct? Dont have a solid understanding of this topic quite yet ;)
Post by: Happy Physics Land on July 16, 2016, 10:51:04 pm
Thanks guys. ;)
Another Question, ;D for the dot point 'Identify Einstein contribution to quantum theory and its relation to black body radiation. What is his relation to black body radiation? Didn't he just extend upon Planck's model of energy being quantised and use it to explain the photoelectric effect? With threshold frequency, work function, E = hf and all that stuff? Just confused as to what Einstein had to do with 'Black body radiation'.
Cheers.
Hey Mysterymarker!
If you were at the physics lecture last week you might have heard me talking about this - both einstein and planck agreed on the mechanism to produce light but there differed in their explanation towards the nature of light. They agree that electromagnetic radiation or electromagnetic wave is produced as a result of change in quantised energy levels. But Planck is actually still partly a classical physicist and believed that energy is evenly distributed across wavefronts. On the other hand, Einstein ascertained the particle nature of light through the photoelectric effect and the 1:1 interaction between photons and electrons, hence affirming light as a stream of photons. So there is a fundamental shift from Planck to Einstein in terms of the explanation towards the nature of light. Obviously through photoelectric effect and the stopping voltage experiment, Einstein was able to confirm Planck's second postulate E=hf and this confirms Planck's explanation towards the blackbody radiation in terms of UV catastrophe and peak wavelength radiation. So essentially Planck theorised a relationship between frequency and quantised energy but Einstein used photoelectric effect and the related implications as evidences to E=hf.
Best Regards
Happy Physics Land
Post by: Spencerr on July 16, 2016, 10:53:07 pm
I think this was lost in the sea of posts - can someone validate whether this is correct? Dont have a solid understanding of this topic quite yet ;)
I haven't seen the original post but i'm assuming the question is why the wave model of light couldn't account for the photoelectric effect.
The classical wave model of light proposed that the energy of light was related to its intensity. So pretty much if you shine light onto a photemitting metal (no matter the frequency), after a while, the electrons in the metal would have enough energy and they would be freed. However experiments showed that this was not the case, as only light above a certain frequency (called the threshold frequency) would cause the electrons to be freed.
As a result 3 problems came up
1. The existence of a threshold frequency
2. The absence of a time delay or immediate release
3. How does intensity affect the emission of photoelectrons.
In order to solve these problems, Einstein extended Planck's idea of quantised energy and suggested that light was also quantised into small packets called photons. These photons would have energy equal to E=hf (linking energy with frequency) and would interact with the electrons on the surface of the metal on a 1:1 basis. If the photons had enough energy, it would transfer all of it to the electron and the electron would have the requried energy to free it self from the attractive force of the metal. However if the photon did not have enough energy, it would be absorbed by the metal and then reemitted. This particle nature of light explained the absence of a time delay and the existence of a threshold frequency. Intensity refers to how many photons of light are there in the beam of light and is independent of the energy of the photons.
Post by: Swagadaktal on July 16, 2016, 10:56:00 pm
I haven't seen the original post but i'm assuming the question is why the wave model of light couldn't account for the photoelectric effect.Thank you for the response!
The classical wave model of light proposed that the energy of light was related to its intensity. So pretty much if you shine light onto a photemitting metal (no matter the frequency), after a while, the electrons in the metal would have enough energy and they would be freed. However experiments showed that this was not the case, as only light above a certain frequency (called the threshold frequency) would cause the electrons to be freed.
As a result 3 problems came up
1. The existence of a threshold frequency
2. The absence of a time delay or immediate release
3. How does intensity affect the emission of photoelectrons.
In order to solve these problems, Einstein extended Planck's idea of quantised energy and suggested that light was also quantised into small packets called photons. These photons would have energy equal to E=hf (linking energy with frequency) and would interact with the electrons on the surface of the metal on a 1:1 basis. If the photons had enough energy, it would transfer all of it to the electron and the electron would have the requried energy to free it self from the attractive force of the metal. However if the photon did not have enough energy, it would be absorbed by the metal and then reemitted. This particle nature of light explained the absence of a time delay and the existence of a threshold frequency. Intensity refers to how many photons of light are there in the beam of light and is independent of the energy of the photons.
Um the thing I was inquiring about was in the quote: I'll write it bellow - thanks for the explanation though! I was more specifically asking about whether this point was valid or I've mistaken it with something else:
An observation it could not explain is the photoelectric effect : In theory if it were a wave it would reflect off a surface even at lower frequencies. However, there is a stopping voltage which can overcome the emission of light which suggests that light isn't a wave but rather a particle. - Not sure if I worded this correctly so any help would be appreciated :)
Post by: Spencerr on July 16, 2016, 11:01:48 pm
Thanks guys. ;)
Another Question, ;D for the dot point 'Identify Einstein contribution to quantum theory and its relation to black body radiation. What is his relation to black body radiation? Didn't he just extend upon Planck's model of energy being quantised and use it to explain the photoelectric effect? With threshold frequency, work function, E = hf and all that stuff? Just confused as to what Einstein had to do with 'Black body radiation'.
Cheers.
Happy Physics Land gave an excellent answer but here's my take on how to answer the question if it was in an exam.
Planck had a problem applying classical physics to black body radiation observations (UV catastrophe)
Thus, in order to account for experimental observations, Planck suggested that EMR emitted or absorbed by black bodies were quantised i.e. they existed as discrete separate packets. This explained by the black body radiation curve peaked and then declined as it approached higher frequencies. He proposed that the energy of these quanta would be E=hf, at that time, Planck himself thought this was crazy and it was all theoretical. However Einstein used Planck's idea of quantised energy to explain the photoelectric effect and solved many problems with it such as the threshold frequency etc. Einstein's practical application of Planck's idea of quantised energy in explaining the photoelectric effect greatly validated the theory and laid the foundation for quantum theory.
Thank you for the response!
Um the thing I was inquiring about was in the quote: I'll write it bellow - thanks for the explanation though! I was more specifically asking about whether this point was valid or I've mistaken it with something else:
An observation it could not explain is the photoelectric effect : In theory if it were a wave it would reflect off a surface even at lower frequencies. However, there is a stopping voltage which can overcome the emission of light which suggests that light isn't a wave but rather a particle. - Not sure if I worded this correctly so any help would be appreciated :)
Hey there, to my knowledge I don't know any links between stopping voltage and the emission of light. I think how stopping voltage works is that, its the voltage at which electrons with max KE cannot reach the electrode and as for the first part, I don't think there is a relevance between wave reflections and the photoelectric effect.
If someone knows more, then I want to hear a better explanation but I don't think it's completely valid.
Moderator action: Merged double post. You can edit and include multiple quotes in the same post.
Post by: jamonwindeyer on July 16, 2016, 11:42:16 pm
Thank you for the response!
Um the thing I was inquiring about was in the quote: I'll write it bellow - thanks for the explanation though! I was more specifically asking about whether this point was valid or I've mistaken it with something else:
An observation it could not explain is the photoelectric effect : In theory if it were a wave it would reflect off a surface even at lower frequencies. However, there is a stopping voltage which can overcome the emission of light which suggests that light isn't a wave but rather a particle. - Not sure if I worded this correctly so any help would be appreciated :)
Sorry Swag! Must have gotten lost ;) I'm not 100% sure what you are driving at in that statement, which leads me to think the wording may be a little off, reckon you could explain what you mean in another way? Not quite clicking with it right now that's all ;D
Post by: Happy Physics Land on July 16, 2016, 11:43:09 pm
Happy Physics Land gave an excellent answer but here's my take on how to answer the question if it was in an exam.
Planck had a problem applying classical physics to black body radiation observations (UV catastrophe)
Thus, in order to account for experimental observations, Planck suggested that EMR emitted or absorbed by black bodies were quantised i.e. they existed as discrete separate packets. This explained by the black body radiation curve peaked and then declined as it approached higher frequencies. He proposed that the energy of these quanta would be E=hf, at that time, Planck himself thought this was crazy and it was all theoretical. However Einstein used Planck's idea of quantised energy to explain the photoelectric effect and solved many problems with it such as the threshold frequency etc. Einstein's practical application of Planck's idea of quantised energy in explaining the photoelectric effect greatly validated the theory and laid the foundation for quantum theory.
Hey there, to my knowledge I don't know any links between stopping voltage and the emission of light. I think how stopping voltage works is that, its the voltage at which electrons with max KE cannot reach the electrode and as for the first part, I don't think there is a relevance between wave reflections and the photoelectric effect.
If someone knows more, then I want to hear a better explanation but I don't think it's completely valid.
Moderator action: Merged double post. You can edit and include multiple quotes in the same post.
Yep yep just combine my answer and Diii's answer together you will get an almost perfect response! Nice job! :)
Post by: jamonwindeyer on July 16, 2016, 11:52:18 pm
Thanks
1. Why is it difficult to see newton's first law?
2. What is the definition of a normal force (I tried to research this but don't understand it). Are normal forces the reaction force of gravity?
3. What is the relative velocity formula and what does it tell us?
4. For centripetal force, why do we say that objects travelling in a circular motion are still accelerating?
5. Why are speed limits a good idea
Definitely can help you with Question 2! A normal force does sort of manifest as the reaction force to gravity, but a little more complicated. Basically, objects are accelerated towards the centre of the earth by a gravitational force. Eventually, this means objects will hit the floor. At this point, the gravitational force is still there, and yet the object is now sitting still. There must be another force cancelling the gravitational force! This is the normal force, and it comes from the ground pushing back against the object pushing against it.
For the others, give your textbook a bit of a flick! Us feeding you answers won't benefit you, you should come to us with things you've tried to understand but can't, or are confused about, or want to confirm, etc. We can answer all your questions, but that won't help you.
Incidentally, Question 1 has primarily to do with friction forces. Think of it this way; if you roll a ball along the ground, will the ball keep rolling forever? Given Newton's 1st Law, would you expect it to? Why doesn't it then?
I'll give you this link to help you with Question 3.
Question 4 is a matter of convention. We say that acceleration is any change in velocity. Velocity is a vector, meaning it has a direction and a magnitude. While circular motion may not change the magnitude of an objects velocity, it definitely changes the direction as it moves in a circle, and thus, it is 'accelerating.'
I'll leave Question 5 for you. Have a bit of a think, I reckon you'll come up with some ideas, remember to tie in things like the impulse formula:
Post by: jakesilove on July 17, 2016, 10:43:03 am
Happy Physics Land gave an excellent answer but here's my take on how to answer the question if it was in an exam.
Hey there, to my knowledge I don't know any links between stopping voltage and the emission of light. I think how stopping voltage works is that, its the voltage at which electrons with max KE cannot reach the electrode and as for the first part, I don't think there is a relevance between wave reflections and the photoelectric effect.
If someone knows more, then I want to hear a better explanation but I don't think it's completely valid.
Moderator action: Merged double post. You can edit and include multiple quotes in the same post.
Just quickly, there IS a relationship between the quantisation of light and stopping voltage, although I don't remember how much you needed (if any) to know in the HSC. Basically, there were three key points that suggested the photoelectric effect was caused by a "particle" like light, rather than a wave.
1. Stopping voltage
If light were a wave, there wouldn't be a minimum frequency (ie. work function) for which electrons would be ejected. We know that the energy of a wave is proportional to the square if its amplitude (don't worry about this, not important). But what that means it that, if light were a wave, if would CONTINUOUSLY add energy to the electron (as more amplitudes pass through the electron) until, eventually, it should have enough energy to jump off the surface of the metal. This doesn't happen: you can shine light at a low enough frequency for as long as you want onto a surface, and if it below the work function nothing will happen. Therefore, stopping voltage is related to the photoelectric effect, and it's explanation.
2. Time delay
This is a similar point as above; the only time delay between the light being switched on and the electrons jumping off was the distance between the light and the surface, divided by the speed of light (ie. the light had to GET to the surface, but after that the electrons jumped off instantly). Again, this is indicative of the particle nature of light (think all or nothing principle, rather than a gradual build up).
Frequency, not intensity, dependent
This one is straight from the course, so I'll assume you understand this!
If you want a super comprehensive, impressive answer, you could always just cite the three things talked about above. Nice to understand them though :)
Jake
Post by: MysteryMarker on July 18, 2016, 03:41:47 pm
I dont understand why the answer for Q14 is C, or why the answer for Q19 is D. Any help is appreciated.
Thanks.
Post by: zsteve on July 18, 2016, 03:52:10 pm
Hey guys
I dont understand why the answer for Q14 is C, or why the answer for Q19 is D. Any help is appreciated.
Thanks.
Hey MysteryMarker,
For Q14, the train is turning right, (option C as specified). Try to visualise the situation (this might be hard for some people!)
In the reference frame of the ground, as the train begins to turn right, the ball travels in a straight line. Newton III predicts that the ball will tend to retain its straight path in the inertial frame of the ground. This is achieved more or less, assuming friction between ball and table isn't massive.
In the non-inertial frame of the turning train carriage, the ball appears to bend as shown.
I'm having severe difficulties with explaining this one :P. Best if you can visualise it by putting yourself 'in' the carriage and doing it intuitively.
Q19.
This is rather weird, because from theory we know that their accelerations are the same. However, this comes to mind...
\( a = \dfrac{F_g}{m}\), so we see that acceleration is inversely proportional to mass. (considering only gravity here)
But \(F_g \propto m\) as well, so if we combine this with a, we find that m cancels and acceleration is independent of mass.
Not a well-phrased question in my opinion.
Post by: jakesilove on July 18, 2016, 03:57:25 pm
Hey MysteryMarker,
For Q14, the train is turning right, (option C as specified). Try to visualise the situation (this might be hard for some people!)
In the reference frame of the ground, as the train begins to turn right, the ball travels in a straight line. Newton III predicts that the ball will tend to retain its straight path in the inertial frame of the ground. This is achieved more or less, assuming friction between ball and table isn't massive.
In the non-inertial frame of the turning train carriage, the ball appears to bend as shown.
I'm having severe difficulties with explaining this one :P. Best if you can visualise it by putting yourself 'in' the carriage and doing it intuitively.
Q19.
This is rather weird, because from theory we know that their accelerations are the same. However, this comes to mind...
\( a = \dfrac{F_g}{m}\), so we see that acceleration is inversely proportional to mass. (considering only gravity here)
But \(F_g \propto m\) as well, so if we combine this with a, we find that m cancels and acceleration is independent of mass.
Not a well-phrased question in my opinion.
Thanks for the great response Steve! I totally agree with all of the above. I'll just add that Q19 is a tricky bugger, and I seem to remember getting it wrong myself. Basically, if you go through each answer, you'll find that all but D are wholly incorrect. D is also true so, even if you don't FULLY understand the answer (as the question is badly worded, I agree), you can get the answer correct
Post by: MysteryMarker on July 18, 2016, 04:04:45 pm
Hey MysteryMarker,
For Q14, the train is turning right, (option C as specified). Try to visualise the situation (this might be hard for some people!)
In the reference frame of the ground, as the train begins to turn right, the ball travels in a straight line. Newton III predicts that the ball will tend to retain its straight path in the inertial frame of the ground. This is achieved more or less, assuming friction between ball and table isn't massive.
In the non-inertial frame of the turning train carriage, the ball appears to bend as shown.
I'm having severe difficulties with explaining this one :P. Best if you can visualise it by putting yourself 'in' the carriage and doing it intuitively.
Q19.
This is rather weird, because from theory we know that their accelerations are the same. However, this comes to mind...
\( a = \dfrac{F_g}{m}\), so we see that acceleration is inversely proportional to mass. (considering only gravity here)
But \(F_g \propto m\) as well, so if we combine this with a, we find that m cancels and acceleration is independent of mass.
Not a well-phrased question in my opinion.
Wouldn't the train travelling right or left have the same effect on the motion of the ball as viewed from the sideview?
Sorry, just confused with Q14.
Post by: Spencerr on July 18, 2016, 04:57:51 pm
Imaging that theres a ball hanging on a string attached to to the top of the train and the train accelerates straight forward. Hopefully you can easily see in your mind's eye that the ball will hang at an angle away from the direction of motion. So if the train is moving to the left, the string will look like this: \ with the ball attached on the end. This is because there is a phantom force acting on the ball along with the force of gravity. The phantom force is produced by the acceleration of the train and acts in the opposite direction.
You can apply this scenario to the billiard ball question. The force of gravity can be ignored. you see that the ball has moved to the left, therefore it has experienced a force to the left (doesnt matter if its centripetal or not). Since there is a force to the left acting on the ball, and using the first scenario, we can deduce that there must be a force acting on the train towards the right that produces an acceleration i.e. its turning towards the right.
or you could just visualise it.
Post by: MysteryMarker on July 18, 2016, 08:08:30 pm
For q 14 I'm going to use a different scenario to take a look at forces.
Imaging that theres a ball hanging on a string attached to to the top of the train and the train accelerates straight forward. Hopefully you can easily see in your mind's eye that the ball will hang at an angle away from the direction of motion. So if the train is moving to the left, the string will look like this: \ with the ball attached on the end. This is because there is a phantom force acting on the ball along with the force of gravity. The phantom force is produced by the acceleration of the train and acts in the opposite direction.
You can apply this scenario to the billiard ball question. The force of gravity can be ignored. you see that the ball has moved to the left, therefore it has experienced a force to the left (doesnt matter if its centripetal or not). Since there is a force to the left acting on the ball, and using the first scenario, we can deduce that there must be a force acting on the train towards the right that produces an acceleration i.e. its turning towards the right.
or you could just visualise it.
When you say that the ball has moved to the left, how is this so? isn't the ball going to the back of the carriage, and so from our perspective going to the 'right'?
Sorry for all these posts, i still don't really understand how we can deduce the train to be turning left/right. I understand the pendulum analogy you used but other than that i can't really see how the train is moving to the right.
Post by: conic curve on July 19, 2016, 12:15:58 pm
So I got a group physics presentation coming up and I need someone to give me feedback on the assessment (once the assignment is completed)
Where can I post my physics assignment for feedback?
Thanks ;D
Post by: RuiAce on July 19, 2016, 12:37:36 pm
HeyWatch out for what you post. We can give you tiny feedback but we cannot do your assignment for you or we are breaking the rules of all my own work.
So I got a group physics presentation coming up and I need someone to give me feedback on the assessment (once the assignment is completed)
Where can I post my physics assignment for feedback?
Thanks ;D
Feel free to post it here.
Post by: jamonwindeyer on July 19, 2016, 12:41:10 pm
When you say that the ball has moved to the left, how is this so? isn't the ball going to the back of the carriage, and so from our perspective going to the 'right'?
Sorry for all these posts, i still don't really understand how we can deduce the train to be turning left/right. I understand the pendulum analogy you used but other than that i can't really see how the train is moving to the right.
Let me have a go at explaining, this is definitely tough ;D
Consider the train moving forward and us looking down on it, as per the question.
Now an important concept to understand here is inertia, the tendency for objects to remain in their set path. It's the reason why the tissue box at the back of your car flies forward when you brake suddenly: Because the tissue box wants to keep moving forward, and there is nothing forcing it to stop with the car.
In this example, the train makes a movement halfway through the balls path (when it starts to curve). Now, moving forward or backward would never cause the ball to curve, like braking a car on a straight road would never cause the tissue box to hit the side window. Doesn't make sense. But why is the train turning right not left?
Consider the ball as a separate object, because it is. When the train turns, the ball is going to want to continue to go straight on the initial path. Consider the train turning left, that is, moving down the page if we are looking at the diagram, and the ball staying where it is. Except this isn't the right way to think of it. We want to stick with the train. For this to make sense, as we move down the page with the train, the ball stays where it is, and so moves upwards on the diagram. Not quite what we need.
Consider the train turning right. The train would move up the page, and so as we go up the page with the train, the ball must stay where it is and so move down the page in our reference frame. This corresponds to the turn in the diagram.
I tried to take out all the jargon, does this help at all? I feel like an animation would be the best way to explain this, hopefully this helps a little :P
Post by: conic curve on July 19, 2016, 02:12:36 pm
Watch out for what you post. We can give you tiny feedback but we cannot do your assignment for you or we are breaking the rules of all my own work.
Feel free to post it here.
Yes I know, don't worry. If I make you guys do it for me, then obviously I'm not going to learn anything from it. It's like making you guys write an essay/creative for me, if you write it for me, then I will not benefit from anything hence I'll struggle to write an essay/creative on the day of the exam (I hope I did not confuse anyone here)
Anyways, I was asking for feedback. How much feedback is allowed when it comes to physics assignments here? (i.e. is it extensive like the free essay marking)
Post by: jamonwindeyer on July 19, 2016, 02:19:53 pm
Yes I know, don't worry. If I make you guys do it for me, then obviously I'm not going to learn anything from it. It's like making you guys write an essay/creative for me, if you write it for me, then I will not benefit from anything hence I'll struggle to write an essay/creative on the day of the exam (I hope I did not confuse anyone here)
Anyways, I was asking for feedback. How much feedback is allowed when it comes to physics assignments here? (i.e. is it extensive like the free essay marking)
It will depend on what sort of task it is, but it will probably just be some general recommendations made at the end, it would be hard to give specific feedback throughout a presentation.
You should also post it in a new thread in the Physics section, keep this section for Q+A and physics chat ;D
Post by: conic curve on July 19, 2016, 02:24:22 pm
It will depend on what sort of task it is, but it will probably just be some general recommendations made at the end, it would be hard to give specific feedback throughout a presentation.
You should also post it in a new thread in the Physics section, keep this section for Q+A and physics chat ;D
It's a presentation and speech (in pairs). I will show you the marking criteria and will also be telling you of our teacher's feedback. I think the task is on researching the GPS/internet or doing research on the application of waves, I don't know what it exactly it to be honest
Our teacher was talking about it today and we haven't gotten a copy of the assessment notification yet
Post by: MysteryMarker on July 19, 2016, 05:55:41 pm
Let me have a go at explaining, this is definitely tough ;D
Consider the train moving forward and us looking down on it, as per the question.
Now an important concept to understand here is inertia, the tendency for objects to remain in their set path. It's the reason why the tissue box at the back of your car flies forward when you brake suddenly: Because the tissue box wants to keep moving forward, and there is nothing forcing it to stop with the car.
In this example, the train makes a movement halfway through the balls path (when it starts to curve). Now, moving forward or backward would never cause the ball to curve, like braking a car on a straight road would never cause the tissue box to hit the side window. Doesn't make sense. But why is the train turning right not left?
Consider the ball as a separate object, because it is. When the train turns, the ball is going to want to continue to go straight on the initial path. Consider the train turning left, that is, moving down the page if we are looking at the diagram, and the ball staying where it is. Except this isn't the right way to think of it. We want to stick with the train. For this to make sense, as we move down the page with the train, the ball stays where it is, and so moves upwards on the diagram. Not quite what we need.
Consider the train turning right. The train would move up the page, and so as we go up the page with the train, the ball must stay where it is and so move down the page in our reference frame. This corresponds to the turn in the diagram.
I tried to take out all the jargon, does this help at all? I feel like an animation would be the best way to explain this, hopefully this helps a little :P
Holy crap, i just realised i've been looking at the diagram the bloody wrong way. This whole time i thought we were looking at a side view and not an aerial view! Damn, thanks Jamon that explanation was freakin mint. Totally gonna ace this concept if it comes in any of my future exams. ;D
Post by: jamonwindeyer on July 19, 2016, 06:51:48 pm
Holy crap, i just realised i've been looking at the diagram the bloody wrong way. This whole time i thought we were looking at a side view and not an aerial view! Damn, thanks Jamon that explanation was freakin mint. Totally gonna ace this concept if it comes in any of my future exams. ;D
Nice!! Aha love that click moment 8) awesome work! ;D
Post by: wyzard on July 19, 2016, 06:52:45 pm
It's a presentation and speech (in pairs). I will show you the marking criteria and will also be telling you of our teacher's feedback. I think the task is on researching the GPS/internet or doing research on the application of waves, I don't know what it exactly it to be honest
Our teacher was talking about it today and we haven't gotten a copy of the assessment notification yet
Ah the GPS 8) You might wanna look into how relativistic correction to time measurement is needed due to the gravitational time dilation. This serves a very important application to Einstein's relativity. I think that will be a really cool presentation topic.
Post by: MysteryMarker on July 19, 2016, 08:45:53 pm
'Describe Planck's contribution to the development of the atomic model and explain how his work was critical in allowing the development of the Bohr model' (4 marks)
The marking criteria states that TWO of Bohr's postulates must be linked with Planck's concept of quantisation. I understand how to link Bohr's 2nd postulate to planck, but how would i link any of his others?
Post by: jamonwindeyer on July 19, 2016, 09:38:04 pm
Quanta to Quarks Questions: (That's alotta Q's)
'Describe Planck's contribution to the development of the atomic model and explain how his work was critical in allowing the development of the Bohr model' (4 marks)
The marking criteria states that TWO of Bohr's postulates must be linked with Planck's concept of quantisation. I understand how to link Bohr's 2nd postulate to planck, but how would i link any of his others?
I'm hoping a helpful student reading this will lend a hand with this question, neither Jake, Rui or myself studied Quanta to Quarks, help out a fellow peer! It will help you too!:o
Post by: Spencerr on July 19, 2016, 09:39:02 pm
Planck, in order to explain observations for black body radiation proposed that emr emitted or absorbed by black bodies occurred in discrete packets known as quanta which carried energy equivalent to E = hf. This was the foundation of a new type of physics called quantum physics that was later validated by Einstein in his application of it to explain the photoelectric effect.
Bohr used Planck's idea of quantised energy to develop his atomic model which extended upon rutherford's planetary model of electrons orbiting a central nuclues. Since energy was quantised and existed in discrete packets, that means the energy shells within the atom must also be quantised instead of occuring over a spectrum. This was Bohr's first postulate and a direct consequance of Planck's contribution. Bohr's second postulate was that when an electron jumped from a lower energy shell to a higher energy shell, it would need to absorb a photon of emr that had energy equivalent to E=hf and when an electron jumped from a higher to a lower energy level, it would release a photon with energy equal to E=hf. Bohr's third postulate regarding the quantisation of angular momentum was also a direct consequence of quantum physics but it has more to do with why the electrons are stable in teh orbits than Planck's impact. Therefore, Planck's contribution had a critical impact on the development of Bohr's atomic model.
Post by: Spencerr on July 19, 2016, 09:39:35 pm
I'm hoping a helpful student reading this will lend a hand with this question, neither Jake, Rui or myself studied Quanta to Quarks, help out a fellow peer! It will help you too!:o
Luckily I do q2q !
Post by: jamonwindeyer on July 19, 2016, 10:28:51 pm
Luckily I do q2q !
You are a legend!
Post by: MysteryMarker on July 20, 2016, 05:01:55 pm
'Describe Planck's contribution to the development of the atomic model and explain how his work was critical in allowing the development of the Bohr model' (4 marks)
Planck, in order to explain observations for black body radiation proposed that emr emitted or absorbed by black bodies occurred in discrete packets known as quanta which carried energy equivalent to E = hf. This was the foundation of a new type of physics called quantum physics that was later validated by Einstein in his application of it to explain the photoelectric effect.
Bohr used Planck's idea of quantised energy to develop his atomic model which extended upon rutherford's planetary model of electrons orbiting a central nuclues. Since energy was quantised and existed in discrete packets, that means the energy shells within the atom must also be quantised instead of occuring over a spectrum. This was Bohr's first postulate and a direct consequance of Planck's contribution. Bohr's second postulate was that when an electron jumped from a lower energy shell to a higher energy shell, it would need to absorb a photon of emr that had energy equivalent to E=hf and when an electron jumped from a higher to a lower energy level, it would release a photon with energy equal to E=hf. Bohr's third postulate regarding the quantisation of angular momentum was also a direct consequence of quantum physics but it has more to do with why the electrons are stable in teh orbits than Planck's impact. Therefore, Planck's contribution had a critical impact on the development of Bohr's atomic model.
Thanks man, that answer was mint as. Just got another one on the limitations of Bohr's model:
One of the limititations was that Bohr's model used a mixture of both 'classical' and 'quantum' physics. What is the classical part of Bohr's model?
Post by: Spencerr on July 20, 2016, 05:56:13 pm
Thanks man, that answer was mint as. Just got another one on the limitations of Bohr's model:
One of the limititations was that Bohr's model used a mixture of both 'classical' and 'quantum' physics. What is the classical part of Bohr's model?
In order to derive Balmer's empirical equation regarding the wavelengths of emr emitted when electrons move from a higher energy level to a lower energy level. whcih is the 1 /lamda = ....Bohr had to combine quantum physics mathematics and classical physics mathematics so it was not a purely quantum model itself. If you search up the mathematics for his derivation of the Rhydberg equation, he used classical physics formulas for KE and GPE and then combined it with his quantisation condition L = nh/ 2pi.
This limitation of Bohr's atomic model was solved by Heisenberg's matrix mechanics which was a purely quantum model that described the interactions of subatomic particles and their movements. Since it was purely quantum, matrix mechanics also solved the existence of hyperfine spectra lines, the zeeman effect, and accurately predicted wavelengths for larger atoms. This was Heisenberg's second contribution after his uncertainty principle
Post by: Neutron on July 20, 2016, 07:22:04 pm
'Describe Planck's contribution to the development of the atomic model and explain how his work was critical in allowing the development of the Bohr model' (4 marks)
Planck, in order to explain observations for black body radiation proposed that emr emitted or absorbed by black bodies occurred in discrete packets known as quanta which carried energy equivalent to E = hf. This was the foundation of a new type of physics called quantum physics that was later validated by Einstein in his application of it to explain the photoelectric effect.
Bohr used Planck's idea of quantised energy to develop his atomic model which extended upon rutherford's planetary model of electrons orbiting a central nuclues. Since energy was quantised and existed in discrete packets, that means the energy shells within the atom must also be quantised instead of occuring over a spectrum. This was Bohr's first postulate and a direct consequance of Planck's contribution. Bohr's second postulate was that when an electron jumped from a lower energy shell to a higher energy shell, it would need to absorb a photon of emr that had energy equivalent to E=hf and when an electron jumped from a higher to a lower energy level, it would release a photon with energy equal to E=hf. Bohr's third postulate regarding the quantisation of angular momentum was also a direct consequence of quantum physics but it has more to do with why the electrons are stable in teh orbits than Planck's impact. Therefore, Planck's contribution had a critical impact on the development of Bohr's atomic model.
Nice answer dude! It might also be a nice touch to include that by using Planck's quantisation theory within the Bohr model, the characteristic hydrogen emission spectra could be explained! (One of the primary experimental evidences for the Bohr model)
Post by: Neutron on July 20, 2016, 07:26:31 pm
I was wondering whether there's anything fundamentally wrong with my understanding about Planck and if one of you kind fellows could read my response to this question, that would be amazing :D
Describe the hypothesis proposed by Planck which resolved the problem with black body radiation (This was part 2 of a question, I already explained the UV catastrophe in the first part :P )
Planck's proposed the quantisation theory, which stated that energy is emitted and absorbed in discrete amounts given by E=hf. When an electron absorbs a discrete packet of energy, it will increase to a higher energy state before returning to its original energy state and re-emitting the photon of the specific energy. Therefore, in order for an electron to emit high frequency radiation, it must absorb the high frequency radiation first. However, the energy associated with such high frequency radiation is equal to or greater than the ionising energy. As such, the electron is liberated from the atom (ionised) before it can jump back down and emit the high frequency radiation. Thus, Planck was able to explain the lack of high frequency radiation emitted during blackbody radiation.
Thanks guys!
Neutron
Post by: Spencerr on July 20, 2016, 07:39:55 pm
Here's a bit of I2I to mix up the posts ;)
I was wondering whether there's anything fundamentally wrong with my understanding about Planck and if one of you kind fellows could read my response to this question, that would be amazing :D
Describe the hypothesis proposed by Planck which resolved the problem with black body radiation (This was part 2 of a question, I already explained the UV catastrophe in the first part :P )
Planck's proposed the quantisation theory, which stated that energy is emitted and absorbed in discrete amounts given by E=hf. When an electron absorbs a discrete packet of energy, it will increase to a higher energy state before returning to its original energy state and re-emitting the photon of the specific energy. Therefore, in order for an electron to emit high frequency radiation, it must absorb the high frequency radiation first. However, the energy associated with such high frequency radiation is equal to or greater than the ionising energy. As such, the electron is liberated from the atom (ionised) before it can jump back down and emit the high frequency radiation. Thus, Planck was able to explain the lack of high frequency radiation emitted during blackbody radiation.
Thanks guys!
Neutron
Hey there, I have 3 responses.
Firstly, when planck proposed his quantisation theory, he didn't link it to electrons moving up or down energy levels, this was done much later. What Planck proposed was that the EMR packets were produced by atoms within the black body oscillating at certain discrete frequencies. The source of the EMR were atomic oscillations.
Secondly, the modern day kinda explanation for why the black body radiation curve peaks an then falls down to zero is that there aren't that many electrons making that huge fall from a very high energy level to a low energy level and producing that large frequency within the EMR photons (in a simplistic sense).
'
Thirdly, from what i understand, if a photon of emr hits an electron, if it has sufficient energy, it will propel the electron to the next energy shell or completely remove it. But if it does not have energy to move the electron to a higher stationary state/energy level, it will be absorbed and re-emitted.
This is the extent of what I know, I'd be interested if something else has a more comprehensive answer.
Post by: jamonwindeyer on July 20, 2016, 10:18:16 pm
Hey there, I have 3 responses.
Firstly, when planck proposed his quantisation theory, he didn't link it to electrons moving up or down energy levels, this was done much later. What Planck proposed was that the EMR packets were produced by atoms within the black body oscillating at certain discrete frequencies. The source of the EMR were atomic oscillations.
Secondly, the modern day kinda explanation for why the black body radiation curve peaks an then falls down to zero is that there aren't that many electrons making that huge fall from a very high energy level to a low energy level and producing that large frequency within the EMR photons (in a simplistic sense).
'
Thirdly, from what i understand, if a photon of emr hits an electron, if it has sufficient energy, it will propel the electron to the next energy shell or completely remove it. But if it does not have energy to move the electron to a higher stationary state/energy level, it will be absorbed and re-emitted.
This is the extent of what I know, I'd be interested if something else has a more comprehensive answer.
You are correct on all three counts, though the first response goes a little deeper I think, certainly deeper than HSC Physics requires ;D second response is perfect, the larger energy 'jumps' are just less common, which makes fundamental sense. And your third response is also spot on :D
Here's a bit of I2I to mix up the posts ;)
I was wondering whether there's anything fundamentally wrong with my understanding about Planck and if one of you kind fellows could read my response to this question, that would be amazing :D
Thanks guys!
Neutron
I think your response is great Neutron! I'd pull back a bit though, this is a Describe question after all, in an exam you'd only need roughly half of that word count to get the marks. Perhaps take out the details about ionising radiation and replace it with the more simplistic explanation suggested above, which is also a little more correct in this context :D
Post by: MysteryMarker on July 20, 2016, 10:27:16 pm
'Analyse the ability of the Rutherford-Bohr model to completely explain observed spectral characteristics' - 4 marks
To attain 4 marks two observations it could and could not explain must be analysed. I'm stuck on finding two observations it could analyse, i've found one being: The production of spectral lines due to the electrons in the atom jumping from one energy level to a lower energy level. What is another observation that it could explain?
Cheers.
Post by: Spencerr on July 21, 2016, 07:07:33 am
For my homeboi diiiiiii - quanta to quarks
'Analyse the ability of the Rutherford-Bohr model to completely explain observed spectral characteristics' - 4 marks
To attain 4 marks two observations it could and could not explain must be analysed. I'm stuck on finding two observations it could analyse, i've found one being: The production of spectral lines due to the electrons in the atom jumping from one energy level to a lower energy level. What is another observation that it could explain?
Cheers.
The Rutherford model was the planetary model of random orbiting electrons. Bohr placed those electrons into discrete energy levels and shells and proposed that when an electron moves from a higher to a lower energy she'll, it'll emit emr that correspond to the frequency of the spectra lines e =hf thus explaining the hydrogen spectrum. Bohr used his atomic model to theoretically derive Balmer equation which was completely empirical based on the observations of the spectral line wavelengths thus bohrs model not only explained why the spectral lines were discrete but it also explained the exact wavelengths of those lines. Limitations include zee man effect, hyperfine lines relative intensity, and why it doesn't work for larger atoms. All these limitations exist because it's a simplistic conflation of quantum and classical physics
Post by: conic curve on July 22, 2016, 01:11:13 pm
1. A train of straight waves of frequency 4.0 MHz passing though soft tissue in the human body at 1540 ms^-1 crosses into bone where its speed is 4216 ms^-1. Calculate:
a. The wavelength of the waves in the soft tissue
b. The frequency of the waves in the bone
c. The wavelength of the waves in the bone
d. The relative index of the bone relative to the soft tissue
For a, I got 385m for b I got 1054m, for c I got 1054m and for d I am not sure how to do it :(
Also how do you do Snell's law questions when it has a temperature of some sort in it e.g. determine the speed of waves in the salt water if their speed in fresh water is 1493 ms^-1 at 25 degrees celcius (don't solve this for me, give me a hint)
Thanks
Post by: m_woods on July 23, 2016, 11:34:55 am
Two conductors labelled X and Y, are carrying currents of 10A and 20A as shown. The force experienced by conductor Y is F newtons (refer to diagram attached). The force experienced by conductor X would be:
(A) F ↑
(B) F ↓
(C) 2F ↑
(D) 2F ↓
The answer was (B) F ↓ but I put down (D) 2F. Could someone please explain why it is (B) not (D)
Post by: RuiAce on July 23, 2016, 12:45:17 pm
Hi, I am struggling with the following question:The force experienced between the conductors is definitely equal to each other in magnitude.
Two conductors labelled X and Y, are carrying currents of 10A and 20A as shown. The force experienced by conductor Y is F newtons (refer to diagram attached). The force experienced by conductor X would be:
(A) F ↑
(B) F ↓
(C) 2F ↑
(D) 2F ↓
The answer was (B) F ↓ but I put down (D) 2F. Could someone please explain why it is (B) not (D)
The key thing to know is that all the current do is vary the force on BOTH wires. If you increase the current in just one wire, the force between BOTH wires increases.
This is because the force per metre is an attractive force. Wire X attracts wire Y, as does wire Y attract wire X. The fact that it's equal in magnitude and opposite in direction follows directly from Newton's Third Law of Motion - For every action an object causes, there exists an equal and opposite reaction being exerted onto the object.
Remember: The attractive (or repulsive where appropriate) force is the force BETWEEN two wires. The conductors are exerting the same force on each OTHER.
Post by: FallonXay on July 23, 2016, 02:41:03 pm
Regarding the Idea to Implementation topic - Cathode Rays, how would I assess the validity of the experiments on the nature of cathode rays (i.e paddle wheel, Maltese cross, etc)?
thanks.
Post by: nimasha.w on July 23, 2016, 04:12:36 pm
Post by: jamonwindeyer on July 23, 2016, 11:13:07 pm
Hello,
Regarding the Idea to Implementation topic - Cathode Rays, how would I assess the validity of the experiments on the nature of cathode rays (i.e paddle wheel, Maltese cross, etc)?
thanks.
Hey FallonXay! First and foremost, we must remind ourselves that validity refers to how well variables are controlled in an experiment. The independent variable should be altered systematically, the dependent variable should be measured correctly, and everything else should be kept constant ;D essentially, this amounts to the following question: Does the experiment test what it actually should?
For these experiments it's a little trickier, because the experiments are quantitative, not qualitative. There are no actual measurements made, so we step back a little bit, and just consider in general how well the variables were controlled. Would anything have affected the results of these experiments, and/or were results considered correctly?
My interpretation of this would be to discuss how correctly the results were interpreted. Obviously, some of the Cathode ray experiments suggested that the rays were waves, and other suggested they were particles. The ones suggesting particle behaviour were valid because the results were interpreted correctly, the others were (somewhat) invalid because the results, while controlled, were interpreted incorrectly.
This would be how I would approach it, as well as coming up with variables in the experiments that may not have been controlled correctly (air pressure in tube, voltage, etc.) ;D in saying that, this question is just a tad strange, I personally doubt it would be asked in the HSC ;D hope this helps!
Post by: jamonwindeyer on July 23, 2016, 11:28:34 pm
hiiiii! i don't fully understand the depletion zone in a pn junction :-))
Hey there! So this is a little bit of a topic that you don't need to completely understand for the HSC, it is a reasonably complex piece of Physics, but we can get a rough idea quite effectively and that will do nicely.
I'm going to assume that you know the basics of extrinsic semiconductors: That p-type semiconductors have an excess of holes, and n-type semiconductors have an excess of electrons.
Therefore, when we join a P and N type semiconductor together, we have what is called electron and hole drift. The excess electrons drift from the N type to the P type, and the holes drift from the P type to the N type. On the N side, this therefore leaves behind a positive donor ion, and on the P side, this leaves behind negative acceptor ions. This is shown in the diagram below.
Note that we call this a depletion zone because the diffused electrons and holes are depleted. The electrons come into contact with the holes on the P side and disappear via recombination, and the same thing happens in reverse on the N side. This leaves behind no mobile charge carriers in the depletion zone. What does happen, however, is that the positive/negative ions set up an electric field, as shown below. This electric field is the basis for the modern transistor!
The very astute may notice that this electric field is what eventually prevents further diffusion of electrons and holes. Once this happens, we say the junction is in equilibrium, but this is beyond HSC Physics ;D
I hope this helps!!
(https://upload.wikimedia.org/wikipedia/commons/3/3e/Pn_Junction_Diffusion_and_Drift.svg)
Post by: Aliceyyy98 on July 24, 2016, 12:48:12 am
Was wondering how the pn junction in solar cells create electricity? Confused about the direction of current and electron flow >< and also how does the induction motor - squirrel cage work? Like how does the rotor chase the magnetic field! Thank youuuu
Post by: jamonwindeyer on July 24, 2016, 01:35:27 am
Heyyy~
Was wondering how the pn junction in solar cells create electricity? Confused about the direction of current and electron flow >< and also how does the induction motor - squirrel cage work? Like how does the rotor chase the magnetic field! Thank youuuu
Hey there!! For the PN junction in solar cells: In terms of how a PN junction works, check out the post I made right before your post, that should explain it for you! Now, for solar cells specifically:
The PN junction effectively works to ensure that any charge can only flow in the same direction as the electric field. Well, it can flow the other way if you really want it to, but that isn't relevant here. In solar cells, the photoelectric effect causes electrons to be released from the semiconductors in the junction. A special coating maximises the amount of photoelectric emission occurring. These released electrons are then forced to move according to the electric field in the junction: They move in the opposite direction. Remember, the electric field indicates the direction positive charges will flow: Electrons do the opposite.
So what we've established is that these electrons are forced to move in a single direction by the electric field in the depletion zone caused by the PN junction. The junction therefore becomes a current source! Electrons flow out of the junction, and through an external load or battery, and this process continues. Keep in mind that a complete circuit is required for this to continue; the electrons continually flow around the circuit and back into the PN junction ready to be re-emitted.
Note that current will just be in the opposite direction to electron flow, by definition ;D
Now to the induction squirrel cage rotor. The principle at play here is Lenz's Law, the idea that induced currents will oppose the change that created them. An induction motor is exposed to a rotating magnetic field. Essentially, this means that a North pole (several, but lets simplify) is spinning around the rotor. The changing field induces a current in the rotor, which will cause the rotor to generate its own magnetic field.
According to Lenz's Law, the rotors magnetic field will oppose the change that created it. It doesn't want the North pole to be spinning. So, it gives chase. It spins as well, effectively to 'cancel out' the change.
This is, kind of, like looking at a sideways picture on a piece of paper glued to the table. The paper can't be moved, so YOU turn your head to cancel out the change and look at it the right way up. Same thing above; the rotor spins so it is doing the same thing as the spinning North pole, thus cancelling out the change. Cancelling out the change isn't the appropriate way to think of it as a Physicist, but for getting the intuitive sense of things, it helps!
The rotor wants to stay aligned with the North pole, and so it rotates around to chase it. This seems kind of silly, but it makes sense! The easiest model for this is spinning a permanent magnet below an aluminium disc; the disc will spin around and chase the magnet. Same thing ;D
The rotor is shaped the way it is to maximise induced current flow, thus maximising the response of the rotor and maximising the efficiency of the induction motor as a whole ;D
I hope this helps! Let me know if anything needs clearing up ;D
Post by: FallonXay on July 24, 2016, 07:06:33 am
Hey FallonXay! First and foremost, we must remind ourselves that validity refers to how well variables are controlled in an experiment. The independent variable should be altered systematically, the dependent variable should be measured correctly, and everything else should be kept constant ;D essentially, this amounts to the following question: Does the experiment test what it actually should?
For these experiments it's a little trickier, because the experiments are quantitative, not qualitative. There are no actual measurements made, so we step back a little bit, and just consider in general how well the variables were controlled. Would anything have affected the results of these experiments, and/or were results considered correctly?
My interpretation of this would be to discuss how correctly the results were interpreted. Obviously, some of the Cathode ray experiments suggested that the rays were waves, and other suggested they were particles. The ones suggesting particle behaviour were valid because the results were interpreted correctly, the others were (somewhat) invalid because the results, while controlled, were interpreted incorrectly.
This would be how I would approach it, as well as coming up with variables in the experiments that may not have been controlled correctly (air pressure in tube, voltage, etc.) ;D in saying that, this question is just a tad strange, I personally doubt it would be asked in the HSC ;D hope this helps!
Ok, makes sense - thanks. However, wouldn't the theories that demonstrated the wave nature of cathode rays still be valid due to the wave-particle duality proposal?
Post by: Aliceyyy98 on July 24, 2016, 08:40:45 am
Hey there!! For the PN junction in solar cells: In terms of how a PN junction works, check out the post I made right before your post, that should explain it for you! Now, for solar cells specifically:
The PN junction effectively works to ensure that any charge can only flow in the same direction as the electric field. Well, it can flow the other way if you really want it to, but that isn't relevant here. In solar cells, the photoelectric effect causes electrons to be released from the semiconductors in the junction. A special coating maximises the amount of photoelectric emission occurring. These released electrons are then forced to move according to the electric field in the junction: They move in the opposite direction. Remember, the electric field indicates the direction positive charges will flow: Electrons do the opposite.
So what we've established is that these electrons are forced to move in a single direction by the electric field in the depletion zone caused by the PN junction. The junction therefore becomes a current source! Electrons flow out of the junction, and through an external load or battery, and this process continues. Keep in mind that a complete circuit is required for this to continue; the electrons continually flow around the circuit and back into the PN junction ready to be re-emitted.
Note that current will just be in the opposite direction to electron flow, by definition ;D
Now to the induction squirrel cage rotor. The principle at play here is Lenz's Law, the idea that induced currents will oppose the change that created them. An induction motor is exposed to a rotating magnetic field. Essentially, this means that a North pole (several, but lets simplify) is spinning around the rotor. The changing field induces a current in the rotor, which will cause the rotor to generate its own magnetic field.
According to Lenz's Law, the rotors magnetic field will oppose the change that created it. It doesn't want the North pole to be spinning. So, it gives chase. It spins as well, effectively to 'cancel out' the change.
This is, kind of, like looking at a sideways picture on a piece of paper glued to the table. The paper can't be moved, so YOU turn your head to cancel out the change and look at it the right way up. Same thing above; the rotor spins so it is doing the same thing as the spinning North pole, thus cancelling out the change. Cancelling out the change isn't the appropriate way to think of it as a Physicist, but for getting the intuitive sense of things, it helps!
The rotor wants to stay aligned with the North pole, and so it rotates around to chase it. This seems kind of silly, but it makes sense! The easiest model for this is spinning a permanent magnet below an aluminium disc; the disc will spin around and chase the magnet. Same thing ;D
The rotor is shaped the way it is to maximise induced current flow, thus maximising the response of the rotor and maximising the efficiency of the induction motor as a whole ;D
I hope this helps! Let me know if anything needs clearing up ;D
Post by: Spencerr on July 24, 2016, 10:22:13 am
Ok, makes sense - thanks. However, wouldn't the theories that demonstrated the wave nature of cathode rays still be valid due to the wave-particle duality proposal?
Hey there, as Jamon mentioned validity refers to how well the experiment is set out to test what it intends to test. At the time of the cathode ray debate, the way theory of cathode rays wad proposed because the germans passed a cathode ray in between two electrically charged plates and found that the rays did not deflect. Since charged particles would deflect and waves wouldn't, they concluded that cathode rays were waves. However the experiment was not valid because the experiment had inherent flaws. At that time they did not have the technology to fully pump air out of the vacuum tube therefore when they did the experiment, there were a significant amount of gas in the discharge tube
These gas molecules were ionized by the cathode rays causing them to be attracted to the electric fields. This neutralized the electric fields once equilibrium was reached and so there was no e field acting on the cathode rays and hence no deflection. This explains why the interpretation of the results were invalid.
Referring to the wave particle duality, I believe that is a separate concept that comes up much later in quanta to quarks with de broglie. It's beyond what you need to know in i2i as its a core component of the elective q2q.
As for the other experiments to showboat they were particles you could argue they are valid due to correct experimental procedure, controlled variables etc. One thing I would like to mention that is beyond hsc is that the paddle wheel experiment which showed the cathode rays had momentum is actually invalid. This is because the rotation of the paddle wheels are cause air currents flowing in a certain direction due to the heating produced by the cathode rays. So there was no momentum pushing the wheel but air currents. But this is beyond the syllabus.
I hope this has helped :)
Post by: FallonXay on July 24, 2016, 12:19:14 pm
Hey there, as Jamon mentioned validity refers to how well the experiment is set out to test what it intends to test. At the time of the cathode ray debate, the way theory of cathode rays wad proposed because the germans passed a cathode ray in between two electrically charged plates and found that the rays did not deflect. Since charged particles would deflect and waves wouldn't, they concluded that cathode rays were waves. However the experiment was not valid because the experiment had inherent flaws. At that time they did not have the technology to fully pump air out of the vacuum tube therefore when they did the experiment, there were a significant amount of gas in the discharge tube
These gas molecules were ionized by the cathode rays causing them to be attracted to the electric fields. This neutralized the electric fields once equilibrium was reached and so there was no e field acting on the cathode rays and hence no deflection. This explains why the interpretation of the results were invalid.
Referring to the wave particle duality, I believe that is a separate concept that comes up much later in quanta to quarks with de broglie. It's beyond what you need to know in i2i as its a core component of the elective q2q.
As for the other experiments to showboat they were particles you could argue they are valid due to correct experimental procedure, controlled variables etc. One thing I would like to mention that is beyond hsc is that the paddle wheel experiment which showed the cathode rays had momentum is actually invalid. This is because the rotation of the paddle wheels are cause air currents flowing in a certain direction due to the heating produced by the cathode rays. So there was no momentum pushing the wheel but air currents. But this is beyond the syllabus.
I hope this has helped :)
Yeah, definitely helps, thanks!
Just one thing: Doesn't Einstein's use the wave/particle duality to explain the photoelectric effect in Idea to Implementation?
Post by: Spencerr on July 24, 2016, 12:32:48 pm
Post by: FallonXay on July 24, 2016, 12:41:00 pm
Your right he does but that's after Planck introduced the idea of quantified energy in black bodies so it's still part of quantum theory which is still much later. :)
ok, Cheers ;D
Post by: Aliceyyy98 on July 24, 2016, 02:40:29 pm
Could someone clarify the g-forces during launch and the physics behind slingshot effect for me!
Thankyou~
Post by: ATWalk on July 24, 2016, 03:43:28 pm
I'm struggling with these multiple choice questions from a 2013 Sydney Boys trial and would love an explanation on how they got these answers.
The answers:
14 - D (Wouldn't a back emf be induced because of the relative motion of the coil?)
16 - B
Thank you! :D
Post by: Spencerr on July 24, 2016, 05:04:23 pm
Hey,
I'm struggling with these multiple choice questions from a 2013 Sydney Boys trial and would love an explanation on how they got these answers.
The answers:
14 - D (Wouldn't a back emf be induced because of the relative motion of the coil?)
16 - B
Thank you! :D
Hey there for 14. Note that it's a generator and not a motor. Back emf only occurs in motors due to the relative motion between the rotor and the b field. However in generators although this same relative motion exists the back emf in this case is actually the generated electricity. Whereas the back emf in motors opposed the supply voltage. The back emf in generators is the output so technically there is no back emf.
For the next question note how it days the ammeter voltmeter are connected in series. Remember back to when you did prac in class, you would always have to connect the voltmeter in parallel because the voltmeter has a significant amount of resistance. This explains why there is no current reading and the light bulb does not work as the large resistance nullifies the current flow.
Post by: jamonwindeyer on July 24, 2016, 09:05:26 pm
Hi！
Could someone clarify the g-forces during launch and the physics behind slingshot effect for me!
Thankyou~
Hey there Alice!!
Okay, so G forces. Basically, G forces are a way to measure the forces on astronaut in terms of multiples of the regular gravitational force of 1G. If we were travelling upwards with an acceleration of 9.8 metres per second per second, then we would experience a reaction force of 1G, as well as the regular gravitational force of 1G, meaning a total of 2G's. Astronauts accelerating upwards can experience G forces in excess of 5-6G's (may want to confirm that number, going off memory) ;D
Excessive or prolonged G force can injure or kill, however, we need a lot of acceleration to launch a rocket. So, it becomes a balance between accelerating the rocket, without exposing the astronauts to excess G forces. Acceleration is very controlled throughout the launch, and the moments of separation (when the fuel tanks are separated from the rocket and return to earth) also offer a break for the astronauts from the G forces ;D
Slingshot effect is much easier to picture with an example. Imagine you are riding a push bike down a road. You want to speed up, so you grab on to the back of one of the cars. The car drags you along, and then you let go, and you've sped up with absolutely no extra pedalling, and the car barely knew you were there!
In the slingshot effect, you are the probe, and the car is the planet. The probe is dragged along by the planet's gravitational field. Effectively, some of the momentum of the planet is transferred to the probe. However, since the planets mass is massively larger than the probe's, the massive speed increase of the probe has almost no effect on the motion of the planet ;D
Post by: Aliceyyy98 on July 24, 2016, 11:57:01 pm
Hey there Alice!!
Okay, so G forces. Basically, G forces are a way to measure the forces on astronaut in terms of multiples of the regular gravitational force of 1G. If we were travelling upwards with an acceleration of 9.8 metres per second per second, then we would experience a reaction force of 1G, as well as the regular gravitational force of 1G, meaning a total of 2G's. Astronauts accelerating upwards can experience G forces in excess of 5-6G's (may want to confirm that number, going off memory) ;D
Excessive or prolonged G force can injure or kill, however, we need a lot of acceleration to launch a rocket. So, it becomes a balance between accelerating the rocket, without exposing the astronauts to excess G forces. Acceleration is very controlled throughout the launch, and the moments of separation (when the fuel tanks are separated from the rocket and return to earth) also offer a break for the astronauts from the G forces ;D
Slingshot effect is much easier to picture with an example. Imagine you are riding a push bike down a road. You want to speed up, so you grab on to the back of one of the cars. The car drags you along, and then you let go, and you've sped up with absolutely no extra pedalling, and the car barely knew you were there!
In the slingshot effect, you are the probe, and the car is the planet. The probe is dragged along by the planet's gravitational field. Effectively, some of the momentum of the planet is transferred to the probe. However, since the planets mass is massively larger than the probe's, the massive speed increase of the probe has almost no effect on the motion of the planet ;D
Thank you for your explanations! They are always so easy to understand, saved my life :)
Post by: jamonwindeyer on July 25, 2016, 12:22:16 am
Thank you for your explanations! They are always so easy to understand, saved my life :)
Awesome! Happy to help ;D
Post by: FallonXay on July 25, 2016, 04:19:27 pm
Hey there for 14. Note that it's a generator and not a motor. Back emf only occurs in motors due to the relative motion between the rotor and the b field. However in generators although this same relative motion exists the back emf in this case is actually the generated electricity. Whereas the back emf in motors opposed the supply voltage. The back emf in generators is the output so technically there is no back emf.
For the next question note how it days the ammeter voltmeter are connected in series. Remember back to when you did prac in class, you would always have to connect the voltmeter in parallel because the voltmeter has a significant amount of resistance. This explains why there is no current reading and the light bulb does not work as the large resistance nullifies the current flow.
On the circuit question, does the voltmeter act to prevent the flow of electrons completely (like does it block the potential difference or something)? - Or how come the current isn't registered on the ammeter before being impeded by the voltmeter. How does this work?
Post by: FallonXay on July 25, 2016, 04:54:12 pm
Thanks.
Post by: jamonwindeyer on July 25, 2016, 07:34:20 pm
On the circuit question, does the voltmeter act to prevent the flow of electrons completely (like does it block the potential difference or something)? - Or how come the current isn't registered on the ammeter before being impeded by the voltmeter. How does this work?
Voltmeters have a huge resistance, so when they are connected in parallel, almost no current flows through it. This is to minimise the effect of the measurement on the circuit being measured ;D
In terms of the question, the voltmeter having a huge resistance means that almost no current flows through the circuit. Consider:
The voltage is constant (from the source), so when the resistance is increased massively due to the voltmeter, the current is brought essentially to zero. Thus, the ammeter reads zero and the light bulb doesn't turn on. However, the voltmeter will still read 12 volts ;D
Also, for the Core Topic - Space, do we need to know about sun spots/solar flares/ effects of solar radiation on space flights?
Thanks.
Not really, you need to know ionisation blackout for the problems affecting re-entry dot point, but not solar storms (as to my knowledge) ;D
Post by: Spencerr on July 25, 2016, 09:26:55 pm
Also, for the Core Topic - Space, do we need to know about sun spots/solar flares/ effects of solar radiation on space flights?
Thanks.
Hey there, in response to this, I've encountered a question on the effects of solar flares/sun spots on communication and I think there was another one on space flights. (syllabus might have changed) but heres how I answered the question.
1. On communication
Sunspots are regions of intense magnetic fields that occur beneath the surface of the sun, and is normally cooler than the surrounding areas. Sunspots can interfere with communications as emr (such as microwaves and radio waves used by satellites) are weakened and disrupted by strong magnetic fields. On solar flares and solar radiation, communications satellites are situated at the top of the van allen belt, and emr used in communication has to travel to and from the satellite, passing through the van allen belt to reach earth. Solar flares and solar radiation can ionise the charged particles in the van allen belts and attenuate communication signals. They can also damage sensitive electron equipment and cause interference.
2. On Space travel
Solar flares and solar radiation can be a problem during space travel due to the collision of ionised particles with the spacecraft. These collisions can 1) disrupt and interfere with communication 2) can damage electronic equipment on board the spacecraft 3) post a danger to the occupants inside the spacecraft. however there are measures to mitigate the impact of these charged particles through 1) using a magnetic shield to repel the charged particles from the ship 2) Sending communication signals from the back end of the spacecraft (or the end that is not exposed to the charged particles) 3) Layering the outside of the spacecraft with something (i forgot, a dense material which prevents the charged particles from entering the spacecraft).
hope this helps!
Post by: Spencerr on July 25, 2016, 09:30:20 pm
On the circuit question, does the voltmeter act to prevent the flow of electrons completely (like does it block the potential difference or something)? - Or how come the current isn't registered on the ammeter before being impeded by the voltmeter. How does this work?
Remember potential difference and current are two differnt things. Potential difference is creates the E- field or the force that pushes the current in one direction (high to low potential). So potential difference is NOT the flow of electrons. Jamon explained the rest haha
Post by: Aliceyyy98 on July 27, 2016, 12:54:36 am
Post by: Spencerr on July 27, 2016, 10:16:20 am
Hi! Is anyone doing Quanta to Quarks for their elective able to explain the spectral line series? How they contributed to the atomic theory?
Hey there I do Q2Q. I'm going phrase the question as "Explain the spectral line series and their contribution to atomic theory?"
Atoms when excited emit a characteristic set of wavelengths. The The spectral line series (of Hydrogen) shows the specific wavelengths and thus specific energies (remember E = Hf) emitted by an excited hydrogen atom (i.e. when an eletron jumps from a higher energy level to a lower energy level). The spectral lines were crucial to Bohr's development of his atomic model. Since spectral lines existed as discrete specific values instead of a continuous spectrum, Bohr proposed that electrons orbited the hydrogen atom in discrete energy levels known as stationary states. Any movement of electrons would have to involve a complete transition from one state to the other and all energy would be kept within the state (meaning that there is no emission of emr as the electrons undergo centripetal acceleration in the orbit). This was his first postulate. His second postulate was that when an electron moves from a higher energy level to a lower energy level, it would have to emit a photon of emr equivalent to the energy difference i.e. Change in E = hf, conversely if an electron moves from a lower energy level to a higher energy level it would have to absorb an equivalent photon. This explains why there as multiple lines as it indicated that electrons were falling from different energy levels . His third postulate was the quantisation condition that L=mvr (angular momentum) only exists in integer multiples of h/2pi.
The hydrogen spectral lines allowed Bohr to develop his three postulates and from those postulates, he derived a mathematical equation to predict the wavelengths emitted by hydrogen. This equation was almost identical with the empirical equation derived by Balmer after he observed the spectral lines. Such a coincidence lend support to Bohr's atomic model and thus the spectral lines were crucial to the development and understanding of atomic structure and theory.
the energy levels in the hydrogen atom occurred as discrete
Post by: Aliceyyy98 on July 27, 2016, 10:59:31 am
Hey there I do Q2Q. I'm going phrase the question as "Explain the spectral line series and their contribution to atomic theory?"
Atoms when excited emit a characteristic set of wavelengths. The The spectral line series (of Hydrogen) shows the specific wavelengths and thus specific energies (remember E = Hf) emitted by an excited hydrogen atom (i.e. when an eletron jumps from a higher energy level to a lower energy level). The spectral lines were crucial to Bohr's development of his atomic model. Since spectral lines existed as discrete specific values instead of a continuous spectrum, Bohr proposed that electrons orbited the hydrogen atom in discrete energy levels known as stationary states. Any movement of electrons would have to involve a complete transition from one state to the other and all energy would be kept within the state (meaning that there is no emission of emr as the electrons undergo centripetal acceleration in the orbit). This was his first postulate. His second postulate was that when an electron moves from a higher energy level to a lower energy level, it would have to emit a photon of emr equivalent to the energy difference i.e. Change in E = hf, conversely if an electron moves from a lower energy level to a higher energy level it would have to absorb an equivalent photon. This explains why there as multiple lines as it indicated that electrons were falling from different energy levels . His third postulate was the quantisation condition that L=mvr (angular momentum) only exists in integer multiples of h/2pi.
The hydrogen spectral lines allowed Bohr to develop his three postulates and from those postulates, he derived a mathematical equation to predict the wavelengths emitted by hydrogen. This equation was almost identical with the empirical equation derived by Balmer after he observed the spectral lines. Such a coincidence lend support to Bohr's atomic model and thus the spectral lines were crucial to the development and understanding of atomic structure and theory.
the energy levels in the hydrogen atom occurred as discrete
Wow thanx a bunch, rlly helpful!
Post by: jakesilove on July 27, 2016, 10:55:47 pm
Hey there I do Q2Q. I'm going phrase the question as "Explain the spectral line series and their contribution to atomic theory?"
Atoms when excited emit a characteristic set of wavelengths. The The spectral line series (of Hydrogen) shows the specific wavelengths and thus specific energies (remember E = Hf) emitted by an excited hydrogen atom (i.e. when an eletron jumps from a higher energy level to a lower energy level). The spectral lines were crucial to Bohr's development of his atomic model. Since spectral lines existed as discrete specific values instead of a continuous spectrum, Bohr proposed that electrons orbited the hydrogen atom in discrete energy levels known as stationary states. Any movement of electrons would have to involve a complete transition from one state to the other and all energy would be kept within the state (meaning that there is no emission of emr as the electrons undergo centripetal acceleration in the orbit). This was his first postulate. His second postulate was that when an electron moves from a higher energy level to a lower energy level, it would have to emit a photon of emr equivalent to the energy difference i.e. Change in E = hf, conversely if an electron moves from a lower energy level to a higher energy level it would have to absorb an equivalent photon. This explains why there as multiple lines as it indicated that electrons were falling from different energy levels . His third postulate was the quantisation condition that L=mvr (angular momentum) only exists in integer multiples of h/2pi.
The hydrogen spectral lines allowed Bohr to develop his three postulates and from those postulates, he derived a mathematical equation to predict the wavelengths emitted by hydrogen. This equation was almost identical with the empirical equation derived by Balmer after he observed the spectral lines. Such a coincidence lend support to Bohr's atomic model and thus the spectral lines were crucial to the development and understanding of atomic structure and theory.
the energy levels in the hydrogen atom occurred as discrete
Legendary answer, thank you for helping out so many people!
Post by: Spencerr on July 28, 2016, 12:21:42 am
Legendary answer, thank you for helping out so many people!
I pretend every question is an exam question. Helps with the revision!!
Post by: Spencerr on July 28, 2016, 03:21:30 am
Could someone clarify the concept of mass defect in relation to to nuclear fission.
Like in nuclear fission, an element splits to form two daughter elements with a higher binding energy per nucleon.
Where does this higher binding energy come from? Is it from the mass defect? If it is from the mass defect, then where does the energy that is released as heat etc. come from?
Post by: jakesilove on July 28, 2016, 09:34:04 am
I have my own physics question that I've been mulling over for a long while.
Could someone clarify the concept of mass defect in relation to to nuclear fission.
Like in nuclear fission, an element splits to form two daughter elements with a higher binding energy per nucleon.
Where does this higher binding energy come from? Is it from the mass defect? If it is from the mass defect, then where does the energy that is released as heat etc. come from?
Hey!
As you know, I didn't do Q2Q, but I think I roughly know how this works (although not in any depth). When you add up the mass of the two daughter elements, it will be slightly lower than the original particle. This is the notion of mass defect. This mass has to go SOMEWHERE (it can't just disappear!), so in steps Einstein's famous formula
With just a tiny mass defect, and enormous amount of energy can be released. This explains the higher binding energy, however there is still WAY to much energy for the daughter particles to hold. So, the rest is given off as heat/light/dubstep mashups of Darude Sandstorm.
I hope that makes sense!
Post by: nimasha.w on July 29, 2016, 01:48:24 am
Post by: jamonwindeyer on July 29, 2016, 02:18:57 am
hi!! how would you make an assessment of validity (like what would say/how would you structure) of secondary sources you've used
Hey!! In this context, validity refers to how 'correct' the information you have obtained is, and there are a few ways to assess this.
- Does the source have appropriate citations (citations indicate a well-researched article, and thus, a valid source)
- Is the source credible (a peer-reviewed paper, a government website, etc. are all valid sources)
- Does there exist any bias/self interest on the part of the author? (less a big deal in Physics than it is in other subjects)
Happy for anyone to jump in with extra suggestions ;D
Post by: Spencerr on July 29, 2016, 09:53:13 am
Hey!! In this context, validity refers to how 'correct' the information you have obtained is, and there are a few ways to assess this.
- Does the source have appropriate citations (citations indicate a well-researched article, and thus, a valid source)
- Is the source credible (a peer-reviewed paper, a government website, etc. are all valid sources)
- Does there exist any bias/self interest on the part of the author? (less a big deal in Physics than it is in other subjects)
Happy for anyone to jump in with extra suggestions ;D
Our school had a two mark question similar. On top of what Jamon mentioned the information needs to be CROSSCHECKED to ensure CONSISTENCY across different sources.
The capital words were the necessary ones for each mark awarded
Post by: wyzard on July 29, 2016, 12:19:52 pm
hi!! how would you make an assessment of validity (like what would say/how would you structure) of secondary sources you've used
An unspoken rule of the Information Age: Never believe what's on the internet immediately :P This is why assessing validity is important, to make sure the information squares with reality and real science.
To assess the validity; check to see if the source is credible like where is it being published, is it being peer-reviewed, running a quick background check on the author for his credentials.
Usually scientific articles from Google Scholar, usually published in scientific journals, and textbook citations are fine.
Science articles on news websites or popular science websites generally, on the other hand, are not as they're written by journalist for the general public, so the information in there are usually heavily trimmed down, or made more dramatic to attract attention. They are nonetheless useful as they provide the general ideas, and will usually include links to the actual scientific research papers so use those ;D
On a final thought, sources from sketchy science websites, such as those promoting "New Free Unlimited Energy Generating Device" or "New Quantum Consciousness" is a big no no. For example, a quick google search for "free energy devices" leads me to this site: http://www.free-energy-info.com/, an example of a page where you don't want to cite your sources. I can't even verify the author(s) credentials, and reading the content gives me so much cringe on the misunderstanding of physics behind it. :o
Post by: jamonwindeyer on July 29, 2016, 01:07:09 pm
For example, a quick google search for "free energy devices" leads me to this site: http://www.free-energy-info.com/, an example of a page where you don't want to cite your sources. I can't even verify the author(s) credentials, and reading the content gives me so much cringe on the misunderstanding of physics behind it. :o
Oh my goodness, this site actually says:
Oliver Heaviside's equation E = mC2...
If that whole site isn't actually a giant troll, then I actually don't know what to do right now... :'(
Post by: conic curve on July 29, 2016, 05:35:44 pm
Qualitatively describe the effect on intensity if the distance is tripled and the strength of the source is halved?
Post by: Spencerr on July 29, 2016, 05:47:23 pm
I do not understand this question and how the answer is 1/18
Qualitatively describe the effect on intensity if the distance is tripled and the strength of the source is halved?
Intensity equals strength over distance squared. It's and inverse law. Someone else might be able to explain it more I'm depth but that's the formula used
Post by: wyzard on July 29, 2016, 05:51:53 pm
Oh my goodness, this site actually says:
Oliver Heaviside's equation E = mC2...
If that whole site isn't actually a giant troll, then I actually don't know what to do right now... :'(
hahahaha join me in cringing ;D Yeah judging by the 3000 ebook they have available, I highly doubt they're trolling. I think they truly believe in building a free energy device, and that the government and big oil companies are suppressing such technology :'(
Post by: jakesilove on July 29, 2016, 06:19:34 pm
Intensity equals strength over distance squared. It's and inverse law. Someone else might be able to explain it more I'm depth but that's the formula used
Yep, that's all you need! Sub in your initial values (ie. Strength = A, distance = d) and then try it with some new values (ie. Strength = (1/2)*A, distance = 3*d) and you should get 1/18th of the initial value!
Post by: Loki98 on July 29, 2016, 07:30:50 pm
Post by: conic curve on July 29, 2016, 08:17:05 pm
Post by: jamonwindeyer on July 29, 2016, 08:30:44 pm
Why is it hard to see Newton's first law in real life?
Newton's 1st Law is (roughly stated) that an object at rest stays at rest and an object in motion stays in motion, unless acted upon by an external force.
So, why doesn't a football we kick (for example) just fly upwards forever, why doesn't a car roll forever without accelerating? The answer is friction, in real life there is constantly frictional forces impeding the motion of objects, thus making it very difficult to ever find an actual situation where Newton's 1st Law hold valid ;D
Post by: conic curve on July 29, 2016, 08:33:08 pm
Newton's 1st Law is (roughly stated) that an object at rest stays at rest and an object in motion stays in motion, unless acted upon by an external force.
So, why doesn't a football we kick (for example) just fly upwards forever, why doesn't a car roll forever without accelerating? The answer is friction, in real life there is constantly frictional forces impeding the motion of objects, thus making it very difficult to ever find an actual situation where Newton's 1st Law hold valid ;D
Oh thank you thank you thank you
Another question but what is the action and reaction force between a rocket flying into space and it lifting off from the ground. I heard that there is a misconception where the smoke that the rocket exerts pushes the rocket up into space (apparently this isn't true)
Post by: jamonwindeyer on July 29, 2016, 08:36:18 pm
Could someone please explain this question to me =]
This question shows DC motors where the permanent magnets have been replaced by electromagnets. Further, the same current is used to drive these magnets as is used in the rotor/coil itself.
First, we must determine the polarity of the electromagnets. Remember that current flows from positive to negative. We use the right hand grip rule. Take your right hand and give a thumbs up. If your fingers are wrapping in the direction of the current, then your thumb points in the direction of the North Pole. Obviously, the opposite pole will be the south pole. This should allow you to determine the polarity of the magnets on either side of the coil.
For each option, with the left/right polarity are:
A – South/North (as in, on the left, the south pole is pointed towards the coil, and on the right, the North pole is pointed towards the coil)
B – South/North
C – South/South
D – South/South
In fairness, those diagrams are terrible to see this properly. We can now disregard options C and D, because the two poles are the same. We need one of each for a magnetic field.
Now we see which has the right direction by using the right hand slap rule (or your version). The magnetic field in both options goes from North to South, or, from right to left. Point your fingers in your right hand in this direction. Then, we point our thumb in the direction of current. Pick, for example, the left hand side of coil B. Current is going away from us, magnetic field is going to the left. We then slap in the direction of the force (meaning, which way that side of the coil will move). Using the rule, this is upwards.
If the left hand side of the coil is moving up, the coil is spinning clockwise. So, B is incorrect. We can use the same idea to prove that A is the answer. Look the left hand side of the coil in A. The magnetic field is going to the left, but now the current is going towards us. Use the slap rule here to prove to yourself that the resultant force will thus push upwards, and thus, that the coil will spin clockwise.
Therefore, the answer is A.
Post by: jakesilove on July 29, 2016, 08:37:34 pm
Oh thank you thank you thank you
Another question but what is the action and reaction force between a rocket flying into space and it lifting off from the ground. I heard that there is a misconception where the smoke that the rocket exerts pushes the rocket up into space (apparently this isn't true)
In very basic terms, the action is the gas being pushed DOWN from the rocket. This doesn't exert a force on the ground, thus exerting a force on the rocket, or anything like that; just imagine that the gas expands, and since it can only go down, it does.
However, we know that all momentum must be conserved. Therefore, if there is a spontaneous DOWN force, there has to be an equivalent UP force. This reaction projects the rocket upward!
Post by: jamonwindeyer on July 29, 2016, 08:37:51 pm
Oh thank you thank you thank you
Another question but what is the action and reaction force between a rocket flying into space and it lifting off from the ground. I heard that there is a misconception where the smoke that the rocket exerts pushes the rocket up into space (apparently this isn't true)
No that's pretty much right! The idea is that the rocket exerts a downwards force on the fuel particles, causing them to be accelerated downwards as exhaust. It is the reaction force to this that pushes the rocket upwards, and this is called the thrust force ;D
In very basic terms, the action is the gas being pushed DOWN from the rocket. This doesn't exert a force on the ground, thus exerting a force on the rocket, or anything like that; just imagine that the gas expands, and since it can only go down, it does.
However, we know that all momentum must be conserved. Therefore, if there is a spontaneous DOWN force, there has to be an equivalent UP force. This reaction projects the rocket upward!
Ditto ;D
Post by: Loki98 on July 29, 2016, 09:50:25 pm
Post by: Neutron on July 29, 2016, 11:20:09 pm
Post by: jamonwindeyer on July 29, 2016, 11:34:26 pm
Is Lenz's law basically irrelevant to transformers :O and rip do you guys have some spare motivation lying around
Lenz's Law is relevant everywhere! ;D is there something specifically confusing you? It is a little outside the syllabus but I can try and help out? I did a decent amount on transformers last semester ;D
Edit: Lol, one week into uni and already procrastinating, sorry, don't think I can help there ;)
Post by: Aliceyyy98 on July 29, 2016, 11:37:42 pm
My teacher has been going on about triodes for the ideas to implementation topic, do we need to understand how it works? like the reverse and forward bias things, as well as base current or something!
I was really confused and was wondering if this is required for the syllabus?
Cheers
Post by: Neutron on July 29, 2016, 11:43:22 pm
Lenz's Law is relevant everywhere! ;D is there something specifically confusing you? It is a little outside the syllabus but I can try and help out? I did a decent amount on transformers last semester ;D
Edit: Lol, one week into uni and already procrastinating, sorry, don't think I can help there ;)
Well when you're describing the formation of eddy currents within a transformer's iron core, it's not actually because of Lenz's law is it? Actually, forgive me if this is just a massive brain fart but why do eddy currents form within transformer's cores again? It's not like the eddy current is opposing the rate of change of magnetic flux so i'm like ??? ah i'm tired
Post by: jamonwindeyer on July 30, 2016, 12:25:04 am
Well when you're describing the formation of eddy currents within a transformer's iron core, it's not actually because of Lenz's law is it? Actually, forgive me if this is just a massive brain fart but why do eddy currents form within transformer's cores again? It's not like the eddy current is opposing the rate of change of magnetic flux so i'm like ??? ah i'm tired
Well the currents themselves are formed as a result of Faraday's Law of induction, but yes, Lenz's Law tells us in which direction they will move! Remember, we have a changing magnetic flux due to the AC current in the primary coil, and this changing flux induces eddy currents in the core (and it is this flux which allows the new current to flow in the secondary coil too) ;D of course the laminations in a transformer mean that this doesn't have too much of an impact, but Lenz's Law definitely plays a big role in describing these eddy currents, they are formed to oppose the changing flux from the primary coil! ;D
Post by: jamonwindeyer on July 30, 2016, 12:36:55 am
Hi!
My teacher has been going on about triodes for the ideas to implementation topic, do we need to understand how it works? like the reverse and forward bias things, as well as base current or something!
I was really confused and was wondering if this is required for the syllabus?
Cheers
Hey! This is definitely not something that is super important in the HSC Physics course, it's a little beyond the scope. A basic understanding of forward and reverse bias for PN junctions would be useful; Basically, forward bias allows current to flow, reverse bias makes it really difficult (just depends on which end of the junction is connected to the positive terminal of voltage). Essentially, it means current can only flow in a single direction.
There is a great summary of the forward and reverse bias stuff on HyperPhysics that I recommend you read! It gives a great explanation of the ideas, but it is a little complicated, let me know if you need any help with it ;D
The base current stuff is to do with basic transistor theory. Essentially, there are three terminals: The base, the collector and the emitter. The base current controls the current flow between the collector and the emitter. The specifics of that are definitely unimportant! A very complicated article here if you feel brave, but it is beyond what would be asked in HSC Physics, save that for tertiary electronics ;)
Post by: Loki98 on July 30, 2016, 03:01:19 pm
How would you explain how a transformer works using the conservation of law of energy?
Also, in the syllabus is asks to "compare the structure and function of a generator to an electric motor," when they say electric motor, do they ask for AC and DC motors?
thank you :)
Post by: MysteryMarker on July 30, 2016, 07:37:01 pm
For the fluorescence screen investigation carried out to deterimine the nature of cathode rays, why is it that this experiment can be considered as giving evidence for both the particle and wave nature of cathode rays. Like i understand that the CR's transferred their energy to the fluorescent atoms, but how do they do work?
Thanks Guys.
Post by: RuiAce on July 30, 2016, 07:39:03 pm
Ideas to Implementation Question:Very dumbed down, short and unorthodox answer
For the fluorescence screen investigation carried out to deterimine the nature of cathode rays, why is it that this experiment can be considered as giving evidence for both the particle and wave nature of cathode rays. Like i understand that the CR's transferred their energy to the fluorescent atoms, but how do they do work?
Thanks Guys.
It could be that a transfer of energy is taking place (particle), or it could be that the cathode rays were what's glowing themselves (wave)
Post by: jamonwindeyer on July 30, 2016, 08:57:06 pm
Could someone please help me with these questions.
How would you explain how a transformer works using the conservation of law of energy?
Also, in the syllabus is asks to "compare the structure and function of a generator to an electric motor," when they say electric motor, do they ask for AC and DC motors?
thank you :)
Hey Loki!
Basically with a transformer, you are never going to get an out more energy than you put in (conservation of energy). For a real transformer, the output power will always be ever so slightly less than the input, due to power losses in the transformer. The input power is equal to the output power, plus any losses in the transformer (heating, flux leakage, etc.).
The idea with this formula is that in a step up transformer, where the voltage is increased, that voltage must come from somewhere. There has to be a trade off, even if the losses aren't there. The trade off is that current decreases. The same works in reverse; if we step down the voltage, our current increases!
Summarising, the power into a transformer is close to equal to the power out of a transformer. So, if voltage goes up, current goes down. If voltage goes down, current comes up ;D
And yep, you need to know both AC and DC Motors/Generators for that dot point (excluding three phase induction motors) ;D
Post by: conic curve on July 31, 2016, 08:45:37 am
Explain, in terms of the principle of physics involved, why it is potentially dangerous to leave loose objects on the back shelf of a car
I know it has to do with Newton's law where that if a car comes to a stop, the object starts flying (I think it's the first law) but I'm not sure how to structure this in an extended response format
Thanks
Post by: jamonwindeyer on July 31, 2016, 02:31:00 pm
How do I answer this question
Explain, in terms of the principle of physics involved, why it is potentially dangerous to leave loose objects on the back shelf of a car
I know it has to do with Newton's law where that if a car comes to a stop, the object starts flying (I think it's the first law) but I'm not sure how to structure this in an extended response format
Thanks
Hey conic! Yep, that's exactly it, Newton's 1st Law of Motion (Inertia) ;D in terms of structuring an answer around it, you shouldn't need an extended response, just explain what happens, why it happens (inertia), and how it is potentially dangerous (sharp objects, blunt trauma, etc) ;D physics responses are fairly free-form! Don't think about it too much, just put down what you know :)
Post by: Neutron on July 31, 2016, 03:38:07 pm
For this 6 marker, I was wondering what key points we need to include? Thank you!!
Explain why Uranium ore is enriched for use in nuclear weapons, with reference to critical mass and neutron capture
This is a q2q question but maybe you guys would know a thing or two? :O
Thank you again!
Neutron
Post by: jamonwindeyer on July 31, 2016, 03:41:09 pm
Oh my god sorry Jamon i was thinking of something completely different!! Cause my friend drew the diagram of a transformer with the eddy currents forming the same way as the magnetic flux was travelling so I was like ??? Anyhow, all good now!
For this 6 marker, I was wondering what key points we need to include? Thank you!!
Explain why Uranium ore is enriched for use in nuclear weapons, with reference to critical mass and neutron capture
This is a q2q question but maybe you guys would know a thing or two? :O
Thank you again!
Neutron
All good, no dramas at all no reason to apologise! ;D I'll throw to Jake and/or diiiii for that question ;)
Post by: Neutron on July 31, 2016, 04:09:20 pm
All good, no dramas at all no reason to apologise! ;D I'll throw to Jake and/or diiiii for that question ;)
Okay :D In the meantime, do you think you could help me with this Space calculation question? :O
A special vehicle bound for Mars has a weight of 2.45x10^3 N at rest on the ground on Earth. It is loaded aboard a space craft and launched. The initial acceleration of the craft as it lifts fro the ground is 28.4 ms^-2 . What is the force applied to the vehicle by its mountings at the time?
The answer is 9.55 x 10^3 N and I've tried all the methods I know and could not get it :O
Sorry I have a lot of questions, my trials in like a few days D: pls shoot me
Neutron
Post by: RuiAce on July 31, 2016, 04:26:52 pm
Okay :D In the meantime, do you think you could help me with this Space calculation question? :O
A special vehicle bound for Mars has a weight of 2.45x10^3 N at rest on the ground on Earth. It is loaded aboard a space craft and launched. The initial acceleration of the craft as it lifts fro the ground is 28.4 ms^-2 . What is the force applied to the vehicle by its mountings at the time?
The answer is 9.55 x 10^3 N and I've tried all the methods I know and could not get it :O
Sorry I have a lot of questions, my trials in like a few days D: pls shoot me
Neutron
Post by: conic curve on July 31, 2016, 04:42:17 pm
Hey conic! Yep, that's exactly it, Newton's 1st Law of Motion (Inertia) ;D in terms of structuring an answer around it, you shouldn't need an extended response, just explain what happens, why it happens (inertia), and how it is potentially dangerous (sharp objects, blunt trauma, etc) ;D physics responses are fairly free-form! Don't think about it too much, just put down what you know :)
Thanks Jamon
Another question I find difficulty in answering is this: A man wishes to drill some holes in the paneling of his car, but for the sake of safety, the holes must be well away from any electrical wiring. Describe the procedure he could use to detect the whereabouts of electrical wiring in the cavity, using only a small compass
Like is this even a physics question? lol
Post by: jamonwindeyer on July 31, 2016, 07:05:13 pm
Thanks Jamon
Another question I find difficulty in answering is this: A man wishes to drill some holes in the paneling of his car, but for the sake of safety, the holes must be well away from any electrical wiring. Describe the procedure he could use to detect the whereabouts of electrical wiring in the cavity, using only a small compass
Like is this even a physics question? lol
It definitely is! I'll leave answering the question to you, but the principle lies within the fact that currents have magnetic fields around them. This is caused by the moving electric charges. To explain why it happens is some really complex Physics arising out of special relativity and electrodynamics, so not necessary to go into any more detail than "because it does."
The magnetic field lines form circles around a current, the direction being given by the right hand rule. Give yourself a thumbs up with your right hand. If your thumb is the direction of the current, the magnetic field has a direction indicated by your fingers, the field lines wrap around the current. This isn't strictly necessary for the question, though it might come in handy.
So the electrical wires produce magnetic fields! The compass will always align itself with any magnetic field! Therefore, I'll leave you to come up with a method that uses the compass to find out where the wires are :D
Post by: conic curve on July 31, 2016, 07:16:50 pm
It definitely is! I'll leave answering the question to you, but the principle lies within the fact that currents have magnetic fields around them. This is caused by the moving electric charges. To explain why it happens is some really complex Physics arising out of special relativity and electrodynamics, so not necessary to go into any more detail than "because it does."
The magnetic field lines form circles around a current, the direction being given by the right hand rule. Give yourself a thumbs up with your right hand. If your thumb is the direction of the current, the magnetic field has a direction indicated by your fingers, the field lines wrap around the current. This isn't strictly necessary for the question, though it might come in handy.
So the electrical wires produce magnetic fields! The compass will always align itself with any magnetic field! Therefore, I'll leave you to come up with a method that uses the compass to find out where the wires are :D
By the way this is from the module electrical energy in the home
Do we have to talk about safety though and ways to reduce the chances of getting electrocuted?
Also do we have to talk about (in this question) what happens when we get electrocuted?
Post by: jamonwindeyer on July 31, 2016, 07:33:59 pm
By the way this is from the module electrical energy in the home
Do we have to talk about safety though and ways to reduce the chances of getting electrocuted?
Also do we have to talk about (in this question) what happens when we get electrocuted?
Nope, the question only wants a description of the procedure itself, nothing else is necessary here ;D
Post by: conic curve on August 01, 2016, 08:37:09 am
A stone is dropped from a 220m cliff. Calculate the:
a)time taken to reach the bottom (2 marks)
b) Velocity just before impact (2 marks)
I know that we have 220m as part of our information and we need to let one side be positive but what other information am I missing and how would I execute the question with all the information I have?
Thanks
Post by: jamonwindeyer on August 01, 2016, 11:10:16 am
I don't understand this question, mainly because I'm not given much information:
A stone is dropped from a 220m cliff. Calculate the:
a)time taken to reach the bottom (2 marks)
b) Velocity just before impact (2 marks)
I know that we have 220m as part of our information and we need to let one side be positive but what other information am I missing and how would I execute the question with all the information I have?
Thanks
Howdy, so this is a pretty standard projectile question for Year 12, we'll use our kinematic formulae ;D
For the first one, we use the vertical displacement formula (this formula can be applied on any axis for which there is a constant acceleration). If we consider the initial height as zero, we want to find how long it takes the stone to reach -220 metres (remember acceleration is -9.8, and the initial vertical velocity is zero at the moment it is dropped)
For the second one, we have a time of flight until collision with the ground. To find the speed at this point, we have a simple formula to help:
That is, 67.13 metres per second towards the ground. Make sure you give a direction as is required for a velocity, rather than a speed :D
Post by: wyzard on August 01, 2016, 08:08:13 pm
I don't understand this question, mainly because I'm not given much information:
A stone is dropped from a 220m cliff. Calculate the:
a)time taken to reach the bottom (2 marks)
b) Velocity just before impact (2 marks)
I know that we have 220m as part of our information and we need to let one side be positive but what other information am I missing and how would I execute the question with all the information I have?
Thanks
A word on the signs of acceleration and velocity. Their signs are determined by where you point the positive direction to be, if you decide upwards is the positive direction then acceleration due to gravity will be downwards. Likewise if you decide the positve direction to be downwards, then acceleration due to gravity will be downwards.
This assigning of positive direction is very often overlooked by students, hence the common confusion of when and where the acceleration due to gravity is positive or negative.
The more technical term for doing so is assigning a coordinate system, something they really drill into uni first year physics students doing mechanics :P
Post by: conic curve on August 01, 2016, 08:27:42 pm
A word on the signs of acceleration and velocity. Their signs are determined by where you point the positive direction to be, if you decide upwards is the positive direction then acceleration due to gravity will be downwards. Likewise if you decide the positve direction to be downwards, then acceleration due to gravity will be downwards.
This assigning of positive direction is very often overlooked by students, hence the common confusion of when and where the acceleration due to gravity is positive or negative.
The more technical term for doing so is assigning a coordinate system, something they really drill into uni first year physics students doing mechanics :P
Um...it said calculate it...
Need help with the following
1. A 200kg space probe fires its rocket motor while in deep space, exerting a force always perpendicular to its velocity. This causes it to accelerate at 2.6 ms^-2. The firing of the motor lasts 8.0 seconds
a)Determine the force exerted by the space probe's rocket motor (2 marks)
Is this asking for net force or reaction force or what?
b) Calculate the increase in the space probe's momentum (2 marks)
(the formula isn't clicking into me right now)
c) Determine the acceleration of the vehicle during the first 8.0s of it's motion (1 mark)
Do we use F=ma for this?
24. A street outside a school has its speed limit reduced from 70 kmh^-1 to 40 kmh^-1. Describe, quantitatively and qualitatively, how the stopping distance of a vehicle would differ between these 2 speeds (3 marks)
This question requires us to describe the situation in terms of value and by visual observation but how am I supposed to do that?
Post by: jamonwindeyer on August 01, 2016, 08:40:05 pm
Um...it said calculate it...
wyzard is explaining conceptually the way the positives/negatives are assigned, it's a very important thing to understand :D
1. A 200kg space probe fires its rocket motor while in deep space, exerting a force always perpendicular to its velocity. This causes it to accelerate at 2.6 ms^-2. The firing of the motor lasts 8.0 seconds
a)Determine the force exerted by the space probe's rocket motor (2 marks)
Is this asking for net force or reaction force or what?
This (if I interpret the question correctly) is simply asking for an application of Newton's 2nd Law, F=ma. You have the mass of the probe and its acceleration due to the motor, you can use that to find the force!
b) Calculate the increase in the space probe's momentum (2 marks)
(the formula isn't clicking into me right now)
The formula you want here is that the change in momentum (also called the impulse) is linked to the force applied, and how long it is applied for, in the following way:
You have all the values you need here since you found force in the previous question.
c) Determine the acceleration of the vehicle during the first 8.0s of it's motion (1 mark)
Isn't that in the question?I'd love for someone else to have a look at this question, I think I've misinterpreted it somewhere :P
24. A street outside a school has its speed limit reduced from 70 kmh^-1 to 40 kmh^-1. Describe, quantitatively and qualitatively, how the stopping distance of a vehicle would differ between these 2 speeds (3 marks)
What you are looking for here is the idea of human reaction times. Say I take 1 second to respond to a hazard on the road. That means that my vehicle travels 1 second before I ever apply brake force. Then, when I do apply brake force, decelerating to a stop takes time as well, potentially another whole second. 2 seconds at 70km/h, for example, therefore equates to roughly another 20 metres of distance.
Qualitatively, you'd simply be describing how reducing speed reduces the distance taken to stop, since the distance travelled while the driver reacts is less, and the brake force will stop the vehicle faster. Quantitatively, you'd be doing rough calculations with the formula:
To find the distance travelled at some speed given some reaction time ;D for 3 marks, you'd want a couple of dot points of qualitative description plus the use of a formula (such as above) to estimate the difference in stopping time. It doesn't have to be exact ;D
Post by: wyzard on August 01, 2016, 09:57:56 pm
Um...it said calculate it...
Assigning the sign is part of the calculation, which you'll be using for the equations kinematic later. jamonwindeyer covered the calculation bit really well, was explaining why sometimes acceleration is positive, and why sometimes is negative ;D
Post by: Neutron on August 02, 2016, 04:04:16 pm
A projectile of mass 1150 kg is fired vertically from the surface of an asteroid of mass 1.1x10^20 kg and radius 2.6x10^6m with a speed of 50 ms^-1. Determine the maximum distance the projectile reaches from the centre of the asteroid.
Thank you guys <3
Post by: jakesilove on August 02, 2016, 05:25:40 pm
Heyy another calculations question, for some reason I can never get these :(
A projectile of mass 1150 kg is fired vertically from the surface of an asteroid of mass 1.1x10^20 kg and radius 2.6x10^6m with a speed of 50 ms^-1. Determine the maximum distance the projectile reaches from the centre of the asteroid.
Thank you guys <3
Hey! So, first we have to calculate the value of acceleration due to gravity on this asteroid. The relevant formula is
So, subbing our information in, you get
This seems pretty low; maybe I did the calculation wrong on that? You'll have to check me on that.
Now, we can use the formula
Where the initial velocity is 50m/s^2, the final velocity is zero (as it has to stop), and the acceleration is as described above.
Therefore it will reach a total of 2.6*10^6+1151000m=3751012m
Again, I think I must have calculated acceleration wrong, definitely check me on that one
Post by: Neutron on August 02, 2016, 06:00:55 pm
The solutions found the initial kinetic and gravitational potential energy of the projectile, and then they let that value be equal to -6.67x10^-11 x 1150 x1.1 x 10^20 all over r (think this is like the gravitational potential energy formula but I have no idea why)
Post by: Neutron on August 02, 2016, 06:51:20 pm
Post by: JennHen on August 02, 2016, 08:08:19 pm
Post by: RuiAce on August 02, 2016, 08:19:55 pm
And also, could someone please explain how a coloured TV works? Like is it basically the same as a cathode ray tube only with three electron guns? :O Sorry for the bombardment of questions, have trials in two days ;;From memory the answer to this question is yes. One electron gun for each of the three digital colours. Each gun makes one of the three metals in one pixel glow blue/red/green.
Hey! Just a question about back EMF - i'm a little confused as to how it is caused, what it actually is, and how it affects the motor. Would be awesome if someone could give an explanation. Cheers!So a bit of background information:
We know that by Faraday's law, if an electrical conductor is subject to a change in magnetic flux, some EMF is induced.
We also know that by Lenz's law, the induced EMF always seeks to oppose the cause of induction.
But here's the thing. As a consequence of Faraday's law, it doesn't matter whether we move the conductor or the magnetic field. If the magnetic field is moved then obviously an EMF will be induced. However, if a conductor is moved through a magnetic field, that also triggers electromagnetic induction.
So keep in mind that for your motor, the armature (coils) are basically conductors moving through a magnetic field! This is the EMF induced into the motor.
Applying Lenz's law, the cause of induction is the current forcing the coils to rotate in the magnetic field. The reason why it's "back" EMF is that well, by Lenz's law it's going to try to make the coils go back the other way.
What does this do? There will be a point where this EMF induced cannot oppose the actual voltage of the circuit anymore. So this effectively puts a limiting (maximum) velocity on what a motor can achieve.
Post by: imtrying on August 02, 2016, 09:03:05 pm
Can't seem to make sense of my notes :-\
Post by: jamonwindeyer on August 02, 2016, 09:10:23 pm
:O (just a bunch of maths errors hahaha)
The solutions found the initial kinetic and gravitational potential energy of the projectile, and then they let that value be equal to -6.67x10^-11 x 1150 x1.1 x 10^20 all over r (think this is like the gravitational potential energy formula but I have no idea why)
Oh that's an interesting method!! Jake's is what I would do, but let me explain this as well. Basically, it is using the conservation of energy. The energy of the projectile must remain conserved (it acts as a closed system in this example, we don't add any energy after launch).
By calculating the initial kinetic and potential energies and summing them, we are calculating the initial total energy of the projectile. This cannot change. How it is distributed, however, can change.
Consider a projectile launched vertically. It will go up, slow down, and at the peak of its motion it will stop. At THIS precise moment, at the highest point, it has no kinetic energy. It has stopped and is about to turn around to plummet back to ground. At this point, we can equate the initial total energy to the gravitational potential energy of the projectile, since this is the only form of energy it has at that point. So, the total initial energy is equated to GPE with the radius unknown: That is what we find to get our answer ;D
I hope that helps!
Post by: jamonwindeyer on August 02, 2016, 09:16:54 pm
And also, could someone please explain how a coloured TV works? Like is it basically the same as a cathode ray tube only with three electron guns? :O Sorry for the bombardment of questions, have trials in two days ;;
Just confirming Rui's answer above! Three separate electron guns, three separate electron beams, and the screen of a colour TV has three different phosphors arranged in groups of three, one for each colour. Each electron beam is focused on illuminating only one colour phosphor ;D
Good luck for your Trials! Be sure to shoot any more questions our way! ;D
Post by: conic curve on August 02, 2016, 09:17:49 pm
Question is
a) If marble B rebounds to the right with a velocity of 3ms^-1, what happens to marble A? (2 marks)
b) Is this an elastic or inelastic collision? Using calculations, justify your answer (2 marks)
Post by: jamonwindeyer on August 02, 2016, 09:20:41 pm
Hey! Just a question about back EMF - i'm a little confused as to how it is caused, what it actually is, and how it affects the motor. Would be awesome if someone could give an explanation. Cheers!
Welcome to the forums JennHen! Hope Rui's answer helped, be sure to let me know if you need any help finding anything ;D
Post by: jakesilove on August 02, 2016, 09:25:14 pm
Hey:) Just doing some summaries for Medical Physics and a little stuck on "Explain that the amplitude of the signal given out when precessing nuclei relax is related to the number of nuclei present"
Can't seem to make sense of my notes :-\
Hey! So basically, when a precessing nucleus is 'given' more energy, it will flip (ie. move to a anti-parallel state). When the energy source stops (the signal stops being sent into the body), the nuclei will want to flip back to a parallel position. This means it has to give off the extra energy it had been given. The more nuclei in a region, the more energy that will be given off. Therefore, the amplitude/amount of signal/energy given out when the nuclei relax is related to the number of nuclei present! Does that make sense?
Jake
Post by: Neutron on August 02, 2016, 09:38:54 pm
Just confirming Rui's answer above! Three separate electron guns, three separate electron beams, and the screen of a colour TV has three different phosphors arranged in groups of three, one for each colour. Each electron beam is focused on illuminating only one colour phosphor ;D
Good luck for your Trials! Be sure to shoot any more questions our way! ;D
Thank you so much Jamon you're too nice D:
This is probably really stupid but can someone please show me what the eddy currents look like in that experiment with the magnet rotating around a non-magnetic metal disk? It's the one with that stimulates an AC induction motor.. Like do two eddy currents form like in electromagnetic braking or only one..? Idk why I keep getting confused with old concepts now, maybe I never understood it properly :/
Post by: jakesilove on August 02, 2016, 09:41:19 pm
Thank you so much Jamon you're too nice D:
This is probably really stupid but can someone please show me what the eddy currents look like in that experiment with the magnet rotating around a non-magnetic metal disk? It's the one with that stimulates an AC induction motor.. Like do two eddy currents form like in electromagnetic braking or only one..? Idk why I keep getting confused with old concepts now, maybe I never understood it properly :/
As far as I remember, the eddy currents formed will just be concentric circles that "chase" the movement of the magnet (or move in the opposite direction, depending on which pole is being held towards the non-magnetic plate!).
Post by: jamonwindeyer on August 02, 2016, 09:45:37 pm
Me and physics maths don't work together :'(
Question is
a) If marble B rebounds to the right with a velocity of 3ms^-1, what happens to marble A? (2 marks)
b) Is this an elastic or inelastic collision? Using calculations, justify your answer (2 marks)
Hey! Okay, so for the first one, we first need to calculate the total initial momentum of the system. Remember that the marbles are moving in opposite directions, so one will have positive momentum, one will have negative. I'll take A's momentum/direction as positive:
So after the collision, this momentum MUST be the same as before the collision. That's the conservation of momentum. So, we know the momentum of B after the collision:
So the momentum of A must be equal to the difference:
So Marble A continues to the right with a reduced velocity of 6 metres per second ;D
For the next question, we need to understand that collisions are elastic if kinetic energy is conserved. The principal here is very similar, we first calculate the kinetic energy initially:
I'll leave you to finish this, calculate the kinetic energy after the collision with the values above!! If it matches, your collision is elastic, if not, you've lost energy and so it is inelastic ;D
Post by: imtrying on August 02, 2016, 09:48:54 pm
Hey! So basically, when a precessing nucleus is 'given' more energy, it will flip (ie. move to a anti-parallel state). When the energy source stops (the signal stops being sent into the body), the nuclei will want to flip back to a parallel position. This means it has to give off the extra energy it had been given. The more nuclei in a region, the more energy that will be given off. Therefore, the amplitude/amount of signal/energy given out when the nuclei relax is related to the number of nuclei present! Does that make sense?
Jake
Thanks so much, that does make sense. One dot point down, only 80 more to go haha :)
Post by: jakesilove on August 02, 2016, 09:51:39 pm
Thanks so much, that does make sense. One dot point down, only 80 more to go haha :)
Feel free to post any other questions here! Been a while since I've been able to go over Medical Physics, but I remember it being a really interesting option
Post by: jamonwindeyer on August 02, 2016, 09:52:08 pm
Thanks so much, that does make sense. One dot point down, only 80 more to go haha :)
Let us know if you need any more help with MRI stuff! Probably the most complex thing I learned in Physics! ;D
Post by: Aliceyyy98 on August 03, 2016, 10:43:55 am
Could someone plz explain how cathode rays work as in like the electron guns and deflection plates etc? The dot pt says in tv and oscilloscopes so do they work differently in different things??
Cheers
Post by: jamonwindeyer on August 03, 2016, 11:11:07 am
Hi guys!
Could someone plz explain how cathode rays work as in like the electron guns and deflection plates etc? The dot pt says in tv and oscilloscopes so do they work differently in different things??
Cheers
Hey Alice!
So they are applied ever so slightly differently. Are you okay with the Physics behind how the deflection plates actually work? Let me know if you aren't, but assuming you are, it's all about what external signal the horizontal and vertical deflection plates are aligned to.
In Oscilloscopes, the horizontal deflection is controlled by a time varying signal. This causes the beam to sweep horizontally across the screen, from left to right, at a constant pace. The vertical deflection is controlled by some other signal, a measurement (for example, a microphone). What this creates is a vertical axis dependent on signal strength of some stimuli, and a horizontal axis dependent on time. Think of your typical heartbeat monitor in the hospital drama shows, that is exactly what I'm talking about ;)
In Cathode Ray Televisions, both the horizontal and vertical deflection plates have time varying strengths. The idea here, put simply, is that it makes the electron beam sweep across the entire television screen. It sweeps along the top row of phosphor dots (kind of like pixels on your laptop), then the second row, then the third row, so quick that you just see the solid image.
There are 3 electron guns in Colour CRT's, each corresponding to either Red, Green, or Blue. Each is tasked with creating a specific colour on the screen (there are different phosphors, one for each colour). The intensity of each is controlled with a separate accelerating anode :D
This explanation is definitely lacking a bit in certain areas, there is some pretty complex stuff involved, but this is a good overview. Is there any specific part of this that is a bit iffy for you?
(http://s.hswstatic.com/gif/tv-cathode.gif)
Post by: Neutron on August 03, 2016, 02:08:24 pm
Hey Alice!
So they are applied ever so slightly differently. Are you okay with the Physics behind how the deflection plates actually work? Let me know if you aren't, but assuming you are, it's all about what external signal the horizontal and vertical deflection plates are aligned to.
In Oscilloscopes, the horizontal deflection is controlled by a time varying signal. This causes the beam to sweep horizontally across the screen, from left to right, at a constant pace. The vertical deflection is controlled by some other signal, a measurement (for example, a microphone). What this creates is a vertical axis dependent on signal strength of some stimuli, and a horizontal axis dependent on time. Think of your typical heartbeat monitor in the hospital drama shows, that is exactly what I'm talking about ;)
In Cathode Ray Televisions, both the horizontal and vertical deflection plates have time varying strengths. The idea here, put simply, is that it makes the electron beam sweep across the entire television screen. It sweeps along the top row of phosphor dots (kind of like pixels on your laptop), then the second row, then the third row, so quick that you just see the solid image.
There are 3 electron guns in Colour CRT's, each corresponding to either Red, Green, or Blue. Each is tasked with creating a specific colour on the screen (there are different phosphors, one for each colour). The intensity of each is controlled with a separate accelerating anode :D
This explanation is definitely lacking a bit in certain areas, there is some pretty complex stuff involved, but this is a good overview. Is there any specific part of this that is a bit iffy for you?
(http://s.hswstatic.com/gif/tv-cathode.gif)
On top of Jamon's flawless answer, it might also be worth mentioning that Cathode Ray Oscilloscopes (CRO) use electric fields as their deflection plates and TVs use magnetic fields. CROs use electric fields as they can switch at higher frequencies than magnetic fields (so it can basically respond to the stimuli faster whereas magnetic fields take a bit longer) and TVs use magnetic fields because when they deflect the electron beam, they deflect it in a circular path (the force is perpendicular to the direction of motion of the electron beam (from right hand palm rule), which means it provides a centripetal force and therefore, circular motion) whereas electric fields deflect in a parabolic path. So then the radius of curvature is greater and therefore, the TV can have a bigger screen (CROs don't need a massive screen) !! But yeah, the electron gun functions the same way for both (i.e. both have heating filament at cathode, which 'boils' off electrons through thermionic emission, which then goes through control electrode, focussing anode and accelerating anode before reaching the deflection plates) and its just the different fields used by the deflection plates :D
If you wanted a few more advantages and disadvantages between electric plates and electromagnetic coils (magnetic fields), here they are:
Electric plates
Advantages
-Can respond at higher frequencies (mentioned above)
-Less power required for deflection
Disadvantages
-Needs to be inside the tube in order for the field to be strong enough (this complicates the tube design)
-Deflects in parabolic path (lower radius of curvature)
-Larger defocussing during deflection (longer tube is required)
Electromagnetic coils
Advantages
-Smaller defocussing of the electron beam when deflecting
-Deflects in a circular path, therefore greater radius of curvature (reduces length of cathode ray tube needed for a bigger screen size)
-Can be positioned outside the tube (simplifies tube design)
Disadvantages
-Requires more power to operate
-Harder to get uniform magnetic field
I was just studying this yesterday so feel free to correct me if I stuffed up somewhere!!
Post by: jakesilove on August 03, 2016, 02:12:08 pm
On top of Jamon's flawless answer, it might also be worth mentioning that Cathode Ray Oscilloscopes (CRO) use electric fields as their deflection plates and TVs use magnetic fields. CROs use electric fields as they can switch at higher frequencies than magnetic fields (so it can basically respond to the stimuli faster whereas magnetic fields take a bit longer) and TVs use magnetic fields because when they deflect the electron beam, they deflect it in a circular path (the force is perpendicular to the direction of motion of the electron beam (from right hand palm rule), which means it provides a centripetal force and therefore, circular motion) whereas electric fields deflect in a parabolic path. So then the radius of curvature is greater and therefore, the TV can have a bigger screen (CROs don't need a massive screen) !! But yeah, the electron gun functions the same way for both (i.e. both have heating filament at cathode, which 'boils' off electrons through thermionic emission, which then goes through control electrode, focussing anode and accelerating anode before reaching the deflection plates) and its just the different fields used by the deflection plates :D
If you wanted a few more advantages and disadvantages between electric plates and electromagnetic coils (magnetic fields), here they are:
Electric plates
Advantages
-Can respond at higher frequencies (mentioned above)
-Less power required for deflection
Disadvantages
-Needs to be inside the tube in order for the field to be strong enough (this complicates the tube design)
-Deflects in parabolic path (lower radius of curvature)
-Larger defocussing during deflection (longer tube is required)
Electromagnetic coils
Advantages
-Smaller defocussing of the electron beam when deflecting
-Deflects in a circular path, therefore greater radius of curvature (reduces length of cathode ray tube needed for a bigger screen size)
-Can be positioned outside the tube (simplifies tube design)
Disadvantages
-Requires more power to operate
-Harder to get uniform magnetic field
I was just studying this yesterday so feel free to correct me if I stuffed up somewhere!!
Neutron mate, you're a star
Post by: jakesilove on August 03, 2016, 02:18:36 pm
Neutron mate, you're a star
See what I did there
Post by: jakesilove on August 03, 2016, 02:19:03 pm
See what I did there
(https://upload.wikimedia.org/wikipedia/commons/e/e1/Neutron_star_illustrated.jpg)
Post by: jamonwindeyer on August 03, 2016, 02:22:43 pm
On top of Jamon's flawless answer, it might also be worth mentioning that Cathode Ray Oscilloscopes (CRO) use electric fields as their deflection plates and TVs use magnetic fields...
Ahh I forgot about this, VERY good point, and probably one of the main things to mention, aha thanks Neutron! ;D
Post by: Loki98 on August 03, 2016, 04:08:16 pm
In the question attached, for part ii) Could someone plz explain why the LED would not light up. Wouldn't there still be a change in flux causing an emf to be produced which should light up the LED but slightly weaker than before since the rate of change in flux is slower than before?
Post by: Spencerr on August 03, 2016, 04:23:13 pm
Hey guys,
In the question attached, for part ii) Could someone plz explain why the LED would not light up. Wouldn't there still be a change in flux causing an emf to be produced which should light up the LED but slightly weaker than before since the rate of change in flux is slower than before?
Hey, my interpretation of it is that the magnitude of the emf induced via Faraday's law is proportional to the rate of change in flux. The question mentions that the change in shape is "much more slowly", I would assume that the emf induced would be too small to cause the light bulb to light up (V = IR), if V is too small, then there will not be enough current to create the light. (Any change in flux will always cause emf to be produced)
Post by: Loki98 on August 03, 2016, 05:33:34 pm
Hey, my interpretation of it is that the magnitude of the emf induced via Faraday's law is proportional to the rate of change in flux. The question mentions that the change in shape is "much more slowly", I would assume that the emf induced would be too small to cause the light bulb to light up (V = IR), if V is too small, then there will not be enough current to create the light. (Any change in flux will always cause emf to be produced)
So the only way it can be justified is by saying the current would be insufficient to turn the LED on?
Post by: jakesilove on August 03, 2016, 05:57:56 pm
So the only way it can be justified is by saying the current would be insufficient to turn the LED on?
I think these two answer cover it pretty well! You're absolutely right, an EMF is produced, however it is much lower. This will be insufficient to turn the LED on. Since the question SAYS that it doesn't turn on, that's really all you can say! Great answers :)
Post by: MysteryMarker on August 03, 2016, 06:16:55 pm
How does electron density affect the resistance of a conductor? Like would an increase in electrons result in more collisions, or allow for the movement of more charge?
Cheers guys.
Post by: conic curve on August 03, 2016, 07:12:55 pm
Does the water affect this at all?
Post by: jamonwindeyer on August 03, 2016, 09:20:44 pm
I think these two answer cover it pretty well! You're absolutely right, an EMF is produced, however it is much lower. This will be insufficient to turn the LED on. Since the question SAYS that it doesn't turn on, that's really all you can say! Great answers :)
So the only way it can be justified is by saying the current would be insufficient to turn the LED on?
That's definitely the answer!! The rate of change of magnetic flux density (magnetic field strength) is much less in the second scenario, and thus, the induced EMF will not be sufficient to light the LED. To give an electrical engineers perspective, an LED requires a minimum forward operating current to turn on, anything below that can't induce electroluminescence in the junction (an LED is, speaking very roughly, essentially the reverse of a simple solar cell, converting electrical energy into visible light ;D)
Post by: jamonwindeyer on August 03, 2016, 09:41:13 pm
Hey guys,
How does electron density affect the resistance of a conductor? Like would an increase in electrons result in more collisions, or allow for the movement of more charge?
Cheers guys.
Hey there!! The mathematics behind proving this is a little bit intense (even I don't quite get it tbh) but we can express the electrical resistance of something in the following way:
The 'n' in the denominator is your free electron density. The free electron density being larger increases your conductivity, and thus decreases resistance. More electrons equals more current flow ;D
Post by: jamonwindeyer on August 03, 2016, 10:06:58 pm
What is the action and reaction force(s) when it comes to running, swimming and rowing a boat?
Does the water affect this at all?
It's all pretty much the same I think! You apply a force on the ground/water, the reaction force propels you forward! ;D the difference being I suppose that for the water, it is the buoyancy of the water rather than the normal reaction force at play :)
Post by: MysteryMarker on August 03, 2016, 10:27:37 pm
Could someone explain this question to me and how the answer is B? Wouldn't it be A as I = V/R and therefore the current is directly proportional to an increase in voltage?
Cheers.
Post by: jamonwindeyer on August 03, 2016, 10:43:20 pm
Hey guys,
Could someone explain this question to me and how the answer is B? Wouldn't it be A as I = V/R and therefore the current is directly proportional to an increase in voltage?
Cheers.
Hey there! I'm very happy you asked this question because it is an important principle to remember: Cathode/anode setups do not obey Ohm's Law! The relationship is more complicated, Ohm's Law only applies in certain situations and this isn't one of them ;D
That being said, we can approach this question by process of elimination.
Graph D is probably the most incorrect, since it implies that the current is infinite with no voltage applied. That is wrong on many levels, so no go.
Graph A is linear, which we would not expect in this situation.
We can explain that the answer is B pretty simply even without knowing that though. If there were no voltage source (that is, V=0), we would still expect some current flow. Why? Because of the photoelectric effect. The photoelectric effect causes a small photocurrent to flow through the circuit, completely independent of the applied voltage. The only graph that has a current flowing when V=0 is B, so, the answer must be B by default ;D
Post by: Aliceyyy98 on August 04, 2016, 12:16:10 am
Hey Alice!
So they are applied ever so slightly differently. Are you okay with the Physics behind how the deflection plates actually work? Let me know if you aren't, but assuming you are, it's all about what external signal the horizontal and vertical deflection plates are aligned to.
In Oscilloscopes, the horizontal deflection is controlled by a time varying signal. This causes the beam to sweep horizontally across the screen, from left to right, at a constant pace. The vertical deflection is controlled by some other signal, a measurement (for example, a microphone). What this creates is a vertical axis dependent on signal strength of some stimuli, and a horizontal axis dependent on time. Think of your typical heartbeat monitor in the hospital drama shows, that is exactly what I'm talking about ;)
In Cathode Ray Televisions, both the horizontal and vertical deflection plates have time varying strengths. The idea here, put simply, is that it makes the electron beam sweep across the entire television screen. It sweeps along the top row of phosphor dots (kind of like pixels on your laptop), then the second row, then the third row, so quick that you just see the solid image.
There are 3 electron guns in Colour CRT's, each corresponding to either Red, Green, or Blue. Each is tasked with creating a specific colour on the screen (there are different phosphors, one for each colour). The intensity of each is controlled with a separate accelerating anode :D
This explanation is definitely lacking a bit in certain areas, there is some pretty complex stuff involved, but this is a good overview. Is there any specific part of this that is a bit iffy for you?
(http://s.hswstatic.com/gif/tv-cathode.gif)
Thank you Jamon! I'm not quite sure how the signals in deflection plates work actually... :'( could you clarify for me!
Cheers!
Post by: jamonwindeyer on August 04, 2016, 09:19:27 am
Thank you Jamon! I'm not quite sure how the signals in deflection plates work actually... :'( could you clarify for me!
Cheers!
Sure!
Okay, so let's just assume we are working with electric field deflection (you can apply the same idea to magnetic). The idea is that by controlling the electric field strength and direction, we can change the path of the electron beam. In this case, this would be done by applying different voltages to the deflection plates :)
Let's say we can apply a voltage between 100V and -100V to the plates (same max magnitude, just opposite polarity) to a pair of plates controlling vertical deflection. 0V would mean the electron beam is unaffected, and travels straight. If we apply 100V, the beam might (for example) be deflected such that it hits the very top of the screen. 99V means it hits JUST below that. 98V, just below that again. All the way to zero where we are back in the centre. Then, -1V, just below the centre. And the process continues, all the way to -100V where the beam hits the bottom of the screen.
If we do this for both horizontal and vertical deflection, we can hit ANY point on our two dimensional screen. The top right corner might be 100V applied to vertical plates, 100V applied to horizontal plates. A point in the bottom right would be -100V vertical, but still 100V horizontal. We adjust the voltages of each pair to move up/down, or left/right.
We can do creative things with this ideas (again, just assume electrical is used for everything to keep things numerically simple). For example, in an Oscilloscope, the horizontal plates have a time varying voltage. We start by applying -100V, and the beam is at the very left. We gradually increase this to 0, and then to 100, to cause the beam to sweep across the screen. Then, when we hit 100V, we immediately reset to -100V and start the sweep again (for prospective electrical engineers, this signal would be called a sawtooth wave).
So we have an electron beam sweeping across the middle of the screen (maybe once a second). If we then connect the vertical deflection plates to some source we want to measure (for example, an electrode connected to the chest to measure heartbeat), that means the electron beams vertical position is dictated by what we are measuring. Thus, we then have your typical cardiograph, with the heartbeat visible as it fluctuates over time! The beam sweeps across, and jumps up and down as it sweeps in correspondence with the heartbeat.
Television screens are more complex, but basically, both the vertical and horizontal deflection plates are time varying. The periods are set up such that the beam does a sweep across the top row of the screen, then the next row in the reverse direction, then the next row, etc etc. Remember Donkey Kong? How you'd go all the way to the right, then jump, then all the way to the left, then jump, etc? That's what happens here (roughly speaking). It's a little tougher to picture, but let's say we start in the top left corner:
Horizontal = 100V, Vertical = 100V
Then we want the 'phosphor dot' to its right:
Horizontal = 99V, Vertical = 100V
We keep sweeping to the right until we get to the top right corner:
Horizontal = -100V, Vertical = 100V
Then we shift down:
Horizontal = -100V, Vertical = 99V
Then we begin sweeping left:
Horizontal = -99V, Vertical = 99V
And the process repeats in this fashion ;D
Post by: Neutron on August 04, 2016, 10:01:06 am
Hey there! I'm very happy you asked this question because it is an important principle to remember: Cathode/anode setups do not obey Ohm's Law! The relationship is more complicated, Ohm's Law only applies in certain situations and this isn't one of them ;D
That being said, we can approach this question by process of elimination.
Graph D is probably the most incorrect, since it implies that the current is infinite with no voltage applied. That is wrong on many levels, so no go.
Graph A is linear, which we would not expect in this situation.
We can explain that the answer is B pretty simply even without knowing that though. If there were no voltage source (that is, V=0), we would still expect some current flow. Why? Because of the photoelectric effect. The photoelectric effect causes a small photocurrent to flow through the circuit, completely independent of the applied voltage. The only graph that has a current flowing when V=0 is B, so, the answer must be B by default ;D
Wait wait but just with this question, when the voltage is equal to the stopping voltage, shouldn't the current be zero since no photoelectrons actually reach the cathode? Why do none of the graphs account for this D:
Post by: Neutron on August 04, 2016, 10:13:19 am
Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)
Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)
Post by: Neutron on August 04, 2016, 10:20:58 am
Adios amigos
Rip me HAHAHA
Thank you, thank god my exams in the afternoon
7.The Earth’s radius at the equator is close to 6 380km, and its rotational period is equal to the rotational period of the Earth. If a new satellite launching station were established at Woomera in South Australia, which of the following would best describe the launch velocity advantage, given its intended orbit?
(A) 464 m s-1 Intended orbit from east → west
(B) more than 464 m s-1 Intended orbit from south → north
(C) 464 m s-1 Intended orbit from west → east
(D) less than 464 m s-1 Intended orbit from west → east
When you have a current carrying conductor that's travelling diagnoally to the right through a magnetic field coming out of a page, can it experience a force to the bottom right? Or only to the bottom? (won't elt me screenshot)
Ahh apparently the file size is too big, will just have to bathe in this ignorance hahah
Post by: jamonwindeyer on August 04, 2016, 10:38:56 am
Ahh it's the morning of the exam and I still have a crap tonne of questions (not a good sign, rip)
Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)
Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)
You'll be fine Neutron!! Just relax mate, you'll be sweet ;D
First question, leave it with me, nothing strikes me immediately.
Next, the answer is definitely D. Orbital decay causes the velocity of the satellite to decrease, thus causing its orbital radius to drop. However, the drop in orbital radius causes an increase in orbital velocity (smaller orbit, needs to move faster to stay in orbit), so it actually ends up travelling faster than it did before the drop. The net effect? Orbital decay causes satellites to gradually spiral inwards towards earth, travelling faster and faster as it goes.
Next, there is definitely changing flux in the answer. However, the changing flux on opposite sides of the loop cancel each other out. One side moves down, the other moves up,, the net effect is no change in magnetic field strength and thus no induced current. Every change in flux on one side is opposite in sign and equal in magnitude to the change in flux directly opposite. Everything cancels, no induced current. The answer is therefore B. Every other example, there is some change in magnetic field strength in some part of the loop. The diagram is pretty bad, the magnet is at the dead centre of the loop, which is why this occurs. If the magnet was somewhere closer to one side of the loop the outcome would be different.
Question 11, I agree with you, I would say the answer is C not D (any takers? Anything I missed?)
Question 7, you would normally need to calculate the velocity of the earth's surface at the equator. We can do this by considering the distance travelled (one circumference, the length of the equator) divided by the time taken (24 hours). However, the question doesn't really need it because it's the same number everywhere. If we launch from Woomera, the surface of the earth isn't moving as quickly as the equator.
To demonstrate this, pick up a ball of any kind. Put black dot on the very edge of the ball (where the equator would be), and then put one lower down. If you spin it, the black dot on the equator rotates faster. Why? Because it has a greater distance to travel in the same amount of time. The 'circumference' circumnavigated from Woomera is lower. Thus, the velocity is lower. We won't get as much benefit there. Therefore, the answer must be D.
Post by: jamonwindeyer on August 04, 2016, 10:42:44 am
Post by: jakesilove on August 04, 2016, 10:46:32 am
Ahh it's the morning of the exam and I still have a crap tonne of questions (not a good sign, rip)
Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)
Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)
Do I have a story about that first question: Gather round children.
I topped my Physics Trial by one mark, and it was because of this multiple choice question. How do you answer it? It's something about calculating the change in potential and kinetic energy, and subtracting terms. Just calculate the initial potential energy, the final kinetic+potential energy, and find that some energy is missing. By conservation of energy, that had to go somewhere (ie. heat and light).
However, Jake wasn't that smart in the exam. Instead, I noticed that D-A=C. By Multiple Choice theory, I assumed that D and A were tricks, something that could be calculated, but that the LOSS would be some sort of 'difference'. So, I put C, the correct answer. Round of applause.
What should you take from this? Be smart with you multiple choice questions. Even if an answer doesn't come to mind, you can logic it out almost every time. You'll be fine mate
Post by: jamonwindeyer on August 04, 2016, 10:48:11 am
Work is equal to the force applied over the distance. We need the gravitational force on the daredevil (which for this question, we can approximate with W=mg) applied over 10km, to figure out how much work is done accelerating the guy.
However, if we calculate the kinetic energy of the daredevil:
Edit: Indeed Jake! I think the calculation above is close, because the difference between the two is the same as C, off by a factor of 1000.
Post by: jakesilove on August 04, 2016, 10:53:39 am
Question 15, I have no idea how a fluorescent screen can be a particle? That makes no sense to me aha ;D not sure about this one either :P
For 15, I think the answer has to be A right? First of all, it is clear that the difference between X and Y is either than one is more charged than the other, or one is faster than the other. That already only really leaves us with A. Gamma particles have no charge, so Z makes sense, and Beta particles will experience a force in the opposite direction to the Alpha particles (X,Y) so that makes sense to.
Post by: jamonwindeyer on August 04, 2016, 10:57:22 am
Now we take the potential energy and kinetic energy after the 10km drop:
The difference between those two results gives the answer of C, that's better!! ;D
Ignore the above working, this is more correct and yields the correct answer, but if you did happen to do the above then that could also lead you to a correct guess! ;D
Post by: jamonwindeyer on August 04, 2016, 10:59:05 am
For 15, I think the answer has to be A right? First of all, it is clear that the difference between X and Y is either than one is more charged than the other, or one is faster than the other. That already only really leaves us with A. Gamma particles have no charge, so Z makes sense, and Beta particles will experience a force in the opposite direction to the Alpha particles (X,Y) so that makes sense to.
I definitely agree!! Not sure why the answer would read D, or, how the hell a fluorescent screen is a particle?
I'll add that these are tricky questions Neutron, I don't blame you for getting stuck! :o
Post by: Neutron on August 04, 2016, 11:29:33 am
You'll be fine Neutron!! Just relax mate, you'll be sweet ;D
First question, leave it with me, nothing strikes me immediately.
Next, the answer is definitely D. Orbital decay causes the velocity of the satellite to decrease, thus causing its orbital radius to drop. However, the drop in orbital radius causes an increase in orbital velocity (smaller orbit, needs to move faster to stay in orbit), so it actually ends up travelling faster than it did before the drop. The net effect? Orbital decay causes satellites to gradually spiral inwards towards earth, travelling faster and faster as it goes.
Next, there is definitely changing flux in the answer. However, the changing flux on opposite sides of the loop cancel each other out. One side moves down, the other moves up,, the net effect is no change in magnetic field strength and thus no induced current. Every change in flux on one side is opposite in sign and equal in magnitude to the change in flux directly opposite. Everything cancels, no induced current. The answer is therefore B. Every other example, there is some change in magnetic field strength in some part of the loop. The diagram is pretty bad, the magnet is at the dead centre of the loop, which is why this occurs. If the magnet was somewhere closer to one side of the loop the outcome would be different.
Question 11, I agree with you, I would say the answer is C not D (any takers? Anything I missed?)
Question 7, you would normally need to calculate the velocity of the earth's surface at the equator. We can do this by considering the distance travelled (one circumference, the length of the equator) divided by the time taken (24 hours). However, the question doesn't really need it because it's the same number everywhere. If we launch from Woomera, the surface of the earth isn't moving as quickly as the equator.
To demonstrate this, pick up a ball of any kind. Put black dot on the very edge of the ball (where the equator would be), and then put one lower down. If you spin it, the black dot on the equator rotates faster. Why? Because it has a greater distance to travel in the same amount of time. The 'circumference' circumnavigated from Woomera is lower. Thus, the velocity is lower. We won't get as much benefit there. Therefore, the answer must be D.
omg you guys are actual legends, but just with the satellite one, if the surface velocity is lower in Woomera, how is that an advantage? And since all satellites have to have their centre of orbit pass through the centre of the Earth (this is going to be really dumb) why does that satellite go west to east? Like I know normally satellites go that way because that's the way the Earth rotates and therefore rotational velocity helps blah blah but if it starts off in Australia, doesn't it have to go up and around the Earth? So like South to North in order to sustain its orbit?If that does happen, and it just has an orbital inclination, won't it still be travelling at the same speed as a geostationary orbit? Or are you allowed to just have a tiny orbit around the bottom of the Earth that's not passing through the centre?? Thank you so much
Post by: jamonwindeyer on August 04, 2016, 01:23:32 pm
omg you guys are actual legends, but just with the satellite one, if the surface velocity is lower in Woomera, how is that an advantage? And since all satellites have to have their centre of orbit pass through the centre of the Earth (this is going to be really dumb) why does that satellite go west to east? Like I know normally satellites go that way because that's the way the Earth rotates and therefore rotational velocity helps blah blah but if it starts off in Australia, doesn't it have to go up and around the Earth? So like South to North in order to sustain its orbit?If that does happen, and it just has an orbital inclination, won't it still be travelling at the same speed as a geostationary orbit? Or are you allowed to just have a tiny orbit around the bottom of the Earth that's not passing through the centre?? Thank you so much
So launching from Woomera is less advantageous than launching from the equator, but it is still advantageous. You still get a boost, just not as much ;D and you are correct, we can't just have a random orbit that doesn't have the centre of the earth at its focus. However, the idea would be launching the satellite from Woomera, then adjusting once beyond the atmosphere to place the satellite in a proper orbit, probably just by bending the trajectory downwards until it is wrapping diagonally (picture a line with gradient -1). The point being, the orbit can be established later, we still like the speed boost ;D
Post by: Aliceyyy98 on August 04, 2016, 03:39:39 pm
Sure!
Okay, so let's just assume we are working with electric field deflection (you can apply the same idea to magnetic). The idea is that by controlling the electric field strength and direction, we can change the path of the electron beam. In this case, this would be done by applying different voltages to the deflection plates :)
Let's say we can apply a voltage between 100V and -100V to the plates (same max magnitude, just opposite polarity) to a pair of plates controlling vertical deflection. 0V would mean the electron beam is unaffected, and travels straight. If we apply 100V, the beam might (for example) be deflected such that it hits the very top of the screen. 99V means it hits JUST below that. 98V, just below that again. All the way to zero where we are back in the centre. Then, -1V, just below the centre. And the process continues, all the way to -100V where the beam hits the bottom of the screen.
If we do this for both horizontal and vertical deflection, we can hit ANY point on our two dimensional screen. The top right corner might be 100V applied to vertical plates, 100V applied to horizontal plates. A point in the bottom right would be -100V vertical, but still 100V horizontal. We adjust the voltages of each pair to move up/down, or left/right.
We can do creative things with this ideas (again, just assume electrical is used for everything to keep things numerically simple). For example, in an Oscilloscope, the horizontal plates have a time varying voltage. We start by applying -100V, and the beam is at the very left. We gradually increase this to 0, and then to 100, to cause the beam to sweep across the screen. Then, when we hit 100V, we immediately reset to -100V and start the sweep again (for prospective electrical engineers, this signal would be called a sawtooth wave).
So we have an electron beam sweeping across the middle of the screen (maybe once a second). If we then connect the vertical deflection plates to some source we want to measure (for example, an electrode connected to the chest to measure heartbeat), that means the electron beams vertical position is dictated by what we are measuring. Thus, we then have your typical cardiograph, with the heartbeat visible as it fluctuates over time! The beam sweeps across, and jumps up and down as it sweeps in correspondence with the heartbeat.
Television screens are more complex, but basically, both the vertical and horizontal deflection plates are time varying. The periods are set up such that the beam does a sweep across the top row of the screen, then the next row in the reverse direction, then the next row, etc etc. Remember Donkey Kong? How you'd go all the way to the right, then jump, then all the way to the left, then jump, etc? That's what happens here (roughly speaking). It's a little tougher to picture, but let's say we start in the top left corner:
Horizontal = 100V, Vertical = 100V
Then we want the 'phosphor dot' to its right:
Horizontal = 99V, Vertical = 100V
We keep sweeping to the right until we get to the top right corner:
Horizontal = -100V, Vertical = 100V
Then we shift down:
Horizontal = -100V, Vertical = 99V
Then we begin sweeping left:
Horizontal = -99V, Vertical = 99V
And the process repeats in this fashion ;D
Thanks heaps! Just one more question sorry my trials is tmr :( what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!
Post by: jamonwindeyer on August 04, 2016, 03:53:48 pm
Thanks heaps! Just one more question sorry my trials is tmr :( what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!
I think that it would not have too much effect! What have you read? :o
Post by: Spencerr on August 04, 2016, 05:32:42 pm
Thanks heaps! Just one more question sorry my trials is tmr :( what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!
Hey there, so what the Hertz radio wave experiment did is that it (serendipitously) discovered the photoelectric effect (Which was the emission of photoelectrons from the surface of the electrodes when exposed to UV light). Hertz decided to test this and so he placed a glass block between an emitter and receiver. He observed that the maximum spark length in the receiving loop DECREASED and when the glass block was removed, it was INCREASED. This suggested that the glass block in fact blocked or absorbed the UV whilst allowing the radio wave to pass through. He then did a similar experiment with quartz but found that quartz did nothing as it allowed the UV rays to pass.
Post by: MysteryMarker on August 04, 2016, 09:38:36 pm
Been stuck on this question for quite some time now. I've tried GPE and KE formulas and i seem to be either getting A or C, but the answer is B.
Cheers Guys.
Post by: conic curve on August 04, 2016, 09:40:33 pm
Post by: Aliceyyy98 on August 04, 2016, 11:35:24 pm
Hey there, so what the Hertz radio wave experiment did is that it (serendipitously) discovered the photoelectric effect (Which was the emission of photoelectrons from the surface of the electrodes when exposed to UV light). Hertz decided to test this and so he placed a glass block between an emitter and receiver. He observed that the maximum spark length in the receiving loop DECREASED and when the glass block was removed, it was INCREASED. This suggested that the glass block in fact blocked or absorbed the UV whilst allowing the radio wave to pass through. He then did a similar experiment with quartz but found that quartz did nothing as it allowed the UV rays to pass.
Thank you!
________________
Hi guys,
Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??
Thankyou!
Moderator action: In these situations, please try to merge your posts :)
Post by: RuiAce on August 04, 2016, 11:56:38 pm
Hi guys,A loudspeaker has one of those E-shaped magnets, where the two outer prongs are of the same pole and the inner prong is of a different pole. I.e.
Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??
Thankyou!
----N ----S
----S or ----N
----N ----S
A paper diaphragm is connected to the ends of the magnets. The diaphragm is essentially what will provide the sound waves.
An image from Physics in Focus is attached.
How are the sound waves generated? The coil is connected to the rest of the circuit. Now, the electrical signals are sent such that the nature of the current is AC, not DC. Note that from the diagram, because of how the coils are wrapped one end shows current going into the page, whereas the other end shows current going out of the page. (Do you see why? You have to analyse how the current is travelling in a loop going away from you.)
The motor effect states that a current carrying conductor experiences a force when placed in an external magnetic field. The right-hand push (aka palm) rule is demonstrated in the diagram. Overall, the wire will be pushed off the magnet.
But recall that this is AC current, not DC. When the direction of the current reverses, the right-hand push rule now predicts that the wire will be pulled back in again! So what should happen is that this wire is moving in and away from the magnet recursively.
But there's a problem - the wire is too tightly wounded. This is the point - because the wire is so tightly wounded, all it will do is vibrate on the magnet. Because we now have these vibrations, that's how the sound gets produced!
As for the nature of the sound:
If the frequency of the electrical signals increase, the pitch goes up.
If the amplitude of the electrical signals increase, the volume goes up.
(I'll let someone else do the galvanometer. Lost experience to word it properly.)
Post by: jamonwindeyer on August 05, 2016, 12:41:22 am
Hey Guys
Been stuck on this question for quite some time now. I've tried GPE and KE formulas and i seem to be either getting A or C, but the answer is B.
Cheers Guys.
Hey! Let me have a go, that's the approach I'd take!
First, let's consider the loss of potential energy between the two positions (difference of 1 metre). In this case we approximate with the following formula:
Now let's see how much kinetic energy it has by the end of that drop!
We have a difference in energy here of 0.26 Joules, and this is what I'd say would be the energy converted to other forms in the bulb. So I actually get D.
Hmm, it's late, I'll have another look in the morning, any ideas people? ;D
Post by: jakesilove on August 05, 2016, 10:45:38 am
Hey! Let me have a go, that's the approach I'd take!
First, let's consider the loss of potential energy between the two positions (difference of 1 metre). In this case we approximate with the following formula:
Now let's see how much kinetic energy it has by the end of that drop!
We have a difference in energy here of 0.26 Joules, and this is what I'd say would be the energy converted to other forms in the bulb. So I actually get D.
Hmm, it's late, I'll have another look in the morning, any ideas people? ;D
I continuously got that answer, no matter what method I used (even tried a projectile motion version!) but figured that it was late, so I was getting something wrong. If Jamon got the same answer, I'd say that D is correct!
Post by: jamonwindeyer on August 05, 2016, 11:26:01 am
I continuously got that answer, no matter what method I used (even tried a projectile motion version!) but figured that it was late, so I was getting something wrong. If Jamon got the same answer, I'd say that D is correct!
Really? Awesome, I just did it again and I keep getting D as well. B seems dubious to me because it is more than the kinetic energy of the magnet at the end, it seems a little out of the ballpark ;D
Post by: Spencerr on August 05, 2016, 11:28:30 am
Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??
Moderator action: In these situations, please try to merge your posts :)
For the Galvanometer.
There are loops of coils wrapped around a soft iron core (soft iron core is used to concentrate the B field). The coils are connected to an external circuit which provides an input of DC current.
The soft iron core is placed inside a radial magnetic field (which ensures that the Magnetic field strength is always at a maximum no matter how much the coil turns).
Now remember in DC motors, that if you input current into a coil, that is in a B-field, the coil will experience a force (Via the Motor effect) and this force will produce Torque (T = FD) which causes the coil to turn.
Same situation here, when current passes through the coils wrapped around the soft iron core, the coils will rotate in one direction, however it is counterbalanced by a torsional spring which produces a retarding force on the rotational motion. This counterbalancing spring is the reason why the reading at 0 when current is 0.
Since F = B I L sine theta, sine theta equals 1 as there is a radial magnetic field, L is the same as length of the coil does not change, and B is the same as the magnetic field strength does not change. So the only thing Force is dependent and thus Torque is dependent on is I which is current.
A pointer is attached to the centre of the coil, so as the coil rotates it also moves linearly. A calibrated scale is used in conjunction with the pointer to provide measurements/readings of the magnitude of the current.
P.S. I also got D as well for the other question.
EDITED: Added a schematic diagram of a galvanometer
Post by: RuiAce on August 05, 2016, 11:31:52 am
You either upload it to imgur or something and copy/paste the [img] code, or you upload it where you write the post - attachments are uploaded below the box you type your message
P.S. I'm not sure how to insert diagrams but a quick google search will show you all the different components of the galvanometer.
Post by: jamonwindeyer on August 05, 2016, 11:34:50 am
Thank you!
________________
Hi guys,
Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??
Thankyou!
I'll take Rui's tag in ;)
Okay, so a galvanometer basically consists of a coil inside a radial magnetic field. What this means is that, while regularly the torque produced by a current carrying coil varies based on its angle with the field, the torque produced here is constant. This is because, if the field is radial, it is always parallel to an equal amount of the field.
Now, the current we are measuring with a galvanometer is fed through this coil. This causes it to rotate, just like a motor, due to the Motor Effect. However, the coil is kept in place with a spring, that provides an opposing torque. Basically, it keeps it from spinning too far. The spring can stretch, but it doesn't let it spin.
The result is that the coil will rotate ever so slightly, until the opposing torque provided by the string equals the torque due to the motor effect. The coil is attached to the needle on your galvanometer, and so the slight rotation causes the needle to move up the scale to read 1mA, or whatever. If you have more current, you have more deflection, and so the needle will move higher up the scale ;D
Of course, there are digital ammeters now which do fulfil this function with transistors and such, but this covers the basic principles you'll need to answer HSC questions! ;D
Edit: Combine my response with Spencer's more mathematical one for a complete picture ;D
Post by: jamonwindeyer on August 05, 2016, 12:00:52 pm
As a result of being hit from behind by a toy truck, a 500g toycar, initially at rest, rolls 12.0m across a floor that applies a constant retarding force of 1.2N to it. The car stops 2.0s after being hit. If the truck was in contact with the car for 0.12s calculate the impulse given to the car.
Hey! So this isn't the typical momentum question, because we have retarding forces involved. Therefore we can't apply the conservation of momentum (we don't have enough info anyway), so let's try something else:
We can calculate the work done on the toy car by the floor due to the retarding force:
This must have been the initial kinetic energy of the toy car, which was then taken away by friction. So, we can use that to find the initial velocity.
Meaning the initial momentum of the car was:
This is actually our answer! The impulse given to the car is equal to the change in momentum, and since the car started from rest, the change in momentum is its initial momentum!
There are definitely more ways to tackle this, but this is the way I'd choose ;D
Post by: brontem on August 05, 2016, 01:41:58 pm
Post by: RuiAce on August 05, 2016, 01:43:04 pm
Can someone really simply sum up the components of an AC motor in as little words as possible (brief enough to fit on a flashcard :D) Thank you ;DComponents? It's the exact same as for a DC motor with one exception
No split-ring commutator - there's a slip ring commutator.
Or did you mean the AC induction motor?
Post by: brontem on August 05, 2016, 01:44:32 pm
Or did you mean the AC induction motor?
woops yes I meant induction motor :)
Post by: RuiAce on August 05, 2016, 01:49:31 pm
The stator for the AC induction motor is a solenoid that produces a rotating magnetic field. The currents are indeed AC currents to ensure this happens.
The rotor looks like a series of bars that form what we call a "squirrel cage". In principle, due to the rotating magnetic field it is always subject to changing magnetic flux. The fact that it's rotational will cause, as a result of Lenz's law, the squirrel cage to rotate in the same direction the magnetic field is applied.
(Will let anyone else fill in more if they have something to say.)
Post by: Spencerr on August 05, 2016, 02:11:19 pm
From all I remember:
The stator for the AC induction motor is a solenoid that produces a rotating magnetic field. The currents are indeed AC currents to ensure this happens.
The rotor looks like a series of bars that form what we call a "squirrel cage". In principle, due to the rotating magnetic field it is always subject to changing magnetic flux. The fact that it's rotational will cause, as a result of Lenz's law, the squirrel cage to rotate in the same direction the magnetic field is applied.
(Will let anyone else fill in more if they have something to say.)
The stator consists of coil windings that are fed with an AC current to produce a rotating magnetic field. The rotor is a squirrel cage made from copper bars attached to a ring on each end. Faraday's law induces an emf in the copper bars and a current (as it is a conductor). The current flows in a direction to oppose the cause of motion i.e. the rotating B-field. In order to do so, the rotor rotates in the same direction, "chasing" the B-field. The speed of the B-field is always greater than the speed of the rotor and this difference is known as the "slip speed", if the speed were the same there would be no change in flux. The rotor is attached to a load to produce a torque and do work. There is no physical contact between the stator and the rotor so the induction motor is much more compact and reliant. However, it is low power.
Post by: conic curve on August 05, 2016, 03:22:58 pm
http://imgur.com/a/S4S24
http://imgur.com/a/pSih9
Thanks
Post by: jakesilove on August 05, 2016, 03:29:57 pm
Can someone here please help me with this:
http://imgur.com/a/S4S24
http://imgur.com/a/pSih9
Thanks
For the first question, you need to calculate the angle of incidence (ie. angle between the entrance-beam of light into the water and the central line). You can do this using Trig, knowing the two sides of that triangle. Then, you use Snell's law to calculate the angle of refraction (knowing the refractive index of air and water). Finally, having the angle in the bottom triangle, you can use trigonometry to figure out the bottom line of the Triangle. I won't do the maths here, because I'm sure you can do it; have a go and show me your working if you need further support!
Post by: levendibigd on August 05, 2016, 06:57:55 pm
At what angle to the horizontal was it launched?
Post by: jamonwindeyer on August 05, 2016, 07:36:14 pm
A projectile was launched from the ground. It had a range of 70 metres and was in the air for 3.5 seconds.
At what angle to the horizontal was it launched?
Hey there, welcome to the forums!! A bit of a mean question this one, they don't give much away ;D
Let's consider the horizontal component of velocity first. We know the range and time of flight, so it is actually a straightforward equation:
For vertical, we need to find the velocity required for the vertical displacement to be zero at t=3.5:
Now we form a right angled triangle with these values, and the angle we want is just the inverse tangent of the vertical and horizontal velocities:
This last bit might be a bit hard to picture, but remember that we can break a launch velocity into its vertical and horizontal components! This forms a right angled triangle, with the vertical side corresponding to vertical velocity, and the base to horizontal velocity ;D let me know it this makes sense! ;D
Post by: conic curve on August 05, 2016, 10:13:58 pm
Thanks
http://imgur.com/a/QVIou
http://imgur.com/a/uZb55
Post by: RuiAce on August 06, 2016, 08:42:20 am
Hey GuysP.S. I am absolutely convinced that the answer is D. May I please see the source of where you got this question from.
Been stuck on this question for quite some time now. I've tried GPE and KE formulas and i seem to be either getting A or C, but the answer is B.
Cheers Guys.
Post by: jamonwindeyer on August 06, 2016, 02:18:17 pm
P.S. I am absolutely convinced that the answer is D. May I please see the source of where you got this question from.
With 4 people (Jake, Spencerr, you and myself) I think that is virtually a definite ;D these sorts of errors are really annoying, imagine all the students who kick themselves over the question thinking they can't get it right? :P
Post by: RuiAce on August 06, 2016, 06:41:49 pm
With 4 people (Jake, Spencerr, you and myself) I think that is virtually a definite ;D these sorts of errors are really annoying, imagine all the students who kick themselves over the question thinking they can't get it right? :PI just got word that it came from the 2013 Exam Choice M/C and the correct answer WAS in fact D.
Post by: jamgoesbam on August 07, 2016, 08:31:18 am
(Answers in order of attachments: A, D, C)
Post by: RuiAce on August 07, 2016, 09:01:30 am
Could someone please explain the answer to these multiple choice questions? Would be much appreciated! Thanks :)
(Answers in order of attachments: A, D, C)
___________________________
Post by: jamonwindeyer on August 07, 2016, 10:39:02 am
Could someone please explain the answer to these multiple choice questions? Would be much appreciated! Thanks :)
(Answers in order of attachments: A, D, C)
That last one is nasty! Okay, so basically, the golf ball is travelling at 0.1c, and if we didn't take special relativity into account, we would say 0.7c is the answer, since the ship is travelling at 0.6c. However, we must remember that:
- The length of the ship as measured from earth (also the distance travelled by the golf ball as measured from earth) is shorter due to length contraction
- The time taken for the ball to reach the other side of the ship (as measured from earth) is longer due to time dilation.
If we look at the equation for speed:
Then clearly, the speed of the golf ball as viewed from earth will be less than what is measured by the astronauts. However, it will still add to the 0.6c, just not as much as we would expect.
This gives the answer of C ;D let me know if this makes sense!
Post by: jakesilove on August 07, 2016, 11:23:13 am
That last one is nasty! Okay, so basically, the golf ball is travelling at 0.1c, and if we didn't take special relativity into account, we would say 0.7c is the answer, since the ship is travelling at 0.6c. However, we must remember that:
- The length of the ship as measured from earth (also the distance travelled by the golf ball as measured from earth) is shorter due to length contraction
- The time taken for the ball to reach the other side of the ship (as measured from earth) is longer due to time dilation.
If we look at the equation for speed:
Then clearly, the speed of the golf ball as viewed from earth will be less than what is measured by the astronauts. However, it will still add to the 0.6c, just not as much as we would expect.
This gives the answer of C ;D let me know if this makes sense!
Just going to add a bit to this answer, because I've been asked about it a couple to times during the Lectures and private messages.
Jamon is totally right, and his method is how you get to the correct solution. However, I think what throws most students is that this LOOKS like a straight up maths question; they should be able to sub numbers into a formula somewhere, and a number should pop out. However, this isn't the case. There IS a formula for exactly this scenario, but it isn't part of the syllabus (except in Western Australia for some reason!).
So, if you ever get stuck and genuinely can't figure out how to approach a maths question, you can generally logic it out. I would have done this as Jamon did, or by elimination; we can easily eliminate the other three multiple choice answers, as they don't make any logical sense. Either way, just remember that this is a legit method of answering what seems to be a Maths question!
Post by: Skidous on August 07, 2016, 12:32:33 pm
Post by: RuiAce on August 07, 2016, 12:34:55 pm
Just a question about content required. Do I need to know Maxwell's equation for the exam, or is it alright just to know that Hertz verified Maxwell's equation?Just that Hertz verified them.
Maxwell's equations have gone way outside of the syllabus
Post by: Skidous on August 07, 2016, 12:36:56 pm
Just that Hertz verified them.
Maxwell's equations have gone way outside of the syllabus
Ok just verified, thanks Rui
Post by: Skidous on August 07, 2016, 01:30:43 pm
Also is it good to have an example of developments into improvements of maglev trains?
Post by: jakesilove on August 07, 2016, 01:47:09 pm
Another question, when talking about Maglev trains, do we need a specific example of one currently in use?
Also is it good to have an example of developments into improvements of maglev trains?
I don't believe so; just understand how they 'technically' work, and that's all you need. From memory, I contrasted two models of Maglev (German and Japanese I think?) which essentially just switched where the superconductors/electromagnets were in the construction.
Post by: Skidous on August 07, 2016, 01:48:06 pm
Post by: jakesilove on August 07, 2016, 01:49:44 pm
Ah ok, but would it be ok to add them in an answer to help get marks?
Oh absolutely! If you already have the knowledge, definitely use it. For instance, in a standard "Discuss uses of Superconductors", an exemplar answer would discuss MagLev, issues with it, and developments in the field. If you have the knowledge, use it whereever you can! I was just trying to say that you don't need to learn it, necessarily.
Post by: Skidous on August 07, 2016, 02:22:22 pm
Post by: jakesilove on August 07, 2016, 02:26:23 pm
Ok, so not for the syllabus but for band 5-6 answers will have that knowledge?
It will basically secure you full marks if you're bordering on a 5-6 out of six :)
Post by: FallonXay on August 07, 2016, 02:32:16 pm
Could someone please explain how to answer these questions?
(The Answer is A for Q9; and B for Q8; and C for Q14)
Thanks.
Post by: jakesilove on August 07, 2016, 02:44:36 pm
Hi!
Could someone please explain how to answer these questions?
(The Answer is A for Q9; and B for Q8; and C for Q14)
Thanks.
Hey!
For Q9, the way I understand it, the Earth's magnetic field will be strongest right at the pole, and then become weaker (in, like, concentric circles) as the distance from the pole increases. So, the original loop contains all of the strongest magnetic field lines. However, the final loop will not. It may still contain the very strongest point (the pole), but it no longer contains field lines radiating from areas just outside the pole. Thus, the answer is A. I hope this makes sense, just imagine lines coming out of the earth, that get weaker as they move away from the pole. The first loop contains more lines!
For Q8, I think of it like this. We know that Lenz's law states that a current will be induced that resists whatever change has been made. In this case, MORE field lines have gone through the loop, into the page. Therefore, Lenz's law will 'want' LESS field lines to go through the loop into the page, by CREATING field lines pointing out the the page. By the right hand rule, this will create an anti-clockwise current, thus making the answer B.
Post by: FallonXay on August 07, 2016, 03:13:35 pm
Hey!
For Q9, the way I understand it, the Earth's magnetic field will be strongest right at the pole, and then become weaker (in, like, concentric circles) as the distance from the pole increases. So, the original loop contains all of the strongest magnetic field lines. However, the final loop will not. It may still contain the very strongest point (the pole), but it no longer contains field lines radiating from areas just outside the pole. Thus, the answer is A. I hope this makes sense, just imagine lines coming out of the earth, that get weaker as they move away from the pole. The first loop contains more lines!
For Q8, I think of it like this. We know that Lenz's law states that a current will be induced that resists whatever change has been made. In this case, MORE field lines have gone through the loop, into the page. Therefore, Lenz's law will 'want' LESS field lines to go through the loop into the page, by CREATING field lines pointing out the the page. By the right hand rule, this will create an anti-clockwise current, thus making the answer B.
Thanks, Q8 makes sense. But for Q9 wouldn't there be an EMF induced due to the changing magnetic flux on the wire to vectorially add magnetic field lines? So how come it's detecting a decrease?
Post by: jakesilove on August 07, 2016, 03:20:11 pm
Thanks, Q8 makes sense. But for Q9 wouldn't there be an EMF induced due to the changing magnetic flux on the wire to vectorially add magnetic field lines? So how come it's detecting a decrease?
I don't think the question is asking about the CURRENT going through the wire, just the field WITHIN the wire. Note that they haven't added an Ammeter, but some sort of 'field detector'. Since none of the answers have to do with current, we know that it's not really a Lenz's law question
Post by: FallonXay on August 07, 2016, 03:23:42 pm
I don't think the question is asking about the CURRENT going through the wire, just the field WITHIN the wire. Note that they haven't added an Ammeter, but some sort of 'field detector'. Since none of the answers have to do with current, we know that it's not really a Lenz's law question
Yes, but wouldn't there be an induced current which creates extra magnetic field lines around/ within the wire?
Post by: jakesilove on August 07, 2016, 03:26:57 pm
Yes, but wouldn't there be an induced current which creates extra magnetic field lines around/ within the wire?
Ah okay, I see your point. There was a decrease (purely from the Magnetic field within the wire changing), however Lenz's law should result in an increase, to counteract the change!
The thing to know is that the field induced by Lenz's law will not make up for the original change. I'm not sure if you do Chemistry, but it's sort of like Le Chatelier's principle if you do. If you change the flux through a loop, the current will try to reverse that change, but not so strongly as to actually reverse it. Sorry, I understand your point now, and I hope you understand my explanation (well, it's less of an explanation, more of a 'this is just how it is')!
Jake
Post by: FallonXay on August 07, 2016, 03:34:49 pm
Ah okay, I see your point. There was a decrease (purely from the Magnetic field within the wire changing), however Lenz's law should result in an increase, to counteract the change!
The thing to know is that the field induced by Lenz's law will not make up for the original change. I'm not sure if you do Chemistry, but it's sort of like Le Chatelier's principle if you do. If you change the flux through a loop, the current will try to reverse that change, but not so strongly as to actually reverse it. Sorry, I understand your point now, and I hope you understand my explanation (well, it's less of an explanation, more of a 'this is just how it is')!
Jake
ahh ok, makes sense - thanks!
Post by: jakesilove on August 07, 2016, 03:35:39 pm
Hi!
Could someone please explain how to answer these questions?
(The Answer is A for Q9; and B for Q8; and C for Q14)
Thanks.
For your final question, I'm getting the answer of C when I divide the energy of the photoelectron by the distance between the plates. This would almost suggest that
Where E is the energy of the field required to exactly oppose the energy of the photo electron. This doesn't make any sense though, so hopefully someone else can step in and help us out! Either the multiple choice is wrong, or I'm not thinking of it correctly.
Post by: jakesilove on August 07, 2016, 03:36:16 pm
ahh ok, makes sense - thanks!
No worries! Sorry that I can't take a proper crack at that last one!
Post by: jamonwindeyer on August 07, 2016, 04:00:02 pm
Hi!
Could someone please explain how to answer these questions?
(The Answer is A for Q9; and B for Q8; and C for Q14)
Thanks.
Hey! Your last question is deceptive, you'd think it wanted Photoelectric Effect stuff, but do you remember this little formula from Year 11?
That's what we need, work is equivalent to the force applied over distance. The amount of work the voltage does will need to be equal to the kinetic energy of the photoelectrons (given in the last column. We know the distance over which it is done, 5mm. So:
Nasty question! Goes to show, it is important to know your Prelim results (this formula is on your reference sheet as well) ;D
Edit: So Jake's calculation was actually inadvertently correct ;)
Post by: FallonXay on August 07, 2016, 04:21:18 pm
Hey! Your last question is deceptive, you'd think it wanted Photoelectric Effect stuff, but do you remember this little formula from Year 11?
That's what we need, work is equivalent to the force applied over distance. The amount of work the voltage does will need to be equal to the kinetic energy of the photoelectrons (given in the last column. We know the distance over which it is done, 5mm. So:
Nasty question! Goes to show, it is important to know your Prelim results (this formula is on your reference sheet as well) ;D
Edit: So Jake's calculation was actually inadvertently correct ;)
No worries! Sorry that I can't take a proper crack at that last one!
mm, I see. Thanks a ton, your answers are very appreciated! :)
Post by: FallonXay on August 07, 2016, 07:35:51 pm
I've got a few more questions:
With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)
and I don't understand how to derive the answer for Q3.
Thanks once again.
Post by: Alexander23 on August 07, 2016, 07:37:09 pm
Thanks!
Post by: RuiAce on August 07, 2016, 07:43:06 pm
Hello, again!
I've got a few more questions:
With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)
and I don't understand how to derive the answer for Q3.
Thanks once again.
That generator question is very dodgy. My guess would be that if we incorporate Lenz's Law in, since the induced EMF always seeks to oppose the cause of induction using -sin(x) is more appropriate to +sin(x). It does say 'most accurately' represented.
Post by: jamonwindeyer on August 07, 2016, 07:55:36 pm
Hello, again!
I've got a few more questions:
With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)
and I don't understand how to derive the answer for Q3.
Thanks once again.
Hey! Normally, we'd use the right hand rule to determine the direction of induced current in the wire, and that would tell us whether it was 'positive' or 'negative.'
However, that relies on the question defining what positive and negative actually is. This question doesn't do that! To me, either A or B could be correct and I don't see a way of distinguishing, you don't have enough information, you need to be told which way they define as positive current flow ;D
Post by: Spencerr on August 07, 2016, 09:53:37 pm
Anyone know why some lines in the hydrogen spectrum brighter than others? And why couldn't Bohr explain this?
Thanks!
Bohr's atomic model consisted of electrons orbiting around the nucleus of an atom is discrete energy levels or shells that were a specific distance away from the nucleous. (this is his first postulate that electrons orbited in "stationary states"). His second postulate was that if an electron moved between discrete energy levels, (either moving to a higher one or to a lower one), an emission or absorption of a photon (E=hf) that has an equivalent energy to the energy difference between the two levels, is required. An emission of a photon of emr occurs when electrons fall from a higher energy level to a lower energy level. Since E=hf, we can determine the frequency of the photons. The hydrogen spectra line series shows exactly this, the falling of electrons from different higher energy levels to the second lowest energy level. Their relative intensities suggest that some electron falls or electron jumps are more favoured than other electron jumps. (as more electron jumps = more identical photons=brighter/sharper lines on the hydrogen spectrum). Now this was a probabilistic nature describing the behaviour of electrons and Bohr's model did not factor this in. Hence he could not explain the relative intensities :)
Post by: FallonXay on August 08, 2016, 12:04:57 pm
Post by: Spencerr on August 08, 2016, 12:12:52 pm
Weight is defined as the force experienced by an object when placed inside a gravitational field. As you know large celestial bodies like the moon and the earth have their own gravitational field that is directed towards the centre of the body. Midway between the earth and the moon, an object will experience both a force of attraction towards the centre of earth from earth's gravity and another force of attraction towards the centre of the moon from the moons gravity however these two forces oppose each other, acting in separate directions. Hence the gravitational fields cancel out leaving the object near weightlessness (due to the absence of a NET FORCE). The other answer that might seem right is if the object is in orbit but remeber that at that point the object is still being subjected to a weight force or attractive force which forms the centripetal force keeping it in orbit. Feel free to ask any questions if you need me to clear things up
Ps is this a hsc paper? Or a schools paper. I've done this question before..
Post by: FallonXay on August 08, 2016, 03:48:24 pm
Hey there, since I'm online I'll give you a quick explanation.
Weight is defined as the force experienced by an object when placed inside a gravitational field. As you know large celestial bodies like the moon and the earth have their own gravitational field that is directed towards the centre of the body. Midway between the earth and the moon, an object will experience both a force of attraction towards the centre of earth from earth's gravity and another force of attraction towards the centre of the moon from the moons gravity however these two forces oppose each other, acting in separate directions. Hence the gravitational fields cancel out leaving the object near weightlessness (due to the absence of a NET FORCE). The other answer that might seem right is if the object is in orbit but remeber that at that point the object is still being subjected to a weight force or attractive force which forms the centripetal force keeping it in orbit. Feel free to ask any questions if you need me to clear things up
Ps is this a hsc paper? Or a schools paper. I've done this question before..
Thanks for the prompt answer!
Yeah, this was from the 2011 Exam Choice Trial HSC Physics paper.
Just with the question: Wouldn't the strength of the gravitational field exerted by the Earth be stronger than that of the Moon. So at the halfway point, wouldn't the net force experienced still be acceleration due to gravity on Earth?
Post by: Spencerr on August 08, 2016, 04:07:52 pm
Thanks for the prompt answer!
Yeah, this was from the 2011 Exam Choice Trial HSC Physics paper.
Just with the question: Wouldn't the strength of the gravitational field exerted by the Earth be stronger than that of the Moon. So at the halfway point, wouldn't the net force experienced still be acceleration due to gravity on Earth?
In terms of it simply being a MC question, using the values for the mass of the Earth and the mass of the Moon and Newton's Law of Gravitational Attraction would be too much. I would say it is the most correct answer out of the four.
Post by: FallonXay on August 08, 2016, 04:40:29 pm
In terms of it simply being a MC question, using the values for the mass of the Earth and the mass of the Moon and Newton's Law of Gravitational Attraction would be too much. I would say it is the most correct answer out of the four.
ok, fair enough. Thanks! :)
Post by: Klexos on August 08, 2016, 09:44:05 pm
This is from the 2014 CSSA and I need help because my answer doesn't agree with the MC, rather it agrees with parts of MC answers
Post by: jamonwindeyer on August 08, 2016, 11:58:01 pm
(http://uploads.tapatalk-cdn.com/20160808/ebf1141d9e73da955dfec816c4841992.jpg)
This is from the 2014 CSSA and I need help because my answer doesn't agree with the MC, rather it agrees with parts of MC answers
Hey Klexos! Subtle trick here, this is a circular magnetic field. That means that you don't need the cos30 in your calculation, the coil is always parallel to the field lines! ;D if you take that out you should get one of the answers in the options ;D
Post by: Klexos on August 09, 2016, 07:14:10 am
Hey Klexos! Subtle trick here, this is a circular magnetic field. That means that you don't need the cos30 in your calculation, the coil is always parallel to the field lines! ;D if you take that out you should get one of the answers in the options ;D
Oooh as in the radial electromagnets?
I'm so done ;___;
Post by: RuiAce on August 09, 2016, 08:14:47 am
Hey Klexos! Subtle trick here, this is a circular magnetic field. That means that you don't need the cos30 in your calculation, the coil is always parallel to the field lines! ;D if you take that out you should get one of the answers in the options ;DAh shit, completely overlooked that one
Oooh as in the radial electromagnets?Well they're not "electromagnets" necessarily though
I'm so done ;___;
Post by: jamonwindeyer on August 09, 2016, 10:41:25 am
Oooh as in the radial electromagnets?
I'm so done ;___;
They might be! The diagram doesn't suggest that though, and it doesn't make a difference for the question, you can have permanent magnets setting up a radial magnetic field ;D
Post by: conic curve on August 09, 2016, 09:23:14 pm
1. Satellites send microwaves into the area where the receiver is shown to be
2. When the satellite signals are sent out, the satellite themselves record the exact time and position this occurs and sends this information along
3. Now as the receiver receives the signal, it compares the time at which it receives the signal and the when it was sent out
4. This time frame tells us an approximation to were you can be in terms of regions, because light travels at 3 time 10^8
Is this sufficient enough?
Post by: jamonwindeyer on August 09, 2016, 10:00:43 pm
What other information do I need for GPS right now (for an assignment other than the information I have down here)
1. Satellites send microwaves into the area where the receiver is shown to be
2. When the satellite signals are sent out, the satellite themselves record the exact time and position this occurs and sends this information along
3. Now as the receiver receives the signal, it compares the time at which it receives the signal and the when it was sent out
4. This time frame tells us an approximation to were you can be in terms of regions, because light travels at 3 time 10^8
Is this sufficient enough?
You've got the rough idea, but remember this happens for 3 or 4 different satellites at once! This allows the GPS receiver to triangulate its exact position on the earth. Note that 3 satellites are required for a precise 2 dimensional position (latitude/longitude), and a fourth is required to accurately determine height ;D you may want to do some research into the specifics of how that works, and further (if you want), into the extra considerations required to account for the effects of special/general relativity (which can throw things off a little) ;D
Post by: Loki98 on August 11, 2016, 05:15:45 pm
Could someone please explain how an AC induction motor works and what the practical was to demonstrate the principle of an induction motor.
Thanks =]
Post by: jamonwindeyer on August 11, 2016, 05:42:21 pm
Hey guys,
Could someone please explain how an AC induction motor works and what the practical was to demonstrate the principle of an induction motor.
Thanks =]
Hey Loki! So the practical you did to model this was something like the magnet rotating under the disc!
So, we hang a conducting disc on a string and rotate a magnet beneath it. Due to the changing magnetic field, eddy currents form in the disc. Their direction is determined by Lenz's Law, and so we know that the induced current will generate a new magnetic field that opposes the change that created it. But how do we oppose a rotating magnet?
This is a little tricky to explain, but consider some point marked X on the side of the disc, just above the rotating north pole. The north pole rotates away. That specific point X doesn't want the north pole to leave (oppose the change), and so it wants to pull it back. The net effect of this is that the disc chases the magnet.
So, an AC induction motor. These motors rely on eddy currents to spin the rotor, and it works in the exact same way as above. The stator consists of six or more field coils in pairs, fed a special version of an AC current (it's 3 Phase) which essentially creates a rotating magnetic field. The rotor spins in response to induced eddy currents. The rotor is shaped a bit like one of those hamster wheels that you see the little hamsters running on, which allows eddy currents to flow and maximise the torque provided by the motor (details on that shape not important).
You could go into more detail here, but this covers you for the sorts of questions asked in the HSC ;D does that make sense?
Post by: MysteryMarker on August 11, 2016, 06:08:23 pm
For Mass defect calculation questions, when do you include the mass of the electrons? Different examples include the mass of the electrons or just use the mass of the constituents of the nucleus.
Cheers.
Post by: jakesilove on August 11, 2016, 06:52:28 pm
Quanta to Quarks Question:
For Mass defect calculation questions, when do you include the mass of the electrons? Different examples include the mass of the electrons or just use the mass of the constituents of the nucleus.
Cheers.
Hey! I didn't do Quanta, so maybe someone that did can help you out! Otherwise, if you post one question where they DID include the electron, and one where they DIDN'T, I can take a look and try to discern a pattern? I would generally say that, if it specifically tells you the NUCLEUS lost mass, you would ignore electrons. If it said something like the 'element' lost mass, then use the electrons. That being said, totally ignoring electrons shouldn't be an issue; they're freaking small. If you completely convert an electron into energy, you get 8*10^(-14) Joules of energy out; negligable!
Post by: Loki98 on August 11, 2016, 07:59:23 pm
Also, what would be a suitable description and diagram to explain how a solar cell works and its applications?
Post by: jamonwindeyer on August 11, 2016, 11:28:10 pm
Thx for the help Jamon,
Also, what would be a suitable description and diagram to explain how a solar cell works and its applications?
You're welcome! So the diagram can be very simple, a quick something like this would suffice! Label the elements of the PN junction, show an external circuit, and give some indication of the release of an electron and it's subsequent movement.
(http://dc.edu.au/wp-content/uploads/how-solar-cells-work.png)
In terms of a description, you wouldn't want more than a few sentences, I'd do something like:
- Sentence One: Identify the PN junction and the setup of the depletion zone
- Sentence Two: Describe the photoelectric effect releasing an electron
- Sentence Three: Identify that the electron moves through the external circuit due to the electric field set up by the depletion zone
No more detail than that for a describe question, then just delve in a bit more for explain ;D let me know if you need any of these points clarified :)
Post by: Spencerr on August 12, 2016, 12:08:07 am
Quanta to Quarks Question:
For Mass defect calculation questions, when do you include the mass of the electrons? Different examples include the mass of the electrons or just use the mass of the constituents of the nucleus.
Cheers.
Hey there I've encountered questions where I have both included and not included the mass of the electrons. However, in most of the questions i DID NOT include the mass of the electrons. This is because during a nuclear reaction (as is the case of an alpha decay or a beta decay), it is NUCLEUS undergoing fission (or fusion). The nucleus breaking apart into daughter nuclei is what produces the mass defect and the electrons do not play a part in this. Furthermore, mass defect is linked with binding energy (using Einstein's mass energy equivalence E = mc squared). Binding energy holds the nucleons together in the nucleus and also does not deal with electrons. So in most cases, I would not include the mass of the electron :)
Post by: Phillorsm on August 12, 2016, 04:41:22 pm
Post by: jamonwindeyer on August 12, 2016, 07:56:38 pm
Hey guys, I need some help understanding Eddy currents. In particular, how to determine the direction of the Eddy current in terms of the physics principals involved when moving a sheet of metal through a changing magnetic field.
Hey Phillorsm! I'd be happy to help here, the rule I recommend is the right hand grip rule! That said, it's best to show you with an example, do you have an example of one of the questions you are struggling with?
Post by: Phillorsm on August 13, 2016, 11:09:09 am
Post by: jamonwindeyer on August 13, 2016, 01:00:59 pm
Yep, sure. This one still makes no sense haha
Okay cool! So we know that the eddy currents are only being produced at A and C, because those are the only locations where the magnetic field is changing. Let me know if you need help with that bit, otherwise, let's focus on direction ;D
First, let's look at Point A. We are introducing a magnetic field going out of the page. Thus, Lenz's Law says we will want to create a magnetic field going into the page that counteracts it. How do we create such a field?
Pretend that the metal plate is a solenoid, and we can use the right hand grip rule. The right hand grip rule is as follows. Hold your right hand out in a thumbs up position. In that position, if your thumb is pointing in the direction of the North pole you want to create with a solenoid, then your fingers wrap in the direction the current should flow. Although this is a rule used mostly for solenoids, it also works here!
At Point A, we want a field going INTO the page. Thus, we need to set the plate up so that the field lines are going INTO it on the side closest to us. This means, the side closest to us needs to be set up as a south pole. This means the other side of the plate is a north pole, that is, your thumb points into the page with the right hand grip rule.
If you use your right hand grip rule with your thumb pointing into the page, you'll find your fingers wrap clockwise. Thus, the eddy currents also wrap clockwise.
The exact same analysis works in reverse at Point C, where we are removing some of the field out of the page. Lenz's Law says we want to introduce a new field out of the page to counteract that change. Thus, our same analysis applies with everything reversed, and you'll get eddy currents flowing anti-clockwise ;D
Read this over a few times, it's a tad confusing, but this is how you determine the direction of eddy current flow. Now, in the questions, you wouldn't go into depth about the right hand grip rule, you would just explain that the eddy currents need to oppose the change that created them. Less field out of the page? We need to introduce more field out of the page to compensate for that. By right hand grip rule, this requires an anti-clockwise eddy current flow ;D
Does that make sense at all? ;D
Post by: Phillorsm on August 13, 2016, 01:10:58 pm
I'll get back to you if I need help with any other concepts ;D
Post by: conic curve on August 14, 2016, 08:25:07 pm
Please do not involve any scientific principles into this and the wave properties associated with as that is another dotpoint
Post by: Happy Physics Land on August 14, 2016, 10:06:24 pm
When it says "describe GPS" (5 marks) what should I write?
Please do not involve any scientific principles into this and the wave properties associated with as that is another dotpoint
This is a bit absurd to describe GPS without referring to scientific principles.
But if you do have to refer to scientific principles please refer to wave transmission and explain what triangulation/trilateration is
I will give you a few dotpoints to consider:
- GPS = Global Positioning System
- GPS is a constellation of 24 low-earth orbit satellites
- Each GPS satellite weighs 1 tonne with about 5 metres span
- GPS works based upon the relay of radio waves between satellites
- GPS uses the principle of triangulation (latitude, longitude, altitude), and the time difference between when the one satellite receives the signal from a transmitting antenna and when the receiver antenna receives the same signal from the satellite in order to accurately spot the location of the destination
- GPS works in any weather conditions since all satellites are in the external universe
- GPS satellites make 2 orbits around the Earth in less than 24 hours
- GPS softwares are often seen on cars and mobile phones
- Google Maps is a similar form of GPS
Post by: conic curve on August 15, 2016, 08:44:05 am
This is a bit absurd to describe GPS without referring to scientific principles.
But if you do have to refer to scientific principles please refer to wave transmission and explain what triangulation/trilateration is
I will give you a few dotpoints to consider:
- GPS = Global Positioning System
- GPS is a constellation of 24 low-earth orbit satellites
- Each GPS satellite weighs 1 tonne with about 5 metres span
- GPS works based upon the relay of radio waves between satellites
- GPS uses the principle of triangulation (latitude, longitude, altitude), and the time difference between when the one satellite receives the signal from a transmitting antenna and when the receiver antenna receives the same signal from the satellite in order to accurately spot the location of the destination
- GPS works in any weather conditions since all satellites are in the external universe
- GPS satellites make 2 orbits around the Earth in less than 24 hours
- GPS softwares are often seen on cars and mobile phones
- Google Maps is a similar form of GPS
Yeah I will be referring to the scientific principles as well but that's another thing I need to do for a presentation
I need hints for the following questions I have attached
Post by: RuiAce on August 15, 2016, 09:11:54 am
Yeah I will be referring to the scientific principles as well but that's another thing I need to do for a presentationAll formula work.
I need hints for the following questions I have attached
For some questions, figure out what is constant (e.g. in F=mg, g and F will vary on different planets but the mass of the object is still unchanged) and carry through.
When doing calculations, always figure out what you actually KNOW, before selecting a formula to try out.
Post by: conic curve on August 15, 2016, 10:46:16 am
All formula work.
For some questions, figure out what is constant (e.g. in F=mg, g and F will vary on different planets but the mass of the object is still unchanged) and carry through.
When doing calculations, always figure out what you actually KNOW, before selecting a formula to try out.
wait what do you mean, I'm still confused
Care to elaborate further?
Post by: RuiAce on August 15, 2016, 10:48:25 am
wait what do you mean, I'm still confusedI mean exactly that. Use your formulas as appropriate.
Care to elaborate further?
And as for what's in my brackets, the acceleration on each planet is different but the mass of the object is not.
Post by: bethjomay on August 17, 2016, 05:03:34 pm
Post by: conic curve on August 20, 2016, 09:31:17 pm
The radius of earth's orbit is 1.49 times 10^11m and that if Jupiter is 7.783 times 10^11. What is the period in seconds of Jupiter's orbit around the sun?
(note: the period of earth around the sun is 365.25 days) (2 marks)
Thanks
Post by: RuiAce on August 20, 2016, 09:48:24 pm
How do you do this question
The radius of earth's orbit is 1.49 times 10^11m and that if Jupiter is 7.783 times 10^11. What is the period in seconds of Jupiter's orbit around the sun?
(note: the period of earth around the sun is 365.25 days) (2 marks)
Thanks
Post by: jamonwindeyer on August 21, 2016, 12:16:19 am
Hiya! I'm just wondering if anyone know whether we need to know Bohrs derivation of the Rydberg constant (for quanta to quarks)? Thank you!
^^^
Just bringing this back to people's attention in case someone who did Quanta happens to know the answer to this one ;D
--
And conic, what Rui used up there was an example of Kepler's Law of Periods. It's taught in the HSC course, just so you know it isn't a Prelim question ;D
Post by: jakesilove on August 21, 2016, 11:41:58 am
^^^
Just bringing this back to people's attention in case someone who did Quanta happens to know the answer to this one ;D
--
And conic, what Rui used up there was an example of Kepler's Law of Periods. It's taught in the HSC course, just so you know it isn't a Prelim question ;D
In response to the Bohr derivation; I can say with fair certainty that you don't!
Post by: Spencerr on August 22, 2016, 12:23:25 am
Hiya! I'm just wondering if anyone know whether we need to know Bohrs derivation of the Rydberg constant (for quanta to quarks)? Thank you!Hey there,
You don't need to know the exact derivation, I've tried memorising it a few times but it's not worth it haha, but you just need to know a few points about the derivation.
-Bohr used a mixture of classical mechanics and quantum mechanics (i.e. he used classical formulas for kinetic energy and gravitational potential energy) and combined them with quantum mechanics (like his quantisation condition L = nh/2 (pi) ). This was a limitation of Bohr's model as the mixture of classical and quantum mechanics is a problem in itself
- Bohr's derivation of the Balmer's equation and the Rhydberg constant (which were both previously empirically based on the hydrogen spectra) validated his atomic model. Essentially, he won over alot of support from the scientific community.
But for the direct derivation it's not needed, make sure you're well acquainted with how to use the formula to find wavelengths though!
-
Post by: amina_98 on August 24, 2016, 05:50:34 pm
Post by: RuiAce on August 24, 2016, 06:09:36 pm
i know this may sound very silly, but does anyone have tips on answering multiple choice? im always loosing atleast 10-12 marks in that section. :(Well is there any particular reason that multiple choice seems to be a weakness?
Multiple choice is attacked through either calculations (alike short response), process of elimination or immediate deduction. The third option is just something that comes with practice.
If it's a calculation, then the worst thing they can do is throw you a twist. Figure out what the twist is. Not simple? You have to do even more past papers. You only learn tricks by being exposed to them (unless you're that capable).
Otherwise, process of elimination is one of the most common things in multiple choice. Eliminate what is wrong so that you narrow down your opinions. Whilst occasionally this narrows you all the way down to one correct answer, generally you only have two to choose which you just use your brain power to figure out which is right.
Multiple choice is generally more applied (calculations) or logical (reasoning). Practice more of these questions to get better at the style. Board of Studies Multiple Choice can be used if you wish. Also, keep in mind textbooks such as Excel Success One explain the answers to multiple choice as well.
Post by: imtrying on August 24, 2016, 08:31:00 pm
Post by: jakesilove on August 24, 2016, 08:34:59 pm
Hey just a question on the semiconductors topic: I get that n-type semiconductors become better conductors due to the extra electron available for conduction (due to the extra one from the Group 5 impurity added) but how does having more positive 'holes' make p-types better conductors?
Basically, because there are more 'holes' (ie. spaces with a LESS negative charge) it is easier for electrons to move into those regions, as they are repelled less. The hole moves along, and each adjacent electron can move into the space with more ease. Therefore, the semiconductor will conduct more easily, simply because electrons can move in one direction without being impeded as much! Does that make sense? It's a bit of a conceptual leap, definitely a tough topic to understand. I hope my explanation made sense!
Post by: RuiAce on August 24, 2016, 08:38:46 pm
When electrons go from the valence band to the conduction band, they move along and then fill a hole ahead of them. But the thing is, because the electron left the original atom it just left a hole behind. So the hole basically moved in the opposite direction of the electron.
Since holes carry a +'ve charge, you're basically causing a lot of positive charge to go in the opposite direction of electrons. Hence conducting electricity.
Post by: imtrying on August 24, 2016, 08:42:12 pm
Basically, because there are more 'holes' (ie. spaces with a LESS negative charge) it is easier for electrons to move into those regions, as they are repelled less. The hole moves along, and each adjacent electron can move into the space with more ease. Therefore, the semiconductor will conduct more easily, simply because electrons can move in one direction without being impeded as much! Does that make sense? It's a bit of a conceptual leap, definitely a tough topic to understand. I hope my explanation made sense!Thank you, your explanation does makes sense! This topic is so hard to get my head around, I probably be spamming this thread until my exam haha
Post by: jakesilove on August 24, 2016, 08:48:14 pm
Thank you, your explanation does makes sense! This topic is so hard to get my head around, I probably be spamming this thread until my exam haha
Then I look forward to plenty more great questions :) Good luck with your study!
Post by: jamonwindeyer on August 24, 2016, 11:27:26 pm
i know this may sound very silly, but does anyone have tips on answering multiple choice? im always loosing atleast 10-12 marks in that section. :(
I agree with Rui, unless you are super confident; answer all MC questions by eliminating 3 incorrect answers, not picking 1 correct one. This doesn't quite work for calculations, but for others, process of elimination is a great strategy ;D
Post by: imtrying on August 25, 2016, 09:48:56 am
With the one with the car speedometer, I assumed eddy currents would oppose the motion of the magnet and go in the opposite direction, but the answer was actually C.
With the other question, I honestly just have no idea :-\
Post by: jamonwindeyer on August 25, 2016, 10:25:24 am
Hi would someone be able to explain the answers to these multiple choice questions to me?
With the one with the car speedometer, I assumed eddy currents would oppose the motion of the magnet and go in the opposite direction, but the answer was actually C.
With the other question, I honestly just have no idea :-\
Hey! I can't seem to open the attachments, just me guys or? :)
Post by: imtrying on August 25, 2016, 10:31:57 am
Post by: jakesilove on August 25, 2016, 11:04:37 am
Ive checked and think theres something wrong with my attachments:( Hopefully they work now?
Hey! So for the first question, you just need to think about what Lenz's law actually means. We know that, initially, the disc is within some sort of magnetic field (it doesn't really matter how much). Now, when the magnet starts to rotate, the magnetic field lines going through the disc will change. According to Lenz's law, a current will be induced in a changing magnetic field that attempts to minimise/remedy any change. So, we know that some sort of force will be induced on the disc. It's going to want to maintain the same field as if the disc was still directly below the magnet, so it's going to turn towards the magnet! Then, as the magnet continues to move, it will continue to try to 'chase' the magnet so that the field lines go back to how it's 'used to'. So, the answer is C.
For the second one, a diagram is absolutely necessary. Draw the current carrying wire, and the magnetic field caused by in. I'm going to assume the electron is above the wire, and so the field lines (when the current is moving to the right) are coming out of the page. Therefore, when using the right hand rule for the electron, the magnetic field is also going out of the page. So, we've got the magnetic field, and the direction of force that we want (to the right). Aligning our hands properly, we see that the 'current' of the electron must go up the page. So, the electron itself (moving in the opposite direction of the current) is going down the page, towards the wire, and perpendicular. Thus, the answer is also C
That last one is hard to explain without a diagram, so try drawing one!
Post by: imtrying on August 25, 2016, 09:56:10 pm
Thanks so much
Lets hope the attachments work this time :-\
Post by: RuiAce on August 25, 2016, 10:15:00 pm
Hi me again with another multiple choice question. Would someone please explain why the answer is C?Many students struggle to understand the difference between the graph of the "force", and the "torque", when it comes to a motor.
Thanks so much
Lets hope the attachments work this time :-\
Why does this happen? Because theta is DIFFERENT depending on whether we talk about FORCE, or TORQUE.
Force: We need to look at the TOP view.
If you look at the top view, you will find the direction of the current and the magnetic field being perpendicular to each other. As the motor spins, the current and magnetic field are STILL perpendicular to each other. Thus, in the formula F = B I L sin(theta), we have (theta) = 90 degrees ALWAYS.
Torque: We need to look at the FRONT view.
The angle is made between the entire PLANE (rectangle) of the wire, and the magnetic field. As the motor spins, this new theta does go from 0 degrees to 90 degrees. This is why TORQUE is not constant, and changes.
Because force IS constant but torque is NOT, REGARDLESS of the magnetic field we use, the graph will always be straight. That is why the answer is C.
Let me know if anything does not make sense. A diagram does help.
Post by: amina_98 on August 28, 2016, 03:05:56 pm
Answers:
3. B
6. A
7. C
15. B
Post by: jakesilove on August 28, 2016, 03:23:19 pm
My physics trial is not this monday, but the monday after and ive started to hit trial papers and im freaking out. i have a bunch of multiple questions i need explanations to :-[ :'(
Answers:
3. B
6. A
7. C
15. B
For your first question, we know from your formula sheet that
We also know that
And so the answer is B, as required. The trick to questions like this is looking for your formula sheet, figuring out what you know and what you can find out, and the brute forcing the rest!
Post by: jakesilove on August 28, 2016, 03:26:52 pm
My physics trial is not this monday, but the monday after and ive started to hit trial papers and im freaking out. i have a bunch of multiple questions i need explanations to :-[ :'(
Answers:
3. B
6. A
7. C
15. B
For your second question, we can quickly figure out the time taken for the projectile to fall to the ground. This is because the initial vertical velocity is zero, and the only force acting downwards is gravity.
Now, we know that horizontal velocity doesn't change over time, as there is no force acting upon it in that direction. Therefore, from our formula sheet,
Making the answer C as required. My comment for HOW to do this question is the same as above
Post by: RuiAce on August 28, 2016, 05:07:58 pm
My physics trial is not this monday, but the monday after and ive started to hit trial papers and im freaking out. i have a bunch of multiple questions i need explanations to :-[ :'(Refer to post #820 for your fourth question. This is Q3.
Answers:
3. B
6. A
7. C
15. B
Post by: RuiAce on August 28, 2016, 05:15:32 pm
My physics trial is not this monday, but the monday after and ive started to hit trial papers and im freaking out. i have a bunch of multiple questions i need explanations to :-[ :'(
Answers:
3. B
6. A
7. C
15. B
Post by: amina_98 on August 28, 2016, 05:36:16 pm
Post by: amina_98 on August 29, 2016, 08:26:55 pm
however this sample answer i was reviewing, it mentioned that hertz placed a sheet of copper between the coils and he was able to show that the sparks exhibited reflecting and refraction properties.
i just wanted to know, are both of these statements correct? did hertz use both glass and copper sheets? or are my notes wrong with the information about the glass?
Post by: Jakeybaby on August 29, 2016, 08:58:26 pm
So I'm currently completing my physics prac report on the diameter of hair.
To do this we used a laser and measured the diffraction pattern which occurred when the laser had been shone through the hair.
I was just wondering if there were any suggestions on systematic errors which could have occurred? We did not set up this equipment either, UniSA did it for us.
Post by: znaser on August 29, 2016, 09:01:53 pm
A photon is incident on a hydrogen atom in the ground state. Explain, using de Broglie’s hypothesis, why the photon is not absorbed by the hydrogen atom.
So I know de Broglie proposed that electrons were standing waves to explain stability of electron orbits. These are quantised as the electron orbit must have a circumference equal to the multiple of the wavelength of the electron (2.pi.r = n.lambda) to allow for a standing wave (otherwise the beginning and end points are out of phase, leading to destructive interference and the wave losing energy). Therefore, the electrons only absorb photons if its energy (e=hc/lamda) allowed it to jump to one of these fixed energy states. But there are no numbers in the question so I'm not sure what the explanation is as to why the photon won't be absorbed by the electron in the hydrogen atom and the bos sample answer didn't rlly specify this. Thanks :)
Post by: jamonwindeyer on August 29, 2016, 09:02:02 pm
Hey! so im reading over my physics notes from the ideas to implementation chapter and was also reviewing a sample answer for a 5 marker on heinrich hertz. my notes say heinrich hertz placed a glass sheet in between the transmitter coil and receiver coil which shielded the spark so no effect was produced at the receiver. removing the glass resulted in the spark returning.
however this sample answer i was reviewing, it mentioned that hertz placed a sheet of copper between the coils and he was able to show that the sparks exhibited reflecting and refraction properties.
i just wanted to know, are both of these statements correct? did hertz use both glass and copper sheets? or are my notes wrong with the information about the glass?
Hey! Hertz definitely used glass sheets and they definitely have the effect you described. So yes, I would deduce that Hertz used both! I mean why not, if you wanted to be thorough, you'd definitely test multiple materials :)
Post by: jamonwindeyer on August 29, 2016, 09:26:47 pm
hey guys,
So I'm currently completing my physics prac report on the diameter of hair.
To do this we used a laser and measured the diffraction pattern which occurred when the laser had been shone through the hair.
I was just wondering if there were any suggestions on systematic errors which could have occurred? We did not set up this equipment either, UniSA did it for us.
Hey guys. So this is a question from 2014 hsc that's kind of confusing me:
A photon is incident on a hydrogen atom in the ground state. Explain, using de Broglie’s hypothesis, why the photon is not absorbed by the hydrogen atom.
Hey guys! Sorry, I'm not sure how to answer either of these :( hopefully someone else can lend a hand?
Although Jake, systematic errors are almost always due to the measuring equipment, perhaps inaccuracies in that regard? Not sure, I've not heard of that experiment before :P
Post by: Jakeybaby on August 29, 2016, 09:33:19 pm
Hey guys! Sorry, I'm not sure how to answer either of these :( hopefully someone else can lend a hand?To measure the diffraction, we used a ruler to measure the distance between the minima.
Although Jake, systematic errors are almost always due to the measuring equipment, perhaps inaccuracies in that regard? Not sure, I've not heard of that experiment before :P
I'll link a video which shows how the experiment was completed.
https://www.youtube.com/watch?v=kpsN78mQ6YY
Post by: jamonwindeyer on August 29, 2016, 09:39:48 pm
To measure the diffraction, we used a ruler to measure the distance between the minima.
I'll link a video which shows how the experiment was completed.
https://www.youtube.com/watch?v=kpsN78mQ6YY
Oh that's really cool!
Okay, well definitely:
- Inaccuracy in wavelength of laser (often varies by a small percentage, to my knowledge)
- Inaccuracy of measuring tool (EG - if using a ruler, it is only accurate to nearest millimetre, even if used perfectly)
And perhaps some others! Anyone want to tag in?
Post by: jakesilove on August 29, 2016, 09:45:23 pm
Hey guys. So this is a question from 2014 hsc that's kind of confusing me:
A photon is incident on a hydrogen atom in the ground state. Explain, using de Broglie’s hypothesis, why the photon is not absorbed by the hydrogen atom.
So I know de Broglie proposed that electrons were standing waves to explain stability of electron orbits. These are quantised as the electron orbit must have a circumference equal to the multiple of the wavelength of the electron (2.pi.r = n.lambda) to allow for a standing wave (otherwise the beginning and end points are out of phase, leading to destructive interference and the wave losing energy). Therefore, the electrons only absorb photons if its energy (e=hc/lamda) allowed it to jump to one of these fixed energy states. But there are no numbers in the question so I'm not sure what the explanation is as to why the photon won't be absorbed by the electron in the hydrogen atom and the bos sample answer didn't rlly specify this. Thanks :)
Hey! I would say exactly what you've written there; that only a very specific TYPE of photon could be absorbed by the electrons, for all the reasons outlined above. I would definitely give you full marks; you've explained the concepts, included relevant formulas, and done it all succinctly. Well done; the question is certainly worded badly, but I don't see any other way of answering it.
Post by: jakesilove on August 29, 2016, 09:49:28 pm
To measure the diffraction, we used a ruler to measure the distance between the minima.
I'll link a video which shows how the experiment was completed.
https://www.youtube.com/watch?v=kpsN78mQ6YY
Lol this experiment is hilarious. There should definitely be an error in wavelength (usually in the order of 5nm). The biggest error is using a ruler, as Jamon pointed out. If you want to go into a bit more depth, you could discuss the approximation used (what they did is that they approximated sin(theta)=tan(theta). Draw a diagram to understand what I mean!). If you don't quite understand, happy to draw up a diagram. This approximation is good we D is much bigger than d, and in this case it sort of is, but still; as with any approximation, there will be an associated error.
Jake
Post by: jakesilove on August 29, 2016, 09:51:56 pm
Lol this experiment is hilarious. There should definitely be an error in wavelength (usually in the order of 5nm). The biggest error is using a ruler, as Jamon pointed out. If you want to go into a bit more depth, you could discuss the approximation used (what they did is that they approximated sin(theta)=tan(theta). Draw a diagram to understand what I mean!). If you don't quite understand, happy to draw up a diagram. This approximation is good we D is much bigger than d, and in this case it sort of is, but still; as with any approximation, there will be an associated error.
Jake
I also don't mean to be demeaning about the experiment in any way; it's just similar to Young's single and double slit experiment, and it's like they need to find a legit application to explain what's going on the students (by measuring a human hair), rather than just explaining the theory to you, which you could definitely understand.
Post by: Jakeybaby on August 29, 2016, 09:56:18 pm
I also don't mean to be demeaning about the experiment in any way; it's just similar to Young's single and double slit experiment, and it's like they need to find a legit application to explain what's going on the students (by measuring a human hair), rather than just explaining the theory to you, which you could definitely understand.Instead of using the equation which they used, we were instructed to use the following equation:
Bandwidth = (Wavelength * L) / a
Where L: Distance from hair to screen
a: width of hair sample
Post by: znaser on August 29, 2016, 10:01:43 pm
Hey! I would say exactly what you've written there; that only a very specific TYPE of photon could be absorbed by the electrons, for all the reasons outlined above. I would definitely give you full marks; you've explained the concepts, included relevant formulas, and done it all succinctly. Well done; the question is certainly worded badly, but I don't see any other way of answering it.
Lol that explains a lot. Thought there was some weird fact about the ground state or something. Thank you rlly helped clear things ;D
Post by: jakesilove on August 29, 2016, 10:17:41 pm
Instead of using the equation which they used, we were instructed to use the following equation:
Bandwidth = (Wavelength * L) / a
Where L: Distance from hair to screen
a: width of hair sample
I would probably just focus on the errors in those variables then; Wavelength, and use of a ruler. I can't imagine you need much more depth than that!
Post by: Jakeybaby on August 29, 2016, 10:31:42 pm
Post by: Neutron on August 31, 2016, 10:50:46 pm
Neutron
Post by: jamonwindeyer on August 31, 2016, 11:06:17 pm
I'm back! Still drained from trials though rip it's been 2 weeks :/ Anyhow, first off, I just wanted to thank all of you for helping me throughout the year! This forum has been so so amazing and ahh I can't believe only the actual hsc is left (yay :/) heh okay so, with p-type semiconductors, you know how the primary mobile charge carriers are holes in the valence band? In this case, does it not matter that the free charge carriers are not in the conduction band? Cause I'm doing an HSC question and the solutions said "These extra holes lower the energy required for charge to be mobilised to the conduction band and move as current". I'm just a bit confused cause I thought the acceptor level formed was so that the electrons could easily jump from the valence band into the acceptor level, meaning it left behind more holes in the valence? Nothing to do with the conduction band? Thank you y'alllll
Neutron
Hey Neutron! ;D welcome back, thank you so much for your kind words :)
That wording does seem a tad strange to me, is it from a sample solution from BOSTES or? :)
Post by: Neutron on September 01, 2016, 01:55:17 pm
Hey Neutron! ;D welcome back, thank you so much for your kind words :)
That wording does seem a tad strange to me, is it from a sample solution from BOSTES or? :)
Yeah it was, from the success one books :O Are we not supposed to follow those??
Post by: jakesilove on September 01, 2016, 02:26:02 pm
Yeah it was, from the success one books :O Are we not supposed to follow those??
Basically, the answer given above is correct, and so is your understanding of it. The acceptor/donor bands are just representations of the fact that it is easier for electrons to be mobilised to form current. However you want to say that is up to you, because in reality the reasons ISN'T taught at an HSC level. So, saying anything along the lines of 'doping the semi-conductor increases the ease with which current can flow. This is because of excess electrons/holes etc. etc.' will definitely get you the marks!
Post by: jamonwindeyer on September 01, 2016, 06:37:55 pm
Yeah it was, from the success one books :O Are we not supposed to follow those??
So the Success One answers are correct (they are BOSTES, after all), but they aren't always 100% clear. I don't like the way that answer was worded, for example ;)
But here is an explanation: Remember that as Jake said, the valence and conduction bands are representations/models of reality. The conduction band represents a set of energies that an electron/hole must attain to break free from a covalent bond and travel freely (within reason) through the lattice. The conduction band isn't actually 'above' the valence band; that is just the way we represent it. There is an energy level separating them, not physical space :)
Essentially, if an electron is moving through the atom, then it is in the conduction band (for our understanding, at least). It must be, because if it wasn't, it would still be forming part of a covalent bond. The acceptor impurity atoms that are introduced generate additional holes in the lattice; an adjacent electron may therefore jump to fill the hole. It had the energy to break its covalent bond to do so, and therefore, in the conduction band :)
Kinda technical, but does that make sense? :)
Post by: conic curve on September 06, 2016, 02:31:08 pm
can anyone tell me the expected way of doing a question like this for year 11 physics? i looked it up and found ways which correlate to actual echo sounders, but realise i've never done this stuff in class.
any help is greatly appreciated!
Post by: jamonwindeyer on September 06, 2016, 02:39:08 pm
The captain of a fishing boat uses an echo sounder to determine the depth of a school of fish below the boat. The captain finds that the reflected waves return after o.15s and o.20s. The captain believes the first reflectin to be from the fish while the seconds is from the ocean floor. If the speed of sound in seawater is 1440m/s, determine the depth of the sea floor. (3 marks)
can anyone tell me the expected way of doing a question like this for year 11 physics? i looked it up and found ways which correlate to actual echo sounders, but realise i've never done this stuff in class.
any help is greatly appreciated!
Hey conic! You'd just be using \(\text{Speed}=\frac{\text{Distance}}{\text{Time}}\) formulae. Since the time taken to return to the boat is 0.20 seconds, and the speed of sounds is 1440m/s:
However, note that this is the distance travelled to the floor and back, so the actual answer is:
Post by: jakesilove on September 06, 2016, 02:39:31 pm
The captain of a fishing boat uses an echo sounder to determine the depth of a school of fish below the boat. The captain finds that the reflected waves return after o.15s and o.20s. The captain believes the first reflectin to be from the fish while the seconds is from the ocean floor. If the speed of sound in seawater is 1440m/s, determine the depth of the sea floor. (3 marks)
can anyone tell me the expected way of doing a question like this for year 11 physics? i looked it up and found ways which correlate to actual echo sounders, but realise i've never done this stuff in class.
any help is greatly appreciated!
If something travels for x m/s for y seconds, then it has traveled xy meters.
The sound took 0.20s to go from the boat, to the sea floor, and back. That means the sea floor is 0.10s away (half the time will be spent going down, half will be spent going up). Therefore, given the speed of sound in water, it is
away
Post by: levendibigd on September 13, 2016, 11:39:08 pm
Bohr’s model of the atom'.
Post by: jakesilove on September 14, 2016, 12:06:21 pm
Hi there! Could I have a hand with this syllabus dot point 'Analyse the significance of the hydrogen spectrum in the development of
Bohr’s model of the atom'.
Hey! So, before Bohr, electrons just sort of floated around in a cloud and did their own thing (in fact, they weren't even electrons; just a negative cloud!). However, some interesting observations of the hydrogen spectrum lead to a change in the theory. EM waves were directed towards Hydrogen atoms at varying frequencies, and then turned off. Light was emitted from the Hydrogen atoms, but only ever at very specific wavelengths based on the incident frequencies! If you put in 100Hz, you'd always get the same colour out! Weird, if the cloud was just random. Instead, Bohr suggested that what was happening was that electrons could only occupy very specific orbitals. When the frequency of EM radiation was being directed towards the Hydrogen, electrons had the ability to 'jump up' energy levels. When the frequency stopped, they would 'jump down', emitting light. However, since there was only specific energy levels they could occupy, there was only specific wavelengths of light that could be emitted! This explains the specificity of the Hydrogen spectrum. Let me know if you need me to clarify any of this!
Jake
Post by: conic curve on September 15, 2016, 05:21:23 pm
A spaceship in deep space fires its rocket motor that exerts a force of 8.00 x 10^4 N. The rocket continues to fire while the spaceship moves through a distance of 1.0 x 10^7 m. If the 6.0 x 104 kg spaceship had been moving at 4.5 x 10^3 m s^-1 before the rocket motor was fired, how fast is it moving now?
Method 1:
Change in kinetic energy = Fs = 8x10^11 J
Therefore, (Δv)^2=(8x10^11)/(m/2) = 2.67x10^7, Δv = 5163 ms^-1
Therefore, final v=9663ms^-1
Method 2:
Total kinetic energy = 8x10^11+3x10^4x(4.5x10^3)^2 = 1.41x10^12 J
v = (1.41x10^12/(3x10^4))^(1/2) = 6855 ms^-1
Post by: jamonwindeyer on September 15, 2016, 06:05:11 pm
Hey, I've done a question two ways and I have no idea which one is right, can someone explain why one is wrong.
Hey conic! Subtle thing here, the issue is that in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer! ;D
Had to think about that one for a tad, happy to be corrected if someone spots a misinterpretation :)
Post by: jakesilove on September 15, 2016, 06:11:32 pm
Hey conic! Subtle thing here, the issue is that in Method 1, you are assuming that the mass of the rocket is constant. It is not, it is losing fuel, and so its mass is actually decreasing. Therefore, you cannot say the change in kinetic energy is purely due to the change in velocity. Your second method is more appropriate and should be giving you the correct answer! ;D
Had to think about that one for a tad, happy to be corrected if someone spots a misinterpretation :)
I'm not entirely sure whether the second method makes up for this; I agree that the 'total kinetic energy' is correct (as it is just initial kinetic energy plus the work done, all of which would go towards increasing the speed of the shuttle), however when determining the final speed, the initial mass of the shuttle is used. Therefore, this would be the final speed if the shuttle doesn't lose mass. Which is the same assumption made in the first calculation. So, to be perfectly honest, I'm not entirely sure what the problem is here!
Post by: jamonwindeyer on September 15, 2016, 06:14:30 pm
I'm not entirely sure whether the second method makes up for this; I agree that the 'total kinetic energy' is correct (as it is just initial kinetic energy plus the work done, all of which would go towards increasing the speed of the shuttle), however when determining the final speed, the initial mass of the shuttle is used. Therefore, this would be the final speed if the shuttle doesn't lose mass. Which is the same assumption made in the first calculation. So, to be perfectly honest, I'm not entirely sure what the problem is here!
Ahhh good point, I shall have another look after my lab ;)
Post by: jakesilove on September 15, 2016, 06:17:40 pm
Ahhh good point, I shall have another look after my lab ;)
I can't figure out why this happens, but basically the two answers are different because, when you deconstruct them, it gives you
(ie. one calculation uses the left, one uses the right, and so obvious as the above relations is false, the answers are different.)
Not sure why at the moment!
Post by: jamonwindeyer on September 16, 2016, 01:05:53 am
So I'm now even more sure that the second answer is the correct one, but I still can't figure out why the first one doesn't work....
From Rui: Fixed up mistake in LaTeX leading to a wrong equation :P
Post by: RuiAce on September 16, 2016, 06:51:21 am
Difference is a linear operator. Not quadratic.
Post by: jamonwindeyer on September 16, 2016, 09:18:15 am
Literally just read Jake's incorrect equation twice lol
Difference is a linear operator. Not quadratic.
Ahhh, right you are, how subtle! I've never actually seen that pop up in HSC Physics, which is surprising, because I bet this would come up a fair bit :P
Post by: jakesilove on September 16, 2016, 12:04:13 pm
Literally just read Jake's incorrect equation twice lol
Difference is a linear operator. Not quadratic.
Yeah sorry that's what I was getting at (wasn't sure how to write not equal to in Latex, so just explained it in words beneath and didn't have time to apply it).
Post by: Adriaclya on September 17, 2016, 09:16:56 pm
I'm stumped. Is trajectory and projectile motion the same thing? 0.o
Post by: RuiAce on September 17, 2016, 09:52:18 pm
Hi!Trajectory is just the actual path the projectile moves through.
I'm stumped. Is trajectory and projectile motion the same thing? 0.o
Post by: Klexos on September 17, 2016, 09:54:21 pm
Can anyone distinguish this for me?
Post by: jamonwindeyer on September 17, 2016, 11:46:52 pm
I don't know if anyone did Medical Physics but I'm having a hard time understanding the difference between T1 & T2 weighted for MRI.
Can anyone distinguish this for me?
Hey! It was my option (Jake's too), happy to lend a hand! This was probably the most difficult concept in the course for me, so I hope I can explain it in a way that is beneficial ;)
Okay, so we know that when we expose nuclei to radio waves of an appropriate frequency, they resonate and precess, and this causes change to the net magnetisation vector. Relaxation is related to this vector.
T1 Relaxation: As nuclei are exposed to radio waves, they flip into anti-parallel alignments. This shrinks the component of the net magnetisation vector in the direction of the file. T1 Relaxation concerns this vector returning to its initial value (specifically, the T1 relaxation time is when the vector returns to 63% of the initial value), as nuclei dissipate their energy into the surrounding lattice. This diagram does a pretty good job showing what this looks like (ignore the equation, it's beyond this course):
(http://mriquestions.com/uploads/3/4/5/7/34572113/2140077_orig.gif)
T2 Relaxation: For reasons slightly beyond the scope of the courses, exposing precessing nuclei to radio frequencies causes the net magnetisation vector to shift into the transverse plane, perpendicular to the applied field. Remember that previously, there was no component of the vector in this direction, it was only parallel to the field. The result is that the precession of the nuclei as a whole loses coherency and phase. As the nuclei relax, those precessing about this transverse axis transfer their energy to those precessing parallel to the field, and the component of the vector perpendicular to the field shrinks. T2 Relaxation Time is when the vector shrinks back to 37% of its initial. Again, see diagram:
(http://mriquestions.com/uploads/3/4/5/7/34572113/2581257_orig.gif)
That is the complicated version: If you want simple, go for this instead. Applying a magnetic field rotates the net magnetisation vector 90 degrees, so it ends up perpendicular to the applied field. T1 Relaxation time measure the parallel component increasing to normal, T2 relaxation time concerns the perpendicular component shrinking away. This is kind of why the two numbers (63% and 37%) are related (adding to 100%), because they concern the same vector ;D
I hope this makes sense! Understanding this demands a rock solid knowledge of everything else in this section; and then a lot of dedication and research to really wrap your head around it!
Oh! So for actually weighting to detect these relaxation times, a few things to note about MRI imaging radio waves first.
Repetition Time (TR): Elapsed time between pulses of radio waves
Echo Delay Time (TE): Time delay between the sending of radio waves and measurement of emitted signals
For T1 weighted images, we want to emphasise areas with short T1 relaxation time (meaning, they dissipate energy very quickly). This is achieved by minimising the repetition time, meaning that only nuclei with short T1 will have the time to dissipate their energy before the next pulse.
For T2 weighted images, we want to emphasise areas with a long T2 relaxation time (meaning, it takes a long (relative) time for them to return to their initial coherency, and thus, will emit MR signals for longer). We do this by maximising echo delay time, so that by the time we take the measurement, only nuclei with long T2 will still be emitting MR signals ;D
Post by: jamonwindeyer on September 17, 2016, 11:47:31 pm
Hi!
I'm stumped. Is trajectory and projectile motion the same thing? 0.o
Hey! Essentially, like analysing the trajectory of a projectile is what we do in projectile motion ;)
Post by: conic curve on September 18, 2016, 12:11:03 am
Post by: jamonwindeyer on September 18, 2016, 12:16:35 am
Damm, no moderators did quanta to quarks did they?
Spencerr did I believe, and Jake has a good knowledge of that Option from his own studies! Definitely ask any questions you have ;D
Post by: conic curve on September 18, 2016, 12:22:56 am
Spencerr did I believe, and Jake has a good knowledge of that Option from his own studies! Definitely ask any questions you have ;D
Oh fantastic, no need to worry
Post by: Klexos on September 18, 2016, 04:01:17 pm
Hey! It was my option (Jake's too), happy to lend a hand! This was probably the most difficult concept in the course for me, so I hope I can explain it in a way that is beneficial ;)
Okay, so we know that when we expose nuclei to radio waves of an appropriate frequency, they resonate and precess, and this causes change to the net magnetisation vector. Relaxation is related to this vector.
T1 Relaxation: As nuclei are exposed to radio waves, they flip into anti-parallel alignments. This shrinks the component of the net magnetisation vector in the direction of the file. T1 Relaxation concerns this vector returning to its initial value (specifically, the T1 relaxation time is when the vector returns to 63% of the initial value), as nuclei dissipate their energy into the surrounding lattice. This diagram does a pretty good job showing what this looks like (ignore the equation, it's beyond this course):
(http://mriquestions.com/uploads/3/4/5/7/34572113/2140077_orig.gif)
T2 Relaxation: For reasons slightly beyond the scope of the courses, exposing precessing nuclei to radio frequencies causes the net magnetisation vector to shift into the transverse plane, perpendicular to the applied field. Remember that previously, there was no component of the vector in this direction, it was only parallel to the field. The result is that the precession of the nuclei as a whole loses coherency and phase. As the nuclei relax, those precessing about this transverse axis transfer their energy to those precessing parallel to the field, and the component of the vector perpendicular to the field shrinks. T2 Relaxation Time is when the vector shrinks back to 37% of its initial. Again, see diagram:
(http://mriquestions.com/uploads/3/4/5/7/34572113/2581257_orig.gif)
That is the complicated version: If you want simple, go for this instead. Applying a magnetic field rotates the net magnetisation vector 90 degrees, so it ends up perpendicular to the applied field. T1 Relaxation time measure the parallel component increasing to normal, T2 relaxation time concerns the perpendicular component shrinking away. This is kind of why the two numbers (63% and 37%) are related (adding to 100%), because they concern the same vector ;D
I hope this makes sense! Understanding this demands a rock solid knowledge of everything else in this section; and then a lot of dedication and research to really wrap your head around it!
Oh! So for actually weighting to detect these relaxation times, a few things to note about MRI imaging radio waves first.
Repetition Time (TR): Elapsed time between pulses of radio waves
Echo Delay Time (TE): Time delay between the sending of radio waves and measurement of emitted signals
For T1 weighted images, we want to emphasise areas with short T1 relaxation time (meaning, they dissipate energy very quickly). This is achieved by minimising the repetition time, meaning that only nuclei with short T1 will have the time to dissipate their energy before the next pulse.
For T2 weighted images, we want to emphasise areas with a long T2 relaxation time (meaning, it takes a long (relative) time for them to return to their initial coherency, and thus, will emit MR signals for longer). We do this by maximising echo delay time, so that by the time we take the measurement, only nuclei with long T2 will still be emitting MR signals ;D
Wow .____, My teacher legit said a couple of sentences and pointed at the PIF, I'm going to have to try understand this and get back to you XD
Thank you, Jamon!
Post by: jamonwindeyer on September 18, 2016, 04:34:56 pm
Wow .____, My teacher legit said a couple of sentences and pointed at the PIF, I'm going to have to try understand this and get back to you XD
Thank you, Jamon!
Ahaha yep, my teacher was a legend and taught this extremely well, I got a lot of detail! You'll never be asked for this much in a HSC Exam, so take as much or little of it as you need to understand what is happening!
Feel free to quote specific bits you don't get and I can help clarify it for you, I did a bit of a content dump on you ;)
Post by: conic curve on September 18, 2016, 05:43:04 pm
Ahaha yep, my teacher was a legend and taught this extremely well, I got a lot of detail! You'll never be asked for this much in a HSC Exam, so take as much or little of it as you need to understand what is happening!
Feel free to quote specific bits you don't get and I can help clarify it for you, I did a bit of a content dump on you ;)
Loving your indepth explanations there Jamon ;D
Post by: Neutron on September 21, 2016, 10:34:43 pm
A single turn coil is positioned in a region of uniform magnetic field with a strength of 0.2 T. The plane of the coil is at 45 degrees to the magnetic field. The coil is a square with 5cm sides and carries a current of 10.0A. Explain why the net force produced by the magnetic field on the coil is zero.
I didn't think the net force would be zero cause doesn't the coil rotate haha thank you!
Post by: jakesilove on September 21, 2016, 10:38:27 pm
Hey guys! Having trouble with this 2011 HSC question (even though i really shouldn't):
A single turn coil is positioned in a region of uniform magnetic field with a strength of 0.2 T. The plane of the coil is at 45 degrees to the magnetic field. The coil is a square with 5cm sides and carries a current of 10.0A. Explain why the net force produced by the magnetic field on the coil is zero.
I didn't think the net force would be zero cause doesn't the coil rotate haha thank you!
That's a totally understandable mistake to make! From memory, one side was labelled AB, and the other was labelled CD. Now, we know that the force on each size is going to be F=BILsin(theta). That will cause AB to 'rotate' in a particular direction, right? However, the force on CD will be exactly the same as the force on AB (same current, length etc.) but in the opposite direction! Therefore, the two forces EXACTLY cancel out. This means that the NET force on the coil is zero, but there IS a force on each side (which causes the rotor to rotate). I hope this makes sense; let me know if I can clarify further!
Jake
Post by: brontem on September 23, 2016, 12:47:25 pm
Post by: jakesilove on September 23, 2016, 03:57:48 pm
Hey can someone help me with part (ii) of this question? :D :D
Hey!
I would structure an answer to include the following. First, we recall that moving particles have a wavelength (the de Broglie wavelength, with formula lambda=h/mv). Now, for orbiting electrons, there are only certain set 'distances' (ie. quantum numbers) that allow the electrons to orbit WITHOUT causing destructive interference with their own waves. This explains why there are set orbits around the nucleus. When a photon is incident on a Hydrogen atom, it is really going to hit an electron. If the photon has the exact right amount of energy to 'jump' the electron to another allowed energy level, it will be absorbed. However, most photons will have a different energy (calculated by E=hf), and so cannot be absorbed by the electron. Thus, they are reflected!
I hope that makes sense! Let me know if you need me to clarify anything.
Jake
Post by: yaboiaderler on September 26, 2016, 09:59:04 pm
1. Topic: Space
In special relativity, in regards to calculations of mass dilation, length contraction and time dilation, I have some difficulty in determining which values are which - how do you know/determine, for example in length contraction, which value is lo and lv?
2. Topic: Motors and Generators
In some multiple choice questions, I have come across questions that ask you to determine the direction of an eddy current when say, a metal sheet is pulled in/out of a magnetic field. How do you actually determine the direction? (Lenz's Law is confusing)
Thanks!
Post by: RuiAce on September 26, 2016, 10:02:46 pm
Hiya, I have two content based questions:The way I learnt how to distinguish between lo and lv is that instead of considering rest frame and moving frame, I started considering two things:
1. Topic: Space
In special relativity, in regards to calculations of mass dilation, length contraction and time dilation, I have some difficulty in determining which values are which - how do you know/determine, for example in length contraction, which value is lo and lv?
2. Topic: Motors and Generators
In some multiple choice questions, I have come across questions that ask you to determine the direction of an eddy current when it its pulled in/out of a magnetic field. How do you actually determine the direction? (Lenz's Law is confusing)
Thanks!
1) Which frame is your point of view. Your point of view is the lo
2) Use logic. Length contracts, time and mass dilate. So if the wrong thing happens (say, your length got longer when it should've been shorter), just swap them around lol.
And for Lenz's law, I think about what's going on. Say a magnet with north pole is travelling downwards. The way I think about it is that it's introducing "too much north". So (not literally, but) by Lenz's law it wants to get rid of the north. How do we get rid of the north? Put a south back in there.
So I figure out the polarity first. THEN I choose to use the grip rule (or push rule where applicable).
The other mods can give more tips here. This was just me.
Post by: jamonwindeyer on September 26, 2016, 10:21:46 pm
Hiya, I have two content based questions:
1. Topic: Space
In special relativity, in regards to calculations of mass dilation, length contraction and time dilation, I have some difficulty in determining which values are which - how do you know/determine, for example in length contraction, which value is lo and lv?
2. Topic: Motors and Generators
In some multiple choice questions, I have come across questions that ask you to determine the direction of an eddy current when say, a metal sheet is pulled in/out of a magnetic field. How do you actually determine the direction? (Lenz's Law is confusing)
Thanks!
Ditto with Rui above!! Common sense is the best way to check which is which in those equations; since it will always change depending on frame of reference as to which is which. Remember, as Rui said, that lengths get shorter as you watch something fly past, times/masses get larger as you watch something fly past ;D
I'd love to give you an example of how to figure out the direction, but I'd prefer it to be something you are specifically having trouble with. Reckon you could snap a picture of an example? ;D
Post by: MysteryMarker on September 27, 2016, 11:11:55 am
Can someone please explain to me how the answer to this question is A? I thought that by using np/ns = Is/Ip, Is = (np x Ip)/ns, thus by decreasing the number of turns in the secondary coil, the current in the secondary coil would increase and so too would the deflection underwent by the pointer in the galvanometer.
Post by: jamonwindeyer on September 27, 2016, 03:13:16 pm
Hey guys
Can someone please explain to me how the answer to this question is A? I thought that by using np/ns = Is/Ip, Is = (np x Ip)/ns, thus by decreasing the number of turns in the secondary coil, the current in the secondary coil would increase and so too would the deflection underwent by the pointer in the galvanometer.
I saw this in the 2015 Paper; it's a trick, and I reckon a HEAP of people would have put B as an answer.
Your reasoning is correct, but with a minor oversight. Let me first explain why A is correct.
We have a voltage source attached to the primary coil. The voltage, current and resistance of this coil are related by Ohms Law:
But the resistance in the primary coil is dependent on the number of turns in the coil. So, we could say \(V=kIn\), where \(k\) is just some constant (perhaps, the resistivity of the wire). So if we decrease the amount of turns in the primary coil, without decreasing the voltage, that MUST mean that the current in the primary coil increases as a result.
By the law of conservation of energy, the power into a transformer is equal to the power out of a transformer (ignoring losses). We've maintained the same voltage at the input, but increased current, so the input power has increased. At the other end, our current increases to compensate for this. For totality, the effects at the output (in the secondary coil) are:
- The voltage increases due to the changed turn relationship between the coils, \(V_s=\frac{V_pN_s}{N_p}\)
- The current increases due to the increased amount of power going INTO the transformer
So the question remains, why is B incorrect? Well this is where we get tricky, because well, I'm not sure. It is, in general correct, at least to my understanding. My only explanation for why A is chosen would be that the deflection is likely greater. I also notice that the voltage source is DC (not sure if that is intentional), which could mean that the deflection is only brief, not constant, while the switch is flicked. However, that shouldn't change the reasoning in my opinion, all should still work as you'd expect.
I'm yet to find a good reason why A is chosen over B. I know why A is correct, but not why B isn't, ahaha :P
Post by: RuiAce on September 27, 2016, 07:04:06 pm
Which would mean somehow I'd have to contradict off Jamon's argument.
Generally with these questions regarding transformers, I've found that the only thing we can assume constancy is power. For some reason resistance cannot be taken as constant. So since Jamon substituted resistance for n*resistivity, since n is obviously constant it must be that the resistivity changes somehow.
In other words, k is a variable, not a constant.
Post by: pels on September 27, 2016, 07:41:23 pm
Got some questions for physics.
For this specific dotpoint,
process information to identify some of the metals, metal alloys and compounds that have been
identified as exhibiting the property of superconductivity and their critical temperatures
What exactly are we required to know here? In my notes I have Aluminium and Tin and their corresponding Critical Temperatures, but is that all we need to know for that point?
Additionally, what are the differences between AC Synchronous Motors and AC Induction motors, in terms of the principles used and components/structure?
Also, what is the difference between magnetic flux and magnetic flux density. Is it simply that flux density is B X A?
Also one last question. What year 11 material in important to know for physics, even though most of it is directly linked to year 12 content.
Thanks.
Post by: RuiAce on September 27, 2016, 07:57:53 pm
Also, what is the difference between magnetic flux and magnetic flux density. Is it simply that flux density is B X A?
Also one last question. What year 11 material in important to know for physics, even though most of it is directly linked to year 12 content.
Thanks.
________________________
Nothing much. Although I found that remembering how to do prelim calculations helped me here and there (e.g. just simple things such as q = I * t, V = W/q)
Also with those superconductors I think just memorised their names and critical temperatures like you suggested. Pick one of each though, so for the allow you could do yttrium barium cuprate. Letting someone else take over from here though
Post by: pels on September 27, 2016, 08:43:30 pm
Still need that explanation of motors and generators.
Other random question I thought of:
Is the only difference in motors and generators the fact that they convert energy differently (e.g. electrical to mechanical and mechanical to electrical), as well as the split ring commutator vs the slip rings?
Cheers
Post by: RuiAce on September 27, 2016, 08:52:54 pm
The split and slip rings - not quite. But yeah the energy conversion is just reversed between a motor and a generator.
The split and slip rings distinguish between DC and AC
Post by: jamonwindeyer on September 27, 2016, 10:49:34 pm
Thanks heaps.
Still need that explanation of motors and generators.
Other random question I thought of:
Is the only difference in motors and generators the fact that they convert energy differently (e.g. electrical to mechanical and mechanical to electrical), as well as the split ring commutator vs the slip rings?
Cheers
I got you! So an AC Synchronous Motor is just your typical motor structure; slip rings, rotor, stator, etc. Exactly as you learn for DC Motors ;D
An AC Induction Motor is a different animal entirely. The rotation is a consequence of Lenz's Law, induced by a rotating magnetic field. There is a section on it in the second post in this thread! Let me know if I can explain it in a little more depth for you though :)
Post by: pels on September 27, 2016, 11:40:30 pm
:)
Cheers
Post by: jakesilove on September 28, 2016, 08:44:51 am
Yeah i might need an explanation on it still.
:)
Cheers
In my HSC year, I used this video to understand Induction motors. Don't worry if you don't perfectly get what's going on; you just need to be able to describe it!
Post by: pels on September 28, 2016, 01:37:31 pm
:) ;D
Post by: dylan862 on September 28, 2016, 09:37:09 pm
Post by: jamonwindeyer on September 28, 2016, 09:59:35 pm
Best way to study for astrophysics??
Speaking in general for Physics rather than for astrophysics particularly - Past papers! Practice makes perfect for a subject like Physics, do as many practice questions as you can :) you might want to read this article I wrote that taks about this a tad :)
You might also want to try things like dot point brainstorms; brainstorming everything you remember about a specific dot point. Palm cards for terminology work too. Lots of cool stuff to try! But past papers/practice questions should take precedence ;D
Post by: Neutron on September 29, 2016, 05:24:29 pm
Can someone please explain to me how a transistor operates as a switch or amplifier? I've never properly understood it :/ and also what are transistor radios and how do they work?
Cheers!
Post by: jamonwindeyer on September 29, 2016, 07:32:12 pm
Yo!
Can someone please explain to me how a transistor operates as a switch or amplifier? I've never properly understood it :/ and also what are transistor radios and how do they work?
Cheers!
Hey Neutron! To properly understand the workings of a transistor is beyond the HSC course. I'm studying transistors now; definitely nothing nice ;) you need to understand PN junctions, but not how they are used in a transistor, so you are all set! I can help you understand PN junctions if that is something you need help with :)
Transistor Radio = Radio that uses transistors in its circuitry (again, no more detail required) ;D
It seems strange that the operation of a transistor isn't covered, but tbh, it's a little too complex to properly understand at Year 12 level :P
Post by: RuiAce on September 29, 2016, 07:35:34 pm
Hey Neutron! To properly understand the workings of a transistor is beyond the HSC course. I'm studying transistors now; definitely nothing nice ;) you need to understand PN junctions, but not how they are used in a transistor, so you are all set! I can help you understand PN junctions if that is something you need help with :)Can confirm. My teacher explicitly told us that whilst we needed to know everything about diodes, by contrast virtually nothing about how transistors work
Transistor Radio = Radio that uses transistors in its circuitry (again, no more detail required) ;D
It seems strange that the operation of a transistor isn't covered, but tbh, it's a little too complex to properly understand at Year 12 level :P
Post by: jakesilove on September 29, 2016, 09:31:56 pm
Can confirm. My teacher explicitly told us that whilst we needed to know everything about diodes, by contrast virtually nothing about how transistors work
I spent ages before Trials trying to figure out how transistors work, only to find out explicitly that we can't be asked about them! There's enough complicated stuff in the syllabus to keep you occupied; why make things harder on yourself?
Post by: RuiAce on September 29, 2016, 09:33:51 pm
I spent ages before Trials trying to figure out how transistors work, only to find out explicitly that we can't be asked about them! There's enough complicated stuff in the syllabus to keep you occupied; why make things harder on yourself?On my part, it was because our physics teacher gave us an assignment on transistors before he had taught us the material (purposefully...) and we had to ask him explicitly to end our confusion lel
Post by: samuels1999 on September 30, 2016, 12:11:38 am
Really appreciate what you do!
I am moving into year 12 next term, and with regard to Physics, do you have any advice on which elective topic is the best to do between "From Quanta to Quarks" and "Medical Physics"?
Thanks
Samuel
Post by: jamonwindeyer on September 30, 2016, 01:40:15 am
Hi Jamon and Jake
Really appreciate what you do!
I am moving into year 12 next term, and with regard to Physics, do you have any advice on which elective topic is the best to do between "From Quanta to Quarks" and "Medical Physics"?
Thanks
Samuel
Hey Sam! Thanks for your kind words ;D
I'm pretty sure Jake and I would both recommend Medical Physics, we both did it as our Option, and speaking personally it was seriously interesting. More interesting than most of the core (at least for me), and not too difficult either! Quanta to Quarks is more popular though, and so probably has more resources about :)
It doesn't really matter either way; do you get to pick your Option do you? ;D
Post by: cherryred on September 30, 2016, 08:17:50 am
How would you suggest I draw my diagrams for this question:
A student claims that a DC generator is an ‘electric motor in reverse’.
Analyse this claim with reference to the structure and function of a simple DC
generator and an electric motor.
Include diagrams in your answer.
THANK YOU :)
Post by: RuiAce on September 30, 2016, 08:45:03 am
Hey,I'll assume you can handle the written part of the response.
How would you suggest I draw my diagrams for this question:
A student claims that a DC generator is an ‘electric motor in reverse’.
Analyse this claim with reference to the structure and function of a simple DC
generator and an electric motor.
Include diagrams in your answer.
THANK YOU :)
For the diagrams, when you draw a DC motor and a DC generator, the bulk of the two diagrams should be identical. You will need an armature, commutator, brushes and the magnetic field in both cases.
The differences lie in thus:
In a motor, we convert electrical energy to kinetic energy. Because electrical energy is what's being converted, we need to draw a battery (cell) connected in a closed circuit to the brushes.
In a generator, we convert kinetic energy to electrical energy. Somehow we must provide the (rotational) kinetic energy, so it really helps to draw an axle for us to rotate. Also, because we usually use a generator to power something up, instead of a battery we will draw a load (this could be a light bulb or many other things).
Post by: Spencerr on September 30, 2016, 11:28:20 am
A student claims that a DC generator is an ‘electric motor in reverse’.
Analyse this claim with reference to the structure and function of a simple DC
generator and an electric motor.
Adding on to Rui's answer, I would suggest that the DC Motor has the Load (where the motor does work as it rotates) and the DC Generators has a handle/ crank which can be rotated to input Mechanical energy. I would say that both the motors and generators use the axle for rotation. Rui's covered the rest :)
I am moving into year 12 next term, and with regard to Physics, do you have any advice on which elective topic is the best to do between "From Quanta to Quarks" and "Medical Physics"?
Samuel
Since I do Q2Q, i'll provide an opposing view :P Q2Q is quite an interesting topic. It's hard to learn at first but afterwards everything fits together neatly and perfectly. Since we study Ideas to Implementation, and we learn about photons etc., Q2Q is just an extension of all that and an understanding of Q2Q really helps mix in with I2I. At this point, I find it difficult to even separate which topics are in which elective because they're so related :)
Post by: RuiAce on September 30, 2016, 11:38:03 am
Adding on to Rui's answer, I would suggest that the DC Motor has the Load (where the motor does work as it rotates) and the DC Generators has a handle/ crank which can be rotated to input Mechanical energy. I would say that both the motors and generators use the axle for rotation. Rui's covered the rest :)Oh my bad. When I meant axle I implicitly included the handle. Thanks for making it more explicit there
Since I do Q2Q, i'll provide an opposing view :P Q2Q is quite an interesting topic. It's hard to learn at first but afterwards everything fits together neatly and perfectly. Since we study Ideas to Implementation, and we learn about photons etc., Q2Q is just an extension of all that and an understanding of Q2Q really helps mix in with I2I. At this point, I find it difficult to even separate which topics are in which elective because they're so related :)
But I don't believe an axle is necessary for a motor; in practice I'm pretty sure they are there but for a "simple" motor it's not necessary; the armature can rotate without the motor.
Don't think a simple motor needs to do work either :P
Post by: amina_98 on September 30, 2016, 07:21:51 pm
thanks
Post by: ml125 on September 30, 2016, 07:41:26 pm
Hi guys so my school decided to do the quanta to quarks option this year and a group of classmates found it challenging and decided to study and teach themselves the medical physics option. in our trials however they didnt have a medical phyiscs section so it was left. but obviously in the hsc exam the medical physics questions are going to be there. but the question is will it be marked? will they have to show some sort of "proof" that they did 30 hours of medical physics study since the course requires 120 hours?Whatever option you answer in the hsc will be the one that's marked. It doesn't matter which option you did at school.
thanks
Post by: conic curve on September 30, 2016, 08:22:24 pm
Post by: ml125 on September 30, 2016, 09:25:45 pm
Just curious but did an of you find the option harder than the core?Found all of it to be of pretty similar difficulty tbh. Doing Quanta though :D
Post by: conic curve on October 01, 2016, 01:11:49 pm
Found all of it to be of pretty similar difficulty tbh. Doing Quanta though :D
Yay another person doing Quanta (I'm doing it next year though) :D
Post by: RuiAce on October 01, 2016, 01:12:36 pm
Post by: conic curve on October 01, 2016, 07:47:34 pm
Also when they ask for "change in GPE" how do you do those types of questions, I never got them
Post by: ml125 on October 02, 2016, 01:27:08 am
In GPE questions when it asks "calculate the energy________" what are they asking?If it was something like "Calculate the energy required to lift an object from height x to height y," you would need to calculate GPE at points x and y, then find the difference between these two values. This is basically a "change in GPE" question - just need to find two separate values for GPE depending on the scenario given in the question then subtract. It would help if there was a specific question you had in mind so we could explain it more clearly.
Also when they ask for "change in GPE" how do you do those types of questions, I never got them
Post by: jakesilove on October 02, 2016, 09:37:01 am
If it was something like "Calculate the energy required to lift an object from height x to height y," you would need to calculate GPE at points x and y, then find the difference between these two values. This is basically a "change in GPE" question - just need to find two separate values for GPE depending on the scenario given in the question then subtract. It would help if there was a specific question you had in mind so we could explain it more clearly.
Spot on
Post by: samuels1999 on October 02, 2016, 05:28:34 pm
Regarding year the 12 Physics course, I just wanted to know: is a student that does Chemistry advantaged over a student who doesn't in certain topic areas?
Thanks,
Samuel
Post by: RuiAce on October 02, 2016, 05:33:17 pm
Hi Jake,Chemistry and physics are 97% distinct in what is taught. The advantage that a chemistry student has over a physics student is minimal. In fact, for the crossover between galvanic cells and cathode ray tubes, it causes more confusion arguably.
Regarding year the 12 Physics course, I just wanted to know: is a student that does Chemistry advantaged over a student who doesn't in certain topic areas?
Thanks,
Samuel
There is a tiny benefit in that a chemist will have an immediate understanding of valence and conduction bands in semiconductors from considering valence electrons in Year 11. It also helps at the point of some options, where chemists will be better adapted to the nuclear chemical equations.
Post by: jakesilove on October 03, 2016, 11:47:39 am
Hi Jake,
Regarding year the 12 Physics course, I just wanted to know: is a student that does Chemistry advantaged over a student who doesn't in certain topic areas?
Thanks,
Samuel
Very little advantage, and definitely nothing to worry about :)
Post by: pels on October 04, 2016, 12:00:48 pm
Very little advantage, and definitely nothing to worry about :)
As both a Chemistry and Physics student this year, it was at first confusing as in Physics its the positive current flow that starts from the positive ANODE, whereas in Chemistry its the negative current flow starting from the negative ANODE, but there is not much advantage I would say, other than if you do the Option Topic: From Quanta To Quarks as I currently am, and a small amount in From Ideas To Implementation, as you go over emission spectra and the atom, which is closely studied in Chemistry :)
Post by: WyattRey on October 04, 2016, 12:07:07 pm
Could someone explain to me why the relaxation time for hydrogen in different molecules differs? For example in water it would be longer than fat, etc. An answer I'm looking at says "Hydrogen in water has a longer relaxation time because the interaction with its surroundings (water) is small. The relaxation time of hydrogen in other molecules is shorter because it has greater interactions with its surroundings". I feel this is quite vague, even if its probably all you need to write.
Thanks :)
Post by: pels on October 04, 2016, 12:29:42 pm
Hi everyone,
Could someone explain to me why the relaxation time for hydrogen in different molecules differs? For example in water it would be longer than fat, etc. An answer I'm looking at says "Hydrogen in water has a longer relaxation time because the interaction with its surroundings (water) is small. The relaxation time of hydrogen in other molecules is shorter because it has greater interactions with its surroundings". I feel this is quite vague, even if its probably all you need to write.
Thanks :)
What sort of question are you answering?
Post by: jakesilove on October 04, 2016, 12:54:36 pm
Hi everyone,
Could someone explain to me why the relaxation time for hydrogen in different molecules differs? For example in water it would be longer than fat, etc. An answer I'm looking at says "Hydrogen in water has a longer relaxation time because the interaction with its surroundings (water) is small. The relaxation time of hydrogen in other molecules is shorter because it has greater interactions with its surroundings". I feel this is quite vague, even if its probably all you need to write.
Thanks :)
Hey! To be honest, I would probably stick to the answer you've described, because I'm not able to explain it in any more real detail. Personally, I always understood the relaxation time as dependent on the number of Hydrogen atoms present in a given area, rather than the way they interact with their surroundings. Needless to say, this is an extremely complicated area of Physics, and not one you need to know it great depth (other than in an explanatory sense). Sorry that I can't give you more detail!
Jake
Post by: Neutron on October 06, 2016, 05:17:46 pm
A scientist at a particle accelerator laboratory observes the lifetime of a particular subatomic particle to be 1.0 x 10^-6 s when it is travelling at 0.9999c. What would the lifetime of the particle be if it were stationary in the laboratory?
Cheers
Post by: RuiAce on October 06, 2016, 05:38:03 pm
That feel when you still don't really understand how you know which time is the dilated one and which time is the proper one.. Pls help with this question :DMight wanna look at posts #928 and #929.
A scientist at a particle accelerator laboratory observes the lifetime of a particular subatomic particle to be 1.0 x 10^-6 s when it is travelling at 0.9999c. What would the lifetime of the particle be if it were stationary in the laboratory?
Cheers
So using "common sense and logic" as we call it, I'd say that time dilates when the particle is moving, so it's lifetime is longer then. When it's at rest, it's lifetime should be shorter.
So we just sub the one in that will make it smaller
Post by: Neutron on October 06, 2016, 07:35:57 pm
Might wanna look at posts #928 and #929.
So using "common sense and logic" as we call it, I'd say that time dilates when the particle is moving, so it's lifetime is longer then. When it's at rest, it's lifetime should be shorter.
So we just sub the one in that will make it smaller
Yep that definitely makes sense but I think all this fatigue has gotten to my head. So with the particle moving, isn't it only in its reference frame where time dilates? So I'm using the example of like a spacecraft flying past the Earth, only the time on the spacecraft has dilated, the time on Earth is still proper. So using that logic, wouldn't the scientist's time be proper? Sorry if this is hella dumb :/
Neutron
Post by: Neutron on October 06, 2016, 07:38:22 pm
Two significant problems that will affect a manned spaceflight to Mars are:
-the changes in gravitational energy
-protecting the space vehicle from high-speed electrically charged particles from the Sun
Use your understanding of physics to analyse each of these problems.
Thankfully this isn't in the 2016 paper otherwise it would be 8 marks out the window! Rip
Neutron
Post by: Spencerr on October 06, 2016, 09:54:31 pm
One more thing, with this 8marker from 2010, I have absolutely no idea how to approach it, if you guys could perhaps give a dot point outline (or if you have time, a response :o ) on what I'm supposed to cover, that would be absolutely amazing!
Two significant problems that will affect a manned spaceflight to Mars are:
-the changes in gravitational energy
-protecting the space vehicle from high-speed electrically charged particles from the Sun
Use your understanding of physics to analyse each of these problems.
Thankfully this isn't in the 2016 paper otherwise it would be 8 marks out the window! Rip
Neutron
Hi, I'll try answer this the best I can, splitting the question into 2 parts for 4 marks each.
Changes in gravitational energy
- Gravitational potential energy is defined as -GmM/R, therefore as the rocket LEAVES Earth, it's gravitational energy increases. However, the law of conservation of energy (dictating that the total energy in a closed system is constant), suggests that energy must come from another source. This other source is the Kinetic energy / chemical energy within the fuel of the rocket. As the rocket leaves Earth, the KE of the rocket decreases (increasing GPE). This is a problem. In order to maintain the level of KE and increase the velocity of the rocket, more fuel must be burnt and used up. This involves the firing of rocket boosters to provide a constant thrust for the rocket.
- When the rocket enters the G field of Mars, the reverse situation happens. It begins to accelerate, being attracted by the gravity of the planet, towards the surface. As a result of its acceleration, it converts GPE to KE very quickly, losing altitude but gaining heat (from friction) and velocity. (Law of conservation of energy). This again is a problem as a velocity too high is difficult to control, and may be fatal for the probe/occupants inside. In order to solve this problem, rocket boosters are fired in the opposite direction to slow the descent of the probe, reducing its KE as it's GPE decreases.
High Speed Electrical charged Particles
- High speed electrically charged particles coming from the sun are often come from solar flares (where the sun emits a burst of solar radiation). They are quite problematic as they can damage sensitive equipment within the spacecraft and also penetrate through the walls of the spacecraft, damaging other parts inside and may be harmful for the occupants. Alongside causing damage, these charged particles interfere with communication (much like the situation of an ionisation blackout in re-entry), which is another problem.
- In order to solve this problem, the outer layer of the spacecraft can be coated with a material which can be electrically charged to repel these high speed charged particles. Insulating material which prevents these particles from penetrating into the spacecraft can also be applied. Furthermore, communication signals can be sent from the backend or the spacecraft (or whichever end isn't being exposed to these high speed charged particles).
I hope thats enough to get 7 marks at least haha but that's my attempt :)
Post by: jamonwindeyer on October 06, 2016, 11:31:04 pm
Hi, I'll try answer this the best I can, splitting the question into 2 parts for 4 marks each.
Changes in gravitational energy
- Gravitational potential energy is defined as -GmM/R, therefore as the rocket LEAVES Earth, it's gravitational energy increases. However, the law of conservation of energy (dictating that the total energy in a closed system is constant), suggests that energy must come from another source. This other source is the Kinetic energy / chemical energy within the fuel of the rocket. As the rocket leaves Earth, the KE of the rocket decreases (increasing GPE). This is a problem. In order to maintain the level of KE and increase the velocity of the rocket, more fuel must be burnt and used up. This involves the firing of rocket boosters to provide a constant thrust for the rocket.
- When the rocket enters the G field of Mars, the reverse situation happens. It begins to accelerate, being attracted by the gravity of the planet, towards the surface. As a result of its acceleration, it converts GPE to KE very quickly, losing altitude but gaining heat (from friction) and velocity. (Law of conservation of energy). This again is a problem as a velocity too high is difficult to control, and may be fatal for the probe/occupants inside. In order to solve this problem, rocket boosters are fired in the opposite direction to slow the descent of the probe, reducing its KE as it's GPE decreases.
High Speed Electrical charged Particles
- High speed electrically charged particles coming from the sun are often come from solar flares (where the sun emits a burst of solar radiation). They are quite problematic as they can damage sensitive equipment within the spacecraft and also penetrate through the walls of the spacecraft, damaging other parts inside and may be harmful for the occupants. Alongside causing damage, these charged particles interfere with communication (much like the situation of an ionisation blackout in re-entry), which is another problem.
- In order to solve this problem, the outer layer of the spacecraft can be coated with a material which can be electrically charged to repel these high speed charged particles. Insulating material which prevents these particles from penetrating into the spacecraft can also be applied. Furthermore, communication signals can be sent from the backend or the spacecraft (or whichever end isn't being exposed to these high speed charged particles).
I hope thats enough to get 7 marks at least haha but that's my attempt :)
This is a fantastic answer to what was a really difficult question!! ;D
This would almost definitely get the full 8 marks (though it is probably too long to replicate in exam conditions), some other things you could discuss:
- The specific issue of fuel consumption when discussing the changes in GPE
- Use of magnetic fields to deflect the charged particles
With these problems (where you are asked to analyse) be sure to relate to their real world implications! Why does the Physics pose a real world issue? That is what they are looking for in these sorts of questions :)
PS - These sorts of questions are why I can't wait for the new HSC syllabus! ;)
Post by: Neutron on October 07, 2016, 12:56:01 am
This is a fantastic answer to what was a really difficult question!! ;D
This would almost definitely get the full 8 marks (though it is probably too long to replicate in exam conditions), some other things you could discuss:
- The specific issue of fuel consumption when discussing the changes in GPE
- Use of magnetic fields to deflect the charged particles
With these problems (where you are asked to analyse) be sure to relate to their real world implications! Why does the Physics pose a real world issue? That is what they are looking for in these sorts of questions :)
PS - These sorts of questions are why I can't wait for the new HSC syllabus! ;)
Thank you Spencer and Jamon! Your responses have been extremely helpful :D Just with the fuel consumption thing though, doesn't the mass of the rocket decrease as it increases in altitude? Meaning the acceleration should increase? Why would the overall kinetic energy be decreasing (and thus why would more fuel need to be used?) Thank you!
Post by: jamonwindeyer on October 07, 2016, 11:30:05 am
Thank you Spencer and Jamon! Your responses have been extremely helpful :D Just with the fuel consumption thing though, doesn't the mass of the rocket decrease as it increases in altitude? Meaning the acceleration should increase? Why would the overall kinetic energy be decreasing (and thus why would more fuel need to be used?) Thank you!
Ohhh yep good catch Neutron! Missed that in my midnight reading, your interpretation is 100% correct! The overall kinetic energy would increase, assuming that the change in mass causes a reduction that is lower in magnitude than the increase of velocity squared! Which is almost certain ;D
Post by: dylan862 on October 07, 2016, 02:20:05 pm
Post by: jakesilove on October 07, 2016, 02:22:04 pm
What exactly do the donor and acceptor levels do in semiconductors?
All you need to know is where they are in the conduction/valence band structure, which type is which (ie. which type of semi-conductor goes with each band) and be able to draw them. They basically make it 'easier' for electrons to jump from valence to conduction bands!
That's really the greatest complexity at which you need to understand this section.
Jake
Post by: Neutron on October 07, 2016, 04:11:04 pm
Yep that definitely makes sense but I think all this fatigue has gotten to my head. So with the particle moving, isn't it only in its reference frame where time dilates? So I'm using the example of like a spacecraft flying past the Earth, only the time on the spacecraft has dilated, the time on Earth is still proper. So using that logic, wouldn't the scientist's time be proper? Sorry if this is hella dumb :/
Neutron
I think this got lost in the jumble but could someone please explain this to me D: sorry
Post by: jamonwindeyer on October 07, 2016, 04:36:41 pm
I think this got lost in the jumble but could someone please explain this to me D: sorry
Sorry Neutron! So you are sort of right. Time dilates for the particles reference frame, but only with respect to another reference frame. If we were sitting on the particle we'd notice nothing at all, that is the tricky bit to understand. When we talk about time dilation, we talk about time in one reference frame moving more slowly than another. So when we measure the life span of the particle, it lives longer in our reference frame. If we measured the life span of the particle, from the reference frame of the particle, it would be the normal value at rest! That's why we say that our time is screwed up, and then when we bring the particle to rest, our two reference frames sync up again.
Does that sort of make sense? Might have to read it twice ;)
Post by: Neutron on October 07, 2016, 05:48:25 pm
Sorry Neutron! So you are sort of right. Time dilates for the particles reference frame, but only with respect to another reference frame. If we were sitting on the particle we'd notice nothing at all, that is the tricky bit to understand. When we talk about time dilation, we talk about time in one reference frame moving more slowly than another. So when we measure the life span of the particle, it lives longer in our reference frame. If we measured the life span of the particle, from the reference frame of the particle, it would be the normal value at rest! That's why we say that our time is screwed up, and then when we bring the particle to rest, our two reference frames sync up again.
Does that sort of make sense? Might have to read it twice ;)
So how come in this case, it isn't the reference frame of us sitting on the particle moving more slowly than the scientist observing the particle? I just keep envisioning the particle as a spacecraft instead (just because it's slightly easier to grasp) and isn't it normally the other way around? i.e. the thing that is moving has time travelling more slowly and the thing in the external frame of reference has the 'faster' time? Ah sorry Jamon
Post by: jamonwindeyer on October 07, 2016, 08:26:53 pm
So how come in this case, it isn't the reference frame of us sitting on the particle moving more slowly than the scientist observing the particle? I just keep envisioning the particle as a spacecraft instead (just because it's slightly easier to grasp) and isn't it normally the other way around? i.e. the thing that is moving has time travelling more slowly and the thing in the external frame of reference has the 'faster' time? Ah sorry Jamon
Okay let's spaceship it! So the spaceship is travelling at speed, and let's say we are watching a clock tick inside the spaceship (as a substitute for measuring the lifespan). So, we measure the clock on the spaceship from our frame of reference. Remember, there is no absolute reference frame for time, so we have to just compare it to our own. The clock on the ship will be slower than ours. Why? Because time slows down at speed. So when we look at the spaceships clock, it has slowed down, because it is moving at relativistic speed.
Keep in mind that on the spaceship, we could watch a clock on earth. If we are in the spaceship, the clock on Earth will be slower than our clock on the ship. What the hell is that about, it's backwards to what we just said?
That's relativity. Because in our frame of reference, the spaceship is moving at speed. In the spaceships frame of reference, the spaceship isn't moving at all, the Earth is! And thus, we see the clock moving at speed and thus, the clock moving slower.
So we will always view a moving reference frame as having slower time than ours, because the frame of reference we are in is always considered as stationary, and we compare to it.
So when the spaceship comes back to earth, its clock is no longer moving slower. So, it will tick through a minute faster, and so the lifetime (as equivalent) will be shorter.
I kind of feel like you understand this Neutron! Because the expression:
i.e. the thing that is moving has time travelling more slowly and the thing in the external frame of reference has the 'faster' time?
That is correct, the thing we observe has time travelling more slowly. So you get this. I think there is some subtle thing that's throwing you off (which is common with relativity). Read this a few times, does it help? I'm waiting for the "aha" moment ;)
Post by: Neutron on October 07, 2016, 11:03:40 pm
Okay let's spaceship it! So the spaceship is travelling at speed, and let's say we are watching a clock tick inside the spaceship (as a substitute for measuring the lifespan). So, we measure the clock on the spaceship from our frame of reference. Remember, there is no absolute reference frame for time, so we have to just compare it to our own. The clock on the ship will be slower than ours. Why? Because time slows down at speed. So when we look at the spaceships clock, it has slowed down, because it is moving at relativistic speed.
Keep in mind that on the spaceship, we could watch a clock on earth. If we are in the spaceship, the clock on Earth will be slower than our clock on the ship. What the hell is that about, it's backwards to what we just said?
That's relativity. Because in our frame of reference, the spaceship is moving at speed. In the spaceships frame of reference, the spaceship isn't moving at all, the Earth is! And thus, we see the clock moving at speed and thus, the clock moving slower.
So we will always view a moving reference frame as having slower time than ours, because the frame of reference we are in is always considered as stationary, and we compare to it.
So when the spaceship comes back to earth, its clock is no longer moving slower. So, it will tick through a minute faster, and so the lifetime (as equivalent) will be shorter.
I kind of feel like you understand this Neutron! Because the expression:
i.e. the thing that is moving has time travelling more slowly and the thing in the external frame of reference has the 'faster' time?
That is correct, the thing we observe has time travelling more slowly. So you get this. I think there is some subtle thing that's throwing you off (which is common with relativity). Read this a few times, does it help? I'm waiting for the "aha" moment ;)
Oh my god I get it :O Aha! hahaha i think it was the 'lifespan' bit which threw me off completely cause now that i re-read the question, it's quite obvious that the time given was the dilated one! I feel kinda dumb now cause lifespan is kinda like a clock ticking aye hahah thank you Jamon!!
Post by: jamonwindeyer on October 07, 2016, 11:24:25 pm
Oh my god I get it :O Aha! hahaha i think it was the 'lifespan' bit which threw me off completely cause now that i re-read the question, it's quite obvious that the time given was the dilated one! I feel kinda dumb now cause lifespan is kinda like a clock ticking aye hahah thank you Jamon!!
Ayyy I knew we'd get there! Yep I totally get it, definitely don't feel dumb, it took one of the greatest Physicists of all time to understand this at first ;) you are most welcome!
Post by: GeekModeEngaged on October 09, 2016, 03:32:35 pm
The question should be there via attachment. Any help will be greatly appreciated!
Post by: jakesilove on October 09, 2016, 05:18:41 pm
Could you please check if I got this wrong or not? I though it was a relatively simple question and then BOSTES said I got it wrong. #storyofmylife
The question should be there via attachment. Any help will be greatly appreciated!
Hey! So for this question, you basically just need to use the right hand rule. The thing is moving down, so by lenz's law the appropriate force will push it BACK UP to counteract the change. So, force is up. Then, the field is into the page. By the right hand rule, the current should thus be flowing to the right. However, remember that electrons move in the opposite direction to current! Therefore, they will build up on the left hand side, making the answer A :)
Post by: RuiAce on October 09, 2016, 05:21:07 pm
You need to make sure that you've incorporated both Lenz's law, AND the fact we're interested in the flow of electrons. It becomes A because we double reversed the right-hand push rule
Post by: GeekModeEngaged on October 09, 2016, 05:56:43 pm
Sorry, but you're probably about to get a bunch of questions from me that I should probably find easy! I hope you don't mind too much as I am very grateful Thanks Again!
Post by: jakesilove on October 09, 2016, 06:06:31 pm
Another Question: not sure how to approach this one. I tried using Keplars' Law but it didn't get me anywhere because I don't know the mass of the star.
Sorry, but you're probably about to get a bunch of questions from me that I should probably find easy! I hope you don't mind too much as I am very grateful Thanks Again!
Hey! I'll just give you a hint, because I'm sure you can get this one out. Keplar's law, equation the ratio of radius and period to the mass of the central object, implies something interesting. For any two objects orbiting the same central star, the value of r^3/T^2 will be a constant. It doesn't actually matter what the mass of the star is, because
So, we can sub in the information known about Alif, to get information out about Ba!
If you can't get an answer, let me know, and I'll post up a solution :)
Jake
Post by: RuiAce on October 09, 2016, 06:06:44 pm
Another Question: not sure how to approach this one. I tried using Keplars' Law but it didn't get me anywhere because I don't know the mass of the star.
Sorry, but you're probably about to get a bunch of questions from me that I should probably find easy! I hope you don't mind too much as I am very grateful Thanks Again!
Post by: GeekModeEngaged on October 09, 2016, 06:22:40 pm
Post by: jakesilove on October 09, 2016, 06:25:01 pm
Okay, next up:
This you just need to know. P-type semi-conductors have an 'acceptor level' close to the valence band. Thus, this is a P-type semi conductor, and must be doped with a group 3 metal.
Note: N-type semi-conductors have a 'donor level' close to the conduction band.
Post by: GeekModeEngaged on October 09, 2016, 07:18:13 pm
a P-type semi conductor must be doped with a group 3 metal.
Note: N-type semi-conductors have a 'donor level' close to the conduction band.
Sorry, could you please explain what a "group 3 metal" is. I don't do chemistry and I don't recall my teacher explaining it.
Thanks
Post by: RuiAce on October 09, 2016, 07:19:32 pm
Sorry, could you please explain what a "group 3 metal" is. I don't do chemistry and I don't recall my teacher explaining it.Look for III A and V A. (III and V are Roman numerals for 3 and 5.)
Thanks
(In fact, ignore everything with B in it.)
https://www.google.com.au/search?q=periodic+table&espv=2&biw=1745&bih=885&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjbsLfOpM3PAhWFjZQKHZRvCF4Q_AUIBigB#imgrc=ARVfilyMkimROM%3A
If your teacher didn't explain this then he/she forgot something quite important in the physics course...
Post by: GeekModeEngaged on October 09, 2016, 07:34:51 pm
Quote
If your teacher didn't explain this then he/she forgot something quite important in the physics course...
It would not surprise me if he forgot something. Am I supposed to use IIIA for p-type and VA for n-type?
Post by: RuiAce on October 09, 2016, 07:37:00 pm
If your teacher didn't explain this then he/she forgot something quite important in the physics course...Yes.
It would not surprise me if he forgot something. Am I supposed to use IIIA for p-type and VA for n-type?
Also: Note that boron belongs in group 3 (just call it that, don't have to call it IIIA in the HSC). And note that phosphorus belongs in group 5. These are the standard elements used for doping.
Post by: GeekModeEngaged on October 09, 2016, 09:18:38 pm
Post by: jakesilove on October 09, 2016, 09:20:29 pm
I know this should be relatively simple, but when I tried to use F = BILsin(x), I got F=0.3 N (where l=0.4sin(45), I = 3A and B=0.5T) I'm sure I'll just be missing something simple (either that, or i"m doing it completely wrong), but could you help anyway please :)
Looks like you got the length wrong!
Post by: GeekModeEngaged on October 09, 2016, 09:22:41 pm
Looks like you got the length wrong!
Oh my god that is so embarrasing. I think it's time to stop trying to do maths for the night :'( :o
Post by: katnisschung on October 15, 2016, 04:28:08 pm
figure 5.2 shows how the Ep of an object changes with distance either side from planet X.
such a graph is called a gravitational potential energy well or, more simply a gravity
well.
a) add to it a gravity well for a planet which is much more massive.
b) account fro the shape of the graph you have drawn
the new graph would have a well with more depth because the mass would mean
more GPE according to the absolute GPE formula
but wouldn't it also be skinnier and its curve approach zero GPE more quickly
because of its larger mass?
Post by: jakesilove on October 15, 2016, 04:32:23 pm
GPE
figure 5.2 shows how the Ep of an object changes with distance either side from planet X.
such a graph is called a gravitational potential energy well or, more simply a gravity
well.
a) add to it a gravity well for a planet which is much more massive.
b) account fro the shape of the graph you have drawn
the new graph would have a well with more depth because the mass would mean
more GPE according to the absolute GPE formula
but wouldn't it also be skinnier and its curve approach zero GPE more quickly
because of its larger mass?
I think you're right there! It wouldn't matter as much as your first point, because the 'skinniness' of a gravity well is less obvious upon comparison than the depth of the well. Worth mentioning, I think, but your first point should get you the marks.
Post by: jamonwindeyer on October 15, 2016, 04:45:55 pm
GPE
figure 5.2 shows how the Ep of an object changes with distance either side from planet X.
such a graph is called a gravitational potential energy well or, more simply a gravity
well.
a) add to it a gravity well for a planet which is much more massive.
b) account fro the shape of the graph you have drawn
the new graph would have a well with more depth because the mass would mean
more GPE according to the absolute GPE formula
but wouldn't it also be skinnier and its curve approach zero GPE more quickly
because of its larger mass?
Hey! So if we have our function of GPE with respect to distance as:
That's really just a graph of the form \(y=\frac{k}{x}\). By increasing the mass of our planet, we are just increasing our value of \(k\). Let's say we take something 10 times greater, so \(10k\). Then compare the graphs (the green graph is the larger planet):
(http://i.imgur.com/EHyQLAQ.png)
So based on that, I'd actually say that the well is wider! I don't like this question though, because it doesn't specify that the mass is larger specifically, it just says "more massive." Also, I'm not sure whether you are drawing gravity wells that don't tend to negative infinity in the middle, that would alter this slightly and definitely result in a deeper well as you've described :)
Welcome to the forums Katniss! :)
Post by: jakesilove on October 15, 2016, 04:48:09 pm
Hey! So if we have our function of GPE with respect to distance as:
That's really just a graph of the form \(y=\frac{k}{x}\). By increasing the mass of our planet, we are just increasing our value of \(k\). Let's say we take something 10 times greater, so \(10k\). Then compare the graphs (the green graph is the larger planet):
(http://i.imgur.com/EHyQLAQ.png)
So based on that, I'd actually say that the well is wider! I don't like this question though, because it doesn't specify that the mass is larger specifically, it just says "more massive." Also, I'm not sure whether you are drawing gravity wells that don't tend to negative infinity in the middle, that would alter this slightly and definitely result in a deeper well as you've described :)
Welcome to the forums Katniss! :)
I sort of assumed this was part of an option that I didn't do, where you do proper three-dimensional-like gravity wells? Maybe Astrophysics? In any case, none of us did that topic, and I was definitely sort of making stuff up.
Post by: jamonwindeyer on October 15, 2016, 04:49:20 pm
I sort of assumed this was part of an option that I didn't do, where you do proper three-dimensional-like gravity wells? Maybe Astrophysics? In any case, none of us did that topic, and I was definitely sort of making stuff up.
Oh maybe? It says "either side," so I'm thinking a two dimensional thing? Is that right Katniss? ;D
Post by: RuiAce on October 15, 2016, 04:53:23 pm
I sort of assumed this was part of an option that I didn't do, where you do proper three-dimensional-like gravity wells? Maybe Astrophysics? In any case, none of us did that topic, and I was definitely sort of making stuff up.I did astro and it doesn't look like astro
Post by: katnisschung on October 15, 2016, 05:29:58 pm
yes its a 2 dimensional graph
weirdly its part of space i just started my hsc :)
ill try posting a lower quality photo becos the other one was too large
Post by: katnisschung on October 15, 2016, 05:43:46 pm
yuna
Post by: jakesilove on October 15, 2016, 05:47:22 pm
this is the graph i am referring to
yuna
Basically, whilst it's sort of good to understand this, it's far outside the curriculum! So don't worry too much :)
Post by: FallonXay on October 18, 2016, 09:02:08 am
How would you solve this graph question? (The answer is C)
Thanks :)
Post by: FallonXay on October 18, 2016, 09:07:38 am
Thank you!!!
Post by: jamonwindeyer on October 18, 2016, 11:06:14 am
Hi!
How would you solve this graph question? (The answer is C)
Thanks :)
Oh this is a tough one! Okay, so remember Faraday's Law:
This question requires applying the law above in reverse. Or in words, the induced EMF is proportional to the rate of change of voltage.
This question is made easier if you do 2U and have a basic understanding of Calculus, but think of it this way. The EMF will exist as long as the voltage is either increasing or decreasing, and it will be zero when the voltage flattens out. Further, when the voltage is increasing, the EMF will be negative. When the voltage is decreasing, it will be positive. This is due to the negative sign in the above formula (think Lenz's Law) ;D
This yields C as our answer :)
Post by: jamonwindeyer on October 18, 2016, 11:14:45 am
Oh, sorry, and this question (The answer is A) ;D
Thank you!!!
And this one we can answer with a bit of formula. The centripetal acceleration:
However, we can obtain an expression for the velocity:
Combine THAT with a rearranged version of Kepler's Law:
Edit by Rui: Typo spotted
So that yields \(v=\frac{GM}{R}\), and we can substitute that back:
We can substitute our varying values of R (remember that the radius of the earth must be taken into account) to obtain our numerical answers. Or, we can just notice that a higher orbital radius will lower our centripetal acceleration, and so A must be true by default (indeed, this second approach is more time efficient, and may not even require all the working above) ;D
Post by: FallonXay on October 18, 2016, 11:24:33 am
Oh this is a tough one! Okay, so remember Faraday's Law:
This question requires applying the law above in reverse. Or in words, the induced EMF is proportional to the rate of change of voltage.
This question is made easier if you do 2U and have a basic understanding of Calculus, but think of it this way. The EMF will exist as long as the voltage is either increasing or decreasing, and it will be zero when the voltage flattens out. Further, when the voltage is increasing, the EMF will be negative. When the voltage is decreasing, it will be positive. This is due to the negative sign in the above formula (think Lenz's Law) ;D
This yields C as our answer :)
And this one we can answer with a bit of formula. The centripetal acceleration:
However, we can obtain an expression for the velocity:
Combine THAT with a rearranged version of Lenz's Law:
So that yields \(v=\frac{GM}{R}\), and we can substitute that back:
We can substitute our varying values of R (remember that the radius of the earth must be taken into account) to obtain our numerical answers. Or, we can just notice that a higher orbital radius will lower our centripetal acceleration, and so A must be true by default (indeed, this second approach is more time efficient, and may not even require all the working above) ;D
Awesome!!! Thanks a ton!!! ;D
Post by: teapancakes08 on October 18, 2016, 09:30:17 pm
I probably should be able to know how to answer this question but I keep getting an unknown when using formula "s = ut + 1/2at^2"...is there a better way to solve it or did I probably just substitute in the wrong values?
Post by: jakesilove on October 18, 2016, 09:49:56 pm
"A parachutist descending at a constant 4.9m/s dropped is keys when he was 98.0m above the ground. Calculate the time it took for the keys to fall to the ground."
I probably should be able to know how to answer this question but I keep getting an unknown when using formula "s = ut + 1/2at^2"...is there a better way to solve it or did I probably just substitute in the wrong values?
Hey! I think there may be a better way to do this. Using
We get
So, using
Beautiful! When a method looks too difficult, try something else :)
Jake
Post by: teapancakes08 on October 18, 2016, 10:44:16 pm
Hey! I think there may be a better way to do this. Using
We get
So, using
Beautiful! When a method looks too difficult, try something else :)
Jake
Oh, thank you so much! ;D
This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?
Post by: RuiAce on October 18, 2016, 11:05:03 pm
Oh, thank you so much! ;DAny extra bits of information you find out you highlight.
This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?
As in, over the black print they gave you. Highlight any bit of information that you suspect may be useful. Or even go that extra mile and use highlighting for suspicions and a tick for actual usage of it.
You can also choose to make use of the side of the paper in writing any things you need to sub into the formulae. Some people use this because the markers may or may not care about what's to the left of the lines they give you
Also as obvious as it seems break down any formulas you use. Make sure that what you used makes sense. With practice you should be able to see when things go missing.
Post by: jamonwindeyer on October 18, 2016, 11:09:34 pm
Oh, thank you so much! ;D
This is kind of redundant, but I tried using the other methods after I posted this question and got an answer was a few values off...and then found out I forgot to sub something in after seeing this. Any tips on how to avoid doing silly mistakes like that?
Lots and lots of practice, so that when you are doing it you flash back to your earlier work and go, "Yo, this seems a bit off." It's all about experience! You might also want to check out this article! ;D
Post by: jamgoesbam on October 19, 2016, 12:55:20 pm
Could someone please explain these questions? GPE confuses me :-\
Answers:
1.6.6 A --> for this question I thought that it went into a lower orbit as 3E = 3 x -GMm/r (since it is more negative, it would be closer to Earth!)
1.6.11 C (could you also please show working, thanks!)
1.6.12 B
Post by: bethjomay on October 19, 2016, 05:56:52 pm
Post by: nimasha.w on October 19, 2016, 06:18:56 pm
Could someone remind me why AC is better for use in transformers? Thank you!Hey! sorry if this explanation is pretty basic, just trying to test my knowledge
From what i understand a transformer requires an alternating current to create a changing magnetic field in the iron core. It is this changing magnetic field that induces an emf in the secondary coil, hence, DC can't be used in a transformer because it the current produced is only in one direction so it can't create a changing magnetic field
Post by: Spencerr on October 19, 2016, 06:34:18 pm
Could someone remind me why AC is better for use in transformers? Thank you!
Hey there, Nimasha gave a solid answer! Here's my 2cents. In an exam, i think this would be 3 marks. I've tried to allocate the marks below in an explanation.
Transformers are electrical devices which are used to step up or step down the voltage of electricity.
They operate via the principle of electromagnetic induction and Faraday's Law (quote his equation) whereby a changing magnetic flux produced by an oscillating magnetic field in a primary coil induces an emf and current in a secondary coil. (1) In order to continuously produce this change in magnetic flux, an oscillating magnetic field is required. The constant switching in direction of the AC current produces the oscillating magnetic field required. (1) In contrast, the unidirectional current of DC does not produce an oscillating B field, and thus does not subject the secondary coil to a changing magnetic flux. Hence DC cannot be used (unless by switching the current on and off to produce the change in flux), making AC much better for use in transformers. (1)
Post by: bethjomay on October 19, 2016, 06:59:00 pm
Post by: teapancakes08 on October 19, 2016, 07:00:06 pm
Any extra bits of information you find out you highlight.
As in, over the black print they gave you. Highlight any bit of information that you suspect may be useful. Or even go that extra mile and use highlighting for suspicions and a tick for actual usage of it.
You can also choose to make use of the side of the paper in writing any things you need to sub into the formulae. Some people use this because the markers may or may not care about what's to the left of the lines they give you
Also as obvious as it seems break down any formulas you use. Make sure that what you used makes sense. With practice you should be able to see when things go missing.
Ah, noted. Thanks for the advice. ;D
Post by: teapancakes08 on October 19, 2016, 07:01:59 pm
Lots and lots of practice, so that when you are doing it you flash back to your earlier work and go, "Yo, this seems a bit off." It's all about experience! You might also want to check out this article! ;D
Thanks, I'll keep that in mind ;D
Post by: jamonwindeyer on October 19, 2016, 10:34:05 pm
Hi!
Could someone please explain these questions? GPE confuses me :-\
Hey! So for your first question, I agree with your interpretation! I think the question is a little off in that way, but that said, if you swap back to your more simplistic version of \(GPE=mgh\), that gives you the correct answer. This little subtlety comes up a lot, and I don't they press on it like this in HSC Exams! :)
The second question, I can't match any of the answers actually! Anyone else who wants to have an attempt please feel free :P
Your last question is best answered by considering what is wrong with the other answers.
- A is incorrect, air resistance would need to be overcome either way
- C is incorrect, centripetal force IS our gravitational force, they are the same thing
- D is incorrect, doesn't really say anything relevant
B is the only remaining answer, we require extra energy to give the satellite enough horizontal velocity to enter an orbit ;D
Post by: katnisschung on October 21, 2016, 05:16:32 pm
Object A of mass M sits on the surface of planet α
and experiences a force of X due to gravity.
Object B, also of mass M, sits on the surface of planet β
also experiences a force of X due to gravity.
Planet α is twice as massive as planet β
calculate the radius of planet β in terms of planet α.
seems like nothing i do gets this answer
rβ= √2 times rα
Post by: jakesilove on October 21, 2016, 05:22:53 pm
IM STUCK!!!
Object A of mass M sits on the surface of planet α
and experiences a force of X due to gravity.
Object B, also of mass M, sits on the surface of planet β
also experiences a force of X due to gravity.
Planet α is twice as massive as planet β
calculate the radius of planet β in terms of planet α.
seems like nothing i do gets this answer
rβ= √2 times rα
Hey! We can calculate g using the equation
M is the mass of the planet, r is the radius of the planet, G is a constant.
You need to be able to derive this; equate gravitational force and force due to gravity!
Subbing in our values, we get
From this,
As required! I definitely skipped lots of steps here, so let me know if you need any clarification :)
Post by: teapancakes08 on October 21, 2016, 06:21:20 pm
"A satellite has 4000J of work done on it. Does it move to a higher or lower altitude? Explain your answer."
....which is also related to this question:
"Explain the relationship between the work done on an object which changes its positon in a gravitational field and its gravitational potential energy (GPE)"
My understanding of the latter is that the change of GPE is related to the work done on an object in that it is converted into kinetic energy as is falls back to the object/body (if I worded it correctly that is...), although I can't quite grasp how it works or even if I got the concept. Actually, I'm not even sure if I'm making any sense...
If anyone can help explain that'll be great. Sorry for the long winded question ^^;;
Post by: jakesilove on October 22, 2016, 11:58:58 am
I'm slightly confused as how to answer this question:
"A satellite has 4000J of work done on it. Does it move to a higher or lower altitude? Explain your answer."
....which is also related to this question:
"Explain the relationship between the work done on an object which changes its positon in a gravitational field and its gravitational potential energy (GPE)"
My understanding of the latter is that the change of GPE is related to the work done on an object in that it is converted into kinetic energy as is falls back to the object/body (if I worded it correctly that is...), although I can't quite grasp how it works or even if I got the concept. Actually, I'm not even sure if I'm making any sense...
If anyone can help explain that'll be great. Sorry for the long winded question ^^;;
Hey! If an object DOES WORK, it's going to have to work AGAINST SOMETHING. If an object has work DONE ON IT, that work has to be produced by SOMETHING. So, if a satellite has work DONE ON IT, how could that happen? Well, the gravitational field will do the work, thus PUSHING the satellite downwards! So, it moves to a lower orbit.
It sounds like you actually do have a good understanding of the topic! I think it's easiest to use a diagram in situations like this; hopefully the below gives you all the information you need! Let me know if I can clarify anything.
(http://i.imgur.com/MeF64Vk.png)
Post by: Rikahs on October 22, 2016, 03:03:00 pm
Post by: jakesilove on October 22, 2016, 03:17:50 pm
I realise that when a DC motor is running at full speed, the back emf induced due to the combination of Faraday's Law and Len'z law will oppose the supply emf. I was wondering whether they complete negate each other i.e. the net current in the coils would be 0? or is the back emf just a little under the supply emf i.e. the net current would be slight over zero?
No, the back EMF is substantially smaller than the supplied current. This results in a net current that is less than the supplied current (due to the Back EMF/Induced current), however there is still current flowing! Otherwise, the motor wouldn't turn.
Post by: student123456 on October 22, 2016, 04:13:14 pm
Hoping somebody may be able to help me out with solving these problems. I never really know how to handle them. Do you it mathematically? Thanks :D
A spaceship at a distance r metres from the centre of a star experiences a gravitational force of x newtons. The spaceship moves a distance of r/2 metres towards the star. What is the gravitational force acting on the spaceship in this new location?
Post by: Sanaz on October 22, 2016, 04:26:00 pm
Post by: RuiAce on October 22, 2016, 04:27:52 pm
Hey guys!
Hoping somebody may be able to help me out with solving these problems. I never really know how to handle them. Do you it mathematically? Thanks :D
A spaceship at a distance r metres from the centre of a star experiences a gravitational force of x newtons. The spaceship moves a distance of r/2 metres towards the star. What is the gravitational force acting on the spaceship in this new location?
Post by: RuiAce on October 22, 2016, 04:40:04 pm
Hey can anyone help me do this?
Post by: Sanaz on October 22, 2016, 04:51:47 pm
yea thank you <3
Post by: Cindy2k16 on October 23, 2016, 12:51:05 pm
Post by: student123456 on October 23, 2016, 12:53:48 pm
Thank you, RuiAce
Post by: RuiAce on October 23, 2016, 01:02:24 pm
Hi could someone please explain the answers to questions 12 and 18 from the 2014 HSC? The answers are A for q12 and D for q18Attack Q18 by a step by step analysis. Incidentally, the right column of the answers is what we find before the left column.
Recall that a cathode ray is an electron. An electron carries NEGATIVE charge. Hence, apply the reverse right-hand rule (or left-hand rule) to determine the magnetic deflection. You will find that the electron is deflected DOWNWARDS by the magnetic field.
Then, if the electron is deflected downwards, we need the electric field to counteract it by deflecting it UPWARDS.
If we want the electron to go up, it must be that the POSITIVE plate is at the top, whereas the negative plate is at the bottom.
Finally, recall that electric field lines go from + to -. Hence, they go from the top to the bottom, i.e. downwards
Hence D.
An explanation on Q12 is found in post #872
Post by: jamonwindeyer on October 23, 2016, 01:22:17 pm
Hi could someone please explain the answers to questions 12 and 18 from the 2014 HSC? The answers are A for q12 and D for q18
Welcome to the forums Cindy! ;D
Post by: Cindy2k16 on October 23, 2016, 05:28:09 pm
Welcome to the forums Cindy! ;D
Thanks :) I've been lurking for a while now but finally signed up today!
Post by: Cindy2k16 on October 23, 2016, 05:32:59 pm
Post by: jakesilove on October 23, 2016, 05:50:41 pm
Hi another question- the 2014 HSC sample answer for q22 said that if the angle of re-entry for a spacecraft is too steep, then the "spacecraft will decelerate too quickly" I'm a bit confused by this. I thought that if a spacecraft enters at too steep an angle, then it wont be able to slow down fast enough - and so deceleration is too little?
The answer given by BOSTES is definitely a bit confusing. It's not so much that the ship won't have time to decelerate quickly enough; it's more than the deceleration caused by interaction between the atmosphere and the ship will be so great that huge amounts of heat will be released, and the G-force experienced by the astronauts will be deadly. The more quickly the ship goes straight downwards, the more it has to deal with the effects of friction. This will heat the ship up, causing it the burn up in the atmosphere. You can really write whatever you want in this section (ie. G-forces, heat, etc. etc.) as long as what you're saying makes sense. Don't always trust the BOSTES answers, because they often DON'T make sense.
Post by: ml125 on October 23, 2016, 05:56:43 pm
Hi another question- the 2014 HSC sample answer for q22 said that if the angle of re-entry for a spacecraft is too steep, then the "spacecraft will decelerate too quickly" I'm a bit confused by this. I thought that if a spacecraft enters at too steep an angle, then it wont be able to slow down fast enough - and so deceleration is too little?"Decelerate" in this sense, refers to a downwards acceleration. So the magnitude of the acceleration is increasing, but the direction of the movement is downwards - the value for acceleration is becoming increasingly negative and is therefore classified as a deceleration.
Post by: RuiAce on October 23, 2016, 06:05:43 pm
"Decelerate" in this sense, refers to a downwards acceleration. So the magnitude of the acceleration is increasing, but the direction of the movement is downwards - the value for acceleration is becoming increasingly negative and is therefore classified as a deceleration.I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...
Post by: ml125 on October 23, 2016, 06:09:06 pm
I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...This is how my physics teacher explained this specific question to us - not 100% sure though ._.
Post by: jakesilove on October 23, 2016, 06:10:15 pm
I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...
'Deceleration' is a stupid term that means nothing in Physics for fairly obvious reasons, however generally it does only refer to the magnitude. Ie. 'slowing down' as opposed to 'becoming less positive'. Let's ignore the term altogether though, if possible
Post by: jakesilove on October 23, 2016, 06:11:46 pm
This is how my physics teacher explained this specific question to us - not 100% sure though ._.
Aha yeah it's totally fair enough, talking about forces is complicated and teachers tend to over simplify to the point of not really making sense. Which is why you can basically write whatever and not get penalised for it, because people teach in so many different ways! I just personally hate pseudo-Physics words that make Physics even more complicated than it needs to be (ie. slowing down is accelerating, speeding up is accelerating, we don't need two different words aha)
Post by: teapancakes08 on October 23, 2016, 06:34:10 pm
Hey! If an object DOES WORK, it's going to have to work AGAINST SOMETHING. If an object has work DONE ON IT, that work has to be produced by SOMETHING. So, if a satellite has work DONE ON IT, how could that happen? Well, the gravitational field will do the work, thus PUSHING the satellite downwards! So, it moves to a lower orbit.
It sounds like you actually do have a good understanding of the topic! I think it's easiest to use a diagram in situations like this; hopefully the below gives you all the information you need! Let me know if I can clarify anything.
(http://i.imgur.com/MeF64Vk.png)
Oh, I see now. So if the object is moved away from the other body/object then it is working against the gravitational field, hence it moves to a high orbit. And then if it falls back down the field is doing the work on the object, thus the GPE is being converted into kinetic energy, hence it moves to a lower orbit. ;D
Thanks so much for the explanation ^^
Post by: Cindy2k16 on October 23, 2016, 06:54:03 pm
Thanks in advance! (also how do you upload images on posts? do you have to upload them to imgur first?)
Post by: Happy Physics Land on October 23, 2016, 07:30:13 pm
Hi could someone explain how to 2014 HSC Q26 b)? I completely don't understand the working out provided in the answer especially since I've never done calculations where there's a non-zero voltage across the electrodes for the photoelectric effect.
Thanks in advance! (also how do you upload images on posts? do you have to upload them to imgur first?)
Hey Cindy!
No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:
1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV
Here at Atarnotes we want the best for our students, so I will hold your hand through this.
The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.
I will prove to you why this is the case:
Maximum kinetic energy = hf - work function _____ equation 1
This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.
But since stopping voltage works against the motion of electrons:
Maximum kinetic energy = hf - qV ________ equation 2
Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV
Good good, now the difficult part is over, congratulations.
We need to do two things for this question
1. Draw the line on the graph for the experiment done with 0.0V
To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:
Work function = planck's constant x threshold frequency
We know work function = 4.1eV = 4.1 x (1.602 x 10-19)J
We know planck's constant = 6.632 x 10-34
Substitute in the values:
4.1 x (1.602 x 10-19)J = 6.632 x 10-34 x threshold frequency
Hence threshold frequency of the metal = 9.9 x 1014 Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!
Moving onto the next part of the question:
2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV
Recall our formula:
Maximum kinetic energy = hf - work function
We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10-19)J
And finally we know the value for h, which is Planck's constant
So now substitute all your values into the formula, and you obtain an answer of 1.3 x 1015Hz as the radiation frequency
Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!
Best Regards
Happy Physics Land
Post by: jakesilove on October 23, 2016, 07:31:47 pm
Hey Cindy!
No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:
1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV
Here at Atarnotes we want the best for our students, so I will hold your hand through this.
The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.
I will prove to you why this is the case:
Maximum kinetic energy = hf - work function _____ equation 1
This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.
But since stopping voltage works against the motion of electrons:
Maximum kinetic energy = hf - qV ________ equation 2
Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV
Good good, now the difficult part is over, congratulations.
We need to do two things for this question
1. Draw the line on the graph for the experiment done with 0.0V
To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:
Work function = planck's constant x threshold frequency
We know work function = 4.1eV = 4.1 x (1.602 x 10-19)J
We know planck's constant = 6.632 x 10-34
Substitute in the values:
4.1 x (1.602 x 10-19)J = 6.632 x 10-34 x threshold frequency
Hence threshold frequency of the metal = 9.9 x 1014 Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!
Moving onto the next part of the question:
2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV
Recall our formula:
Maximum kinetic energy = hf - work function
We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10-19)J
And finally we know the value for h, which is Planck's constant
So now substitute all your values into the formula, and you obtain an answer of 1.3 x 1015Hz as the radiation frequency
Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!
Best Regards
Happy Physics Land
THE PRODIGAL SON RETURNS!
Post by: Happy Physics Land on October 23, 2016, 07:34:33 pm
THE PRODIGAL SON RETURNS!
Who said you cant write an essay in physics? ;)
Post by: Cindy2k16 on October 23, 2016, 10:03:24 pm
Hey Cindy!
No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:
1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV
Here at Atarnotes we want the best for our students, so I will hold your hand through this.
The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.
I will prove to you why this is the case:
Maximum kinetic energy = hf - work function _____ equation 1
This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.
But since stopping voltage works against the motion of electrons:
Maximum kinetic energy = hf - qV ________ equation 2
Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV
Good good, now the difficult part is over, congratulations.
We need to do two things for this question
1. Draw the line on the graph for the experiment done with 0.0V
To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:
Work function = planck's constant x threshold frequency
We know work function = 4.1eV = 4.1 x (1.602 x 10-19)J
We know planck's constant = 6.632 x 10-34
Substitute in the values:
4.1 x (1.602 x 10-19)J = 6.632 x 10-34 x threshold frequency
Hence threshold frequency of the metal = 9.9 x 1014 Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!
Moving onto the next part of the question:
2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV
Recall our formula:
Maximum kinetic energy = hf - work function
We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10-19)J
And finally we know the value for h, which is Planck's constant
So now substitute all your values into the formula, and you obtain an answer of 1.3 x 1015Hz as the radiation frequency
Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!
Best Regards
Happy Physics Land
Thank you for your explanation it did help!!
I have a few more questions if thats okay. How do you know that 4.1V is the stopping voltage? Do you interpret it from the graph because I'm having trouble understanding the graph (not the one to draw, but the one provided)- I get that its graphing E=hf but how does it relate to the experiment with 4.1V (or does it not relate at all?). I thought that the x-intercept of max KE vs frequency graphs was only dependent on the work function of the metal but does it change when there is a voltage applied across the electrodes? I hope my questions make sense?
It seems today's a bad day for studying physics I've never been so confused about it before :-\
Thanks in advance!
Post by: jamonwindeyer on October 24, 2016, 12:32:57 am
Thank you for your explanation it did help!!
I have a few more questions if thats okay. How do you know that 4.1V is the stopping voltage? Do you interpret it from the graph because I'm having trouble understanding the graph (not the one to draw, but the one provided)- I get that its graphing E=hf but how does it relate to the experiment with 4.1V (or does it not relate at all?). I thought that the x-intercept of max KE vs frequency graphs was only dependent on the work function of the metal but does it change when there is a voltage applied across the electrodes? I hope my questions make sense?
It seems today's a bad day for studying physics I've never been so confused about it before :-\
Thanks in advance!
Hey your question definitely makes sense! Don't worry, we all have those days ;)
This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.
In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.
So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2 ;)
Post by: Happy Physics Land on October 24, 2016, 09:49:14 am
Hey your question definitely makes sense! Don't worry, we all have those days ;)
This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.
In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.
So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2 ;)
HAPPY BIRTHDAY JAMON!!! <3 :D :D
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Post by: Cindy2k16 on October 24, 2016, 11:37:45 am
Hey your question definitely makes sense! Don't worry, we all have those days ;)
This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.
In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.
So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2 ;)
Hi thanks for your help! I understand this idea of pulling the electrons away but i still have a few questions (sorry)
1. Just to clarify, are the electrons being pulled away because the emitter electrode is negative and the opposite electrode is positive? So this pulling away won't happen if the voltage switched to the emitter being positive etc?
2. I forgot to ask this earlier, but if the plate opposite the emitter is positive as shown in the diagram, wouldn't it be attracting the photoelectrons? If so, then how can the 4.1V be the stopping voltage if its not repelling the electrons at all and rather, is pulling them towards the opposite electrode? (I looked at a few websites and I think they all showed the opposite electrode to be negative when the stopping voltage is applied- unless i misunderstood something?)
Thank you! (sorry I'm taking so long to get this..)
Post by: Happy Physics Land on October 24, 2016, 04:57:06 pm
Hi thanks for your help! I understand this idea of pulling the electrons away but i still have a few questions (sorry)
1. Just to clarify, are the electrons being pulled away because the emitter electrode is negative and the opposite electrode is positive? So this pulling away won't happen if the voltage switched to the emitter being positive etc?
2. I forgot to ask this earlier, but if the plate opposite the emitter is positive as shown in the diagram, wouldn't it be attracting the photoelectrons? If so, then how can the 4.1V be the stopping voltage if its not repelling the electrons at all and rather, is pulling them towards the opposite electrode? (I looked at a few websites and I think they all showed the opposite electrode to be negative when the stopping voltage is applied- unless i misunderstood something?)
Thank you! (sorry I'm taking so long to get this..)
Hey Cindy!
Its alright! It is a very tough question to understand. Your first dotpoint is right on the money. If the collector is positive, then electrons would travel towards it spontaneously due to electrostatic attraction (i.e. negative is attracted to positive). If the collector is negative then electrons are repelled and the only way the electrons can ever reach the collector is if they get given enough kinetic energy from the incident light ray.
So yeah with regards to your second question, the only reason why we supply a voltage to the emitter and collector is so that we can repel the electrons up to a stopping voltage value where electrons cannot reach the collector anymore and so we can work out work function. Theoretically if the collector is just positive then yes we do not need any voltage input because electrons will simply be attracted to collector. But experimentally, we still need some voltage supply purely just to bring charge to the collector (i.e. make it positive), but this voltage wouldnt determine anything valuable for us.
Best Regards
Happy Physics Land
Post by: student123456 on October 24, 2016, 06:08:04 pm
Hoping someone can help me out with this multiple choice question please ::::)
Post by: RuiAce on October 24, 2016, 06:13:37 pm
Hola guys and gals!Addressed by Jake in post #987
Hoping someone can help me out with this multiple choice question please ::::)
Post by: jakesilove on October 24, 2016, 07:11:39 pm
Addressed by Jake in post #987
Rui you are a machine, efficient af
Post by: jamonwindeyer on October 24, 2016, 07:19:51 pm
Rui you are a machine, efficient af
I feel like he has a bloody spreadsheet in his head that he just closes his eyes and navigates through 8)
Post by: Cindy2k16 on October 24, 2016, 07:55:45 pm
(Also thanks for the replies to the photoelectric question!)
Post by: RuiAce on October 24, 2016, 07:57:38 pm
Not sure where to put this question but I was just wondering approximately what raw marks you need to get in the HSC exam for a band 5 or a band 6 in physics?Consider the raw marks database
(Also thanks for the replies to the photoelectric question!)
Post by: Cindy2k16 on October 25, 2016, 10:55:38 am
Post by: jakesilove on October 25, 2016, 11:15:18 am
When plotting a graph in the exam, if the points provided don't include ones close to/at the x/y axis, do we extend our line down so that it does cut the x/y axis or only draw the line within the points? I've heard conflicting opinions about this from people of different schools so I'm not sure what to think. (also does the same apply in chemistry? i dont want to post the same question again in chem)
Hey! It genuinely doesn't matter. Personally, I would only ever plot within the range of the given values. If you're asked to estimate based on the plot, it will either only ask you for a point within that range, or explicitly tell you to extend your graph to estimate a higher/lower value. This applies to Chem and Physics; it just doesn't matter! I think it's 'better science' not to extend your graph beyond the data points you're given, though, unless they tell you to.
Post by: RuiAce on October 25, 2016, 11:17:30 am
When plotting a graph in the exam, if the points provided don't include ones close to/at the x/y axis, do we extend our line down so that it does cut the x/y axis or only draw the line within the points? I've heard conflicting opinions about this from people of different schools so I'm not sure what to think. (also does the same apply in chemistry? i dont want to post the same question again in chem)I'm interested in what Jamon and Jake have to say as well so I won't give a concrete answer. I only say yes to within the boundaries of the HSC as if your point extends too far, you can't do the question otherwise. Fortunately, the likelihood of being given a point out of the domain is unlikely in HSC physics and chemistry due to realistic absurdity.
I want Jamon/Jake to read this to input opinions but everyone else up to you
The absurdity arises from something that I only learnt recently. It is the difference between interpolation and extrapolation.
Given a set of data, we want to fit a polynomial of degree ?? to satisfy as many points as possible. In general, for the sake of the HSC course we're interested in fitting a linear function (a degree 1 polynomial, I.e. Straight line) through the points.
The line that we fit is, in general, good for points within the domain. If our data ranges from, say, a range of currents from 1 amp to 10 amps, then it's good to find something for 3 amps. This is interpolation - taking something out from within the boundaries.
Extrapolation is different and is what's considered here. Extrapolation is the act of relying on your polynomial to determine something OUTSIDE the boundaries.
In general, this causes havoc. Because the behaviour of a substance might change outside of reasonable boundaries and you actually need a completely new model for it. An example is in how germanium starts becoming a conductor and not a semiconductor.
That being said, the questions in the HSC are usually designed so that the region of extrapolation is not massive enough to cause problems. These issues would've been monitored in advance.
Given a set of data, we want to fit a polynomial of degree ?? to satisfy as many points as possible. In general, for the sake of the HSC course we're interested in fitting a linear function (a degree 1 polynomial, I.e. Straight line) through the points.
The line that we fit is, in general, good for points within the domain. If our data ranges from, say, a range of currents from 1 amp to 10 amps, then it's good to find something for 3 amps. This is interpolation - taking something out from within the boundaries.
Extrapolation is different and is what's considered here. Extrapolation is the act of relying on your polynomial to determine something OUTSIDE the boundaries.
In general, this causes havoc. Because the behaviour of a substance might change outside of reasonable boundaries and you actually need a completely new model for it. An example is in how germanium starts becoming a conductor and not a semiconductor.
That being said, the questions in the HSC are usually designed so that the region of extrapolation is not massive enough to cause problems. These issues would've been monitored in advance.
Post by: jakesilove on October 25, 2016, 11:25:23 am
I'm interested in what Jamon and Jake have to say as well so I won't give a concrete answer. I only say yes to within the boundaries of the HSC as if your point extends too far, you can't do the question otherwise. Fortunately, the likelihood of being given a point out of the domain is unlikely in HSC physics and chemistry due to realistic absurdity.I want Jamon/Jake to read this to input opinions but everyone else up to youThe absurdity arises from something that I only learnt recently. It is the difference between interpolation and extrapolation.
Given a set of data, we want to fit a polynomial of degree ?? to satisfy as many points as possible. In general, for the sake of the HSC course we're interested in fitting a linear function (a degree 1 polynomial, I.e. Straight line) through the points.
The line that we fit is, in general, good for points within the domain. If our data ranges from, say, a range of currents from 1 amp to 10 amps, then it's good to find something for 3 amps. This is interpolation - taking something out from within the boundaries.
Extrapolation is different and is what's considered here. Extrapolation is the act of relying on your polynomial to determine something OUTSIDE the boundaries.
In general, this causes havoc. Because the behaviour of a substance might change outside of reasonable boundaries and you actually need a completely new model for it. An example is in how germanium starts becoming a conductor and not a semiconductor.
That being said, the questions in the HSC are usually designed so that the region of extrapolation is not massive enough to cause problems. These issues would've been monitored in advance.
Rui, your explanation is essentially why I said it isn't 'good science' to continue your sketch beyond the data points given. Consider Hooke's law, which looks like this over large enough delta x.
(http://i.imgur.com/1oI3QWT.png)
Most experiments only let you take measurements in the first region. So, should you draw a straight line into infinity? Obviously not; that doesn't represent the physical phenomena. This occurs in probably every non-electrodynamic experiment (I believe) due to relativistic effects. So, if it's not a good idea some of the time, it isn't a good idea any of the time. Obviously this strays beyond HSC physics, but does explain why I wouldn't recommend extrapolating unless you're asked to in an exam.
Post by: RuiAce on October 25, 2016, 11:28:44 am
Rui, your explanation is essentially why I said it isn't 'good science' to continue your sketch beyond the data points given. Consider Hooke's law, which looks like this over large enough delta x.They do say that in the real world extrapolation is one of the most dangerous things you can do.
(http://i.imgur.com/1oI3QWT.png)
Most experiments only let you take measurements in the first region. So, should you draw a straight line into infinity? Obviously not; that doesn't represent the physical phenomena. This occurs in probably every non-electrodynamic experiment (I believe) due to relativistic effects. So, if it's not a good idea some of the time, it isn't a good idea any of the time. Obviously this strays beyond HSC physics, but does explain why I wouldn't recommend extrapolating unless you're asked to in an exam.
Post by: jakesilove on October 25, 2016, 11:29:53 am
They do say that in the real world extrapolation is one of the most dangerous things you can do.
They don't call it extrapolation for nothing
Post by: Cindy2k16 on October 25, 2016, 11:33:00 am
Post by: jakesilove on October 25, 2016, 11:33:59 am
Thanks for the replies! It occurred to me because in the biology exam they didn't give us plots near the x/y axis and I only plotted the line within the points but it suddenly occurred to me that what if i should've extended it? I've heard teachers say extending it isnt good science but some teachers say that extending it is what the HSC markers want, even though it's not necessarily 'good science'. I've heard teachers who do HSC marking both say not to extend and also the opposite :/
Yeah look, you're absolutely right, and different teachers have different preferences. The fact is that you'll get full marks no matter what you do, because teachers teach differently.
Post by: Happy Physics Land on October 25, 2016, 02:55:09 pm
Rui, your explanation is essentially why I said it isn't 'good science' to continue your sketch beyond the data points given. Consider Hooke's law, which looks like this over large enough delta x.
(http://i.imgur.com/1oI3QWT.png)
Most experiments only let you take measurements in the first region. So, should you draw a straight line into infinity? Obviously not; that doesn't represent the physical phenomena. This occurs in probably every non-electrodynamic experiment (I believe) due to relativistic effects. So, if it's not a good idea some of the time, it isn't a good idea any of the time. Obviously this strays beyond HSC physics, but does explain why I wouldn't recommend extrapolating unless you're asked to in an exam.
Oh no material testing experiments and Hooke's law ... XD where are my yield stress maximum tensile stress and necking point?????
Post by: jakesilove on October 25, 2016, 02:56:57 pm
Oh no material testing experiments and Hooke's law ... XD where are my yield stress maximum tensile stress and necking point?????
.........
Post by: nay103 on October 25, 2016, 04:25:33 pm
I was doing 2009 HSC and came across this question:
31) a) ii) Describe TWO problems associated with Rutherford’s model and how these were explained by Bohr’s model of the hydrogen atom. (4)
I could only think of the fact that Rutherford's model didn't configure electrons into orbits, so I looked at the textbook answer. It said that Rutherford didn't explain how negative electrons weren't pulled into the positive nucleus, and Bohr solved this by saying that an electron can orbit a nucleus like a planet around the sun, where angular momentum is quantised. But isn't this wrong, since electrons are actually just standing waves?
How else would I be able to answer this question?
Thanks!
Post by: Cindy2k16 on October 25, 2016, 09:08:33 pm
Post by: Cindy2k16 on October 25, 2016, 09:17:23 pm
In the sample answer it mentions Newton's 3rd Law- is that necessary to mention? If so, how does the reaction force show that gravity is stimulated because I don't really understand that.
Thanks in advance :)
Post by: jamonwindeyer on October 25, 2016, 09:40:28 pm
Hi could someone clarify how to ensure/determine the reliability, accuracy and validity of secondary sources? I think reliability is consistency in information across multiple reputable sources but I'm less sure about the difference between accuracy and validity for secondary sources.
Hey Cindy! For reliability, you are spot on.
Validity for an experiment is controlling variables, and for secondary sources, it is similar. Does this source answer the questions you were asking adequately? Is there any bias or error that could impact on its validity? Is the source credible, and current?
We assess accuracy by comparing the source to something like a scientiffic journal. We have established information that is guaranteed accurate; accurate secondary sources will match. For example, a source which says \(g=7ms^{-1}\) is not considered accurate, because it doesn't match the established information. So, ALL information in that source is therefore probably inaccurate, or at least needs to be taken with caution :)
Post by: jamonwindeyer on October 25, 2016, 09:49:08 pm
For this question I wrote that the rotation of the space station would provide a centripetal force that acts on the astronaut causing the astronaut to accelerate towards the centre and thus stimulates gravity. Is that enough to answer the question?
In the sample answer it mentions Newton's 3rd Law- is that necessary to mention? If so, how does the reaction force show that gravity is stimulated because I don't really understand that.
Thanks in advance :)
For this, you would need about 1 more sentence, maybe 2. You've started excellently.
To understand what is happening here, think of a heavy brick in a bucket. If you swing that bucket around at speed, you have a centripetal force applied by your arm which keeps the bucket in circular motion. However, provided you keep the bucket spinning, the brick stays in the bucket! Why? There is a reaction force at play here (that's where Newton's 3rd Law comes in), an equal and opposite force to the centripetal force. We colloquially call this the centrifugal force, and THIS is what keeps the brick in place. Just the same, this is what simulates gravity for your astronaut.
So you need 2 more bits of info:
- There is an equal and opposite force that acts on the astronaut due to Newton's 3rd Law
- This 'centrifugal force' is experienced as a downwards force by the astronaut, similar to gravity
Does that make sense? The analogy is the key here; but not everyone clicks with it (I guess I'm the only one swinging bricks around in buckets ;)) ;D
Post by: Cindy2k16 on October 25, 2016, 10:09:05 pm
For this, you would need about 1 more sentence, maybe 2. You've started excellently.
To understand what is happening here, think of a heavy brick in a bucket. If you swing that bucket around at speed, you have a centripetal force applied by your arm which keeps the bucket in circular motion. However, provided you keep the bucket spinning, the brick stays in the bucket! Why? There is a reaction force at play here (that's where Newton's 3rd Law comes in), an equal and opposite force to the centripetal force. We colloquially call this the centrifugal force, and THIS is what keeps the brick in place. Just the same, this is what simulates gravity for your astronaut.
So you need 2 more bits of info:
- There is an equal and opposite force that acts on the astronaut due to Newton's 3rd Law
- This 'centrifugal force' is experienced as a downwards force by the astronaut, similar to gravity
Does that make sense? The analogy is the key here; but not everyone clicks with it (I guess I'm the only one swinging bricks around in buckets ;)) ;D
Hi thanks for the explanation! I'm still a bit confused though sorry. so in the sample answer it says the astronaut "reacts against the force". is this the same as "an equal and opposite force acts on the astronaut"? One sounds to me like its the astronaut providing the reaction force and the other sounds like something else is providing a reaction force-onto the astronaut, so I'm a bit confused. Also how is the astronaut/brick staying in place similar to how gravity works?
Thanks!
Post by: jamonwindeyer on October 26, 2016, 09:12:38 am
Hi thanks for the explanation! I'm still a bit confused though sorry. so in the sample answer it says the astronaut "reacts against the force". is this the same as "an equal and opposite force acts on the astronaut"? One sounds to me like its the astronaut providing the reaction force and the other sounds like something else is providing a reaction force-onto the astronaut, so I'm a bit confused. Also how is the astronaut/brick staying in place similar to how gravity works?
Thanks!
Ehh it's a little bit the same really; there is a reaction force experienced by the astronaut is what I would say. It's the least confusing and answers the question directly ;D
If the astronaut was standing on flat ground on earth, they would expect to stay in place and not float anywhere. Why? Gravity. On the space station, the reaction/centrifugal force is what keeps the astronaut in place and keeps them from floating off. Just like gravity does on earth; and thus, the reaction force keeping them in place simulates gravity. More specifically, the reaction force that the astronaut experiences (and exerts on the space station) is kind of like the weight force that one experiences due to gravity :)
Post by: aaron_solomon on October 27, 2016, 10:59:32 am
Post by: jamonwindeyer on October 27, 2016, 11:36:57 am
Hey could someone explain what magnetic flux is? i can seem to wrap my head around it
Hey! Magnetic flux is magnetic field lines; it is a magnetic field! So saying that we have a changing magnetic flux, means we have a changing magnetic field!
As to what a magnetic field actually is; that's gross. It's what you get when you mix electrodynamics and special relativity; definitely not necessary to go any further than just "magnetic field lines" ;D
Post by: imtrying on October 27, 2016, 11:41:45 am
Post by: RuiAce on October 27, 2016, 11:46:11 am
Hey could someone explain what magnetic flux is? i can seem to wrap my head around it
Hey! Magnetic flux is magnetic field lines; it is a magnetic field! So saying that we have a changing magnetic flux, means we have a changing magnetic field!In the HSC course, they colloquially define flux as "the number of field lines passing through an imaginary area". I remember in my trials I actually got examined on phi = BA
As to what a magnetic field actually is; that's gross. It's what you get when you mix electrodynamics and special relativity; definitely not necessary to go any further than just "magnetic field lines" ;D
If I were to use calculus to solve a projectile motion question in physics, would I still be able to get full marks? Or do I need to use the physics formulas?If you want to use it to check then go for it, but please do your actual working using the physics formulae. They're designed to not need any calculus at all. That is a risk and depending on your marker you may not be awarded the marks.
Post by: cherryred on October 27, 2016, 12:48:55 pm
I am still getting confused on lv/lo, mv/mo and tv/to. Any ways to understand whether a numerical value is given for the v or the o part would be appreciated.
Thank you
Post by: RuiAce on October 27, 2016, 12:57:21 pm
Hey,See a mention on posts #928 and #929
I am still getting confused on lv/lo, mv/mo and tv/to. Any ways to understand whether a numerical value is given for the v or the o part would be appreciated.
Thank you
Post by: Cindy2k16 on October 27, 2016, 03:11:43 pm
Post by: Cindy2k16 on October 27, 2016, 03:17:25 pm
TIA
Post by: jamonwindeyer on October 27, 2016, 03:17:44 pm
Hi could someone explain why the answer to this is C? Don't things with high resistance tend to heat up faster?
This was like the trickiest question ever. Remember that the rate of change of magnetic flux is a constant (presumably), and this means that the induced EMF or voltage is also a constant. Because the induced voltage is a constant, a high resistance means you get less current, and so get less heating.
Big R means little I, little I means no heating :P this was a weird question though, don't stress too much about it, it fooled everyone :)
The issue is that you want a large R, but not so large that not enough current flows ;D
Post by: Cindy2k16 on October 27, 2016, 03:19:53 pm
This was like the trickiest question ever. Remember that the rate of change of magnetic flux is a constant (presumably), and this means that the induced EMF or voltage is also a constant. Because the induced voltage is a constant, a high resistance means you get less current, and so get less heating.
Big R means little I, little I means no heating :P this was a weird question though, don't stress too much about it, it fooled everyone :)
The issue is that you want a large R, but not so large that not enough current flows ;D
oh ok thank you :)
Post by: jamonwindeyer on October 27, 2016, 03:23:58 pm
Also why is the answer for this one B? I calculated that the energy of the peak wavelength is 0.15eV and so that only overcomes the energy gap of A. Did I calculate it wrong?
TIA
What did you use for your peak wavelength? If you use something just a TOUCH lower, then you will get to the value required for B. To effectively detect human radiation, we'd want an energy gap close to the energy in the typical human-emitted photon. The value you calculated is closest to B (even if it is slightly below), so we'd prefer B over A in that circumstance. A's band gap is too low; so we'd presumably pick up radiation from other stuff! ;D
Post by: Cindy2k16 on October 27, 2016, 03:27:55 pm
What did you use for your peak wavelength? If you use something just a TOUCH lower, then you will get to the value required for B. To effectively detect human radiation, we'd want an energy gap close to the energy in the typical human-emitted photon. The value you calculated is closest to B (even if it is slightly below), so we'd prefer B over A in that circumstance. A's band gap is too low; so we'd presumably pick up radiation from other stuff! ;D
I used 8micrometres for the peak wavelength. But I see why the answer is B. Thank you!
Post by: Cindy2k16 on October 27, 2016, 03:38:03 pm
Post by: imtrying on October 27, 2016, 04:04:37 pm
Post by: jamonwindeyer on October 27, 2016, 04:09:29 pm
For the dot point 'discuss the concept that length standards are defined in terms of time in contrast to then original metre standard" what is it that we discuss? I understand that it was a change from 1 ten millionth of the distance between the equator along the meridian to the distance travelled by light in a vacuum in a fraction of a second, but how exactly does this relate to relativity?
Hey! Well basically the only thing to discuss is that we are now defining the metre based on a more absolute standard. The speed of light is constant! Length is not constant according to special relativity. Therefore, if we use light to define the metre, then we are more accurate! :)
Post by: jamonwindeyer on October 27, 2016, 04:10:39 pm
Hi for this question I used the formula starting with vy2 and presumed that it would hit the base of the house at max height (so i didn't work out the time) I got the required answer of 30m/s. but in the answers they found the time it would take and subbed it into the other formula. So would the way I worked it out be invalid since I had assumed it would hit at max height even though the questions didn't say it did?
Hey Cindy! Unfortunately it would be; unless you could somehow proved that it had to strike at the point of maximum height! :) you'd probably just lose a mark :P
Post by: znaser on October 27, 2016, 04:17:43 pm
After outlining his proposal and Davission and Germer's nickel experiment, is it enough to say that the the treatment of electrons as waves resulted in the explanation of the shortcomings of Bohr's model (stability of electron orbits) and hence, the move from classical to quantum physics, or is there more to it? Thanks :)
Post by: bethjomay on October 27, 2016, 04:43:07 pm
Help! Is this a syllabus point I'm missing or are we meant to just figure it out?
Post by: Goodwil on October 27, 2016, 04:50:38 pm
Post by: jakesilove on October 27, 2016, 05:09:44 pm
Hi. So this is a 7 marker from the 2007 hsc: Analyse how de Broglie's proposal and supporting experimental evidence led to the move from classical physics to quantum physics.
After outlining his proposal and Davission and Germer's nickel experiment, is it enough to say that the the treatment of electrons as waves resulted in the explanation of the shortcomings of Bohr's model (stability of electron orbits) and hence, the move from classical to quantum physics, or is there more to it? Thanks :)
I think that's a completely sufficient answer! Go into a bit of depth with each of the points you've outlined, but I wouldn't recommend more :)
Post by: jakesilove on October 27, 2016, 05:13:33 pm
(http://uploads.tapatalk-cdn.com/20161026/16ba9879960544c60141cd917ffae2f8.jpg)
Help! Is this a syllabus point I'm missing or are we meant to just figure it out?
Hey! This is a really tough question, really really tough. The motor part is just like usual; when we want to car to go forward, we pump energy into a motor which causes shit to rotate etc. Standard stuff. The important part of this question is to relate the fact that a motor is a generator, but in reverse. So, the car can convert electrical energy into kinetic energy, but it can also convert kinetic energy into electrical energy! By making the generator turn, energy is being sucked away, thus decreasing the overall speed of the vehicle. Throw in the term 'Lenz's law' etc., and describe how a motor/generator works, and you should get full marks :)
Post by: jakesilove on October 27, 2016, 05:15:17 pm
Hi there. Why is the answer B and not C in this question?
Because we're talking about WAVELENGTH, not frequency! Recall that when determining the maximum kinetic energy of an electron, it will have to do with the frequency of the photon that is incident. Frequency is proportional to 1/wavelength, so the function should look hyperbolic :)
Post by: bethjomay on October 27, 2016, 05:49:34 pm
Hey! This is a really tough question, really really tough. The motor part is just like usual; when we want to car to go forward, we pump energy into a motor which causes shit to rotate etc. Standard stuff. The important part of this question is to relate the fact that a motor is a generator, but in reverse. So, the car can convert electrical energy into kinetic energy, but it can also convert kinetic energy into electrical energy! By making the generator turn, energy is being sucked away, thus decreasing the overall speed of the vehicle. Throw in the term 'Lenz's law' etc., and describe how a motor/generator works, and you should get full marks :)
Ok, thank you! So the generator 'effect' is happening in the original electric motor, simultaneously? Both are happening at once?
Post by: jakesilove on October 27, 2016, 06:00:40 pm
Ok, thank you! So the generator 'effect' is happening in the original electric motor, simultaneously? Both are happening at once?
No, not quite; when the car 'speeds up', the apparatus will act as a motor. Electrical energy will be converted into kinetic energy, propelling the car forward. However, when the car needs to 'slow down', the apparatus begins to work as a generator. It will draw kinetic energy away, converting it back to electrical energy! So, whether it acts as a generator or a motor will depend on what actually needs to happen/what sort of outcome we want :)
Post by: FallonXay on October 27, 2016, 06:01:53 pm
How would you do the calculations for the graph in part b?
Thanks~
Post by: jakesilove on October 27, 2016, 06:07:52 pm
Hi!
How would you do the calculations for the graph in part b?
Thanks~
Hey! Such a weird question, hey? It was from my paper, so I know it well :D You just need to make sure that the horizontal displacement of Ball Q stays consistent (ie 3 blocks). Then, you know that acceleration due to gravity adds a consistent amount of velocity per second. It starts by 'moving down' 3 blocks, and then 9 blocks, which means each second ADDS 6 blocks worth of down. So, the next vertical position should be 9+6=15 blocks further down! Absolutely stupid question, but it does make you think about the actual physics behind projectile motion :) You can also check the 'actual' maths on the BOSTES answer page, here
Post by: znaser on October 27, 2016, 06:11:47 pm
I think that's a completely sufficient answer! Go into a bit of depth with each of the points you've outlined, but I wouldn't recommend more :)
Thanks Jake!
Post by: bethjomay on October 27, 2016, 06:15:40 pm
No, not quite; when the car 'speeds up', the apparatus will act as a motor. Electrical energy will be converted into kinetic energy, propelling the car forward. However, when the car needs to 'slow down', the apparatus begins to work as a generator. It will draw kinetic energy away, converting it back to electrical energy! So, whether it acts as a generator or a motor will depend on what actually needs to happen/what sort of outcome we want :)
Ah ok, that makes sense! Thank you!
Post by: wyzard on October 27, 2016, 06:18:46 pm
Hi!Basically break it down into horizontal and vertical components and examine them separately.
How would you do the calculations for the graph in part b?
Thanks~
For the horizontal bit, the speed is uniform, so for every second the x-coordinate increases by the same amount.
For the vertical bit it's a more challenging as there's acceleration due to gravity to account for. Fortunately since both the cannon balls have the same vertical speed, they will be aligned at every second and you can use that. For the remaining time however, you can use the following equation:
Post by: RuiAce on October 27, 2016, 06:27:30 pm
Post by: FallonXay on October 27, 2016, 06:29:21 pm
Basically break it down into horizontal and vertical components and examine them separately.
For the horizontal bit, the speed is uniform, so for every second the x-coordinate increases by the same amount.
For the vertical bit it's a more challenging as there's acceleration due to gravity to account for. Fortunately since both the cannon balls have the same vertical speed, they will be aligned at every second and you can use that. For the remaining time however, you can use the following equation:
Note that, on the HSC formula sheet, the above equation is
Hey! Such a weird question, hey? It was from my paper, so I know it well :D You just need to make sure that the horizontal displacement of Ball Q stays consistent (ie 3 blocks). Then, you know that acceleration due to gravity adds a consistent amount of velocity per second. It starts by 'moving down' 3 blocks, and then 9 blocks, which means each second ADDS 6 blocks worth of down. So, the next vertical position should be 9+6=15 blocks further down! Absolutely stupid question, but it does make you think about the actual physics behind projectile motion :) You can also check the 'actual' maths on the BOSTES answer page, here
Got it - Thanks :)
Post by: Cindy2k16 on October 27, 2016, 09:20:38 pm
Post by: RuiAce on October 27, 2016, 09:24:57 pm
Hi so I understand that they needed the aether since they thought light needed a medium to propagate (is this right?) and the MM experiment was designed to test the speed of the earth through the aether. But I don't understand how Einstein's special theory of relativity meant the aether was no longer needed- why did light being constant in all frames of reference mean that light didn't need a medium to travel in? Could someone explain? Thanks in advance :)The fact that the speed of light was constant in all frames of reference, and that all inertial frames of reference are relative to each other (these are the two statements of special relativity), meant that there was no longer a need for some:
a) absolute frame of reference
b) medium of propagation of light
as if the above were true, then technically the speed of light would be affected. If the aether existed, then the existence of an aether wind would've naturally affected the speed of light. But of course, MM showed that the aether may as well not exist, so the model can be discarded as unnecessary.
Post by: Cindy2k16 on October 27, 2016, 09:30:30 pm
The fact that the speed of light was constant in all frames of reference, and that all inertial frames of reference are relative to each other (these are the two statements of special relativity), meant that there was no longer a need for some:
a) absolute frame of reference
b) medium of propagation of light
as if the above were true, then technically the speed of light would be affected. If the aether existed, then the existence of an aether wind would've naturally affected the speed of light. But of course, MM showed that the aether may as well not exist, so the model can be discarded as unnecessary.
Hi thanks for the reply I understand now :)
Also is the speed of light constant in all frames of reference- both inertial and non-inertial- or is it all inertial frames of reference?
TIA
Post by: RuiAce on October 27, 2016, 09:31:34 pm
Hi thanks for the reply I understand now :)Whoops forgot to say that above. But I believe inertial only.
Also is the speed of light constant in all frames of reference- both inertial and non-inertial- or is it all inertial frames of reference?
TIA
Weird things happen in non-inertial frames of reference.
Post by: Cindy2k16 on October 27, 2016, 09:34:05 pm
Whoops forgot to say that above. But I believe inertial only.
Weird things happen in non-inertial frames of reference.
ok cool thanks. much appreciated!
Post by: jamonwindeyer on October 27, 2016, 11:10:45 pm
Whoops forgot to say that above. But I believe inertial only.
Weird things happen in non-inertial frames of reference.
Confirm inertial only. If we are in a non-inertial reference frame we can use the altered behaviour of light to calculate our acceleration ;D
Post by: znaser on October 28, 2016, 10:03:21 am
Post by: RuiAce on October 28, 2016, 10:12:50 am
I'm not exactly sure as to why a step-down transformer is used to transfer energy from a regional to local sub-station? and for the laptop or pretty much any household appliance, I always thought a step-up transformer is used so there isn't too much current flowing through the wires of the appliance that would cause it to burn. But according to the answers (c), a step-down transformer is also actually used. Why is this the case? Thanks :)To bring it to the regional substation we step it up so that we minimise the current flow. This minimises power losses.
But once it reaches the regional substation, we need to bring it to houses. At the substation, the voltages have reached things such as 1M V. If we keep stepping it up, it will probably reach 10M V. Obviously, things in the household do NOT need so much voltage; a powerpoint is only about 240V. So we need to step it down or else there will be too much voltage for things in the house.
Whereas something like a TV requires high voltages, we need the opposite for a laptop. Once we are down at 240V, there is no longer a direct risk of wires burning, or else we wouldn't have electricity in our homes. Laptops and etc. only need about 10V to power, otherwise the higher voltages are what damage the delicate appliances.
Post by: znaser on October 28, 2016, 10:23:40 am
To bring it to the regional substation we step it up so that we minimise the current flow. This minimises power losses.
But once it reaches the regional substation, we need to bring it to houses. At the substation, the voltages have reached things such as 1M V. If we keep stepping it up, it will probably reach 10M V. Obviously, things in the household do NOT need so much voltage; a powerpoint is only about 240V. So we need to step it down or else there will be too much voltage for things in the house.
Whereas something like a TV requires high voltages, we need the opposite for a laptop. Once we are down at 240V, there is no longer a direct risk of wires burning, or else we wouldn't have electricity in our homes. Laptops and etc. only need about 10V to power, otherwise the higher voltages are what damage the delicate appliances.
That makes much more sense now. Thanks Rui!
Post by: MysteryMarker on October 28, 2016, 10:37:10 am
Recently the weapons unit of the UN discovered the components for a 100m long cannon barrel in an unspecified country. The country's defence agencies have denied that this is an offensive weapon to be used against neighbouring countries. They have explained that is is to be used to launch communication satellites into orbit. Explain why this is not considered to be a plausible explanation. (3 marks)
Post by: RuiAce on October 28, 2016, 10:38:30 am
Don't quite understand how to go about this question.Hint: Think about accelerating too quickly causing the satellite to burn, and excessive g-forces if people are meant to be on the satellite.
Recently the weapons unit of the UN discovered the components for a 100m long cannon barrel in an unspecified country. The country's defence agencies have denied that this is an offensive weapon to be used against neighbouring countries. They have explained that is is to be used to launch communication satellites into orbit. Explain why this is not considered to be a plausible explanation. (3 marks)
Post by: Neutron on October 28, 2016, 11:19:35 am
Yo! Could someone please explain these two multiple choices for me? For the first one, if back emf was equal to the applied voltage wouldn't that mean the total voltage is zero so the coil wouldn't even spin? (No voltage=no current) what ahah and for the second one, the heat loss within metal is due to the current's interaction with the resistance right? So I get that with lower resistance you can get more current (and since its I^2 then that will increase power loss) but if you had high resistance, wouldn't the ultimately heat loss be greater? Like if there was no resistance but the current was really high, there'd still be no power loss so idk in this case does a higher current win out? :/ cheers
Neutron
Post by: Happy Physics Land on October 28, 2016, 11:50:21 am
(http://uploads.tapatalk-cdn.com/20161027/c9b2de4f2f31c82b9cc8fab8464fae57.jpg) (http://uploads.tapatalk-cdn.com/20161027/8831eaf356615c8295645d046a6e6fcc.jpg)
Yo! Could someone please explain these two multiple choices for me? For the first one, if back emf was equal to the applied voltage wouldn't that mean the total voltage is zero so the coil wouldn't even spin? (No voltage=no current) what ahah and for the second one, the heat loss within metal is due to the current's interaction with the resistance right? So I get that with lower resistance you can get more current (and since its I^2 then that will increase power loss) but if you had high resistance, wouldn't the ultimately heat loss be greater? Like if there was no resistance but the current was really high, there'd still be no power loss so idk in this case does a higher current win out? :/ cheers
Neutron
Hey Neutron!
I will start with question 19. This question used to always bother me, because my reasoning when answering this question is the same as you (i.e. P=IR^2 and high resistance would result in more power loss and heat dissipation etc.). I reckon it's a badly written question and the question should really specify "the metallic base of the cooking pot".
The base of the pot must be a metal in order for eddy current to be induced and flow. So compared to all the other materials such as plastic which has high resistance, all the materials used to make the base of the pot would have low resistance since they are metals. Really dodgy question, I honestly agree with your reasoning but the answer is C as per BOSTES.
Question 16 was a bit difficult, and I almost ignored the minute detail as well until when Jamon Windeyer (Credit :)) pointed out to me that it is an IDEAL ELECTRIC MOTOR. This means that there will be NO FRICTION and hence the motor will retain its angular momentum and no extra torque is needed for the motor to continue rotating. In real life situations when friction is present, option A would be correct because we need the supply current to create torque to overcome the effect of friction (T = nBIAcostheta). But since here we have an ideal electric motor with no friction, nothing needs to be overcome and no work is done since no force (F=BIL) is applied to help it rotate. Hence back emf = applied voltage (i.e. no net emf, no current) and hence the answer is C. Tricky!
Best Regards
Happy Physics Land
Post by: and1_98 on October 28, 2016, 12:43:16 pm
Post by: student123456 on October 28, 2016, 01:14:17 pm
A photon is incident on a hydrogen atom in the ground state.
Explain, using de Broglie’s hypothesis, why the photon is not absorbed
by the hydrogen atom. (3 marks)
Post by: imtrying on October 28, 2016, 01:40:09 pm
Under the dot point about Planck and Einsteins differing views about whether science research is removed from political and social forces, I definitely see the political forces,like the use of scientific research for use by military and governments, but what were the social forces? Would it be things like WW2 German anti-Semitism impacting on Jewish scientists?
Also, a bit confused on how Hertzs experiment relates to the photoelectric effect.
Post by: Neutron on October 28, 2016, 01:49:29 pm
I initially thought C... Then turned to A... Answer says A. Anyone care to explain?
Hah yew this is something I can answer
so in order for diffraction to occur, the wavelength must be similar to the slit size right? Cause if the slight was too big the wave would just pass right through (so in this case, if the wavelength was smaller than the slit, it would just pass right through with no diffraction). So in order for the Braggs to observe a diffraction pattern, the wavelength of the X Ray's would have to be on the same order of magnitude (fancy way of saying same size) as the atomic spacings within the crystal! Also, loving the creativity of option D heh, hope this helps!Neutron
Post by: jakesilove on October 28, 2016, 01:50:25 pm
I initially thought C... Then turned to A... Answer says A. Anyone care to explain?
Brilliant answer Neutron, nothing to add except that this is just something you need to know, but not necessarily understand.
Post by: jakesilove on October 28, 2016, 01:52:06 pm
Hi. Would anyone be able to help me answer this question please?
A photon is incident on a hydrogen atom in the ground state.
Explain, using de Broglie’s hypothesis, why the photon is not absorbed
by the hydrogen atom. (3 marks)
Hey! Check out posts HERE and HERE for an answer :)
Post by: jakesilove on October 28, 2016, 01:53:57 pm
Hey😄
Under the dot point about Planck and Einsteins differing views about whether science research is removed from political and social forces, I definitely see the political forces,like the use of scientific research for use by military and governments, but what were the social forces? Would it be things like WW2 German anti-Semitism impacting on Jewish scientists?
Also, a bit confused on how Hertzs experiment relates to the photoelectric effect.
In terms of social impacts, you're absolutely right! There really isn't much else to talk about.
Hertz's experiments relates to the photoelectric effect as it was the first observation of it. He observed that EM radiation could cause 'sparking' in wires, which we now realise is due to the photoelectric effect. He didn't do any research into this, or discover the cause (he left that the Planck/Einstein), so the Hertz experiment is significant purely as a starting point :)
Post by: imtrying on October 28, 2016, 01:55:56 pm
In terms of social impacts, you're absolutely right! There really isn't much else to talk about.
Hertz's experiments relates to the photoelectric effect as it was the first observation of it. He observed that EM radiation could cause 'sparking' in wires, which we now realise is due to the photoelectric effect. He didn't do any research into this, or discover the cause (he left that the Planck/Einstein), so the Hertz experiment is significant purely as a starting point :)
Thanks :)
Post by: Neutron on October 28, 2016, 01:57:43 pm
Hey Neutron!
I will start with question 19. This question used to always bother me, because my reasoning when answering this question is the same as you (i.e. P=IR^2 and high resistance would result in more power loss and heat dissipation etc.). I reckon it's a badly written question and the question should really specify "the metallic base of the cooking pot".
The base of the pot must be a metal in order for eddy current to be induced and flow. So compared to all the other materials such as plastic which has high resistance, all the materials used to make the base of the pot would have low resistance since they are metals. Really dodgy question, I honestly agree with your reasoning but the answer is C as per BOSTES.
Question 16 was a bit difficult, and I almost ignored the minute detail as well until when Jamon Windeyer (Credit :)) pointed out to me that it is an IDEAL ELECTRIC MOTOR. This means that there will be NO FRICTION and hence the motor will retain its angular momentum and no extra torque is needed for the motor to continue rotating. In real life situations when friction is present, option A would be correct because we need the supply current to create torque to overcome the effect of friction (T = nBIAcostheta). But since here we have an ideal electric motor with no friction, nothing needs to be overcome and no work is done since no force (F=BIL) is applied to help it rotate. Hence back emf = applied voltage (i.e. no net emf, no current) and hence the answer is C. Tricky!
Best Regards
Happy Physics Land
Aye happy physics land it's been a while hahaha
Yeah yeah that pot thing makes sense now heh
But with the ideal motor one (does ideal just mean we ignore everything in the real world lmao) I get that if the net voltage is zero and the current is therefore zero that coil will keep spinning anyway, but while the back emf is being induced, won't it produce a force that opposes the rotation of the coil (and thus slowing it down?) like I thought the point of back emf never equaling the supply emf being that because when they're equal, the relative motion between the coil and the magnet will be zero (force produced by back=force produced by supply) and therefore, no rate of change of flux=no emf induced.. Idk sorry i think I'm just being dumb haha
Neutron
Post by: jamonwindeyer on October 28, 2016, 02:02:06 pm
Aye happy physics land it's been a while hahaha
Yeah yeah that pot thing makes sense now heh
But with the ideal motor one (does ideal just mean we ignore everything in the real world lmao) I get that if the net voltage is zero and the current is therefore zero that coil will keep spinning anyway, but while the back emf is being induced, won't it produce a force that opposes the rotation of the coil (and thus slowing it down?) like I thought the point of back emf never equaling the supply emf being that because when they're equal, the relative motion between the coil and the magnet will be zero (force produced by back=force produced by supply) and therefore, no rate of change of flux=no emf induced.. Idk sorry i think I'm just being dumb haha
Neutron
No no you've actually got it and don't realise!
I thought the point of back emf never equaling the supply emf being that because when they're equal, the relative motion between the coil and the magnet will be zero (force produced by back=force produced by supply)
You are right! Torque produced by supply equals torque produced by back-emf, so the net torque is zero. What happens to an object where we have a zero net force? It just keeps doing whatever it was doing. Thus, the motor keeps spinning! It's the inertia of the coil that keeps it going ;D
Post by: Neutron on October 28, 2016, 02:54:35 pm
No no you've actually got it and don't realise!
I thought the point of back emf never equaling the supply emf being that because when they're equal, the relative motion between the coil and the magnet will be zero (force produced by back=force produced by supply)
You are right! Torque produced by supply equals torque produced by back-emf, so the net torque is zero. What happens to an object where we have a zero net force? It just keeps doing whatever it was doing. Thus, the motor keeps spinning! It's the inertia of the coil that keeps it going ;D
Ah idek what it is I don't get I think I just need time to wrap my head around this lmao I'll keep thinking about it and hopefully I eventually work out what I don't get. But just with another question, what actually is the neutrino? Like it's not found it the nucleus is it? So during beta decay, won't some of the "kinetic energy the electron is supposed to have but doesn't" be converted to create the mass of the neutrino and then the neutrino's kinetic energy? Cheers
Neutron
Post by: imtrying on October 28, 2016, 03:11:46 pm
Doping a semiconductor with a group 3 impurity increases its conductivity because there is one less electron in the valence band so electrons are not so tightly held in covalent bonds so they are more free to move to the conduction band, meaning there are more positive holes created in the valence band to carry positive charge in the opposite direction.
Doping with a group 5 impurity just means there are more electrons available to move in to the conduction band.
Haha I don't know if this is right, this part of the topic always confuses me.
Post by: and1_98 on October 28, 2016, 03:27:28 pm
I mean, it would definitely come under E = mc^2, mass dilation etc.; just haven't seen anything else like it before.
From the 2012 Girraween 2012 Trial.
Post by: wyzard on October 28, 2016, 03:46:20 pm
I got the answer (C), just curious as to whether or not something like this (Relativistic KE) is assessable...?Not too sure about HSC, but it's always better to know about it just in case it might come up. It's better to know a bit more than what the syllabus expects from you.
I mean, it would definitely come under E = mc^2, mass dilation etc.; just haven't seen anything else like it before.
From the 2012 Girraween 2012 Trial.
To evaluate the kinetic energy you can use the following formula:
Where
Post by: jakesilove on October 28, 2016, 03:47:22 pm
I got the answer (C), just curious as to whether or not something like this (Relativistic KE) is assessable...?
I mean, it would definitely come under E = mc^2, mass dilation etc.; just haven't seen anything else like it before.
From the 2012 Girraween 2012 Trial.
Definitely not able to be assessed. Don't worry about it!
Post by: Goodwil on October 28, 2016, 03:55:36 pm
Post by: imtrying on October 28, 2016, 03:57:57 pm
Also, do I need to know about forward and reverse bias?
Post by: jakesilove on October 28, 2016, 04:14:41 pm
Hi there, just having a bit of trouble with these questions (5 is D, 16 is A)Hey! For the first one, you can figure out the speed of the thing by going
Then, sub straight into our centripetal force formula!
For your second question, the current is essentially going 'out' of the page (opposite direction to electron flow), and your thumb points in the direction of the north pole (to the left). Doing that, your palm should point up! This makes the answer A)
Post by: jakesilove on October 28, 2016, 04:16:20 pm
What are the differences between microchips and microprocessors, and other than their impacts on society, how much do we need to know about them?
Also, do I need to know about forward and reverse bias?
Is this as a part of Ideas to Implementation? If so, you don't need the answer to answer of your questions :) It can't be assessed!
Post by: imtrying on October 28, 2016, 04:19:40 pm
Is this as a part of Ideas to Implementation? If so, you don't need the answer to answer of your questions :) It can't be assessed!
Yep it's from ideas to implementation. I think the textbook I'm using is a little too detailed, glad I don't have to know this!
Post by: and1_98 on October 28, 2016, 04:23:42 pm
That 2010 HSC Cathode Ray multiple choice is ambig af. Would've gone C purely because I thought 'P' was directed 'into the page'...
Is the, I guess, 'hint' regarding direction just in the fact that there is a fat tube coming from 'into the page'; so you can sort of assume that's where the CR is emitted from (therefore ray is out of page)...?
Thanks man
Post by: RuiAce on October 28, 2016, 04:25:43 pm
Yep it's from ideas to implementation. I think the textbook I'm using is a little too detailed, glad I don't have to know this!You can think of a microprocessor as a more powerful microchip. A microprocessor is like a CPU if i recall.
Post by: maksim on October 28, 2016, 04:27:48 pm
I was just wanting to get some clarification as to what the characteristics of an inertial frame of reference are.
I'm aware that it is a non-accelerated frame of reference, but am unsure of whether it is a minimum of the first, or all of Newton's Laws of Motion that must apply for a frame of reference to be inertial. Any additional features that should also be included in a description of an inertial frame of reference would be greatly appreciated.
Post by: imtrying on October 28, 2016, 05:21:46 pm
You can think of a microprocessor as a more powerful microchip. A microprocessor is like a CPU if i recall.
Thanks:)
Post by: imtrying on October 28, 2016, 06:09:54 pm
Doping a semiconductor with a group 3 impurity increases its conductivity because there is one less electron in the valence band so electrons are not so tightly held in covalent bonds so they are more free to move to the conduction band, meaning there are more positive holes created in the valence band to carry positive charge in the opposite direction.
Doping with a group 5 impurity just means there are more electrons available to move in to the conduction band.
Haha I don't know if this is right, this part of the topic always confuses me.
Post by: $Billi$ on October 28, 2016, 06:17:25 pm
need help with these three questions. 1 is A (although I really think it's D) 6 is D 7 is C.
thank you
Post by: jakesilove on October 28, 2016, 06:28:08 pm
Hi,
I was just wanting to get some clarification as to what the characteristics of an inertial frame of reference are.
I'm aware that it is a non-accelerated frame of reference, but am unsure of whether it is a minimum of the first, or all of Newton's Laws of Motion that must apply for a frame of reference to be inertial. Any additional features that should also be included in a description of an inertial frame of reference would be greatly appreciated.
Hey! An inertial frame is one travelling with a constant velocity. That's it; that's the only descriptor necessary, because that's the only definition. Now, the important ASPECT of an inertial frame is that all laws of Physics operate identically, regardless of the inertial frame. If you want to get a bit more technical here, we're actually saying that Maxwell's equations are abided by in all inertial frames. However, that's all you need!
Post by: jakesilove on October 28, 2016, 06:32:51 pm
Would I be correct in saying this?
Doping a semiconductor with a group 3 impurity increases its conductivity because there is one less electron in the valence band so electrons are not so tightly held in covalent bonds so they are more free to move to the conduction band, meaning there are more positive holes created in the valence band to carry positive charge in the opposite direction.
Doping with a group 5 impurity just means there are more electrons available to move in to the conduction band.
Haha I don't know if this is right, this part of the topic always confuses me.
I think that it's important that you get a bit more technical in your response. You're concepts are great, but there's some important jargon that you're missing, and some holes (lol) in your explanation.
P-type
A P-type semiconductor is one doped with a Group 3 metal. This creates an extra 'hole' in the lattice structure, which can be described as positive simply because it is 'less negative' that surrounding portions. The hole allows electrons to more easily flow, as they will 'jump' to fill the holes etc. P-type semiconductors create an accepter level above the Valence band, which allows electrons to more easily reach the Conduction band (Draw diagram)
N-type
A P-type semiconductor is one doped with a Group 5 metal. This creates an extra electron in the lattice structure. The excess electrons allows electrons to more easily flow, as they can be thought of as being placed 'above' lattice electrons. N-type semiconductors create a donor level below the Conduction band, which allows electrons to more easily reach the Conduction band (Draw diagram).
And that's it! Let me know if any of that is confusing
Post by: imtrying on October 28, 2016, 06:41:01 pm
I think that it's important that you get a bit more technical in your response. You're concepts are great, but there's some important jargon that you're missing, and some holes (lol) in your explanation.
P-type
A P-type semiconductor is one doped with a Group 3 metal. This creates an extra 'hole' in the lattice structure, which can be described as positive simply because it is 'less negative' that surrounding portions. The hole allows electrons to more easily flow, as they will 'jump' to fill the holes etc. P-type semiconductors create an accepter level above the Valence band, which allows electrons to more easily reach the Conduction band (Draw diagram)
N-type
A P-type semiconductor is one doped with a Group 5 metal. This creates an extra electron in the lattice structure. The excess electrons allows electrons to more easily flow, as they can be thought of as being placed 'above' lattice electrons. N-type semiconductors create a donor level below the Conduction band, which allows electrons to more easily reach the Conduction band (Draw diagram).
And that's it! Let me know if any of that is confusing
Very clear explanation, thank you!
Post by: anuu.j on October 28, 2016, 06:41:48 pm
Could someone please explain to me about lenz's law and faradays law
Post by: jakesilove on October 28, 2016, 06:42:28 pm
Hey guys,
need help with these three questions. 1 is A (although I really think it's D) 6 is D 7 is C.
thank you
Hey! So, starting with your first question, we know that there has to be some sort of accelerating force to change your weight, right? So, at the very least, the answer can't be C. Now, think about what is actually required; you need to feel less weight that normal! If it were moving up, and speeding up, you would feel 'pressed' into the ground (that's easy to understand). However, if it were moving down and slowing down, you would also feel pressed into the ground! This would make you heavier. The only solution is if it were moving up, and slowing down; you would feel weightless (Imagine if it were moving up really quickly, and then just stopped. You'd be thrown into the roof!).
For your second question, you first need to identify that you need to DO WORK on the objects to pull them apart. They will naturally move together, due to gravity, so you must do POSITIVE work (limiting our answers to C and D). Now, the gravitational ENERGY is the gravitational FORCE times a distance. Therefore, the required force will be 4.0J. Somehow, that gets us to 4.0J as an answer. Hmm... Not sure
Your last question is super hard. In fact, you can't actually get a numerical answer. The answer is 0.66c, and you need to be able to logic that out. Can you see why?
Post by: jakesilove on October 28, 2016, 06:44:20 pm
Heyy Guys
Could someone please explain to me about lenz's law and faradays law
Check out Jamon's complete guide here!
Post by: proficles on October 28, 2016, 06:49:13 pm
The cathode ray tube and transistor circuits in a conventional television rely on transformers.
What transformation of the 240 V AC input voltage do these components require?
Cathode ray tube: step up or down?
Transistor circuits: step up or down?
Post by: Cindy2k16 on October 28, 2016, 07:17:17 pm
Hi how would you arrive at the answer for the 2008 HSC Physics exam multiple choice question 8?
The cathode ray tube and transistor circuits in a conventional television rely on transformers.
What transformation of the 240 V AC input voltage do these components require?
Cathode ray tube: step up or down?
Transistor circuits: step up or down?
Cathode ray tubes require high voltages to cause the emission of electrons, so step up
transistor circuits are delicate and operate at very low voltages so step down. So A
hope this helps :)
Post by: proficles on October 28, 2016, 07:41:03 pm
Cathode ray tubes require high voltages to cause the emission of electrons, so step up
transistor circuits are delicate and operate at very low voltages so step down. So A
hope this helps :)
oh thank you makes sense!
Post by: maksim on October 28, 2016, 07:44:34 pm
Hey! An inertial frame is one travelling with a constant velocity. That's it; that's the only descriptor necessary, because that's the only definition. Now, the important ASPECT of an inertial frame is that all laws of Physics operate identically, regardless of the inertial frame. If you want to get a bit more technical here, we're actually saying that Maxwell's equations are abided by in all inertial frames. However, that's all you need!
Thank you. :)
Post by: maksim on October 28, 2016, 07:50:43 pm
Post by: RuiAce on October 28, 2016, 07:54:13 pm
I was also wondering what the primary types of heat shielding are for spacecrafts re-entering the atmosphere, and the level of detail we are required to know about each in terms of what they are made of and how they absorb/dissipate heat.Pretty sure you need literally no detail about that whatsoever.
Post by: maksim on October 28, 2016, 07:59:15 pm
Pretty sure you need literally no detail about that whatsoever.
Really? Because I've seen questions about the three issues related to safe re-entry which require additional information about overheating as an issue. So I wanted to got some advice about the depth of knowledge I would be required to know.
Post by: RuiAce on October 28, 2016, 08:01:13 pm
Really? Because I've seen questions about the three issues related to safe re-entry which require additional information about overheating as an issue. So I wanted to got some advice about the depth of knowledge I would be required to know.When it comes to overheating I just mention
- Heat shields
- Air conditioners
- As you said, just the physics behind re-entry
I've never seen overheating singled out as its own question. It's only under the umbrella of reentry.
(unless the other two have more to say)
Post by: maksim on October 28, 2016, 08:07:43 pm
When it comes to overheating I just mention
- Heat shields
- Air conditioners
- As you said, just the physics behind re-entry
I've never seen overheating singled out as its own question. It's only under the umbrella of reentry.
(unless the other two have more to say)
So would you say that it would be too much unnecessary information to talk about thermal soak and ablative heat shields. I completely forgot about air conditioning, so thanks for that. :)
Post by: RuiAce on October 28, 2016, 08:09:22 pm
So would you say that it would be too much unnecessary information to talk about thermal soak and ablative heat shields. I completely forgot about air conditioning, so thanks for that. :)I do remember the name "ablative heat shields" but I have never heard of thermal soak before. Nor did I know how heat shields actually worked.
If you want, stay tuned for what the other moderators have to say. But I personally think that's all unnecessary
Post by: zoe_rammie on October 28, 2016, 08:27:12 pm
But does a galvanometer measure EMF or Electric current?
(I know that EMF gives rise to electric current, but still...sometimes even the teenies details makes all the difference)
Thanks to whoever replies to this :D
Post by: imtrying on October 28, 2016, 08:32:25 pm
Post by: RuiAce on October 28, 2016, 08:35:01 pm
Sorry, this is just a simple question,Current
But does a galvanometer measure EMF or Electric current?
(I know that EMF gives rise to electric current, but still...sometimes even the teenies details makes all the difference)
Thanks to whoever replies to this :D
EMF (electromotive force) is equivalent to saying "Voltage" in the HSC
Post by: jakesilove on October 28, 2016, 08:35:17 pm
Sorry, this is just a simple question,
But does a galvanometer measure EMF or Electric current?
(I know that EMF gives rise to electric current, but still...sometimes even the teenies details makes all the difference)
Thanks to whoever replies to this :D
Hey! A galvanometer can be set up to measure both. Most commonly, I think it measures current, but it can be set up to measure voltage as well :)
Jake
Post by: jakesilove on October 28, 2016, 08:36:01 pm
Would it be more correct to refer to ultrasound passing through tissue as being transmitted or as being refracted? My textbook seems to use them interchangeably.
I would probably go with transmitted through tissue. Just seems more scientifically correct, because refracted implies some sort of 'outward' movement, which I guess is also correct... so yeah either, but I would suggest transmitted
Post by: Albertenouttaten on October 28, 2016, 08:39:37 pm
Post by: Cindy2k16 on October 28, 2016, 08:42:53 pm
I can only find the notes from the marking centre for the older papers and they don't give an indication of what the final verdict should be after comparing the structure and function- So I was just wondering what you're meant to end up deciding about the student's claim?
Thanks in advance :)
Post by: Albertenouttaten on October 28, 2016, 08:57:20 pm
Hope this helps
Post by: Goodwil on October 28, 2016, 09:09:44 pm
1. When using the right hand rule for electrons/cathode rays, is the finger that is used for current pointing in the same direction as electron flow, or the opposite direction? (eg if electrons are moving to the left, is the finger for current pointing to the left or the right?)
2. For projectile questions such as the one attached (ie projectiles on uneven ground) should we set acceleration to be positive or negative? And is ΔY the same or opposite sign?
Thanks
Post by: RuiAce on October 28, 2016, 09:12:45 pm
Hi again. I have 2 more questions.If I recall for Q2, in the answers for that one they made acceleration positive.
1. When using the right hand rule for electrons/cathode rays, is the finger that is used for current pointing in the same direction as electron flow, or the opposite direction? (eg if electrons are moving to the left, is the finger for current pointing to the left or the right?)
2. For projectile questions such as the one attached (ie projectiles on uneven ground) should we set acceleration to be positive or negative? And is ΔY the same or opposite sign?
Thanks
But there's a twist. Note how they also made Δy positive. Note that normally we would actually treat Δy as negative. This is cause overall, the particle fell downwards, in the vertical component sense.
Because they made Δy positive, they had to compensate by making ay positive. Here, I would've made ay (gravity) negative anyway, which would mean that Δy was also -150
Post by: bethjomay on October 28, 2016, 09:17:46 pm
Post by: jakesilove on October 28, 2016, 09:19:26 pm
Hey there, could someone help me with the calculations involved in each of these questions? Both from the 2014 paper, I feel sorry for Jamon and Jake now!
Hey! Such a weird question, hey? It was from my paper, so I know it well :D You just need to make sure that the horizontal displacement of Ball Q stays consistent (ie 3 blocks). Then, you know that acceleration due to gravity adds a consistent amount of velocity per second. It starts by 'moving down' 3 blocks, and then 9 blocks, which means each second ADDS 6 blocks worth of down. So, the next vertical position should be 9+6=15 blocks further down! Absolutely stupid question, but it does make you think about the actual physics behind projectile motion :) You can also check the 'actual' maths on the BOSTES answer page, here
Post by: jakesilove on October 28, 2016, 09:25:00 pm
Hey there, could someone help me with the calculations involved in each of these questions? Both from the 2014 paper, I feel sorry for Jamon and Jake now!
As for your second question, also kinda fucked. We need to find the frequency of the first emitted electron, accounting for the 4.1V. We do this as follows
This get's us the frequency at zero volts! So, we plug that in as our x-intercept, and draw a line with the same gradient (always going to be h) from there. From there, you can go on to estimate etc. to get about 12.8*10^14 Hz :)
Post by: bethjomay on October 28, 2016, 09:34:38 pm
As for your second question, also kinda fucked. We need to find the frequency of the first emitted electron, accounting for the 4.1V. We do this as follows
This get's us the frequency at zero volts! So, we plug that in as our x-intercept, and draw a line with the same gradient (always going to be h) from there. From there, you can go on to estimate etc. to get about 12.8*10^14 Hz :)
I'm sorry, I'm not really sure where your first calculation came from?
Post by: jakesilove on October 28, 2016, 09:37:13 pm
I'm sorry, I'm not really sure where your first calculation came from?
Fair enough, I was totally unclear aha, too much trauma involved in this question. The graph is in eV (electron volts). So, I had to convert from volts (given. 4.1V) into electron volts (so you just multiply by e, the elementary charge on the electron). Make sense?
Post by: Happy Physics Land on October 28, 2016, 09:39:33 pm
Hey there, could someone help me with the calculations involved in each of these questions? Both from the 2014 paper, I feel sorry for Jamon and Jake now!
I will just add in my two cents based upon Jake's explanation. Because the falling motion of the projectile is recorded as a stroboscopic photograph, meaning that the time interval between two positions of the projectile are constant, hence the horizontal distance between two consecutive positions should be the same since the horizontal velocity is constant.
I will provide the mathematical approach that I have taken:
According to Galileo's analysis of projectile motion, vertical acceleration of any object under the planet's gravitational field should be the same, meaning that the acceleration of the object that is vertically falling should be the same as the that of the projectile tracing a parabolic path.
For between the 3rd and 4th seconds:
y = ut + 1/2 at2
y = vertical displacement = 3 grids
u = initial vertical velocity = 0 ms-1
t = time of travel = 1 second
hence 3 = 0 + 1/2 a
hence a = 6 ms-2
We know that at the 5th second Ball Q must be in the identical vertical position as Ball P, so we can plot that.
To calculate the position of the ball at the sixth second, relative to its initial position:
y = ut + 1/2 at2
= 0 + 1/2 (6) (3)2
= 27th grid
So for Ball P its next position should be the 27th grid counting down from the top and for Ball Q its next position should also be 27th grid counting down from the top, and 3 grids to the right of its previous position.
Best Regards
Happy Physics Land
Post by: Happy Physics Land on October 28, 2016, 09:44:14 pm
I'm sorry, I'm not really sure where your first calculation came from?
Hey Beth!
If you want more explanation on that second question that you put up, you can have a look at the one I wrote several days ago on the same question!
Physics Question Thread
It tells you step by step what to do
Post by: bethjomay on October 28, 2016, 09:54:50 pm
Hey Beth!
If you want more explanation on that second question that you put up, you can have a look at the one I wrote several days ago on the same question!
Physics Question Thread
It tells you step by step what to do
Oh awesome! Thank you both! I'm still trying to get my head around this question, but I think i'll have another look at it tomorrow. We never really looked at applying electron volts in class, except for pointing it out on the formula sheet :P
Post by: Albertenouttaten on October 28, 2016, 09:55:27 pm
Thanks
Post by: bethjomay on October 28, 2016, 10:02:09 pm
The graphs that show Energy vs frequency (in regards to photoelectric effect). One question asked what is the significance of the gradients of metals being the same? What would be the answer to this kind of question?
Thanks
I think I can have a go at this! Essentially, the gradient on this kind of graph is equal to Planck's constant, so the significance is that is it in fact a constant for all metals. This comes from the equation E = hf, if we take the 'E' as the y value of the graph and 'f' as the x-value (as in y = mx), we can see that the gradient should be 'h' - Planck's constant! I'm sure someone else can give a more comprehensive explanation but hopefully that helps!
Post by: jakesilove on October 28, 2016, 10:08:59 pm
I think I can have a go at this! Essentially, the gradient on this kind of graph is equal to Planck's constant, so the significance is that is it in fact a constant for all metals. This comes from the equation E = hf, if we take the 'E' as the y value of the graph and 'f' as the x-value (as in y = mx), we can see that the gradient should be 'h' - Planck's constant! I'm sure someone else can give a more comprehensive explanation but hopefully that helps!
That's a perfect answer! It just establishes that Planck's relationship, E=hf, holds true for all metals, and is a linear relationship. Thanks Beth!
Post by: maxpedersen on October 29, 2016, 09:47:34 am
Because I've seen conflicting answers on the internet.
Thanks
Post by: proficles on October 29, 2016, 10:07:36 am
A satellite is moved from a geostationary orbit to a higher orbit.
Which statement about the orbit change is correct?
(A) During the move the gravitational potential energy decreases.
(B) The change in gravitational potential energy is independent of the mass of the
satellite.
(C) The work done is the difference between the gravitational potential energy of the
higher orbit and that of the geostationary orbit.
(D) The work done is the energy required to move the satellite, which is in the
gravitational field, from a very large distance away, to the higher orbit.
Post by: FallonXay on October 29, 2016, 10:08:54 am
Post by: Neutron on October 29, 2016, 10:24:57 am
Yo for the AC motor, does the one in the second picture exist? (The one that looks like an AC generator) Or is it just induction, universal and synchronous? Cause Google doesn't seem to think it's a motor but for a question we had to talk about how the AC motor uses the motor effect and the other three don't :o speaking of, what actually is the synchronous motor?
Post by: MysteryMarker on October 29, 2016, 10:44:34 am
Need help with this question, cause i don't have the answers :D. I believe the answer is C because the light from A and B will take the same time to travel to A' and B' but the flashes will be witnessed by O' differently as the observer moves into the flash from B' whereas he moves away from the flash at A'. Is this correct? Appreciate any help given :D
Cheers.
http://prnt.sc/d09xas ( Not letting me upload my 968KB file :/ )
Post by: Cindy2k16 on October 29, 2016, 11:12:29 am
Post by: Happy Physics Land on October 29, 2016, 11:36:24 am
Quick question, just to clarify, but DC motors & generators can never experience eddy currents right, because eddy currents arise due to changing magnetic flux passing through a metal?
Because I've seen conflicting answers on the internet.
Thanks
Hey Max!
Definitely a tricky question to think about, and it is definitely NOT TRUE that eddy currents cant flow in DC motors. The reason why you are perhaps a bit confused is because you didnt recognise that eddy current is not produced by the supply emf, but the induced emf. You are quite right in that DC doesnt really produce a change in magnetic flux, but it has nothing to do with the production of eddy currents, because eddy currents is produced as a result of the rotor spinning in a magnetic field.
So when the rotor rotates in an external magnetic field, it cuts magnetic field lines and hence experiences a change in magnetic flux. According to Faraday's law, this causes an emf to be induced and according to lenz's law this emf will be in a direction such that it opposes the source of change in flux. This is what we otherwise know as back emf or induced emf. This emf creates a current, and this is what we call an eddy current. This eddy current will flow in the rotor to oppose the source of the change in flux which is the supply emf.
Evidently eddy current is always produced in a motor, regardless of the type of supply voltage that the motor is connected to.
Best Regards
Happy Physics Land
Post by: Happy Physics Land on October 29, 2016, 11:47:12 am
Hi for 2009 HSC Physics MC q3, why are A and D wrong? (correct answer is c)
A satellite is moved from a geostationary orbit to a higher orbit.
Which statement about the orbit change is correct?
(A) During the move the gravitational potential energy decreases.
(B) The change in gravitational potential energy is independent of the mass of the
satellite.
(C) The work done is the difference between the gravitational potential energy of the
higher orbit and that of the geostationary orbit.
(D) The work done is the energy required to move the satellite, which is in the
gravitational field, from a very large distance away, to the higher orbit.
Hey Proficles!
We can first eliminate A which is definitely not correct because gravitational potential energy can only be increasing when you move from a lower orbit to a higher orbit. It's like you climbing from the 1st floor of a building to the 40th floor of the building, your gravitational potential energy has increased. Same principle.
C and D can be a bit more confusing and when you look at it both seem viable, but of course its multiple choice and you have to choose the best answer. Indeed it's quite hard to see which one is a better option especially under exam conditions since both options talk about work and energy.
Here is what I think. There are two reasons why I would pick C:
1. The change in gravitational potential energy between two orbits is defined by the formula Ep=-GMm/r (1/rf - 1/ri). Now this explicitly says the same thing as option C, that the change in gravitational potential energy is just the difference between the final orbit and the initial orbit.
2. Option C is more specific about the geostationary orbit part whereas D mentions "a very large distance away” but this doesnt really fit the question, which specifically mentions “geostationary orbit”. You can be 10,000,000,000km from the centre of Earth (hypothetically speaking) and still be in Earth's gravitational field but probably only experiencing a gravitational field strength of 0.001N/kg. So it doesnt quite suit our scenario here.
A tough question, but option C definitely seem more reasonable.
Best Regards
Happy Physics Land
Post by: katnisschung on October 29, 2016, 11:49:04 am
11) A car, travelling 30m/s drives over the edge of a cliff into the water
58m below.
t=3.46s
a) calculate the time it takes the car to hit the water
12) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s
t=4.52s
Post by: Happy Physics Land on October 29, 2016, 11:57:47 am
Hello! How do you do this question? (The Answer is A)
Hey FallonXay!
A difficult question definitely! (A lot of difficult ones today). At the first sight it might seem like a motors problem.But sneaky sneaky it's actually a problem about Faraday and Lenz's law! Let's see how that comes into the question!
Now all the coils are turning in the clockwise direction for all the options, and all the coils are initially placed at a horizontal plane, 90 degrees to the direction of B-field. At this time, it receives the maximum amount of flux - amount of magnetic field lines that penetrates through the area of the coil. (FLUX! NOT CHANGE IN FLUX!) Now as the coil rotates, the plane is no longer 90 degrees to the magnetic field lines and consequently it loses magnetic flux. As it continues to rotate it loses more and more flux and consequently this change in flux results in an emf being induced (Faraday's law) and this causes an eddy current to be induced inside the coil. The eddy current will be a direction such that it opposes the original change in flux (Lenz's law). So, as we lose magnetic flux downwards, the eddy current wants to flow in a direction so that it can produce more magnetic flux downwards to compensate for the loss. Using our Right Hand Coil Rule, our thumb point downwards, our fingers curling in a clockwise direction, hence the answer is A.
Good question.
Best Regards
Happy Physics Land
Post by: znaser on October 29, 2016, 12:03:00 pm
Post by: zoe_rammie on October 29, 2016, 12:08:50 pm
Bostes put the answer down as b and excel put it down as d. I'm pretty sure though that the answer is d, but i could be mistaken. Thanks :)
Hmm that's a bit strange...I would've said D as well, considering the fact that
Induced EMF = - rate of change of Magnetic flux over time
I wonder what other people say.
Post by: Happy Physics Land on October 29, 2016, 12:09:35 pm
(http://uploads.tapatalk-cdn.com/20161028/a1201c77d9545487a686cf05ca68caac.jpg) (http://uploads.tapatalk-cdn.com/20161028/94106ed9de56e970c032e83247e23717.jpg)
Yo for the AC motor, does the one in the second picture exist? (The one that looks like an AC generator) Or is it just induction, universal and synchronous? Cause Google doesn't seem to think it's a motor but for a question we had to talk about how the AC motor uses the motor effect and the other three don't :o speaking of, what actually is the synchronous motor?
Hey Neutron!
Synchronous motor isnt in the HSC syllabus so you dont really need to know about it. Clearly by the way Im speaking here I dont really know what it is (hehe soz soz :) ). Jake and Jamon are a bit busy right now so they can come help later.
In terms of that question on whether or not AC slip ring motors exist, I can tell you that yes it does but not in that form. AC slip ring motors can only work with induction motors (something I learnt from engineering studies, not in the syllabus). And if in the exam they want you to provide a motor that works on AC, just talk about universal or induction motor.
I'm not too capable here, I will inform Jake/Jamon about your question.
Best Regards
Happy Physics Land
Post by: zoe_rammie on October 29, 2016, 12:16:34 pm
Hey Proficles!
We can first eliminate A which is definitely not correct because gravitational potential energy can only be increasing when you move from a lower orbit to a higher orbit. It's like you climbing from the 1st floor of a building to the 40th floor of the building, your gravitational potential energy has increased. Same principle.
C and D can be a bit more confusing and when you look at it both seem viable, but of course its multiple choice and you have to choose the best answer. Indeed it's quite hard to see which one is a better option especially under exam conditions since both options talk about work and energy.
Here is what I think. There are two reasons why I would pick C:
1. The change in gravitational potential energy between two orbits is defined by the formula Ep=-GMm/r (1/rf - 1/ri). Now this explicitly says the same thing as option C, that the change in gravitational potential energy is just the difference between the final orbit and the initial orbit.
2. Option C is more specific about the geostationary orbit part whereas D mentions "a very large distance away” but this doesnt really fit the question, which specifically mentions “geostationary orbit”. You can be 10,000,000,000km from the centre of Earth (hypothetically speaking) and still be in Earth's gravitational field but probably only experiencing a gravitational field strength of 0.001N/kg. So it doesnt quite suit our scenario here.
A tough question, but option C definitely seem more reasonable.
Best Regards
Happy Physics Land
Imma also just add, (And I agree with you that the answer is most likely C)
D says "a very large distance away”, which, for all you know, could mean farther away than the high orbit itself.
So because it is pretty vague as to which direction it'd be moving, I'd choose C.
Post by: Happy Physics Land on October 29, 2016, 12:25:27 pm
Hey guys
Need help with this question, cause i don't have the answers :D. I believe the answer is C because the light from A and B will take the same time to travel to A' and B' but the flashes will be witnessed by O' differently as the observer moves into the flash from B' whereas he moves away from the flash at A'. Is this correct? Appreciate any help given :D
Cheers.
http://prnt.sc/d09xas ( Not letting me upload my 968KB file :/ )
Hey Mystery Marker!
This question is a bit ambiguous, here is why.
If the two charges are in a different inertial frame of reference as the observer, then your reasoning and answer will definitely be correct, no doubt about it.
However, if the two charges are in the same inertial frame of reference as the observer, i.e. they are somehow attached to the train and travel with the train like what the diagram has illustrated, then the answer would be A. Because if the two charges are moving at a speed v with the train, then for the observer O', he isnt approaching any one of the charges as they are all in the same frame of reference and so light would travel equal distance towards O' from point A' and B'.
But yes if its the first scenario where the train is approaching an external point of reference B' then you are definitely correct.
Best Regards
Happy Physics Land
Post by: Happy Physics Land on October 29, 2016, 12:29:58 pm
Hi are hysteresis losses from the constant reversal of magnetisation of the transformer core? TIA
Hey Cindy!
Hysteresis loss is simply the loss of magnetic flux during the transmission process from primary coil to secondary coil. It's not really because of the constant reversal of magnetic polarity, but because the inability of iron core to PERFECTLY transmit the magnetic flux. In other words, you can reverse your magnetic polarity however you like, as long as you have an ideal perfect transmitter of magnetic flux, hysteresis losses cannot occur. Iron core is the best thing we have now and even that isnt perfectly ideal, so hysteresis losses would always occur. And because the equation vp/vs = np/ns relies on the perfect transmission of magnetic flux, energy would be lost because the equation is no longer viable since some magnetic flux is lost.
Best Regards
Happy Physics Land
Post by: Happy Physics Land on October 29, 2016, 12:35:38 pm
Bostes put the answer down as b and excel put it down as d. I'm pretty sure though that the answer is d, but i could be mistaken. Thanks :)
Hey Znaser!
Definitely D. If you've done 2u mathematics you should know that faraday's law is essentially a differentiation of flux. So all you had to do there was just to draw the derivative graph of the original graph and the outcome would look like a cosine graph like the one in B. But because of the implication from lenz's law, there is a negative sign in front of faraday's law and hence the cosine graph should be flipped along the x-axis, producing the answer D. So yeah Im with you there.
Best Regards
Happy Physics Land
Post by: RuiAce on October 29, 2016, 12:40:33 pm
(http://uploads.tapatalk-cdn.com/20161028/a1201c77d9545487a686cf05ca68caac.jpg) (http://uploads.tapatalk-cdn.com/20161028/94106ed9de56e970c032e83247e23717.jpg)
Yo for the AC motor, does the one in the second picture exist? (The one that looks like an AC generator) Or is it just induction, universal and synchronous? Cause Google doesn't seem to think it's a motor but for a question we had to talk about how the AC motor uses the motor effect and the other three don't :o speaking of, what actually is the synchronous motor?
Hey Neutron!Yeah can confirm synchronous motor isn't in the course.
Synchronous motor isnt in the HSC syllabus so you dont really need to know about it. Clearly by the way Im speaking here I dont really know what it is (hehe soz soz :) ). Jake and Jamon are a bit busy right now so they can come help later.
In terms of that question on whether or not AC slip ring motors exist, I can tell you that yes it does but not in that form. AC slip ring motors can only work with induction motors (something I learnt from engineering studies, not in the syllabus). And if in the exam they want you to provide a motor that works on AC, just talk about universal or induction motor.
I'm not too capable here, I will inform Jake/Jamon about your question.
Best Regards
Happy Physics Land
But I am pretty certain, that in your ordinary DC motor if you just
- Swap the split ring commutator with a slip ring commutator
- Swap the DC power source with an AC power source
You still have an AC motor. Just that this is your ordinary motor, not the induction motor which relies on a rotating magnetic field.
The AC currents are monitored such that the frequency of direction reversal will always aid in the running of the motor
In practice, this is probably rarer than the AC induction motor as it's quite obvious that the induction motor has its advantages. But it's definitely a real thing and does exist.
Post by: Happy Physics Land on October 29, 2016, 12:48:29 pm
hullo hullo hullo again.
Got another question aheh.
Could someone please explain the math of this? (2014 q 20)
The answer is D.
tyvm :)
Hey Zoe!
No worries! This question is just a bit difficult to type out by hand, I've written and drawn my solutions, hope you understand!
Don't hesitate to ask if anything's obscure!
(http://i.imgur.com/MCWN21f.jpg)
Best Regards
Happy Physics Land
Post by: zoe_rammie on October 29, 2016, 12:53:01 pm
Hey Zoe!
No worries! This question is just a bit difficult to type out by hand, I've written and drawn my solutions, hope you understand!
Don't hesitate to ask if anything's obscure!
(http://i.imgur.com/MCWN21f.jpg)
Best Regards
Happy Physics Land
oOooOHHH gotcha gotcha. Thank you muchly! :D
Post by: Happy Physics Land on October 29, 2016, 01:02:40 pm
oOooOHHH gotcha gotcha. Thank you muchly! :D
No worries, as long as you got something out of it! :D
Post by: maxpedersen on October 29, 2016, 01:43:44 pm
Hey Max!
Definitely a tricky question to think about, and it is definitely NOT TRUE that eddy currents cant flow in DC motors. The reason why you are perhaps a bit confused is because you didnt recognise that eddy current is not produced by the supply emf, but the induced emf. You are quite right in that DC doesnt really produce a change in magnetic flux, but it has nothing to do with the production of eddy currents, because eddy currents is produced as a result of the rotor spinning in a magnetic field.
So when the rotor rotates in an external magnetic field, it cuts magnetic field lines and hence experiences a change in magnetic flux. According to Faraday's law, this causes an emf to be induced and according to lenz's law this emf will be in a direction such that it opposes the source of change in flux. This is what we otherwise know as back emf or induced emf. This emf creates a current, and this is what we call an eddy current. This eddy current will flow in the rotor to oppose the source of the change in flux which is the supply emf.
Evidently eddy current is always produced in a motor, regardless of the type of supply voltage that the motor is connected to.
Best Regards
Happy Physics Land
Oh right, my bad haha. Thanks for such a comprehensive answer, appreciate it! :)
Post by: Cindy2k16 on October 29, 2016, 01:50:31 pm
Thanks :)
Post by: levendibigd on October 29, 2016, 01:54:42 pm
Post by: JemexR on October 29, 2016, 01:55:08 pm
When drawing the bands of conductors, should we draw the conduction band sitting right on top of the valence band or the two bands overlapping? Or is either fine? Also do we say there is no band gap?
Thanks :)
I would say either as long as you specify that there is no energy gap.
Quick question due to mental blank: What is the term for the minimum amount of energy required for an electron to be released under the photoelectric effect? Thanks!
Post by: Cindy2k16 on October 29, 2016, 02:04:15 pm
I would say either as long as you specify that there is no energy gap.
Quick question due to mental blank: What is the term for the minimum amount of energy required for an electron to be released under the photoelectric effect? Thanks!
Thanks!
Do u mean the work function?
Post by: Cindy2k16 on October 29, 2016, 02:06:33 pm
Post by: MysteryMarker on October 29, 2016, 02:11:42 pm
I keep getting a negative cos curve for the current time graph. Because at its current time it has 0 magnetic flux and hence as it rotates in the direction given it will become max flux and vice versa representing a sin curve. Hence to find the current we just find the negative derivative?
Yea, i'm pretty confused with how to go about this question, pre sure answer is B. Appreciate any help given :D
Cheers.
http://prnt.sc/d0bgpg
Post by: Cindy2k16 on October 29, 2016, 02:13:22 pm
Hey Cindy!
Hysteresis loss is simply the loss of magnetic flux during the transmission process from primary coil to secondary coil. It's not really because of the constant reversal of magnetic polarity, but because the inability of iron core to PERFECTLY transmit the magnetic flux. In other words, you can reverse your magnetic polarity however you like, as long as you have an ideal perfect transmitter of magnetic flux, hysteresis losses cannot occur. Iron core is the best thing we have now and even that isnt perfectly ideal, so hysteresis losses would always occur. And because the equation vp/vs = np/ns relies on the perfect transmission of magnetic flux, energy would be lost because the equation is no longer viable since some magnetic flux is lost.
Best Regards
Happy Physics Land
Hi! Thanks for the answer. I'm still a bit confused through since I've heard somewhere (but ive forgotten) that hysteresis losses are from flux leakage but when I look it up on google I get a lot about reversing the magnetisation of the transformer core? They sound like pretty different things so I'm confused as to what the right one is. Are they both called hysteresis losses?
Post by: jakesilove on October 29, 2016, 02:19:37 pm
Hey! Could I please have a hand with Q20c :) answer a) 20ms^-1 and b)14ms^-1
Hey! Just in future, it'd be super helpful if you post the answers to the first few parts.
So, what do we know about the second part of the motion? Well, actually, we know basically nothing. The ball bounces inelastically, which should mean that neither the horizontal nor vertical components of motion stay the same. So, the first two parts actually won't make a difference to our answer!
Based on the diagram, it looks like the height of motion is at h, when the ball passes through the hole.
Okay, now I'm lost. I actually don't know how to answer this using HSC physics, I feel like there isn't enough information. This is definitely NOT something you can be assessed on in your HSC, so don't stress, but if anyone wants to jump in feel free to answer part c) for me!
Post by: jakesilove on October 29, 2016, 02:20:08 pm
Thanks!
Do u mean the work function?
Yep! All of the above answers are correct
Post by: jakesilove on October 29, 2016, 02:21:54 pm
Hi in 2006 Physics q29 (option) b) i) it says "describe the criteria you would use to determine the reliability of a data source for this purpose"- 3 marks . If reliability is if the information is consistent across many reputable sources, how do you talk about it for 3 marks worth?
Hey! It wants 'criteria' for reliability, so there are probably a few things you can talk about. I don't know what the experiment was on, but the obvious thing to discuss is repeating it a billion times. Then, you could take lots of samples etc. to get a more reliable idea of the results as a whole. Finally, you can draw graphs and take averages to increase reliability. You could also compare to others' results! That's three marks, easily :)
Post by: JemexR on October 29, 2016, 02:22:24 pm
Hey! Just in future, it'd be super helpful if you post the answers to the first few parts.
So, what do we know about the second part of the motion? Well, actually, we know basically nothing. The ball bounces inelastically, which should mean that neither the horizontal nor vertical components of motion stay the same. So, the first two parts actually won't make a difference to our answer!
Based on the diagram, it looks like the height of motion is at h, when the ball passes through the hole.
Okay, now I'm lost. I actually don't know how to answer this using HSC physics, I feel like there isn't enough information. This is definitely NOT something you can be assessed on in your HSC, so don't stress, but if anyone wants to jump in feel free to answer part c) for me!
If it was elastic, could you then infer that V2 = the velocity of the ball when it comes into contact with the ground, and the angles to the normal would be identical?
Post by: jakesilove on October 29, 2016, 02:24:32 pm
If it was elastic, could you then infer that V2 = the velocity of the ball when it comes into contact with the ground, and the angles to the normal would be identical?
Yep absolutely, but the questions specifically says it's inelastic, right?
Post by: jakesilove on October 29, 2016, 02:26:12 pm
Hi! Thanks for the answer. I'm still a bit confused through since I've heard somewhere (but ive forgotten) that hysteresis losses are from flux leakage but when I look it up on google I get a lot about reversing the magnetisation of the transformer core? They sound like pretty different things so I'm confused as to what the right one is. Are they both called hysteresis losses?
If you're a bit confused, then that's good, because hysteresis is super confusing. Don't worry about understanding it at this point, just be able to talk about it for a sentence (ie hysteresis causes energy loss etc.) and move on with your life. There are heaps of things you actually DO need to learn, and this isn't one of them! It's just the icing on the cake
Post by: Cindy2k16 on October 29, 2016, 02:37:58 pm
Hey! It wants 'criteria' for reliability, so there are probably a few things you can talk about. I don't know what the experiment was on, but the obvious thing to discuss is repeating it a billion times. Then, you could take lots of samples etc. to get a more reliable idea of the results as a whole. Finally, you can draw graphs and take averages to increase reliability. You could also compare to others' results! That's three marks, easily :)
Oh sorry I forgot to give more context but It's not about an experiment :/
"During your study of Medical physics you identified data sources, and gathered, processed and presented information to explian why MRI scans can be used to detect abnormalities in the body. Describe the criteria etc"
To answer the question I dont think u need to know about med physics but rather talk about assessing the reliability of secondary sources. But as far as I know that involves comparing information from reputable sources to ensure the info is consistent. But for 3 marks just saying that hardly seems enough. The past papers book I have has a sample answer talking about author credibility, making sure its verified by a scientific institution etc but isnt that validity/accuracy rather than reliability? The book also has some mistakes in it so I'm not sure how much to trust it.
TIA
Post by: FallonXay on October 29, 2016, 02:49:24 pm
Regarding Q8, how can you tell whether the source of power is AC/DC - this one said that it was DC (is there a different symbol used to show AC power sources?)
And lastly, I don't understand the answers for Q15; shouldn't electrons/ holes move in opposite directions? - The Answer is C.
Thanks in advance~ :)
Post by: Cindy2k16 on October 29, 2016, 03:02:03 pm
Hi! For Question 9, the answers say C - how is this true, because shouldn't there still be eddy currents generated in R but just broken - so R will still experience some resistance though just not as much as Q (I've probably asked about this before earlier in the year but I forgot :P)
Regarding Q8, how can you tell whether the source of power is AC/DC - this one said that it was DC (is there a different symbol used to show AC power sources?)
And lastly, I don't understand the answers for Q15; shouldn't electrons/ holes move in opposite directions? - The Answer is C.
Thanks in advance~ :)
Hi for your last question- the electrons and holes move in opposite directions in an electric field- not necessarily in a magnetic field though (though they can). If you use your right hand palm rule for the conventional current (movement of positive charge) youll find there is a downwards force on the 'holes'. Now electron flow is in the OPPOSITE direction to conventional current. So when you use your left hand palm rule (or whatever youve been taught to use for the flow of a negative charge) you point ur thumb in the OPPOSITE direction to the conventional current and youll find that the electrons ALSO experience a downwards force.
Thus both holes and electrons move to the bottom of the rod ie. the same side of the rod.
Therefore C. hope this helps :)
Post by: Yasminpotts1105 on October 29, 2016, 03:34:56 pm
Post by: jamonwindeyer on October 29, 2016, 03:45:49 pm
Hi! For Question 9, the answers say C - how is this true, because shouldn't there still be eddy currents generated in R but just broken - so R will still experience some resistance though just not as much as Q (I've probably asked about this before earlier in the year but I forgot :P)
Regarding Q8, how can you tell whether the source of power is AC/DC - this one said that it was DC (is there a different symbol used to show AC power sources?)
And lastly, I don't understand the answers for Q15; shouldn't electrons/ holes move in opposite directions? - The Answer is C.
Thanks in advance~ :)
Hey! For Q9, the eddy currents would be there, but the effect is thought to be negligible. Thus you can ignore them in this circumstance :)
AC sources normally have a circle with a squiggly line; DC sources are almost always two straight lines in HSC Physics, one longer than the other, with that longer line representing the positive terminal. Otherwise, you look for a '+' or a '-', or both, to indicate the DC source and its direction ;D
Post by: FallonXay on October 29, 2016, 04:20:44 pm
Hi for your last question- the electrons and holes move in opposite directions in an electric field- not necessarily in a magnetic field though (though they can). If you use your right hand palm rule for the conventional current (movement of positive charge) youll find there is a downwards force on the 'holes'. Now electron flow is in the OPPOSITE direction to conventional current. So when you use your left hand palm rule (or whatever youve been taught to use for the flow of a negative charge) you point ur thumb in the OPPOSITE direction to the conventional current and youll find that the electrons ALSO experience a downwards force.
Thus both holes and electrons move to the bottom of the rod ie. the same side of the rod.
Therefore C. hope this helps :)
Thanks - the downward force on the holes makes sense :)
But with the (negative) electrons, I always thought that you still use the right hand palm rule but the thumb just points in the opposite direction to the conventional current - thus giving an upward force? Sorry, I don't quite understand this part, would you be able to elaborate? Thanks a ton!!!
Post by: jamonwindeyer on October 29, 2016, 04:24:56 pm
Silly question, but when using a vector diagram to show the net resultant force should you put in lengths of each side or angles or not??
Hey hey! Not a silly question at all. If it requests it and/or those sides/angles are self apparent or easily found, sure! But if not don't worry. Essentially, don't go into much detail unless you have to. Usually vector diagrams are just testing how well you understand the different forces at play and what direction they act in :)
Post by: jamonwindeyer on October 29, 2016, 04:32:19 pm
Thanks - the downward force on the holes makes sense :)
But with the (negative) electrons, I always thought that you still use the right hand palm rule but the thumb just points in the opposite direction to the conventional current - thus giving an upward force? Sorry, I don't quite understand this part, would you be able to elaborate? Thanks a ton!!!
This is true! With electrons you do point the thumb in the opposite direction, but the issue is that we have TWO things that have swapped. The electron charge is opposite in sign to the holes, AND they are moving in the opposite direction. These two things together mean that the forces actually both end up in the same direction. So you point your thumb in the opposite direction for the charge, and then again because the electrons are moving in the opposite direction. So really, it doesn't switch at all, hence the same force direction ;D
Post by: maksim on October 29, 2016, 04:49:38 pm
Post by: Goodwil on October 29, 2016, 04:51:25 pm
Post by: imtrying on October 29, 2016, 04:53:31 pm
Post by: jakesilove on October 29, 2016, 04:59:54 pm
Hi, I was just wondering what the expected and maximum g-forces would be for an astronaut during a rocket launch.
Hey! Check out the table at the bottom of this page! Although you definitely don't need to know this :)
Post by: jakesilove on October 29, 2016, 05:03:19 pm
Why is the answer A and not C here?
Hey! Check out my answer here :)
Post by: jakesilove on October 29, 2016, 05:06:18 pm
Could someone explain how angles are measured for force vs torque because this always confuses me
Always confused me as well! For force, we're generally talking about a wire, right? So the angle there is easy; just imagine what the field lines look like, and then see how those lines 'interact' with the wire. Usually, it will just be perpendicular, or zero! Simply draw the field lines, and stick the wire through it, and look at the angle made between them :)
For torque, we care about net rotation. So, what's actually rotating? The loop! It wouldn't really make a huge amount of sense to talk about the torque on a single wire, as that wire isn't turning. So, if we care about the loop as a whole, then the 'plane' we look to is the plane of the loop. Imagine a the loop as a surface, and then look at how the field lines interact with that surface. When the loop is at it's starting point, it's like the field lines are running in the same plane as the loop. When the loop is perpendicular to it's starting position, the field lines are perpendicular to the plane of the loop! So, that's the angle we're talking about. It's seriously confusing. I'm still confused.
Post by: imtrying on October 29, 2016, 05:36:03 pm
Always confused me as well! For force, we're generally talking about a wire, right? So the angle there is easy; just imagine what the field lines look like, and then see how those lines 'interact' with the wire. Usually, it will just be perpendicular, or zero! Simply draw the field lines, and stick the wire through it, and look at the angle made between them :)
For torque, we care about net rotation. So, what's actually rotating? The loop! It wouldn't really make a huge amount of sense to talk about the torque on a single wire, as that wire isn't turning. So, if we care about the loop as a whole, then the 'plane' we look to is the plane of the loop. Imagine a the loop as a surface, and then look at how the field lines interact with that surface. When the loop is at it's starting point, it's like the field lines are running in the same plane as the loop. When the loop is perpendicular to it's starting position, the field lines are perpendicular to the plane of the loop! So, that's the angle we're talking about. It's seriously confusing. I'm still confused.
Thank you, less confused now:)
Alas, the confusion continues...how do I do question 18 of this paper https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf (sorry i couldn't upload a photo here's the link to the 2014 HSC)
Post by: JemexR on October 29, 2016, 05:38:31 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf
Post by: JemexR on October 29, 2016, 05:44:45 pm
Alas, the confusion continues...how do I do question 18 of this paper https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf (sorry i couldn't upload a photo here's the link to the 2014 HSC)
Given that cathode rays are electrons, you have to use whichever hand rule you were taught for the electron (motors), from this, you can deduce that the electron will therefore move "down" due to the given magnetic field. Therefore, in order to counteract this movement, the electric field's direction must be "down" as well, given that the electron will be attracted to the positive terminal.
Post by: imtrying on October 29, 2016, 05:46:52 pm
Given that cathode rays are electrons, you have to use whichever hand rule you were taught for the electron (motors), from this, you can deduce that the electron will therefore move "down" due to the given magnetic field. Therefore, in order to counteract this movement, the electric field's direction must be "down" as well, given that the electron will be attracted to the positive terminal.Thanks!
Post by: jakesilove on October 29, 2016, 05:58:56 pm
Thank you, less confused now:)
Alas, the confusion continues...how do I do question 18 of this paper https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf (sorry i couldn't upload a photo here's the link to the 2014 HSC)
So, using the right hand rule, we can find the force on the electron. Make sure to point your thumb to the left, because it should be opposite to electron flow. That should get you a force DOWNWARDS due to the magnetic field.
Now, we what to apply an electric field to counteract the change, so we will put the positive electrode on top (to ATTRACT the electron). Electric field lines are drawn positive to negative, so the direction will be downwards.
Post by: jakesilove on October 29, 2016, 06:01:26 pm
Hi, could someone explain the differences between the four options for the HSC 2010 Q20 Multiple choice? I can understand how to identify which way the wire would turn given both the direction of the current and the direction of the magnetic field, but I cannot differentiate at this point how the diagrams differ. THanks
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf
Yeah, that was a fucking terrible diagram hey! Basically, you need to use the right hand rule to figure out the direction of the electromagnets (follow the current around; will there be a north pole on the right, or the left?). Then, establish the direction of the current flow (again, just follow the positive to the negative, as that will be the direction of current). Then, compare these two things (ie. draw where N and S are, and the direction of current flow), and figure which one will turn as desired! Does that make sense?
Post by: jamonwindeyer on October 29, 2016, 06:02:09 pm
Hi, could someone explain the differences between the four options for the HSC 2010 Q20 Multiple choice? I can understand how to identify which way the wire would turn given both the direction of the current and the direction of the magnetic field, but I cannot differentiate at this point how the diagrams differ. THanks
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf
Hey! It's a tough one to spot hey. What differs is:
a) Which side of the coil that each terminal is connected to. You can see that in half of them, the '+' is connected to the left, and the other half, it is connected to the right. Follow the wire, you'll spot it!
b) The direction of rotation around the pieces of metal. Look carefully, some wrap OVER and so cause what you could call a clockwise rotation, others wrap UNDER and cause a rotation in the opposite direction :)
Post by: katnisschung on October 29, 2016, 06:11:12 pm
11) A car, travelling 30m/s drives over the edge of a cliff into the water
58m below.
t=3.46s
a) calculate the time it takes the car to hit the water
12) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s
t=4.52s
Post by: jakesilove on October 29, 2016, 06:30:00 pm
the surfing answers may be wrong or i am yet again :P
11) A car, travelling 30m/s drives over the edge of a cliff into the water
58m below.
t=3.46s
a) calculate the time it takes the car to hit the water
12) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s
t=4.52s
Hey! For all questions where there is only horizontal velocity, we can completely ignore pretty much all the information except for height, and work it out in the y direction. So, using
Pretty close to the given answer, which means our values are just a little different somewhere! Maybe they used a=10. Can you get the second one using this working?
Post by: katnisschung on October 29, 2016, 06:53:10 pm
but for 12 i got 6.4 secs
Post by: MarkThor on October 29, 2016, 06:58:13 pm
Post by: proficles on October 29, 2016, 07:04:07 pm
Post by: jakesilove on October 29, 2016, 07:13:40 pm
for 11 i got 3.46 thought i got it wrong becos i wrote √12 ::)
but for 12 i got 6.4 secs
Yep, I definitely agree that 12 is 6.4s! Great job, looks like the book is just wrong
Post by: jakesilove on October 29, 2016, 07:16:54 pm
With my logic I got C, but the answer is A. Does this have something to do with the ring being split, or am I just stuffing up induction?
Definitely a seriously tricky question! What you have to remember is that a current will only be induced where there is a change in flux, or basically a 'change in the shit going on between magnetic fields'. In this case, where does the actual CHANGE occur? Only when entering, and leaving, the field! When the magnet swings entirely into the field, nothing is in essence 'changing'. Change only occurs when the number of field lines going through the magnet changes. Thus, there should be two peaks (one for the magnet swinging in, one for the magnet swinging out). Naturally, they will have different signs.
Post by: jakesilove on October 29, 2016, 07:19:07 pm
Hi, why is the answer C?
In this question, the only current is an induced current. Stupid question, but there isn't a supplied current; you're causing the wheel to turn through pedaling, not by supplying current! So, the 'induced' current is the current created by the electromagnets. When that current is a maximum, you've essentially go the maximum back-emf possible, thus reducing the movement of the wheel. Very stupid question, which required you to think about what was actually going on
Post by: MarkThor on October 29, 2016, 07:20:58 pm
Post by: Neutron on October 29, 2016, 07:26:08 pm
Hey Neutron!
Synchronous motor isnt in the HSC syllabus so you dont really need to know about it. Clearly by the way Im speaking here I dont really know what it is (hehe soz soz :) ). Jake and Jamon are a bit busy right now so they can come help later.
In terms of that question on whether or not AC slip ring motors exist, I can tell you that yes it does but not in that form. AC slip ring motors can only work with induction motors (something I learnt from engineering studies, not in the syllabus). And if in the exam they want you to provide a motor that works on AC, just talk about universal or induction motor.
I'm not too capable here, I will inform Jake/Jamon about your question.
Best Regards
Happy Physics Land
Yeah but the thing is they wanted us to talk about how the MOTOR EFFECT (sorry for capitalising, I can't bold on mobile rip) is used in the AC motor haha and if the motor I posted a picture of isnt viable, then what is :O
Post by: Goodwil on October 29, 2016, 07:30:19 pm
Hey! Check out my answer here :)Thanks, makes sense now
Post by: imtrying on October 29, 2016, 07:33:37 pm
Post by: Goodwil on October 29, 2016, 07:36:19 pm
Post by: jakesilove on October 29, 2016, 07:36:57 pm
Hey could I get a hand with question 9 from the 2015 paper? Sorry can't upload the photo but here's the link https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
For questions like this, always look to the formula sheet. You'll find that
You can use that to find the force on R due to Q. The force will be REPULSIVE. Use the formula again to find the force on R due to P. The force will be ATTRACTIVE. Because R due to Q will be greater than R due to P, there will be an overall repulsive force. Subtract the two values, and the force is to the right!
Post by: imtrying on October 29, 2016, 08:32:10 pm
Also thank you so much :)
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Post by: Brenda0708 on October 29, 2016, 09:00:46 pm
Hi, I'm a bit confused with the direction of the magnetic field in this question. If we use the RHP rule then shouldn't the MF be out of the page, not into the page? (HSC 2002 q25c)
Post by: pels on October 29, 2016, 09:17:49 pm
Also, here is the question:
"Discuss the effect of temperature on the conducting properties of both Type I and Type II superconductors with reference to the BCS Theory"
Post by: Cindy2k16 on October 29, 2016, 09:24:35 pm
TIA
Post by: Cindy2k16 on October 29, 2016, 09:37:36 pm
1. electron creates lattice distortion, creating an area of greater positive charge density
2. another electron being attracted to the area of greater positive charge density and thus being indirectly attracted to the electron that created the positive charge dense area
3. Cooper pair is formed and moves through the lattice unobstructed by the lattice and rather, facilitated by it with no energy loss.
Or should we be also talking about electron-phonon exchanges?
Post by: Cindy2k16 on October 29, 2016, 09:47:34 pm
I have a question here. I've never heard of Type 1 and 2 Superconductors, and I don't recall them being in the syllabus, but do we need to know the concept?
Also, here is the question:
"Discuss the effect of temperature on the conducting properties of both Type I and Type II superconductors with reference to the BCS Theory"
Hi Type 1 and 2 semiconductors fall under the syllabus dot point "Process information to identify some of the metals, metal alloys and compounds that have been identified as exhibiting the property of superconductivity and their critical temperatures"
Theres not much you need to know about them though I think:
Type 1: elements (metals). Have very low critical temperatures. Have a sharp transition from normal to superconductivity once critical temperature is reached (not sure if you even need to know this bit)
Type 2: metal alloys/compounds (ceramics). Higher critical temperatures than Type 1 .more gradual transition from normal to superconductivity.
I am not sure if you are required to be able to recall the names of type 1 and 2 superconductors though. I guess if worse comes to worse you can always just recall mercury as a Type 1.
Hope this helps :)
Post by: JemexR on October 29, 2016, 09:59:25 pm
Yeah, that was a fucking terrible diagram hey! Basically, you need to use the right hand rule to figure out the direction of the electromagnets (follow the current around; will there be a north pole on the right, or the left?). Then, establish the direction of the current flow (again, just follow the positive to the negative, as that will be the direction of current). Then, compare these two things (ie. draw where N and S are, and the direction of current flow), and figure which one will turn as desired! Does that make sense?
I did that and it seems I've understood the diagram horribly since I'm coming up with thwo as the exact same T.T
Post by: JemexR on October 29, 2016, 10:03:40 pm
I have a question here. I've never heard of Type 1 and 2 Superconductors, and I don't recall them being in the syllabus, but do we need to know the concept?
Also, here is the question:
"Discuss the effect of temperature on the conducting properties of both Type I and Type II superconductors with reference to the BCS Theory"
Just supporting what Cindy said, but type 2 superconductors don't necessarily hit superconductivity -> in exchange of achieving "superconductivity" at higher temperatures, the substance will exhibit near-superconductivity, but not 0 resistance.
Post by: JemexR on October 29, 2016, 10:16:03 pm
Post by: Cindy2k16 on October 29, 2016, 10:18:34 pm
When calculating torque, does the A stand for the total area or just the half (as in, if it is just one side of the coil; from the edge to the axis)?
The total area of the coil :)
Post by: jakesilove on October 29, 2016, 10:32:02 pm
Question 24c) from the 2015 HSC has got me very confused, could I get a hand with that one please?
Also thank you so much :)
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Hey! This is a seriously tough question, and actually requires you to analyse the motion using projectile formulas! You know the constant accelerating force from b). If anything undergoes a constant acceleration in a single direction, and it's travelling in that direction, it's as though a particle is being acted on by gravity.
Find the acceleration, and then use
to solve for v! We know the electron starts at rest (u=0), we know the acceleration, and we know the distance (delta y) :)
Post by: jakesilove on October 29, 2016, 10:33:48 pm
(http://uploads.tapatalk-cdn.com/20161029/6c0e8031ddd3b35777de37eebf839060.jpg)
Hi, I'm a bit confused with the direction of the magnetic field in this question. If we use the RHP rule then shouldn't the MF be out of the page, not into the page? (HSC 2002 q25c)
So, if the negative plate is the bottom one, it will repel the electrons. We therefore need to ATTRACT the electrons downwards, to cancel out the force. Remember that electrons move in the opposite direction to current, so we stick our thumb to the left, our palm down, and get a MF into the page :)
Post by: jakesilove on October 29, 2016, 10:34:49 pm
I have a question here. I've never heard of Type 1 and 2 Superconductors, and I don't recall them being in the syllabus, but do we need to know the concept?
Also, here is the question:
"Discuss the effect of temperature on the conducting properties of both Type I and Type II superconductors with reference to the BCS Theory"
Type 1 vs Type 2 semiconductors is really not something of great importance in the HSC. They wouldn't ask a question like that anymore, so I genuinely would just ignore that whole area. Not worth your time!
Post by: noonedoesnt on October 29, 2016, 10:35:55 pm
Post by: jakesilove on October 29, 2016, 10:36:29 pm
Hi whats the order of all these events in the reconceptualisation of light like: MM experiment, Special theory of relativity, Maxwell, Hertz, Planck and Einstein's explanation of the photoelectric effect? Should we know it? I only know the order in the separate topics of Space and I2I but not how they overlap.
TIA
Definitely no need to know how they overlap, because in a lot of ways, they don't. Who said that light was an EM wave? How did it travel? Well, none of that is REALLY addressed in the HSC, so you certainly don't need to know it. Keep the sections compartmentalized, but if you do get a really broad question, address Space and I2I separately (ie. Light and the Aether could be a subheading, and the Photoelectric Effect could be another!)
Post by: jakesilove on October 29, 2016, 10:37:52 pm
Hi another question- when talking about the BCS theory, is something along the lines of the following sufficient?:
1. electron creates lattice distortion, creating an area of greater positive charge density
2. another electron being attracted to the area of greater positive charge density and thus being indirectly attracted to the electron that created the positive charge dense area
3. Cooper pair is formed and moves through the lattice unobstructed by the lattice and rather, facilitated by it with no energy loss.
Or should we be also talking about electron-phonon exchanges?
I would definitely throw in the fact that one of the electrons transmits a phonon to another electron (it's cooper pair), but other than that your explanation is perfect! Draw a little diagram of lattice distortion, and that will be full marks for a BCS question :)
Post by: Cindy2k16 on October 29, 2016, 10:38:19 pm
Definitely no need to know how they overlap, because in a lot of ways, they don't. Who said that light was an EM wave? How did it travel? Well, none of that is REALLY addressed in the HSC, so you certainly don't need to know it. Keep the sections compartmentalized, but if you do get a really broad question, address Space and I2I separately (ie. Light and the Aether could be a subheading, and the Photoelectric Effect could be another!)
Thank you for clarifying!
Post by: jakesilove on October 29, 2016, 10:41:51 pm
here is a bit of a challenge i can't work out
This is a FUCKING hard question, and I've never got it before. Let's give it a go.
We know that energy must be conserved. Let's find the initial kinetic and gravitational potential energy.
Obviously, the initial kinetic energy is zero. The initial gravitational potential energy will be
That's sort of cheating a little bit, but let's just assume that the change in height being 1m is sufficient to sub it into this formula. Now, the final gravitational potential energy will be zero (because I've cheated, and defined that the be the floor), but the final kinetic energy will be
UHHA! There's been a loss of energy somewhere, a loss of 0.98-0.722=0.26J. Please tell me that D is the answer :)
Post by: noonedoesnt on October 29, 2016, 10:46:31 pm
Post by: Cindy2k16 on October 29, 2016, 10:47:16 pm
TIA
Post by: RuiAce on October 29, 2016, 10:49:35 pm
Hi does it matter which units we use for electric field ? as in N/C and V/m. Does 2 V/M mean the same as 2 N/C? If not, how do we know when to use either units?That one yes between those two it doesn't matter. It might be worth mentioning though that the standard is N/C.
TIA
Post by: jakesilove on October 29, 2016, 10:49:57 pm
Hi does it matter which units we use for electric field ? as in N/C and V/m. Does 2 V/M mean the same as 2 N/C? If not, how do we know when to use either units?
TIA
The two expressions are 100% identical :) Generally, if you're using E=V/d, you'll stick to V/m, but it genuinely won't matter
Post by: noonedoesnt on October 29, 2016, 10:50:38 pm
Post by: jakesilove on October 29, 2016, 10:51:08 pm
Nope the answer was B!
Nup. I call bullshit. I reckon the answers are wrong aha
Post by: noonedoesnt on October 29, 2016, 10:51:41 pm
Nup. I call bullshit. I reckon the answers are wrong aha
yeah I agree
Post by: jakesilove on October 29, 2016, 10:54:54 pm
Another pickle
We can logic this. If it has the same time of flight, then the ranges will HAVE to be different. This is because they have different Vx values. So, if the flight time was the same, then either A or B would also have to be true. As only one multiple choice answer can be correct, we can eliminate D).
Actually, I think this turns out to be fairly straight forward; draw two triangles, find Vx and Vy, and sub it into formulas. Find values for delta y, and delta x, and compare. Show me some working if you're still confused :)
Post by: noonedoesnt on October 29, 2016, 10:57:59 pm
We can logic this. If it has the same time of flight, then the ranges will HAVE to be different. This is because they have different Vx values. So, if the flight time was the same, then either A or B would also have to be true. As only one multiple choice answer can be correct, we can eliminate D).
Actually, I think this turns out to be fairly straight forward; draw two triangles, find Vx and Vy, and sub it into formulas. Find values for delta y, and delta x, and compare. Show me some working if you're still confused :)
Nor myself or rui could solve it, but we both picked B, however the answer was A.
Post by: jakesilove on October 29, 2016, 11:03:23 pm
Nor myself or rui could solve it, but we both picked B, however the answer was A.
I actually get that the range is exactly the same, which is to be expected, as they're both 5 degrees out from maximum range. Weird Q!
Post by: noonedoesnt on October 29, 2016, 11:14:21 pm
I actually get that the range is exactly the same, which is to be expected, as they're both 5 degrees out from maximum range. Weird Q!
That's exactly what I said! 45 degrees is maximum and calculated 5 degrees either side gives the same range.
The question might take into account air resistance, more air resistance with the increased horizontal speed at 40 degrees than 50 degrees.
Post by: JemexR on October 29, 2016, 11:25:19 pm
Post by: Cindy2k16 on October 29, 2016, 11:32:31 pm
A quick simple question; does HSC specify acceleration due to gravity as 9.8 or 10 ms-2?
9.8 usually. unless the question specifies otherwise
Sometimes the projectile is on a different planet with a different gravitational field. If this is the case they'll usually say what the acceleration due to gravity is, unless the question is to find the acceleration due to gravity.
:)
Post by: RuiAce on October 29, 2016, 11:34:01 pm
A quick simple question; does HSC specify acceleration due to gravity as 9.8 or 10 ms-2?Furthering onto what Cindy said
If you look on your data sheet for physics, it clearly states that g = 9.8
Post by: JemexR on October 30, 2016, 12:23:02 am
Furthering onto what Cindy said
If you look on your data sheet for physics, it clearly states that g = 9.8
hehe. Sorry for that waste of time...
Post by: RuiAce on October 30, 2016, 12:27:27 am
hehe. Sorry for that waste of time...Well if you were genuinely unsure then it's not a waste of time and just clearing up little bits and pieces, so all good
Post by: FallonXay on October 30, 2016, 07:38:19 am
How would you work out this question? (Bostes answers say D; The excel Success One Physics book answers say B)
Thanks!!!
Post by: noonedoesnt on October 30, 2016, 07:52:43 am
Hellloooooooooo! :)
How would you work out this question? (Bostes answers say D; The excel Success One Physics book answers say B)
Thanks!!!
This was an extremely controversial question, and from what I heard the answer D was a violation of conservation of energy, but i'm not too sure about that. Anyways I feel D makes no sense in answering that question, because if he is outside a spacecraft in orbit, both have the same orbital radius and the same period. Thus B would be correct.
Post by: FallonXay on October 30, 2016, 10:02:35 am
This was an extremely controversial question, and from what I heard the answer D was a violation of conservation of energy, but i'm not too sure about that. Anyways I feel D makes no sense in answering that question, because if he is outside a spacecraft in orbit, both have the same orbital radius and the same period. Thus B would be correct.
Ahh k. Cheers :)
Post by: FallonXay on October 30, 2016, 10:03:18 am
How would you answers this question?
Thanks in advance~ :)
Post by: MysteryMarker on October 30, 2016, 10:07:46 am
For Hertz' experiment to demonstrate the production/existence of radio waves, for the transmitter coil I've read that he used a high voltage AC power source connected to an induction coil and so on... But for the 2008 physics Q which asks us to 'describe an investigation used to demonstrate the production and reception of radio waves' they utilised a DC power supply. Does it matter whether AC or DC is used?
Cheers.
Post by: jakesilove on October 30, 2016, 10:30:08 am
Hello again,
How would you answers this question?
Thanks in advance~ :)
Hey! You needed to break down this answer into three parts; launch, slingshot, and orbit.
For launch, discuss conservation of momentum. Explain how we launch rockets, and how we optimise speed etc. You don't need to go into too much depth here, but if you can outline escape velocity that would be ideal.
Then, the slingshot effect. Explain how it works, draw a diagram, and explain why we use it. Simple.
Finally, orbit. Explain orbital velocity, throw in some laws of gravitation, discuss friction etc. and you're golden!
Post by: jakesilove on October 30, 2016, 10:30:53 am
Hey guys,
For Hertz' experiment to demonstrate the production/existence of radio waves, for the transmitter coil I've read that he used a high voltage AC power source connected to an induction coil and so on... But for the 2008 physics Q which asks us to 'describe an investigation used to demonstrate the production and reception of radio waves' they utilised a DC power supply. Does it matter whether AC or DC is used?
Cheers.
Nope :)
Post by: jamonwindeyer on October 30, 2016, 10:50:09 am
Hellloooooooooo! :)
How would you work out this question? (Bostes answers say D; The excel Success One Physics book answers say B)
Thanks!!!
This was an extremely controversial question, and from what I heard the answer D was a violation of conservation of energy, but i'm not too sure about that. Anyways I feel D makes no sense in answering that question, because if he is outside a spacecraft in orbit, both have the same orbital radius and the same period. Thus B would be correct.
When in doubt; BOSTES solution wins. The Success One Physics Book has quite a few errors; my tutoring students and I find them fairly frequently. The answer here is indeed D. But yep, it's a weird one!
The fact that both the spacecraft and astronaut are in orbit with the earth around the sun makes no difference; the effects of the suns gravity are negligible compared to the other forces at play in this scenario. B just doesn't make sense; it's just stating an irrelevant fact. A is incorrect; clearly gravity has to play a significant role if we have an orbit :)
Would C be correct? Well the force of gravity on each is:
The mass of the astronaut is different to the mass of the spacecraft, so the forces are different. So C is incorrect too. That leaves D by process of elimination.
The actual reason is this, consider Newton's 2nd Law:
So the acceleration of the astronaut and the spacecraft is:
Acceleration is inversely proportional to mass, and this is why the accelerations are identical, because substituting the expression for gravitational force yields:
Mass plays no role, so their acceleration will be the same, meaning they'll have the same orbit :)
Post by: imtrying on October 30, 2016, 11:28:05 am
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2011exams/pdf_doc/2011-hsc-exam-physics.pdf
Post by: FallonXay on October 30, 2016, 11:33:03 am
When in doubt; BOSTES solution wins. The Success One Physics Book has quite a few errors; my tutoring students and I find them fairly frequently. The answer here is indeed D. But yep, it's a weird one!
The fact that both the spacecraft and astronaut are in orbit with the earth around the sun makes no difference; the effects of the suns gravity are negligible compared to the other forces at play in this scenario. B just doesn't make sense; it's just stating an irrelevant fact. A is incorrect; clearly gravity has to play a significant role if we have an orbit :)
Would C be correct? Well the force of gravity on each is:
The mass of the astronaut is different to the mass of the spacecraft, so the forces are different. So C is incorrect too. That leaves D by process of elimination.
The actual reason is this, consider Newton's 2nd Law:
So the acceleration of the astronaut and the spacecraft is:
Acceleration is inversely proportional to mass, and this is why the accelerations are identical, because substituting the expression for gravitational force yields:
Mass plays no role, so their acceleration will be the same, meaning they'll have the same orbit :)
Ahh k, gotcha. Thanks :)
Post by: MarkThor on October 30, 2016, 11:34:44 am
Post by: Cindy2k16 on October 30, 2016, 11:45:23 am
Could someone please explain to me how to get the answer for this question?
You estimate the peak weavelength ~7 micrometres from the graph and calculate the energy with E=hf. You'll find the value you get is closest to the band gap of B.
:)
Post by: znaser on October 30, 2016, 11:59:37 am
Post by: jakesilove on October 30, 2016, 12:06:20 pm
A mathematical explanation for the shorter path and lower max. height would just be formulated by manipulating the equations as the acceleration due to gravity will be proportional or inversely proportional to ... But I was wondering in terms of a more scientific approach, would it be because there is more resistance to the movement of the projectile upwards due to a stronger gravitational force acting downwards and as a result, a lower maximum height. and since horizontal velocity is not affected by acceleration, a shorter range would result due to a lower time in the air. or is it something else? Thanks :)
Yep, you're right on the money! I probably wouldn't use the word 'resistance', as that implies air resistance etc; it's about a greater downwards acceleration/force due to gravity! But yeah, you've answered perfectly
Post by: Cindy2k16 on October 30, 2016, 12:13:41 pm
TIA :)
Post by: Cindy2k16 on October 30, 2016, 12:35:41 pm
TIA
When a spacecraft in orbit fires its rockets to leave its orbit, is this just an application of just newton's third law (conservation of momentum) or does it also apply Newton's second law- does the firing of the rockets apply a force on the rocket causing it to accelerate and therefore leave its orbit?
TIA :) (sorry for all these questions!)
Moderator Edit: Posts merged. If you have a few in a row, use the 'Modify' button to add them to your initial question! :)
Post by: jamonwindeyer on October 30, 2016, 12:36:19 pm
Hi for this question how do you work out what the answer is? I can see why each option would affect the amount of current but not which one would affect it the least
TIA :)
Hey! So let's consider each option; the trick here is remembering Faraday's Law for Induced EMF (and thus current):
So we can change the current by increasing how much of a change in flux we expose the wire to.
Making it longer does this, because the length of the wire is cutting through more field lines. Changing your orientation with respect to the earths magnetic field DEFINITELY does this as well. Ditto for the speed of the wire.
Changing the thickness doesn't do much, because it specifies that the wire has a very low resistance. Making something with a low resistivity thicker isn't going to alter things as much as the others. Therefore, the answer (imo) would be B :)
Post by: RuiAce on October 30, 2016, 12:36:41 pm
Another question: for mass dilation, is the mass only observed to increase from an observer in another frame of reference to the object travelling at relativistic speeds? Or does the object travelling at relativistic speeds experience the increase in mass?It's observed to increase. Note that the guy travelling at relativistic speeds actually thinks everyone else is travelling at relativistic speeds (because all inertial frames of reference are relative), so he thinks his mass is fine and observes everyone else getting heavier
TIA
Post by: Albertenouttaten on October 30, 2016, 12:36:48 pm
Hi for this question how do you work out what the answer is? I can see why each option would affect the amount of current but not which one would affect it the least
TIA :)
Hi is the answer C?
Post by: imtrying on October 30, 2016, 12:37:29 pm
You estimate the peak weavelength ~7 micrometres from the graph and calculate the energy with E=hf. You'll find the value you get is closest to the band gap of B.
:)
Hey I was just doing this question as well, and I'm probably missing something obvious, but why would you calculate it from the peak of the graph?
Post by: jamonwindeyer on October 30, 2016, 12:39:15 pm
Hi is the answer C?
Not quite! See my response above :)
Hey I was just doing this question as well, and I'm probably missing something obvious, but why would you calculate it from the peak of the graph?
No no this is actually pretty subtle. If we want to detect human radiation, we would want a work function roughly equal to the peak of our human black body curve. Lower than this, and we would get stuff from other sources and cause interference. Higher, and we'd not detect our radiation entirely :) thus, we use the peak (the answer won't be exact to any of the options, just close) :)
Post by: jamonwindeyer on October 30, 2016, 12:41:30 pm
When a spacecraft in orbit fires its rockets to leave its orbit, is this just an application of just newton's third law (conservation of momentum) or does it also apply Newton's second law- does the firing of the rockets apply a force on the rocket causing it to accelerate and therefore leave its orbit?
TIA :) (sorry for all these questions!)
That's okay! Rui answered your other one; but for this, it's all of the above! Conservation of momentum plays a role, Newton's 2nd Law also plays a role; they are just different ways of analysing the same scenario (and indeed, you should be able to apply both to a rocket launch) ;D
Post by: imtrying on October 30, 2016, 12:43:28 pm
No no this is actually pretty subtle. If we want to detect human radiation, we would want a work function roughly equal to the peak of our human black body curve. Lower than this, and we would get stuff from other sources and cause interference. Higher, and we'd not detect our radiation entirely :) thus, we use the peak (the answer won't be exact to any of the options, just close) :)
Thanks! :)
Post by: Mei2016 on October 30, 2016, 12:51:36 pm
-When Pauli suggested the neutrino, this covers the three types right? (the electron neutrion, tau and muon neutrinos?)
-Also, is it advised to remember the nuclear equation for what was actually happening in Fermi's Uranium Problem(the neutron + 235 U --> 2 daughter products and 3 neutrons)?
-In 2006 paper, in Q9, the answer is B and do they mean clockwise or anticlockwise?
(is it anticlockwise because charges would move clockwise Y to X, however, it's an induced current so due to Lenz's law it would flow the opposite direction, so anticlockwise?)
And for 2013, Q16, why is the answer D and not C?, because isn't it in metals that Valence Band and Conduction Band are both full at the same time. The question doesn't specify what band it is 'electrons in a fully filled band' so I wasn't quite sure on this one.
-Also, for 2011 Q34 c) for forces in the atomic nucleus, is electrostatic only for protons to protons or electrons to electrons? (so it's direction is always 'repulsive'?) or could it be between protons and electrons (then the direction would be 'attractive')
-and isn't the direction for SNF attractive at certain distances and then replusive if the distances are less than 1x10^-15m? ( the answers for the table in Q34c) only had 'attractive' as the direction...)
Post by: Cindy2k16 on October 30, 2016, 12:51:50 pm
Hey! So let's consider each option; the trick here is remembering Faraday's Law for Induced EMF (and thus current):
So we can change the current by increasing how much of a change in flux we expose the wire to.
Making it longer does this, because the length of the wire is cutting through more field lines. Changing your orientation with respect to the earths magnetic field DEFINITELY does this as well. Ditto for the speed of the wire.
Changing the thickness doesn't do much, because it specifies that the wire has a very low resistance. Making something with a low resistivity thicker isn't going to alter things as much as the others. Therefore, the answer (imo) would be B :)
That's okay! Rui answered your other one; but for this, it's all of the above! Conservation of momentum plays a role, Newton's 2nd Law also plays a role; they are just different ways of analysing the same scenario (and indeed, you should be able to apply both to a rocket launch) ;D
It's observed to increase. Note that the guy travelling at relativistic speeds actually thinks everyone else is travelling at relativistic speeds (because all inertial frames of reference are relative), so he thinks his mass is fine and observes everyone else getting heavier
Thank you!
Post by: Brenda0708 on October 30, 2016, 01:15:10 pm
Post by: jakesilove on October 30, 2016, 01:20:23 pm
Hi, for Quanta to quarks do we leave our answer for mass defect in kg or amu? And for the energy that that mass defect is equivalent to, do we calculate in in MeV or in J? Alsoooo how many significant figures should I leave my answer in?
Unless the question specifies, it doesn't matter what units you use! Sig figs do matter though; look at the FEWEST number of sig figs you've got in the question (ie. if mass defect is 1.0g, then you have two sig figs) and use that in your answer.
Post by: Goodwil on October 30, 2016, 01:31:35 pm
Post by: noonedoesnt on October 30, 2016, 01:38:57 pm
How would we do both parts to this question?
for part a)
- By using the Np/Ns = Vp/Vs = Is/Ip we see that the largest voltage is induced when the ratio of primary to secondary coils is low. ie, 5 primary turns to 100 secondary. Thus arrangement B would produce a spark, as a large voltage is required to jump the gap
for part b)
using the same formula from above, by moving different lengths, the induced emf is altered. ie, at arrangement a, ratio of primary to secondary coils is much higher than at arrangement b, thus a lower voltage is induced than that of in arrangement b the more of the primary coil is inside the secondary coil.
Post by: FallonXay on October 30, 2016, 01:39:37 pm
Thanks.
Post by: Phillorsm on October 30, 2016, 01:39:53 pm
I'm getting some conflicting information.
Thanks.
Post by: jakesilove on October 30, 2016, 01:40:57 pm
Just a quick question: Does back EMF occur in both DC AND AC motors or only in DC motors?
Thanks.
We say that it only occurs in DC motors, because of what back EMF really is. An opposing current implies a consistent direction, which wouldn't be the case for any induced back EMF in AC motors. Like, there needs to be a consistent current to oppose, if that makes sense. So, we only talk about Back EMF in DC motors.
Post by: jakesilove on October 30, 2016, 01:42:01 pm
Hey guys, for the medical physics dot point about bone scans vs x rays, are bone scans CAT scans or PET scans?
I'm getting some conflicting information.
Thanks.
Ahahaha you have no idea how close to home this question hits. In my trials, we got a question about a 'Bone scan', which I had never heard of before. I just assumed it was a PET scan, and I got the marks, but others used CAT scan and also got the marks. As far as I understand, you really can use either/or. However, someone is welcome to correct me!
Post by: Phillorsm on October 30, 2016, 01:43:14 pm
Post by: jakesilove on October 30, 2016, 01:46:53 pm
It makes more sense to be a PET scan haha
Yeah I totally agree
Post by: Mei2016 on October 30, 2016, 01:54:49 pm
How would we do both parts to this question?
This is the 2008 Q26 question, and due to Faraday's law of EM induction ϵ=−ndϕ/dt, as Arrangement A has more number of coils inside it, it has a greater induced EMF, therefore producing a spark in Arrangement A.
For part b) again using Faraday's law, different amounts of coil produces different amounts of EMF
(However, I don't get how this works with Np/Ns = Vp/Vs = Is/Ip as the answers to the question is Arrangement A)
I think it's something to do with : in Arrangement B, they're showing you that there's no primary coil within the secondary coil so you get Np=0.
Post by: Brenda0708 on October 30, 2016, 01:58:06 pm
Unless the question specifies, it doesn't matter what units you use! Sig figs do matter though; look at the FEWEST number of sig figs you've got in the question (ie. if mass defect is 1.0g, then you have two sig figs) and use that in your answer.
Oh so even if the fewest number of sig figs is 6 we should leave it to 6? The answer in the book did it to 3
Post by: noonedoesnt on October 30, 2016, 01:58:43 pm
This is the 2008 Q26 question, and due to Faraday's law of EM induction ϵ=−ndϕ/dt, as Arrangement A has more number of coils inside it, it has a greater induced EMF, therefore producing a spark in Arrangement A.
For part b) again using Faraday's law, different amounts of coil produces different amounts of EMF
(However, I don't get how this works with Np/Ns = Vp/Vs = Is/Ip as the answers to the question is Arrangement A)
well shit, that doesn't make any sense. How is it arrangement A?
Jake can you explain?
Post by: jakesilove on October 30, 2016, 02:01:00 pm
Oh so even if the fewest number of sig figs is 6 we should leave it to 6? The answer in the book did it to 3Yep, the book is likely wrong. It's always the fewest number of sig figs in the question!
Post by: jamonwindeyer on October 30, 2016, 02:18:15 pm
well shit, that doesn't make any sense. How is it arrangement A?
Jake can you explain?
It makes perfect sense :) let me explain! In Arrangement B, the two coils are completely separated from each other (remember there is no connection between the two besides flux linkage. Thus, when we pull them apart, the EMF in the secondary coil is tiny (the absence of the core being near the secondary coil in Arrangement B makes this worse).
It's like having a transformer with the coils way too far apart, essentially ;D
Post by: Cindy2k16 on October 30, 2016, 02:20:00 pm
Hey guys, for the medical physics dot point about bone scans vs x rays, are bone scans CAT scans or PET scans?
I'm getting some conflicting information.
Thanks.
Its neither! Bone scans involve injecting a radiopharmaceutical into the body- but the radioisotope used is not a positron emitting one therefore its not PET- Technetium-99m is the radioisotope commonly used for bones scans. Tc-99m emits gamma radiation which is detected with gamma cameras outside the body. It works similar to PET though, as the radioisotope is taken up into the bones and accumulates in areas of high metabolism in the bones and then multiple detections of the radiation from the body allows for the scan to show areas of high concentration of the radioisotope -> shows the presence of bone tumours. Which x-rays can't detect well.
Hope this helps :)
Post by: MarkThor on October 30, 2016, 02:20:37 pm
You estimate the peak weavelength ~7 micrometres from the graph and calculate the energy with E=hf. You'll find the value you get is closest to the band gap of B.
:)
Thank you :)
Post by: Cindy2k16 on October 30, 2016, 02:25:28 pm
Also should we be able to describe thought experiments for time dilation, mass dilation or length contraction? Currently I can only describe a thought experiment for the relativity of simultaneity.
TIA :)
Post by: annabelkaren on October 30, 2016, 02:32:54 pm
Post by: Cindy2k16 on October 30, 2016, 02:37:18 pm
Explain what happens to a rocket's chemical energy, kinetic energy, and gravitational potential energy when it is being launched from the surface of Earth. - 3 Marks
When a rocket burns its fuel during rocket launch, it is converting the chemical energy of the fuel into kinetic energy (the gases released from the combustion of fuel eject out, causing the rocket to move upwards). Thus the rocket's chemical energy decreases and its kinetic energy increases.
When the rocket is launching from the surface of the earth, it is doing work against the gravitational field. Thus, its gravitational potential energy increases.
Hope this helps :)
Post by: Mei2016 on October 30, 2016, 03:14:00 pm
So Pauli predicted the existence of the neutrino after studying beta decays and suggested the existence of them to satisfy the laws of conservation of momentum and energy. However, beta decay (ie beta minus) is due to emission of neutrons, where the neutrons decay into a proton, an electron and an anti-neutrino. So an anti-neutrino is produced by beta decay, so why did Pauli suggest the neutrino?
-When Pauli suggested the neutrino, this covers the three types right? (the electron neutrion, tau and muon neutrinos?)
-Also, is it advised to remember the nuclear equation for what was actually happening in Fermi's Uranium Problem(the neutron + 235 U --> 2 daughter products and 3 neutrons)?
-In 2006 paper, in Q9, the answer is B and do they mean clockwise or anticlockwise?
(is it anticlockwise because charges would move clockwise Y to X, however, it's an induced current so due to Lenz's law it would flow the opposite direction, so anticlockwise?)
And for 2013, Q16, why is the answer D and not C?, because isn't it in metals that Valence Band and Conduction Band are both full at the same time. The question doesn't specify what band it is 'electrons in a fully filled band' so I wasn't quite sure on this one.
-Also, for 2011 Q34 c) for forces in the atomic nucleus, is electrostatic only for protons to protons or electrons to electrons? (so it's direction is always 'repulsive'?) or could it be between protons and electrons (then the direction would be 'attractive')
-and isn't the direction for SNF attractive at certain distances and then repulsive if the distances are less than 1x10^-15m? ( the answers for the table in Q34c) only had 'attractive' as the direction...)
Post by: Cindy2k16 on October 30, 2016, 03:39:50 pm
Hi, I have a few questions:
So Pauli predicted the existence of the neutrino after studying beta decays and suggested the existence of them to satisfy the laws of conservation of momentum and energy. However, beta decay (ie beta minus) is due to emission of neutrons, where the neutrons decay into a proton, an electron and an anti-neutrino. So an anti-neutrino is produced by beta decay, so why did Pauli suggest the neutrino?
-When Pauli suggested the neutrino, this covers the three types right? (the electron neutrion, tau and muon neutrinos?)
-Also, is it advised to remember the nuclear equation for what was actually happening in Fermi's Uranium Problem(the neutron + 235 U --> 2 daughter products and 3 neutrons)?
-In 2006 paper, in Q9, the answer is B and do they mean clockwise or anticlockwise?
(is it anticlockwise because charges would move clockwise Y to X, however, it's an induced current so due to Lenz's law it would flow the opposite direction, so anticlockwise?)
And for 2013, Q16, why is the answer D and not C?, because isn't it in metals that Valence Band and Conduction Band are both full at the same time. The question doesn't specify what band it is 'electrons in a fully filled band' so I wasn't quite sure on this one.
-Also, for 2011 Q34 c) for forces in the atomic nucleus, is electrostatic only for protons to protons or electrons to electrons? (so it's direction is always 'repulsive'?) or could it be between protons and electrons (then the direction would be 'attractive')
-and isn't the direction for SNF attractive at certain distances and then repulsive if the distances are less than 1x10^-15m? ( the answers for the table in Q34c) only had 'attractive' as the direction...)
Hi i can help with some of these questions (i dont do quanta to quarks)
-In 2006 paper, in Q9, the answer is B and do they mean clockwise or anticlockwise?
(is it anticlockwise because charges would move clockwise Y to X, however, it's an induced current so due to Lenz's law it would flow the opposite direction, so anticlockwise?)
It is clockwise from Y to X. When you apply the right hand palm rule, you point the thumb in the direction that the wheel is turning through the magnetic field since that is the direction of movement of positive charge. The movement of the wheel through the magnetic field is basically the movement of the 'positive charge' in the metals as they are being moved with the wheel. Does this make sense? And then you point ur fingers in the direction of the magnetic field and you will get a downwards force on the positive charges. Thus current flows down out of the wheel through Brush Y, through the external circuit, to X.
And for 2013, Q16, why is the answer D and not C?, because isn't it in metals that Valence Band and Conduction Band are both full at the same time. The question doesn't specify what band it is 'electrons in a fully filled band' so I wasn't quite sure on this one.
It isn't true that in conductors both the valence band and conduction band are both full at the same time. In fact too many electrons in the conduction band make it too crowded and in fact makes it harder for current to flow, I think. Sorry I dont know a very scientific explanation.. But in conductors the conduction band is partially filled. So its C.
Hope this helps :)
Post by: jamonwindeyer on October 30, 2016, 03:57:51 pm
Hi when a spacecraft is travelling through space, theoretically (just in terms of what we need to know in the HSC) does it not need to fire any rockets to continue moving? If yes, is this because theres practically no friction in space and unless the spacecraft is passing close to a planet, its not experiencing any (significant) forces and so by Newton's 1st law it'll just continue on its straight line path?
Also should we be able to describe thought experiments for time dilation, mass dilation or length contraction? Currently I can only describe a thought experiment for the relativity of simultaneity.
TIA :)
First point: Yep! You don't need to do any work to keep a craft moving in space (theoretically) :)
Second: The experiments you know are fine, the syllabus only asks you to know about the 'train experiments' and be able to discuss the benefits/drawbacks of thought experiments in general ;D
Ps - You are answering so many questions, you are awesome!! Thank you! ;D
Post by: jamonwindeyer on October 30, 2016, 04:00:54 pm
Explain what happens to a rocket's chemical energy, kinetic energy, and gravitational potential energy when it is being launched from the surface of Earth. - 3 Marks
Welcome to the forums Anna! Let me know if you need a hand finding anything! ;D
Post by: pels on October 30, 2016, 04:04:42 pm
What do we need to know about transistors and diodes?
Post by: Cindy2k16 on October 30, 2016, 04:07:20 pm
First point: Yep! You don't need to do any work to keep a craft moving in space (theoretically) :)
Second: The experiments you know are fine, the syllabus only asks you to know about the 'train experiments' and be able to discuss the benefits/drawbacks of thought experiments in general ;D
Ps - You are answering so many questions, you are awesome!! Thank you! ;D
Thanks for clarifying Jamon! And its no problem :)
Post by: annabelkaren on October 30, 2016, 04:14:43 pm
When a rocket burns its fuel during rocket launch, it is converting the chemical energy of the fuel into kinetic energy (the gases released from the combustion of fuel eject out, causing the rocket to move upwards). Thus the rocket's chemical energy decreases and its kinetic energy increases.Thank you so much!!!
When the rocket is launching from the surface of the earth, it is doing work against the gravitational field. Thus, its gravitational potential energy increases.
Hope this helps :)
Post by: Brenda0708 on October 30, 2016, 04:15:26 pm
Hi I'm confused about the angle used in the equation F=qvBsintheta here. If theta is the angle between the velocity vector and the direction of the MF lines, how does that look for this question?
Post by: Cindy2k16 on October 30, 2016, 04:23:50 pm
(http://uploads.tapatalk-cdn.com/20161029/b7d6362c3926ac5d8b689fa6fa2a55a7.jpg)
Hi I'm confused about the angle used in the equation F=qvBsintheta here. If theta is the angle between the velocity vector and the direction of the MF lines, how does that look for this question?
Hi the way this question is set up is kind of a trick. They give you an angle, but its not needed.. This is because the angle used in F=qvBsintheta is the angle between the velocity vector of the electron and the direction of the magnetic field. If you look at the diagram carefully, you will realise that the electron is in fact travelling perpendicular to the magnetic field and so theta is actually 90 degrees not 40 thus sin90=1. I believe everything else you're doing is fine- just use 90 degrees instead :) (or just use F=qvB, its the same thing as F=qvBsin90)
I've seen a few questions with this trick so always be careful with these questions where they provide an angle!
Hope this helps :)
Post by: imtrying on October 30, 2016, 04:42:15 pm
Post by: jakesilove on October 30, 2016, 05:02:15 pm
With the dotpoint about Hertzs observations with radio waves producing a photoelectric effect, would I be correct in saying that the fact that the spark stopped when UV was blocked by glass showed that it was preventing the photoelectric effect from happening because no UV light was hitting the receiver meaning it was no longer emitting electrons which therefore meant there was no spark?
Actually, the sparking wasn't stopped by the glass, it was only dampened! This showed that whatever was causing the spark (ie. UV rays) could travel through material such as glass.
Post by: Cindy2k16 on October 30, 2016, 05:08:52 pm
Actually, the sparking wasn't stopped by the glass, it was only dampened! This showed that whatever was causing the spark (ie. UV rays) could travel through material such as glass.
Hi Jake I've also heard that Hertz put his receiver coil in a dark container and found that the sparks were dampened. And that when he exposed the receiver to UV light, sparks were more readily induced. Is this also correct to talk about when referring to Hertz's observation of the photoelectric effect? TIA
Post by: MysteryMarker on October 30, 2016, 05:09:11 pm
http://prnt.sc/d0py3e
http://prnt.sc/d0pyhr
Cheers.
Post by: Brenda0708 on October 30, 2016, 05:14:41 pm
Confused as to which l is which??? Doesn't length contract in the stationary observers view not for the person moving??
Post by: jakesilove on October 30, 2016, 05:18:02 pm
Hi Jake I've also heard that Hertz put his receiver coil in a dark container and found that the sparks were dampened. And that when he exposed the receiver to UV light, sparks were more readily induced. Is this also correct to talk about when referring to Hertz's observation of the photoelectric effect? TIA
Absolutely right! He did a few experiments like that, with the results you described, but with no real analysis of what he was observing :)
Post by: Brenda0708 on October 30, 2016, 05:27:16 pm
I keep getting stuck on projectile q LOL please help for 19b) not sure where to start
Post by: Mei2016 on October 30, 2016, 05:30:19 pm
Hi i can help with some of these questions (i dont do quanta to quarks)
-In 2006 paper, in Q9, the answer is B and do they mean clockwise or anticlockwise?
(is it anticlockwise because charges would move clockwise Y to X, however, it's an induced current so due to Lenz's law it would flow the opposite direction, so anticlockwise?)
It is clockwise from Y to X. When you apply the right hand palm rule, you point the thumb in the direction that the wheel is turning through the magnetic field since that is the direction of movement of positive charge. The movement of the wheel through the magnetic field is basically the movement of the 'positive charge' in the metals as they are being moved with the wheel. Does this make sense? And then you point ur fingers in the direction of the magnetic field and you will get a downwards force on the positive charges. Thus current flows down out of the wheel through Brush Y, through the external circuit, to X.
And for 2013, Q16, why is the answer D and not C?, because isn't it in metals that Valence Band and Conduction Band are both full at the same time. The question doesn't specify what band it is 'electrons in a fully filled band' so I wasn't quite sure on this one.
It isn't true that in conductors both the valence band and conduction band are both full at the same time. In fact too many electrons in the conduction band make it too crowded and in fact makes it harder for current to flow, I think. Sorry I dont know a very scientific explanation.. But in conductors the conduction band is partially filled. So its C.
Hope this helps :)
Hi, thanks for the help. It's just that for the 2006 question, the answers mentioned Lenz's law so I got a bit confused there. Thanks :)
Post by: Cindy2k16 on October 30, 2016, 05:33:50 pm
(http://uploads.tapatalk-cdn.com/20161029/f57898b06685cb7eef99a76c9e2f73d9.jpg)
I keep getting stuck on projectile q LOL please help for 19b) not sure where to start
HI I would start by calculating the acceleration the electron experiences inside the electric field. Use F=ma for this, and sub in the force you found in a and the mass of an electron to find a.
Then draw the triangle for the initial velocity to determine the initial horizontal and vertical velocity of the electron at A.
Once you have all these values, you can act as if this is a normal projectile motion question except the acceleration is not 9.8, but the 'a' you calculated.
To find the time taken to travel from A to D, that's basically working out the time of flight. Use the formula starting with 'change in y' (the triangle with 'y' next to it). Sub in the change in y as 0 (since overall, the vertical displacement of the particle from A to D is 0), sub the initial vertical velocity, the acceleration you calculated and then rearrange the equation to find t. One value of t will be 0 (since the particle has 0 vertical displacement at the beginning too) and the other value will be the time it took to reach D.
Hope this helps :)
Post by: Mei2016 on October 30, 2016, 06:04:39 pm
(http://uploads.tapatalk-cdn.com/20161029/f57898b06685cb7eef99a76c9e2f73d9.jpg)
I keep getting stuck on projectile q LOL please help for 19b) not sure where to start
Hi, how I would do this is follow Cindy2k16's way of getting the acceleration from F=ma and getting the y-component of the velocity with the triangle.
But an easier way to do the projectiles part is use v=u+at, where v=-u as this is using the symmetry of the parabolic arc and by using the a from above,
t=-2u/a [ie t=(-2x6x10^6sin60)/(-1.7587x10^14)]
Hope this helps :)
Post by: Mei2016 on October 30, 2016, 06:12:01 pm
(http://uploads.tapatalk-cdn.com/20161029/5c0cf0772e9917aa1883915a3d689a76.jpg)
Confused as to which l is which??? Doesn't length contract in the stationary observers view not for the person moving??
So for these kinds of questions, always think: who's moving at relativistic speeds? When moving at such high speeds, recall that you experience time and mass dilation and length contractions. So since you're moving in this question, the length you measure is always shorter than the real length, so we are trying to calculate Lv. (note the real length is stated as a fact in the first sentence and this is Lo).
Hope this helps. :)
Post by: Albertenouttaten on October 30, 2016, 06:54:17 pm
In a particle accerlator called a synchroton, magnetic fields are used to control the motion of an electron so that it follows a circular path of fixed radius (i know this would probably require me to use either centripetal force of acceleration formula)
Describe the changes required in the magnetic field to accelerate an electron to near the speed of light. Support answer with appropriate mathematical relationships.
Thanks to anyone who is able to put any sort of input at all, any help is greatly appreciated.
Post by: jakesilove on October 30, 2016, 07:00:05 pm
Hi can anyone help me answer this question, I can't think of what to write:
In a particle accerlator called a synchroton, magnetic fields are used to control the motion of an electron so that it follows a circular path of fixed radius (i know this would probably require me to use either centripetal force of acceleration formula)
Describe the changes required in the magnetic field to accelerate an electron to near the speed of light. Support answer with appropriate mathematical relationships.
Thanks to anyone who is able to put any sort of input at all, any help is greatly appreciated.
Hey! So essentially what this question is asking is the method by which we can accelerate electrons, using a magnetic field. We know that the force due to a magnetic field is
Where B is the magnetic field strength. Obviously q is fixed (it's the charge of the electron), so we can only really vary the magnetic field strength. Now, if we apply a constant magnetic field (in terms of direction, at least), it will provide a centripetal force. So, we want the magnetic field to continuously be in the same direction, but we want to keep increasing it's strength.
How do we increase the strength of a magnetic field? Well, assuming that they are electromagnets, we would want to put more and more current through the wiring! Additionally, if we used superconductors, we would get a seriously strong field. You can really talk about whatever you want in a question like this; as you can see, I'm rambling, because I'm making up the answer as I go aha.
You could potentially discuss relativistic effects, such as mass dilation, as problematic, however that's just the icing on the cake
Post by: Cindy2k16 on October 30, 2016, 07:05:11 pm
"Two significant problems that will affect a manned spaceflight to Mars are:
-the changes in gravitational energy
- protecting the space vehicle from high-speed electrically charged particles from the Sun
Use your understanding of physics to analyse these problems"
8 marks
How could you answer this question? is it necessary that we know about the gravitational field/atmosphere etc of Mars?
TIA
Post by: jakesilove on October 30, 2016, 07:08:34 pm
Hi for 2010 q32
"Two significant problems that will affect a manned spaceflight to Mars are:
-the changes in gravitational energy
- protecting the space vehicle from high-speed electrically charged particles from the Sun
Use your understanding of physics to analyse these problems"
8 marks
How could you answer this question? is it necessary that we know about the gravitational field/atmosphere etc of Mars?
TIA
Is this part of an option? Unfortunately, I won't be able to help you out, but hopefully someone else can!
Post by: JemexR on October 30, 2016, 07:08:41 pm
Post by: jakesilove on October 30, 2016, 07:10:49 pm
Hi, for HSC 2005 Q20, when it mentions reliability, in what sense are we meant to respond in? Given that reliability is INCREASED by repetition, it is asking us to "describe how you assessed the reliability of the information you found"; I can only think of accurate source (reliable, i.e. from a scientific institution) at the present moment, but it's worth 6m :/
Reliability of secondary sources is largely about cross-checking your sources; do different sources say the same thing? Does that information correspond with your understanding of the topic area? Go into some depth about sources you used, how you found them, how you analysed them, how many sources you used etc. etc. Basically, make shit up, but I promise you won't get a question that absurd in a modern paper.
Post by: Cindy2k16 on October 30, 2016, 07:11:49 pm
Is this part of an option? Unfortunately, I won't be able to help you out, but hopefully someone else can!
Actually....its in the core :(
Post by: jakesilove on October 30, 2016, 07:15:12 pm
Actually....its in the core :(
Well shit.
No, you definitely don't need to know anything about Mars. The 'Change in gravitational energy' can be understood as ANY movement of the rocket, so potentially just talk about rocket launches in as much detail as possible, as well as layers of protection from solar flares etc. Fucking weird question; basically, just spam information about rockets and hope for the best is my advice!
Post by: Brenda0708 on October 30, 2016, 07:22:04 pm
HI I would start by calculating the acceleration the electron experiences inside the electric field. Use F=ma for this, and sub in the force you found in a and the mass of an electron to find a.
Then draw the triangle for the initial velocity to determine the initial horizontal and vertical velocity of the electron at A.
Once you have all these values, you can act as if this is a normal projectile motion question except the acceleration is not 9.8, but the 'a' you calculated.
To find the time taken to travel from A to D, that's basically working out the time of flight. Use the formula starting with 'change in y' (the triangle with 'y' next to it). Sub in the change in y as 0 (since overall, the vertical displacement of the particle from A to D is 0), sub the initial vertical velocity, the acceleration you calculated and then rearrange the equation to find t. One value of t will be 0 (since the particle has 0 vertical displacement at the beginning too) and the other value will be the time it took to reach D.
Hope this helps :)
Thank you!!!! :)
Post by: Albertenouttaten on October 30, 2016, 07:22:35 pm
1) Earth to Earth Orbit
2) Earth Orbit to Sun Orbit then to Mars Orbit
3) Mars Orbit to Mars Surface
4) Mars surface to Mars Orbit
5) Step 2 but in reverse
6) Lowered down to Earth orbit.
Steps 1,2,4 are increases in GPE and the leftover are decreases in GPE. Also talk about g-forces experienced, the need for retroboosters when entering into Mar's atmosphere (retroboosters would be required in this process as the aim is to slow down the ship as the decrease in GPE results in increased kinetic energy and also the ship will be used again to return to Earth hence cannot simply rely only on the blunt shape of the ship and also another aspect you can talk about is that Mar's atmosphere is not as thick as Earth's so atmospheric friction isn't enough of a force to slow the ship down to safe levels)
Second part of the question refers to the charged particles emited from the Sun (KEY NOTE: they are charged particles and not limited to being +ve or -ve). There are two methods of redirecting or protecting the ship, this is through magnetic fields or electric fields. If we use an electric field, it will only repel one type of charge, hence the key note. Therefore we must use an magnetic field to protect the ships equiment from the high velocity particles. This is a very summarised answer and that question would require you to expand on this but these are the general aspects of the quesiton. Definitely encoporate many aspects of the syllabus.
Post by: Brenda0708 on October 30, 2016, 07:24:34 pm
Hi, how I would do this is follow Cindy2k16's way of getting the acceleration from F=ma and getting the y-component of the velocity with the triangle.
But an easier way to do the projectiles part is use v=u+at, where v=-u as this is using the symmetry of the parabolic arc and by using the a from above,
t=-2u/a [ie t=(-2x6x10^6sin60)/(-1.7587x10^14)]
Hope this helps :)
Thank you for your help. What do you mean by the symmetry of the parabolic arc? V= -u as in the vertical component of the final velocity = the negative of the initial velocity? Why? haha :')
Post by: Cindy2k16 on October 30, 2016, 07:26:36 pm
Hi Cindy, this question requires you to talk about all the changes in potential energy and the consequences of these chagnes. These being:
1) Earth to Earth Orbit
2) Earth Orbit to Sun Orbit then to Mars Orbit
3) Mars Orbit to Mars Surface
4) Mars surface to Mars Orbit
5) Step 2 but in reverse
6) Lowered down to Earth orbit.
Steps 1,2,4 are increases in GPE and the leftover are decreases in GPE. Also talk about g-forces experienced.
Second part of the question refers to the charged particles emited from the Sun (KEY NOTE: they are charged particles and not limited to being +ve or -ve). There are two methods of redirecting or protecting the ship, this is through magnetic fields or electric fields. If we use an electric field, it will only repel one type of charge, hence the key note. Therefore we must use an magnetic field to protect the ships equiment from the high velocity particles. This is a very summarised answer and that question would require you to expand on this but these are the general aspects of the quesiton. Definitely encoporate many aspects of the syllabus.
Well shit.
No, you definitely don't need to know anything about Mars. The 'Change in gravitational energy' can be understood as ANY movement of the rocket, so potentially just talk about rocket launches in as much detail as possible, as well as layers of protection from solar flares etc. Fucking weird question; basically, just spam information about rockets and hope for the best is my advice!
Thank you!
Post by: imtrying on October 30, 2016, 07:26:46 pm
In Medical Physics, is the piezoelectric effect basically when an oscillating potential difference is applied to a piezoelectric crystal it causes vibrations in the lattice of the crystal to give off high frequency sounds, and these are what is used in transducers?
Post by: Cindy2k16 on October 30, 2016, 07:30:09 pm
Hey:)
In Medical Physics, is the piezoelectric effect basically when an oscillating potential difference is applied to a piezoelectric crystal it causes vibrations in the lattice of the crystal to give off high frequency sounds, and these are what is used in transducers?
The piezoelectric effect is how when a potential difference is applied to a piezoelectric crystal, it causes it to deform. And then the other way round too (for ultrasound detection). It doesn't have to be an oscillating potential difference- thats just whats used In ultrasound transducers. It causes the crystal to repeatedly deform at a high frequency so its basically vibrating :)
Post by: Brenda0708 on October 30, 2016, 07:30:34 pm
So for these kinds of questions, always think: who's moving at relativistic speeds? When moving at such high speeds, recall that you experience time and mass dilation and length contractions. So since you're moving in this question, the length you measure is always shorter than the real length, so we are trying to calculate Lv. (note the real length is stated as a fact in the first sentence and this is Lo).
Hope this helps. :)
Waait doesnt the stationary observer see the length of the fast moving object contract in the direction of their velocity? And we measure the real length? but in this case the moving person is measuring the length of something when they are moving so im a bit lost as to whether or not they will see it contract or see its 'real length'
Post by: jakesilove on October 30, 2016, 08:01:50 pm
Waait doesnt the stationary observer see the length of the fast moving object contract in the direction of their velocity? And we measure the real length? but in this case the moving person is measuring the length of something when they are moving so im a bit lost as to whether or not they will see it contract or see its 'real length'
If any object is moving relative to the observer, the observer will measure the relativistic length! So, if I'm running past you, you will measure my relativistic speed. However, I will think YOU'RE running past ME, so I'll actually be measuring your relativistic speed! Does that clarify things?
Post by: Brenda0708 on October 30, 2016, 08:11:47 pm
If any object is moving relative to the observer, the observer will measure the relativistic length! So, if I'm running past you, you will measure my relativistic speed. However, I will think YOU'RE running past ME, so I'll actually be measuring your relativistic speed! Does that clarify things?
ohhhhhh thank you!
Post by: Albertenouttaten on October 30, 2016, 08:13:31 pm
If any object is moving relative to the observer, the observer will measure the relativistic length! So, if I'm running past you, you will measure my relativistic speed. However, I will think YOU'RE running past ME, so I'll actually be measuring your relativistic speed! Does that clarify things?
Lol i just wonder how high Einstein was to come up with these thought experiments.
Post by: Cindy2k16 on October 30, 2016, 08:17:35 pm
Post by: jakesilove on October 30, 2016, 08:19:42 pm
Just wondering to the people who did hsc physics prior to this year and did well, how did you feel when you finished the exam? Did you know that you had aced it? Were there questions you weren't sure of or did you just totally know everything ? ;D
I definitely didn't think I aced the exam. It was a seriously weird paper (2014 was a weird mix of easy and just bloody strange), and so I wasn't hugely confident with my performance. Like, I was sure I did fine, but I definitely didn't think I aced it or anything like that! We're all bad judges of ourselves, so once you walk out of the exam room, you may as well just forget the paper and move on. There's no way to tell what is a good raw mark, and what is a bad raw mark, because you have no idea how the rest of the cohort went. You seem completely ready to smash the exam, and that's all that's important. You'll do great tomorrow :)
Post by: Cindy2k16 on October 30, 2016, 08:25:02 pm
I definitely didn't think I aced the exam. It was a seriously weird paper (2014 was a weird mix of easy and just bloody strange), and so I wasn't hugely confident with my performance. Like, I was sure I did fine, but I definitely didn't think I aced it or anything like that! We're all bad judges of ourselves, so once you walk out of the exam room, you may as well just forget the paper and move on. There's no way to tell what is a good raw mark, and what is a bad raw mark, because you have no idea how the rest of the cohort went. You seem completely ready to smash the exam, and that's all that's important. You'll do great tomorrow :)
Haha I see. That's kind of reassuring because I practically never feel great about my performance after exams. And me? ready? thats the opposite of how I ever feel for exams lol especially this one. I am probably one of the most stressed people at my school. I hope I'll do okay. My school has a pretty inconsistent performance in science subjects in the hsc so that's kind of nerve wracking... I just badly want this all to finish!
Post by: JemexR on October 30, 2016, 08:27:42 pm
Hi for 2010 q32
"Two significant problems that will affect a manned spaceflight to Mars are:
-the changes in gravitational energy
- protecting the space vehicle from high-speed electrically charged particles from the Sun
Use your understanding of physics to analyse these problems"
8 marks
How could you answer this question? is it necessary that we know about the gravitational field/atmosphere etc of Mars?
TIA
Haha, I had the answer for this question for some reason;
Change in GE - humans are essentially designed for gravity on Earth, and given that gravity on Mars is less than that on Earth, and due to the zero gravity experienced during space travel, being in this environment can cause some serious health issues such as muscle weakening and loss of bone density.
High-speed particles - These charged particles are particularly dangerous to humans due to their ability to cause mutations in DNA, hence causing multiple problems such as cancer
Post by: Taylah Jensen on October 30, 2016, 08:48:20 pm
Would someone please be able to tell me the thought process behind this question? I feel like the answer has something to do with the right hand push rule, and there's some key in the fact the magnetic field is "then applied" meaning a change in magnetic flux, but I can't justify the answer ahaha.
Thank you!!
Post by: Cindy2k16 on October 30, 2016, 08:51:31 pm
Hiiiii,
Would someone please be able to tell me the thought process behind this question? I feel like the answer has something to do with the right hand push rule, and there's some key in the fact the magnetic field is "then applied" meaning a change in magnetic flux, but I can't justify the answer ahaha.
Thank you!!
Hi I answered this before so I'll just copy past my answer:
The electrons and holes move in opposite directions in an electric field. If you use your right hand palm rule for the conventional current (movement of positive charge) youll find there is a downwards force on the 'holes'. Now electron flow is in the OPPOSITE direction to conventional current. So when you use your left hand palm rule (or whatever youve been taught to use for the flow of a negative charge) you point ur thumb in the OPPOSITE direction to the conventional current and youll find that the electrons ALSO experience a downwards force.
Thus both holes and electrons move to the bottom of the rod ie. the same side of the rod.
Therefore C.
For this question, induction isn't used to answer it. Its more about the force experienced by charged particles in a magnetic field
hope this helps :)
Post by: Taylah Jensen on October 30, 2016, 08:55:43 pm
Hi I answered this before so I'll just copy past my answer:
The electrons and holes move in opposite directions in an electric field. If you use your right hand palm rule for the conventional current (movement of positive charge) youll find there is a downwards force on the 'holes'. Now electron flow is in the OPPOSITE direction to conventional current. So when you use your left hand palm rule (or whatever youve been taught to use for the flow of a negative charge) you point ur thumb in the OPPOSITE direction to the conventional current and youll find that the electrons ALSO experience a downwards force.
Thus both holes and electrons move to the bottom of the rod ie. the same side of the rod.
Therefore C.
For this question, induction isn't used to answer it. Its more about the force experienced by charged particles in a magnetic field
hope this helps :)
Yes! Thank you so much :)
Post by: Mei2016 on October 30, 2016, 09:15:52 pm
for 2011 Q34 c) for forces in the atomic nucleus, is electrostatic only for protons to protons or electrons to electrons? (so it's direction is always 'repulsive'?) or could it be between protons and electrons (then the direction would be 'attractive')
-and isn't the direction for SNF attractive at certain distances and then repulsive if the distances are less than 1x10^-15m? ( the answers for the table in Q34c) only had 'attractive' as the direction...)
Post by: imtrying on October 30, 2016, 09:26:41 pm
Post by: RuiAce on October 30, 2016, 09:27:22 pm
Just to clarify: when determining the force on a negative charge do I just point my thumb in the direction of negative current using RHP rule or do I have to use my left hand?Either works.
Post by: imtrying on October 30, 2016, 09:30:59 pm
Either works.
Thanks for the quick reply! Having a bit of a last minute panic there haha
Post by: nimasha.w on October 30, 2016, 09:33:23 pm
Post by: MysteryMarker on October 30, 2016, 09:40:05 pm
Hi, for those who have done quanta to quarks and can help out...
for 2011 Q34 c) for forces in the atomic nucleus, is electrostatic only for protons to protons or electrons to electrons? (so it's direction is always 'repulsive'?) or could it be between protons and electrons (then the direction would be 'attractive')
-and isn't the direction for SNF attractive at certain distances and then repulsive if the distances are less than 1x10^-15m? ( the answers for the table in Q34c) only had 'attractive' as the direction...)
Yo, for your first question about the forces in the atomic nucleus, within the nucleus there only exists two particles. These include protons and neutrons. Hence as electrostatic force is only experienced by charged particles the only charged particle within the nucleus is the proton, resulting in the electostatic force between protons to be repulsive as like opposes like. Sure the electrostatic force between a proton and electron is attractive, but within the nucleus these electrostatic forces act to repel each other.
For your second question, yes SNF is repulsive at very short distances and attractive for distances after that. The answers given in Q34c) I believe are very ambiguous because you are just stating their relative strengths. Hence I assume they just wrote attractive as in the second column we wrote that electrostatic forces are repulsive and to counteract this we have an attractive force from the SNF.
This is just my take on these questions, take these answers with a grain of salt! I'm also a year 12 student about to lose his virginity tomorrow (I don't count english cause that was bloody rape), so yeah hope this helps! :D
Post by: imtrying on October 30, 2016, 09:40:06 pm
- extended quantum theory to all forms of EMR not just black body radiation
- proposed light interacted with matter as photons (and hence proposed the photoelectric effect)
(This is for the ideas to implementation topic )
Post by: proficles on October 30, 2016, 09:57:24 pm
Post by: MysteryMarker on October 30, 2016, 10:04:30 pm
hi! 2 things - whats a simple prac to remember for the production and receiving of radio waves and one for the production of alternation current (i don't understand the ones we did in class)
Yo, for your first one I've just attached a screenshot of the 2008 HSC solution for a similar question.
http://prnt.sc/d0rzj3
For your second one to demonstrate the production of AC, this is going to be very simplified and basic, but I reckon something along the lines of moving a magnet in and out of a solenoid connected to a galvanometer should do. Unless the moderators, the actual smart people, have a genius way of producing an alternating current ;).
Post by: Cindy2k16 on October 30, 2016, 10:05:22 pm
But if this is the case, how come current is measured when an ammeter is attached to the motor? Is this because the galvanometer is considered a load?
TIA
Post by: jamonwindeyer on October 30, 2016, 10:32:00 pm
Hi for back emf, so when a motor is running at a constant speed with no load attached, the back emf will equal the supply emf and so the net emf is zero. Thus current in the coil should be zero right? ( http://www.physics.usyd.edu.au/teach_res/hsp/u6/t6_lenz.pdf )
But if this is the case, how come current is measured when an ammeter is attached to the motor? Is this because the galvanometer is considered a load?
TIA
Because in real life, we have friction.
If we want a motor to keep spinning, we actually need a supply EMF slightly larger than the Back EMF, to provide a torque large enough to overcome friction ;D
If anyone has any unanswered questions from above, please let me know, because I can't see if anything got missed!
Post by: Cindy2k16 on October 30, 2016, 10:35:08 pm
Because in real life, we have friction.
If we want a motor to keep spinning, we actually need a supply EMF slightly larger than the Back EMF, to provide a torque large enough to overcome friction ;D
If anyone has any unanswered questions from above, please let me know, because I can't see if anything got missed!
I see thank you!
Post by: Albertenouttaten on October 30, 2016, 10:40:33 pm
1) Back emf equals supply emf, resulting in zero net current in coil ,or
2) Supply emf is slightly greater than the back emf?
Post by: JemexR on October 30, 2016, 10:42:36 pm
Post by: JemexR on October 30, 2016, 10:45:38 pm
Wait so in HSC should we say that:
1) Back emf equals supply emf, resulting in zero net current in coil ,or
2) Supply emf is slightly greater than the back emf?
Back EMF is produced due to lenz's law, and is essentially the force stopping the motor from just going on forever (conservation of energy). Hence supply EMF is greater than the back EMF; otherwise the motor won't run
Post by: Albertenouttaten on October 30, 2016, 10:46:55 pm
Quick last minute quetsion; can someone differentiate Faraday from Lenz for me please?
Faradays law: When a conductor experiences a change in flux, an Emf is induced
Lenz law (sort of like a consequence of lenz law, not sure how to word it): If the circuit is closed, the conductor gives rise to a current which creates a magnetic field to oppose the original change in flux.
So Faraday's just talks about an Emf being induced however Lenz law talks about a current and why it produces a mag field.
Post by: nimasha.w on October 30, 2016, 10:47:17 pm
Quote
For your second one to demonstrate the production of AC, this is going to be very simplified and basic, but I reckon something along the lines of moving a magnet in and out of a solenoid connected to a galvanometer should do. Unless the moderators, the actual smart people, have a genius way of producing an alternating current ;).
could you please expand on this if possible :-)
Post by: jamonwindeyer on October 30, 2016, 10:50:21 pm
Wait so in HSC should we say that:
1) Back emf equals supply emf, resulting in zero net current in coil ,or
2) Supply emf is slightly greater than the back emf?
If they specify a motor with friction, (2), if they specify an ideal motor, (1) ;D you can justify your choice either way, if it is an MC they will alert you to which ;D
Post by: jamonwindeyer on October 30, 2016, 10:53:57 pm
could you please expand on this if possible :-)
If you move a magnet near a solenoid, you expose the solenoid to a changing magnetic flux. This generates an induced EMF (and current) according to Faraday's Law. Moving the magnet in, then out, generates current in opposite directions each time. You've generated AC ;D
Post by: proficles on October 30, 2016, 11:25:22 pm
thanks in advance :)
Post by: $Billi$ on October 30, 2016, 11:33:22 pm
Post by: jamonwindeyer on October 30, 2016, 11:36:59 pm
Hey man can you quickly explain which one is T0 and which one is tv in time dilation formula? Appreciate it bro
Think of T0 as the observer, (o=observer), the thing that ISN'T moving at relativistic speed. Then tv is moving at speed.
Remember though, we could say "An astronaut on a ship moving at 0.9c measures a time of 16 days on his ship. How long is that on earth?"
In that scenario, we have to let the astronaut be the observer (t0), because he's measured it in his own reference frame. Thus, the earth is actually the thing moving past HIM, that's the idea of relativity :)
Try to get a sense of whether your time should get bigger or smaller; that should help prevent errors! :)
Post by: $Billi$ on October 30, 2016, 11:39:54 pm
Think of T0 as the observer, (o=observer), the thing that ISN'T moving at relativistic speed. Then tv is moving at speed.
Remember though, we could say "An astronaut on a ship moving at 0.9c measures a time of 16 days on his ship. How long is that on earth?"
In that scenario, we have to let the astronaut be the observer (t0), because he's measured it in his own reference frame. Thus, the earth is actually the thing moving past HIM, that's the idea of relativity :)
Try to get a sense of whether your time should get bigger or smaller; that should help prevent errors! :)
You're a life saver 🔑!! Thanks alot
Post by: nancy_cc on October 31, 2016, 02:05:14 am
For Einstein's contribution to quantum theory, am I missing anything?
- extended quantum theory to all forms of EMR not just black body radiation
- proposed light interacted with matter as photons (and hence proposed the photoelectric effect)
(This is for the ideas to implementation topic )
Hey i think you might have been missed! If you want to look at Jamon's guide to the photoelectric effect I've attached a link to the forum below, you can check what you have against what he said :)
Physics: A Complete Guide to the Course!
Post by: jamonwindeyer on October 31, 2016, 10:42:43 am
Hey i think you might have been missed! If you want to look at Jamon's guide to the photoelectric effect I've attached a link to the forum below, you can check what you have against what he said :)
Physics: A Complete Guide to the Course!
Legend Nancy! Thanks for the help! ;D
Post by: nancy_cc on October 31, 2016, 01:33:06 pm
Legend Nancy! Thanks for the help! ;D
Hahaha no problem!! :)
Post by: Yasminpotts1105 on November 02, 2016, 06:39:39 pm
Are the only forces I need gravity and tension in the string?
Post by: jakesilove on November 02, 2016, 06:41:28 pm
"Draw all forces acting on the mass of a pendulum, when its velocity is zero, but within a swing."
Are the only forces I need gravity and tension in the string?
Yep! Those are the only two relevant forces :)
Post by: Yasminpotts1105 on November 02, 2016, 07:15:08 pm
So far I just have information about blackouts and red-outs and understand about the g-forces but I am not sure what else I need to say.
Post by: jakesilove on November 02, 2016, 07:19:46 pm
"Discuss safety requirements for astronauts in regard to low gravitational forces in space."
So far I just have information about blackouts and red-outs and understand about the g-forces but I am not sure what else I need to say.
That's probably plenty! If you want to discuss Astronauts whilst in space, think about the need for them to exercise so as not to lose too much muscle mass. But honestly, it sounds like you've got the gist of the dotpoint!
Post by: teapancakes08 on November 03, 2016, 10:15:07 pm
An ‘extreme’ cyclist wants to perform a stunt in which he rides up a ramp, launching himself into the air, then flies through a hoop and lands on another ramp. The angle of each ramp is 30.0° and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 km/h. Calculate:
(a) the maximum height above the ground that the lower edge of the hoop could be placed
(b) how far away the landing ramp should be placed.
I think I can work out the distance okay, but the finding of the max. height concerning the lowest edge of the hoop throws me off a bit. I might be overthinking this though. Does the question just require the usual equations or is there more math to it? If anyone can help me out, thank you so much!
Post by: jamonwindeyer on November 03, 2016, 10:28:35 pm
Having trouble with this question:
An ‘extreme’ cyclist wants to perform a stunt in which he rides up a ramp, launching himself into the air, then flies through a hoop and lands on another ramp. The angle of each ramp is 30.0° and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 km/h. Calculate:
(a) the maximum height above the ground that the lower edge of the hoop could be placed
(b) how far away the landing ramp should be placed.
I think I can work out the distance okay, but the finding of the max. height concerning the lowest edge of the hoop throws me off a bit. I might be overthinking this though. Does the question just require the usual equations or is there more math to it? If anyone can help me out, thank you so much!
Hey there! I'm having a bit of trouble picturing this; do you interpret this as the ramp being 1.5 metres high at the point where the cyclist leaves the ramp? That's how I interpret it, let me know if you think the same.
In that case, just the normal math!! ;D I'm super happy to show you the steps, but just in case you wanted to have a go yourself first, it would just be:
1- Resolve the velocity into horizontal and vertical components
2- Use \(v=u+at\) to find when it reaches the maximum height
3- Use \(\Delta y=ut-\frac{1}{2}at^2\) to find the maximum height itself
Then remember to add 1.5 metres to that number, because we started that high off the ground to begin with :)
If you are having trouble let me know and I'll show you the steps! ;D
Post by: teapancakes08 on November 03, 2016, 11:15:29 pm
Hey there! I'm having a bit of trouble picturing this; do you interpret this as the ramp being 1.5 metres high at the point where the cyclist leaves the ramp? That's how I interpret it, let me know if you think the same.
In that case, just the normal math!! ;D I'm super happy to show you the steps, but just in case you wanted to have a go yourself first, it would just be:
1- Resolve the velocity into horizontal and vertical components
2- Use \(v=u+at\) to find when it reaches the maximum height
3- Use \(\Delta y=ut-\frac{1}{2}at^2\) to find the maximum height itself
Then remember to add 1.5 metres to that number, because we started that high off the ground to begin with :)
If you are having trouble let me know and I'll show you the steps! ;D
Seems to be the case. The answers I got were (a) 2.39m (2.d.p) and (b) 6.14 (2.d.p), which also happen to be the answers when checking the back of the textbook. Thanks for helping me out ;D
Post by: jamonwindeyer on November 03, 2016, 11:23:35 pm
Seems to be the case. The answers I got were (a) 2.39m (2.d.p) and (b) 6.14 (2.d.p), which also happen to be the answers when checking the back of the textbook. Thanks for helping me out ;D
Cool beans!! Not a worry at all :)
Post by: katnisschung on November 05, 2016, 07:23:27 pm
2) A projectile is fired from the top of a 120m high cliff at 25m/s
it lands on the ground 6.4 s after firing
find
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight
thanks!
Post by: jakesilove on November 05, 2016, 07:43:29 pm
stuck again :P
2) A projectile is fired from the top of a 120m high cliff at 25m/s
it lands on the ground 6.4 s after firing
find
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight
thanks!
Hey! This is a behemoth of a question; I've attached my solutions, but I totally understand if anything isn't clear, so if you need clarification just ask!
(http://i.imgur.com/z0l5QEE.jpg?1)
Post by: katnisschung on November 05, 2016, 07:44:03 pm
Post by: katnisschung on November 05, 2016, 07:50:17 pm
but why is change in y considered -120m?
because i keep on picturing the distance from
the peak of the projectile to the bottom
so wouldn't it be >120?
Post by: jakesilove on November 05, 2016, 07:57:20 pm
i can see that the answers are correct
but why is change in y considered -120m?
because i keep on picturing the distance from
the peak of the projectile to the bottom
so wouldn't it be >120?
Basically, you just have to make sure you define a direction, and stick with it. I made down negative, and up positive. That meant that acceleration due to gravity was negative (as it pushes a particle DOWN), and that distance was negative if it traveled in the downwards direction!
Post by: katnisschung on November 05, 2016, 08:09:13 pm
i understand that its negative
let me rephrase
wouldnt change in y be
<-120m becos the projectile travels above the building
then to the ground?
sorry if this isn't clear
Post by: jakesilove on November 05, 2016, 08:29:10 pm
hi jake
i understand that its negative
let me rephrase
wouldnt change in y be
<-120m becos the projectile travels above the building
then to the ground?
sorry if this isn't clear
Ah I see! Remember that the time I used was the 6.4 seconds. The equation asks for the total displacement, at a particular time. If I throw a projectile up, it lands on the ground, the delta y value when it hits the ground is zero! That's just want delta y means; final y minus initial y
Post by: Swagadaktal on November 05, 2016, 10:44:41 pm
With a motor if it is at an angle (say 30 degrees to the vertical) - do you use the formula flux = B*A*cos(thetre) or B*A*sin(thetre) --- if this makes any sense?
Another Q: I'll try explain the scenario if it doesnt work I'll scour the interwebs for a diagram to illustrate my question.
Say there's a generator, with a coil situated within 2 magnets, with the N to the left and S to the right (so b field going to the right), and the coil is currently in its horizontal position, with the 4 corners (where bottom left, top left, top right to bottom right corners are labelled A, B , C and D respectively). And the question asks which way must the coil turn if you want the current to travel through C to D,
how would you answer this question in terms of magnetic flux? And is it appropriate to say that Lenz law informs us that a current will be induced to oppose the motion?
Soz i know this is really convoluted - if you manage to answer this ty xoxo <3 but if you need any further clarification im more than happy to give u guys a better question :D
Post by: jamonwindeyer on November 05, 2016, 11:22:55 pm
Hey guys, tbh only posting here coz i ceebs going to vce and this squad is bomb af soo
With a motor if it is at an angle (say 30 degrees to the vertical) - do you use the formula flux = B*A*cos(thetre) or B*A*sin(thetre) --- if this makes any sense?
Another Q: I'll try explain the scenario if it doesnt work I'll scour the interwebs for a diagram to illustrate my question.
Say there's a generator, with a coil situated within 2 magnets, with the N to the left and S to the right (so b field going to the right), and the coil is currently in its horizontal position, with the 4 corners (where bottom left, top left, top right to bottom right corners are labelled A, B , C and D respectively). And the question asks which way must the coil turn if you want the current to travel through C to D,
how would you answer this question in terms of magnetic flux? And is it appropriate to say that Lenz law informs us that a current will be induced to oppose the motion?
Soz i know this is really convoluted - if you manage to answer this ty xoxo <3 but if you need any further clarification im more than happy to give u guys a better question :D
Hey Swag!
For the first one, I believe you'd use the cosine version, in the HSC we calculate the torque for a motor as \(\tau=BAIn\cos{\theta}\), so I think you're extracting terms from there to calculate the flux in the coil? That would make the most sense to me :)
For your second one, you are spot on the money about Lenz's Law. So it's best to rejig the question a bit, and hopefully it makes sense given Lenz's Law why we'd do this:
If we pretend we are dealing with a motor and send a current through C to D, which way would the coil turn? The generator would need to be turned the OPPOSITE way to that.
The reason for this is (roughly) that when you turn the coil for the generator, the current is induced to oppose it. That current acts to oppose the motion, and so generates a torque in the opposite direction to your turn. So, figure out the direction of the OPPOSING TORQUE like you would a motor, then reverse it ;D
Incidentally, doing this (if I picture your diagram correctly), the answer should be clockwise? C to D would rotate the coil anticlockwise, so we must have turned the coil clockwise to generate that ;D
Post by: Swagadaktal on November 06, 2016, 10:15:52 am
Hey Swag!OOH yeah I understood the reasoning but was having trouble putting it into words properly. Thanks Jamon <3
For the first one, I believe you'd use the cosine version, in the HSC we calculate the torque for a motor as \(\tau=BAIn\cos{\theta}\), so I think you're extracting terms from there to calculate the flux in the coil? That would make the most sense to me :)
For your second one, you are spot on the money about Lenz's Law. So it's best to rejig the question a bit, and hopefully it makes sense given Lenz's Law why we'd do this:
If we pretend we are dealing with a motor and send a current through C to D, which way would the coil turn? The generator would need to be turned the OPPOSITE way to that.
The reason for this is (roughly) that when you turn the coil for the generator, the current is induced to oppose it. That current acts to oppose the motion, and so generates a torque in the opposite direction to your turn. So, figure out the direction of the OPPOSING TORQUE like you would a motor, then reverse it ;D
Incidentally, doing this (if I picture your diagram correctly), the answer should be clockwise? C to D would rotate the coil anticlockwise, so we must have turned the coil clockwise to generate that ;D
- Can't say that it acts like a reverse motor in my exam but I'll include that it must oppose the motion, and using RH slap rule we would want the force being down so generator must be moving up or w/e
- and with my first question, I was asking about which angle we used in the formula. Do we use the angle to the horizontal or angle to the vertical? ty xo
Post by: jamonwindeyer on November 06, 2016, 10:49:33 am
OOH yeah I understood the reasoning but was having trouble putting it into words properly. Thanks Jamon <3
- Can't say that it acts like a reverse motor in my exam but I'll include that it must oppose the motion, and using RH grip we would want the force being down so generator must be moving up or w/e
- and with my first question, I was asking about which angle we used in the formula. Do we use the angle to the horizontal or angle to the vertical? ty xo
Ohh right, angle to the horizontal! Or more specifically, the angle that is formed by the plane of the coil and the field lines ;D
Post by: katnisschung on November 06, 2016, 08:51:23 pm
4) a canon ball is fired at 50m/s from the top of a 200m high cliff so that maximum
range is achieved (presumably delta x... don't know what they mean)
what is the initial vertical velocity
Post by: jakesilove on November 06, 2016, 10:07:21 pm
i got it wrong again :'(
4) a canon ball is fired at 50m/s from the top of a 200m high cliff so that maximum
range is achieved (presumably delta x... don't know what they mean)
what is the initial vertical velocity
For this, you just need to know that the maximum range will occur when theta equals 45 degrees. Then, construct a triangle, and find the vertical velocity!
Post by: jamonwindeyer on November 06, 2016, 10:12:39 pm
For this, you just need to know that the maximum range will occur when theta equals 45 degrees. Then, construct a triangle, and find the vertical velocity!
I did this, but I think the fact that it is on a cliff changes that fact! It will definitely travel further at shallower angles; I'm not sure if I can find a way to tackle this in a way that isn't mathematically rigorous? Seems very MX1 style, where's the question from Katniss? :)
*someone else feel free to have a shot, but I got a few lines in and just thought that it seemed a bit too mathematically rigorous for a HSC Physics problem...
Post by: RuiAce on November 06, 2016, 10:18:08 pm
I did this, but I think the fact that it is on a cliff changes that fact! It will definitely travel further at shallower angles; I'm not sure if I can find a way to tackle this in a way that isn't mathematically rigorous? Seems very MX1 style, where's the question from Katniss? :)Hmm. There are ways of translating an MX1 approach into a physics approach (but I'm a bit too lazy to do the question though so only if you want to put it down aha)
*someone else feel free to have a shot, but I got a few lines in and just thought that it seemed a bit too mathematically rigorous for a HSC Physics problem...
Post by: jamonwindeyer on November 06, 2016, 10:23:47 pm
Hmm. There are ways of translating an MX1 approach into a physics approach (but I'm a bit too lazy to do the question though so only if you want to put it down aha)
Oh I definitely agree, but it gets to a point where it's like "Nah, this doesn't seem right."
Maybe the question is just really bloody tough aha :P but much more likely is I'm missing some approach, because I can't see how to do it without Calculus :P
Post by: katnisschung on November 07, 2016, 05:54:02 pm
the question is from surfing (our teacher likes to bombard us with a lot of q
from it problem is the answers are quite frequently wrong)
the answers says that theta is 45 degrees so maybe they ignored the fact it was on a cliff.
question jake said "the maximum range will occur when theta equals 45 degrees."
im a little confused is this absolute for all types of questions?
Post by: jamonwindeyer on November 07, 2016, 06:37:48 pm
Thanks Jamon and Jake
the question is from surfing (our teacher likes to bombard us with a lot of q
from it problem is the answers are quite frequently wrong)
the answers says that theta is 45 degrees so maybe they ignored the fact it was on a cliff.
question jake said "the maximum range will occur when theta equals 45 degrees."
im a little confused is this absolute for all types of questions?
Hey Katniss! Okay, that's interesting! If it is on a cliff, then it's not an absolute; you can show that you'll have longer ranges for lower angles. You can use this to visualise the scenario! So I think that's another wrong answer to add to the list :P
If you are on flat ground, then yes! Maximum range will always occur (for a given velocity) if you launch the projectile at 45 degrees to the horizontal! :)
Post by: katnisschung on November 09, 2016, 09:16:41 pm
sighs projectiles....
a canon ball is fired at 40 degrees to the horizontal from the top of a 218.7m
cliff and hits a target 300m from the base of the cliff
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight
Post by: jakesilove on November 10, 2016, 12:24:18 pm
i need help again
sighs projectiles....
a canon ball is fired at 40 degrees to the horizontal from the top of a 218.7m
cliff and hits a target 300m from the base of the cliff
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight
Hey! Check out the working I supplied here, and try taking a similar approach! If you're still struggling, show us your working out, and we can talk you through it :)
Post by: katnisschung on November 11, 2016, 09:05:57 pm
Hey! Check out the working I supplied here, and try taking a similar approach! If you're still struggling, show us your working out, and we can talk you through it :)
hi Jake
yeah i did do it just need confirmation for the time.
seems the answers may be wrong again.
my teacher is giving us the old version of surfing which has quite a few mistakes
Post by: RuiAce on November 11, 2016, 09:23:56 pm
hi JakeFor a virtually identical question, if you want checking you should post up your own solutions to be looked at
yeah i did do it just need confirmation for the time.
seems the answers may be wrong again.
my teacher is giving us the old version of surfing which has quite a few mistakes
Post by: katnisschung on November 14, 2016, 04:45:18 pm
For a virtually identical question, if you want checking you should post up your own solutions to be looked at
hi could somebody confirm my answers for ux, uy and t
A cannon ball is fired at 40 degrees to the horizontal fro mthe top of a 218.7 m cliff and hits
a taret 300m from the base of the cliff
surfing says the answers are
ux=30.64 ms-1
uy=25.7ms-1
t=9.8 secs
ignore the massive cross through my answer :P
Post by: and1_98 on November 14, 2016, 06:09:54 pm
hi could somebody confirm my answers for ux, uy and t
A cannon ball is fired at 40 degrees to the horizontal fro mthe top of a 218.7 m cliff and hits
a taret 300m from the base of the cliff
surfing says the answers are
ux=30.64 ms-1
uy=25.7ms-1
t=9.8 secs
ignore the massive cross through my answer :P
Think the answers from surfing are correct.
delta(x) = vtcos(40) = 300 - so, by rearranging, t = 300/(vcos(40)) when the range is achieved (when the ball hits the ground).
Then, sub this value of t into the the quadratic for delta(y) of form - delta(y) = -1/2at^2 + vsin(40) + 218.7
Allowing this quadratic to be = to 0 (as, when the ball has achieved the range of 300m, the vertical displacement of the projectile will be 0 (or -218.7m; depending on how you set up the quadratic initially)) proceed to solve for 'v'.
I had v^2 = 1597.481175... so v = 39.96850228
so uy = vsin(40) = 39.97 x sin(40) = 25.69 m/sec
and ux = vcos(40) = 39.97 x cos(40) = 30.62 m/sec
Then, for the time of flight, I just subbed the aforementioned 'v' back into the initial quadratic for delta(y), letting it = 0 (i.e. delta(y) = 0 since when the time of flight is achieved, the projectile reaches the ground); solving it with the quadratic formula.
You should get two values for t (the positive value being the one you want as t>0), one of them being 9.798270249... secs.
Hope this helps.
Post by: jakesilove on November 14, 2016, 06:20:41 pm
Think the answers from surfing are correct.
delta(x) = vtcos(40) = 300 - so, by rearranging, t = 300/(vcos(40)) when the range is achieved (when the ball hits the ground).
Then, sub this value of t into the the quadratic for delta(y) of form - delta(y) = -1/2at^2 + vsin(40) + 218.7
Allowing this quadratic to be = to 0 (as, when the ball has achieved the range of 300m, the vertical displacement of the projectile will be 0 (or -218.7m; depending on how you set up the quadratic initially)) proceed to solve for 'v'.
I had v^2 = 1597.481175... so v = 39.96850228
so uy = vsin(40) = 39.97 x sin(40) = 25.69 m/sec
and ux = vcos(40) = 39.97 x cos(40) = 30.62 m/sec
Then, for the time of flight, I just subbed the aforementioned 'v' back into the initial quadratic for delta(y), letting it = 0 (i.e. delta(y) = 0 since when the time of flight is achieved, the projectile reaches the ground); solving it with the quadratic formula.
You should get two values for t (the positive value being the one you want as t>0), one of them being 9.798270249... secs.
Hope this helps.
Looks great to me! Thanks for the answer!
Post by: bholenath125 on November 15, 2016, 05:16:17 pm
Post by: RuiAce on November 15, 2016, 05:22:50 pm
Guys i have a lot of trouble with Projectile motion is there someone who could help me outWell what's the primary concern? What exactly troubles you with projectiles?
Some of Jamon's guides in the meantime
Post by: thataveragevegan on November 15, 2016, 06:44:13 pm
9. Explain the reason for the selection of infinity
as the place of zero gravitational potential
energy.
10. Explain how this selection of zero level results
in any point within a gravitational field having
a negative gravitational potential energy.
Post by: jamonwindeyer on November 15, 2016, 07:12:09 pm
So I was wondering if i could get a typical exam response style for these questions? My teachers have verbally explained it in class but i was looking for a specific exam response, so i can compare it to my response.
Hey there! I'll give you how I'd answer these (very rough, by no means Gospel). These are also a little different to what you'd normally get in an exam style scenario! Especially the second one :)
Explain the reason for the selection of infinity as the place of zero gravitational potential energy.
Gravitational fields are infinite in size, and all objects inside a gravitational field, by definition, have GPE. Since they are infinitely large, to have a point with zero GPE, we need to be infinitely far away. Thus, we choose infinity as the place of zero GPE. This is also more convenient on an cosmological scale than using the surface of a planet as the zero point.
Explain how this selection of zero level results in any point within a gravitational field having a negative gravitational potential energy.
It is convention for GPE to increase as our distance from the centre of the gravitational field increases. However, since we have chosen zero as our infinity point, we would need to be increasing towards zero. Therefore, every point within a gravitational field must have a negative GPE, which approaches zero with increasing distance from the centre of the field.
I hope these help! Again, definitely not Gospel, there are many ways to answer questions like this. You could use dot points, or structure it differently, but if I had those questions in front of me that is how I'd respond :)
Post by: thataveragevegan on November 15, 2016, 07:17:18 pm
Hey there! I'll give you how I'd answer these (very rough, by no means Gospel). These are also a little different to what you'd normally get in an exam style scenario! Especially the second one :)
Explain the reason for the selection of infinity as the place of zero gravitational potential energy.
Gravitational fields are infinite in size, and all objects inside a gravitational field, by definition, have GPE. Since they are infinitely large, to have a point with zero GPE, we need to be infinitely far away. Thus, we choose infinity as the place of zero GPE. This is also more convenient on an cosmological scale than using the surface of a planet as the zero point.
Explain how this selection of zero level results in any point within a gravitational field having a negative gravitational potential energy.
It is convention for GPE to increase as our distance from the centre of the gravitational field increases. However, since we have chosen zero as our infinity point, we would need to be increasing towards zero. Therefore, every point within a gravitational field must have a negative GPE, which approaches zero with increasing distance from the centre of the field.
I hope these help! Again, definitely not Gospel, there are many ways to answer questions like this. You could use dot points, or structure it differently, but if I had those questions in front of me that is how I'd respond :)
Thanks heaps!
Post by: Yasminpotts1105 on November 15, 2016, 08:14:50 pm
- They are useful in making predictions.
- They are concepts that lack an experimental basis.
- They are ideas that can't be accepted until they have been tested.
Post by: FallonXay on November 15, 2016, 08:42:25 pm
Which of the following is a true statement about scientific theories, such as Einstein's special theory of relativity?
- They are useful in making predictions.
- They are concepts that lack an experimental basis.
- They are ideas that can't be accepted until they have been tested.
Hello!
"They are concepts that lack and experimental basis" is clearly wrong as Einstein used numerous forms of thought experiments to explain his special theory of relativity i.e looking at a mirror on a train travelling at the speed of light. Furthermore, Einstein also used experiments such as the Michelson-Morley experiment part of the evidence to support his theory. (Remember: The null result of the Michelson-Morley experiment didn't disprove the aether but was useful as partial evidence in supporting Einstein's theory)
"They are ideas that can't be accepted until they have been tested" This is clearly false as Einstein's Special Theory of Relativity was an accepted model before physical tests such as the Hafele-Keating experiment or muon decay in particle accelerators (which exhibit relativistic effects) were applied.
"They are useful in making predictions" is the correct answer as Einstein's theory of special relativity revolves around the constancy of light predicted the relativistic effects (time dilation, mass increase, length contraction). This, although could not be tested at the time, through numerous thought experiments predicted the effects of the constancy of light.
Hope this helped :)
Post by: jamonwindeyer on November 16, 2016, 12:53:01 pm
Hello!
"They are concepts that lack and experimental basis" is clearly wrong as Einstein used numerous forms of thought experiments to explain his special theory of relativity i.e looking at a mirror on a train travelling at the speed of light. Furthermore, Einstein also used experiments such as the Michelson-Morley experiment part of the evidence to support his theory. (Remember: The null result of the Michelson-Morley experiment didn't disprove the aether but was useful as partial evidence in supporting Einstein's theory)
"They are ideas that can't be accepted until they have been tested" This is clearly false as Einstein's Special Theory of Relativity was an accepted model before physical tests such as the Hafele-Keating experiment or muon decay in particle accelerators (which exhibit relativistic effects) were applied.
"They are useful in making predictions" is the correct answer as Einstein's theory of special relativity revolves around the constancy of light predicted the relativistic effects (time dilation, mass increase, length contraction). This, although could not be tested at the time, through numerous thought experiments predicted the effects of the constancy of light.
Hope this helped :)
Legend, thanks for the awesome answer ;D
Post by: thataveragevegan on November 16, 2016, 04:39:19 pm
In my physics notes that i bought from ATAR Notes Newtown's law of Universal Gravitation is defined as
" Newtown's Law of Universal Gravitation determines the gravitational force acing on two objects due to their gravitational attraction.
G is the gravitational constant, m1 and m2 are the two masses and d is the distance between them. Remember this force acts on both objects . The less the mss of the object, the more it is affected. (F=ma).
What exactly do you guys mean by "the more it is affected", affected by what exactly ?
Thanks :)
Post by: jamonwindeyer on November 16, 2016, 08:41:04 pm
hey so i was wondering if i could get some clarification from hopefully Jake but any ATAR notes lecturer will do
In my physics notes that i bought from ATAR Notes Newtown's law of Universal Gravitation is defined as
" Newtown's Law of Universal Gravitation determines the gravitational force acing on two objects due to their gravitational attraction.
G is the gravitational constant, m1 and m2 are the two masses and d is the distance between them. Remember this force acts on both objects . The less the mss of the object, the more it is affected. (F=ma).
What exactly do you guys mean by "the more it is affected", affected by what exactly ?
Thanks :)
Hi! I wrote those notes, so I'd say I'm pretty qualified ;)
So that statement refers to the fact that it is affected more by the gravitational force. In this case, I mean that it will accelerate more.
According to \(F=ma\), because the force is the same, larger masses will accelerate less when experiencing the same force ;D
Reading those sentences in isolation like that, I agree it's a little ambiguous! The notes will be getting a little tidy up soon, so I'll actually make a note of adding a little extra to that bit :) hope this helps!
Post by: Yasminpotts1105 on November 18, 2016, 01:35:03 pm
Post by: RuiAce on November 18, 2016, 01:47:32 pm
Could you please show worked solutions so we know how to actually get the answer, after trying for ages?Refer to posts #990 and #991
Post by: Wales on November 18, 2016, 08:56:05 pm
So throughout the year I've been struggling quite hard in Physics. I've been recommended to drop by my teacher multiple times and yesterday I just received my results for the pendulum test. My result was less then satisfying with a mere 40% in last weighing 20% overall. Should I continue my investment in Physics? Will the failure of this test affect my overall ranking that hard? I came last place.
Is the upcoming physics module extremely challenging? I'm currently on 13 units and enjoying Physics but just can't seem to get it well enough. If I drop I will be left with 11 units. Would it be worth continuing Physics if my mark won't be that great? My teacher tells me the time I need to spend would be far too much and it would be better to drop it and focus on my other subjects. Will post them if needed.
Regards, Wales
Post by: RuiAce on November 18, 2016, 09:07:45 pm
Not sure if this is the appropriate place to post, let me know if it isn't :)The question then becomes what drives you to enjoy physics despite its apparent difficulty?
So throughout the year I've been struggling quite hard in Physics. I've been recommended to drop by my teacher multiple times and yesterday I just received my results for the pendulum test. My result was less then satisfying with a mere 40% in last weighing 20% overall. Should I continue my investment in Physics? Will the failure of this test affect my overall ranking that hard? I came last place.
Is the upcoming physics module extremely challenging? I'm currently on 13 units and enjoying Physics but just can't seem to get it well enough. If I drop I will be left with 11 units. Would it be worth continuing Physics if my mark won't be that great? My teacher tells me the time I need to spend would be far too much and it would be better to drop it and focus on my other subjects. Will post them if needed.
Regards, Wales
Most people are often put off by the difficulty of something (relative to them). The HSC is about doing what you like, however in general (not always) this implies that you do have some kind of a talent for the course.
When teachers suggest you to drop, they're probably right. Teachers cannot ever force you to drop, however if they're providing pressure then it's probably for the better. Of course, that pressure can be rejected though; I know people who have done so.
So the main question is, what is pushing your like for physics? What is this powerful force that pushes for you to not drop? Or is there something about the course itself or how it's taught that you like?
Then, contrast that to both the how and why behind your marks. What do you put into physics, and what do you suspect is causing it to not pay off? A good starting point would be to identify where the marks were lost first, and what you could've done instead.
(Units wise: I did 10 units, yet I know people who did 16. That's your own call.)
Of course, note that it also depends on your performance in physics relative to your performance in everything else as well. If there's some subject creating even more difficulty than this then that should probably be addressed first.
Post by: samuels1999 on November 18, 2016, 10:09:13 pm
I have two questions regarding Mass Dilation in Special Relativity:
Firstly, Does Mass Dilation have anything to do with E=mc^2?
Secondly, When mass increases...according to the definition of mass...there should be more matter in the object. I just want to clarify.. do individual atoms gain more mass? or is there more matter in the object all together at high speeds?
Sorry part 2 is terribly worded.
Thanks
Samuel
Post by: jamonwindeyer on November 18, 2016, 11:11:07 pm
Hi Guys,
I have two questions regarding Mass Dilation in Special Relativity:
Firstly, Does Mass Dilation have anything to do with E=mc^2?
Secondly, When mass increases...according to the definition of mass...there should be more matter in the object. I just want to clarify.. do individual atoms gain more mass? or is there more matter in the object all together at high speeds?
Sorry part 2 is terribly worded.
Thanks
Samuel
Hey Sam!
1. Yes. Consider an object travelling super close to the speed of light. If we keep doing work on that object, it will keep accelerating, and thus eventually reach the speed of light. What if we then did more work on the object? Well, its kinetic energy must increase, we are doing work after all. But the speed cannot exceed the speed of light. Since \(KE=\frac{1}{2}mv^2\), the only option is for the mass to increase. Work (energy) has been converted into mass (!!), and the ratio is given by \(E=mc^2\) ;D
2. The issue here is the difference between invariant mass and relativistic mass. Invariant mass is our typical understanding of mass, the amount of matter in the object. Relativistic mass is to do with the velocity of the observer, it is different and we don't look at it in the same way. So to answer your question, there are no atoms added to our object. We measure the same amount of atoms to be heavier.
Note, the terminology in my second answer is not assessable, and neither is a super precise explanation of the phenomena ;D
Post by: jakesilove on November 19, 2016, 12:22:19 pm
Hey Sam!
1. Yes. Consider an object travelling super close to the speed of light. If we keep doing work on that object, it will keep accelerating, and thus eventually reach the speed of light. What if we then did more work on the object? Well, its kinetic energy must increase, we are doing work after all. But the speed cannot exceed the speed of light. Since \(KE=\frac{1}{2}mv^2\), the only option is for the mass to increase. Work (energy) has been converted into mass (!!), and the ratio is given by \(E=mc^2\) ;D
2. The issue here is the difference between invariant mass and relativistic mass. Invariant mass is our typical understanding of mass, the amount of matter in the object. Relativistic mass is to do with the velocity of the observer, it is different and we don't look at it in the same way. So to answer your question, there are no atoms added to our object. We measure the same amount of atoms to be heavier.
Note, the terminology in my second answer is not assessable, and neither is a super precise explanation of the phenomena ;D
Just to add to your answer to the second question; I think this basically brings to light the typical terminological issue that arises when talking about something like mass. Mass is a descriptor; something has a mass of X. That describes the object, at a particular point in time and space. When something moves at high speeds, it gains mass, because energy is ABSOLUTELY EQUIVALENT to mass. That's what E=mc^2 means; as energy increases, so too does mass. Why do higher speeds cause greater mass? Because energy is mass, so obviously more energy means more mass. It really is a very complicated area of physics, made even more complicated by terminology like 'relativistic mass' etc. If you're interested, check out a great video here. However, to reiterate Jamon's point, none of this is important in your HSC!
Post by: katnisschung on November 19, 2016, 01:05:41 pm
Think the answers from surfing are correct.
delta(x) = vtcos(40) = 300 - so, by rearranging, t = 300/(vcos(40)) when the range is achieved (when the ball hits the ground).
Then, sub this value of t into the the quadratic for delta(y) of form - delta(y) = -1/2at^2 + vsin(40) + 218.7
Allowing this quadratic to be = to 0 (as, when the ball has achieved the range of 300m, the vertical displacement of the projectile will be 0 (or -218.7m; depending on how you set up the quadratic initially)) proceed to solve for 'v'.
I had v^2 = 1597.481175... so v = 39.96850228
so uy = vsin(40) = 39.97 x sin(40) = 25.69 m/sec
and ux = vcos(40) = 39.97 x cos(40) = 30.62 m/sec
Then, for the time of flight, I just subbed the aforementioned 'v' back into the initial quadratic for delta(y), letting it = 0 (i.e. delta(y) = 0 since when the time of flight is achieved, the projectile reaches the ground); solving it with the quadratic formula.
You should get two values for t (the positive value being the one you want as t>0), one of them being 9.798270249... secs.
Hope this helps.
thanks and.....
i'm a little confused...
at the beginning why did u take delta(x) = vtcos(40) = 300 instead of
delta(x) = utcos(40) = 300
Post by: RuiAce on November 19, 2016, 01:23:57 pm
thanks and.....Probably was a typo but it doesn't matter here (fortunately).
i'm a little confused...
at the beginning why did u take delta(x) = vtcos(40) = 300 instead of
delta(x) = utcos(40) = 300
Note that ux=vx
Post by: katnisschung on November 19, 2016, 01:40:28 pm
(im really bad at maths and the quadratics are confusing the hell out of me)
(this is in reference to a question i posted ages answered by and)
Post by: jamonwindeyer on November 19, 2016, 02:37:55 pm
haha can somebody explain the steps to get u
(im really bad at maths and the quadratics are confusing the hell out of me)
(this is in reference to a question i posted ages answered by and)
Hey Katniss! I'll start from this line:
So, we now know how long the projectile was in the air. We also know that AFTER this amount of time, its vertical displacement is 218.7 metres below the origin. So we use \(\Delta y=u_yt+\frac{1}{2}at^2\) from our reference sheet, where \(u_y=u\sin{40}\):
Notice that going into that second line, the \(u\) on the top and bottom of that first fraction cancel out! So the excellent thing about this quadratic is that there is no formula required! Keep rearranging to find:
I know it looks bad, but it is just moving things around in your expression! This formula will let you get an expression of the form \(\frac{1}{u^2}=??\), and then you just rearrange this to get your final answer! :)
If you are having trouble visualising how the algebra itself takes place, try replacing the \(\sin{40}\) and \(\cos{40}\) with numbers earlier in the problem and using a rounded version!
Using the numbers instead, and simplifying and rounding as you go, may help you! It's not as correct as using the real thing all the way through, but it's close! :)
Post by: katnisschung on November 19, 2016, 02:46:25 pm
Post by: Wales on November 19, 2016, 09:27:57 pm
The question then becomes what drives you to enjoy physics despite its apparent difficulty?
Most people are often put off by the difficulty of something (relative to them). The HSC is about doing what you like, however in general (not always) this implies that you do have some kind of a talent for the course.
When teachers suggest you to drop, they're probably right. Teachers cannot ever force you to drop, however if they're providing pressure then it's probably for the better. Of course, that pressure can be rejected though; I know people who have done so.
So the main question is, what is pushing your like for physics? What is this powerful force that pushes for you to not drop? Or is there something about the course itself or how it's taught that you like?
Then, contrast that to both the how and why behind your marks. What do you put into physics, and what do you suspect is causing it to not pay off? A good starting point would be to identify where the marks were lost first, and what you could've done instead.
(Units wise: I did 10 units, yet I know people who did 16. That's your own call.)
Hey Rui :) I didn't expect to see you here. I remember you being part of the accellerated classes at my school a few years back, the Math genius as most people refer to you haha.
Well, I rather enjoy Physics because of the fine detail contained within the course and understanding how, why and what allows something to work. Currently doing the Space module and I'm loving it.
I guess it comes down to whether I'm going to be dedicated enough. When it comes down to raw talent I feel like Physics just doesn't suit me. My 11 Yearly's mark was atrocious and I've only done well in the Practicals up until this point. I feel like what's causing my performance in Physics is the pace the class is going at and I feel like I'm unable to keep up a majority of the time, it's a class of 14 so it's relatively small. A big factor in why I want to drop right now is that I've already botched my Term 1 assessment worth 20% and I know that I won't be able to bring it up due to the nature of the students in the class, at least not to a satisfactory level. I'm taking Ext 2 maths as well so I've considered spending the time freed by Physics in that (Not everyone is as talented as you Rui :( )
I've got a few days to think it over so let me know you're suggestion but basically I'm struggling really hard consistently and I've been told it's not the course for me by my teacher multiple times ( Since term 2 last year ) and I feel that she is right.
Post by: RuiAce on November 19, 2016, 11:30:30 pm
Hey Rui :) I didn't expect to see you here. I remember you being part of the accellerated classes at my school a few years back, the Math genius as most people refer to you haha.Aha hey hey :) Well I'm not even close to being the smartest math person I know anymore. Nice seeing someone from that school here :)
Well, I rather enjoy Physics because of the fine detail contained within the course and understanding how, why and what allows something to work. Currently doing the Space module and I'm loving it.
I guess it comes down to whether I'm going to be dedicated enough. When it comes down to raw talent I feel like Physics just doesn't suit me. My 11 Yearly's mark was atrocious and I've only done well in the Practicals up until this point. I feel like what's causing my performance in Physics is the pace the class is going at and I feel like I'm unable to keep up a majority of the time, it's a class of 14 so it's relatively small. A big factor in why I want to drop right now is that I've already botched my Term 1 assessment worth 20% and I know that I won't be able to bring it up due to the nature of the students in the class, at least not to a satisfactory level. I'm taking Ext 2 maths as well so I've considered spending the time freed by Physics in that (Not everyone is as talented as you Rui :( )
I've got a few days to think it over so let me know you're suggestion but basically I'm struggling really hard consistently and I've been told it's not the course for me by my teacher multiple times ( Since term 2 last year ) and I feel that she is right.
Well that's a good thing to have no doubt about it; an actual interest in what you're learning. But then if you were to keep physics then this interest needs to push you on to actually ensuring you don't fall behind. It's normal for things to not come to you the very instant you're taught it, however if it takes a bit too long for that to happen then something else is up. It could possibly be that your brain isn't well versed against the concepts taught, but that I can't say for sure - only you'd be able to explain how you're falling behind
(I did put heaps of time into MX2 though. I didn't rely on talent 100%)
That being said, if you do start to feel your teacher's right you may want to just drop it then. The HSC is a game; gotta find the option with the better pay-off and judging by what you said, it might be the better choice. So I'm not going to give an instruction but my recommendation would probably be drop as well
Post by: Wales on November 19, 2016, 11:37:52 pm
Aha hey hey :) Well I'm not even close to being the smartest math person I know anymore. Nice seeing someone from that school here :)
Well that's a good thing to have no doubt about it; an actual interest in what you're learning. But then if you were to keep physics then this interest needs to push you on to actually ensuring you don't fall behind. It's normal for things to not come to you the very instant you're taught it, however if it takes a bit too long for that to happen then something else is up. It could possibly be that your brain isn't well versed against the concepts taught, but that I can't say for sure - only you'd be able to explain how you're falling behind
(I did put heaps of time into MX2 though. I didn't rely on talent 100%)
That being said, if you do start to feel your teacher's right you may want to just drop it then. The HSC is a game; gotta find the option with the better pay-off and judging by what you said, it might be the better choice. So I'm not going to give an instruction but my recommendation would probably be drop as well
I'm rather torn. I would love to pursue Physics and get a grasp of what it really is however I feel the investment is far too great for my current position. Maybe I could continue that later in Uni or something along those lines. Right now I think the better choice may be to drop it and dedicate myself to my remaining subjects. Everybody keeps telling me that the teacher cannot force me out and I agree but I think it's at that point where she essentially believes that in my best interests it would be best to drop. I will think over it again but the likely conclusion is that I will indeed drop it.
Thank you for your thoughts and insight. It is greatly appreciated :)
Regards, Wales
Post by: jamonwindeyer on November 20, 2016, 12:00:50 am
I'm rather torn. I would love to pursue Physics and get a grasp of what it really is however I feel the investment is far too great for my current position. Maybe I could continue that later in Uni or something along those lines. Right now I think the better choice may be to drop it and dedicate myself to my remaining subjects. Everybody keeps telling me that the teacher cannot force me out and I agree but I think it's at that point where she essentially believes that in my best interests it would be best to drop. I will think over it again but the likely conclusion is that I will indeed drop it.
Thank you for your thoughts and insight. It is greatly appreciated :)
Regards, Wales
As a side note; HSC Physics is majorly different to what studying Physics at uni is like. At uni it is much more mathematical, a lot more rigorous, much more detailed! That appeals to a lot of people but it turns a lot of people off also ;D
HSC Physics is a fantastic course for a lot of reasons (and a bad one for a few). It's got fascinating content that gives you a cool understanding of many aspects of the world around you. But it is tough, there is no denying that.
I'm always of the belief that you should stick with what you enjoy. That said, this is the HSC, and if you are consistently falling short of your goals then that will eat away at your enthusiasm over time. If you are the kind of person who can stay optimistic and keep working hard, then I'd be keeping Physics around, perhaps even until Half Yearlies (which is the latest anyone should drop anything imo)? :) it just sounds like you really love it, and I hate seeing people drop something they love just because they aren't quite getting where they want to be. I know it sucks, but hey, the HSC isn't just about marks right? It's about learning things that excite and motivate you! :)
Just my two cents. Of course you need to look after yourself and keep your other subjects travelling nicely; only you will ultimately know what is in your best interests ;D
Post by: katnisschung on November 21, 2016, 04:51:01 pm
The line passing through a tube which is moved in such a way as to keep the first mass carrier moving in a circle
Hypothetically if the centripetal force was greater than the weight force would the pass go flying outwards?
Post by: jamonwindeyer on November 21, 2016, 05:34:39 pm
A mass carrier is attached by a length of fishing line to another mass carrier
The line passing through a tube which is moved in such a way as to keep the first mass carrier moving in a circle
Hypothetically if the centripetal force was greater than the weight force would the pass go flying outwards?
Hey Katniss! I'm having a bit of trouble picturing this scenario, is it something from a book? Do you think you could attach a diagram or maybe even draw one to help me understand it a little better? :)
Post by: katnisschung on November 21, 2016, 07:20:37 pm
Hey Katniss! I'm having a bit of trouble picturing this scenario, is it something from a book? Do you think you could attach a diagram or maybe even draw one to help me understand it a little better? :)
Post by: jamonwindeyer on November 21, 2016, 10:33:32 pm
ATTACHMENT
Awesome! Okay, I'm with you now. So the weight force of the larger mass carrier is providing the centripetal force that keeps the smaller mass carrier in uniform circular motion.
Your interpretation (as I interpret it) is 100% spot on. If the mass carrier is in motion with a radius and velocity such that the force required to maintain it is greater than the weight force, then yes, it will fly outwards. Or perhaps just slowly wobble into a wider circle to re-establish the balance? We can't tell which, but yes, the moving mass will definitely move outwards if the weight force isn't sufficient to maintain the motion :)
Post by: katnisschung on November 24, 2016, 10:55:48 am
and the notification gave very limited info regarding what we will be tested on
(it only said the unit of "space")
its worth 25% which is quite a bit, however this is only my second practical test in science.
advice on how to study and does anyone have any practice papers for pracs so i could possibly
carry it out at home? my intuition tells me it could have something to do with projectile motion
Post by: jakesilove on November 24, 2016, 01:04:10 pm
for my first hsc assessment we have a physics prac
and the notification gave very limited info regarding what we will be tested on
(it only said the unit of "space")
its worth 25% which is quite a bit, however this is only my second practical test in science.
advice on how to study and does anyone have any practice papers for pracs so i could possibly
carry it out at home? my intuition tells me it could have something to do with projectile motion
Hey!
There's actually fairly little prep you can do for a practical task like the one you've described. Be familiar with the way in which an experiment is written up (aim, hypothesis, equipment, method, results, discussion, conclusion). Understand how best to write a method (past tense, numbers etc.). Think about the importance of accuracy, reliability, and validity to an experimental design, and feel comfortable in assessing each of these components for a given experiment. That's really the only guidance I can give!
Jake
Post by: jamonwindeyer on November 24, 2016, 01:35:01 pm
for my first hsc assessment we have a physics prac
and the notification gave very limited info regarding what we will be tested on
(it only said the unit of "space")
its worth 25% which is quite a bit, however this is only my second practical test in science.
advice on how to study and does anyone have any practice papers for pracs so i could possibly
carry it out at home? my intuition tells me it could have something to do with projectile motion
Only addition I'd make to above is maybe to smash out a few projectile motion questions? Just a handful of the typical ones to make sure you know the formulae to use for specific situations ;D but yeah, there isn't a whole lot you can do for prep here. That said, if you go here you'll see a few sample assessment tasks from Hurlstone for Term 4; they could match slightly to what you'll get in your Prac! They seem like half Prac half theory though :) good luck!
Post by: FallonXay on November 24, 2016, 03:50:55 pm
for my first hsc assessment we have a physics prac
and the notification gave very limited info regarding what we will be tested on
(it only said the unit of "space")
its worth 25% which is quite a bit, however this is only my second practical test in science.
advice on how to study and does anyone have any practice papers for pracs so i could possibly
carry it out at home? my intuition tells me it could have something to do with projectile motion
My school did a practical exam for physics (we built a ramp, put it on top of a lab bench and had to work out where to place a cup on the ground so that a marble would roll in). The exam involved this practical aspect, then also a short theoretical paper/exam involving analysis of the pendulum experiment that investigates acceleration due to gravity. Unfortunately, I don't have my exam anymore to show you sample questions but things that were tested included ways to improve/check validity, reliability and accuracy and experimental write up - as Jake mentioned. Also, make sure you're able to identify independent, dependent and controlled variables. We were also tested on relevant aspects to the experiment such as Galileo's interpretation of projectile motion (i.e horizontal and vertical components are independent, the horizontal component was constant and the vertical was equal to the acceleration due to gravity, trajectory of a parabola) and there were aspects involving graphical analysis of projectile motion (and the pendulum experiment); testing skills such as interpreting (or drawing) graphs are a popular question - remember things like labelling graphs and line of best fit.
Post by: katnisschung on November 27, 2016, 11:44:18 am
i can't wrap my head around this...
so i was revising "g-forces" and my teacher was like accelerating up
is the same as decelerating down as both result in g-forces more than 1G.
but waht!!? i get the first and the second situation i get (like u feel it in an elevator) but i cannot
see the second situation mathematically...
Post by: jakesilove on November 27, 2016, 11:55:34 am
ok so probably a stupid question and im overthinking it but
i can't wrap my head around this...
so i was revising "g-forces" and my teacher was like accelerating up
is the same as decelerating down as both result in g-forces more than 1G.
but waht!!? i get the first and the second situation i get (like u feel it in an elevator) but i cannot
see the second situation mathematically...
So, if you accelerate up (ie. speed up), it's like when you're in a stationary elevator and it starts going up. There are clear g-forces, because you feel 'pressed' into the ground, and the acceleration is clearly up. How do we tell? Well, initially your velocity is zero, then it's one, then it's 3, then it's 6 etc. Clearly, your change in velocity is increasing in the upwards direction.
If you decelerate down (ie. 'slow down' when you're travelling downwards), it's like when you're in an elevator travelling downwards, and it slows down to stop. There are also clear g-forces, because you feel 'pressed' into the ground again (suggesting that the g-forces will be exactly the same!). But how can we confirm this mathematically? Well, initially you are travelling downward with a velocity of 6. We can say, then, that you're travelling at -6. Then, you're travelling at -3, then -1, then 0. Clearly, your change in velocity is increasing, also in the upward direction, because you're getting less negative!
Does that make sense?
Jake
Post by: katnisschung on November 27, 2016, 12:06:26 pm
i don't know why but i was thinking of velocity and subbing that in instead
of change of acceleration which is positive the last sentence really helped :) :) :)
Thanks!
Post by: jakesilove on November 27, 2016, 12:46:45 pm
I GOT IT!
i don't know why but i was thinking of velocity and subbing that in instead
of change of acceleration which is positive the last sentence really helped :) :) :)
Thanks!
No problem :) It's definitely a difficult concept, and was actually involved in one of the trickier multiple choice questions in the 2016 HSC paper!
Post by: katnisschung on November 28, 2016, 06:06:16 pm
so im pretty hopeless at maths and i was revising Kepler's laws
by deriving them....
i understand the law (equating centripetal force with gravitational force
and i am able to derive v=...and r^3/T^2)
But where the hell did this derivation come from....
would anyone be able to run me through the steps?
i know its probably fairly simple but like i said im terrible at maths
Post by: RuiAce on November 28, 2016, 06:15:33 pm
hi im back again....
so im pretty hopeless at maths and i was revising Kepler's laws
by deriving them....
i understand the law (equating centripetal force with gravitational force
and i am able to derive v=...and r^3/T^2)
But where the hell did this derivation come from....
would anyone be able to run me through the steps?
i know its probably fairly simple but like i said im terrible at maths
Post by: katnisschung on November 28, 2016, 06:20:14 pm
but where did the formula i attached come from?
how do you equate the period of the first planet's orbit to the
period of the second planets orbit...
hope i've expressed myself more clearly
Post by: RuiAce on November 28, 2016, 06:23:28 pm
Post by: katnisschung on November 28, 2016, 06:35:32 pm
Post by: samuels1999 on November 28, 2016, 07:15:16 pm
I need a bit of help with a question dealing with Kepler's Law. The question related to Mars and one of its moons. The question stated that the orbital period of the moon was 1.09x10^5 seconds and that the mass of mars was 7.1x10^23kg. It then asked me to work out the radius of the moon's orbit around Mars.
I know the question relates to Kepler's Law, but I am uncertain how to use it.
Thanks,
Samuel
Post by: RuiAce on November 28, 2016, 07:23:18 pm
Hi Everyone,
I need a bit of help with a question dealing with Kepler's Law. The question related to Mars and one of its moons. The question stated that the orbital period of the moon was 1.09x10^5 seconds and that the mass of mars was 7.1x10^23kg. It then asked me to work out the radius of the moon's orbit around Mars.
I know the question relates to Kepler's Law, but I am uncertain how to use it.
Thanks,
Samuel
Post by: f_tan on November 30, 2016, 09:18:25 pm
Post by: RuiAce on November 30, 2016, 09:19:29 pm
Is angular speed just the rate of rotation?Yes, however angular velocity is not a part of the course.
Post by: jamonwindeyer on November 30, 2016, 10:38:34 pm
Is angular speed just the rate of rotation?
It can come in just implicitly when talking about orbits, perhaps even harnessing the rotational motion of earth (the equator is the point of maximum orbital velocity on earth) ;D but yeah pretty irrelevant for the most part ;D
Post by: f_tan on December 01, 2016, 08:26:56 pm
What g-force is experienced by a 70 kg test pilot pulling out of a downward dive in a circular arc of radius 1000 m at 200 m/s?
A) 42 g
B) 41 g
C) 5.1 g
D) 4.1 g
Thank you! :)
Post by: jakesilove on December 01, 2016, 09:04:42 pm
How do I work out the answer for this question?
What g-force is experienced by a 70 kg test pilot pulling out of a downward dive in a circular arc of radius 1000 m at 200 m/s?
A) 42 g
B) 41 g
C) 5.1 g
D) 4.1 g
Thank you! :)
This is a seriously intense question. You need to equate net force (F=ma) with centripetal force (due to the circular arc, F=mv^2/r) which gets us
This is the 'additional' acceleration. Now, recall that the formula for working out g-forces is
Therefore, I think the answer is C. Does that all make sense?
Post by: f_tan on December 01, 2016, 09:12:14 pm
I have another question:
A spacecraft is rising from the Earth's surface at 49 m/s. At 980 m the booster rocket tanks are jettisoned. Calculate the speed with which the booster tanks crash into the Earth.
Post by: jakesilove on December 01, 2016, 09:16:44 pm
Yep, that makes sense! Thanks so much!
I have another question:
A spacecraft is rising from the Earth's surface at 49 m/s. At 980 m the booster rocket tanks are jettisoned. Calculate the speed with which the booster tanks crash into the Earth.
It's sort of difficult to figure out exactly what the questions asking, but let's make some educated guesses. We can imagine that the booster tank as a projectile, travelling upwards at 49 m/s, 980m above the earth's surface. We want to find out how fast it is going when it crashes into earth.
Turns out, we have a formula we can sub straight into!
Here, it's important that acceleration and displacement have the same sign, as they are both in the same direction.
Since that number pops out perfectly, I'm fairly certain that is the correct answer :)
Post by: katnisschung on December 03, 2016, 05:27:02 pm
(centripetal force formula)
can only be used on problems with a "small" radius correct?
Post by: RuiAce on December 03, 2016, 05:29:17 pm
Fc=mv^2/rI have never heard of this
(centripetal force formula)
can only be used on problems with a "small" radius correct?
Post by: Jakeybaby on December 03, 2016, 05:39:24 pm
Fc=mv^2/rI've only ever seen this formula used within questions regarding the spinning of an object on a string, not sure about larger radii. However, unsure about whether or not it's covered in HSC as I completed SACE this year. I can't see a large radius being used in a question though.
(centripetal force formula)
can only be used on problems with a "small" radius correct?
Post by: jamonwindeyer on December 03, 2016, 05:56:40 pm
Fc=mv^2/r
(centripetal force formula)
can only be used on problems with a "small" radius correct?
As long as the motion is uniformly circular, this formula works. Perhaps you are thinking of the fact that orbits are not actually uniformly circular in the real world, they are elliptical! That said, this formula is still used in the HSC course because it is still very accurate ;D
So, for the HSC course, you can use this formula for any body moving in a circular fashion ;D
Post by: RuiAce on December 03, 2016, 06:03:03 pm
As long as the motion is uniformly circular, this formula works. Perhaps you are thinking of the fact that orbits are not actually uniformly circular in the real world, they are elliptical! That said, this formula is still used in the HSC course because it is still very accurate ;DHey Jamon, whilst we're here :o
So, for the HSC course, you can use this formula for any body moving in a circular fashion ;D
Since the orbits are actually elliptical, does the formula r^3/T^2=GM/(4pi^2) still hold true regardless?
Post by: jamonwindeyer on December 03, 2016, 06:10:08 pm
Hey Jamon, whilst we're here :o
Since the orbits are actually elliptical, does the formula r^3/T^2=GM/(4pi^2) still hold true regardless?
I believe the fact that it is elliptical means that it is more accurate to do this:
That is, the mass of the orbiting body does play a small role. Provided that we are talking about a planet orbiting a star, or a satellite orbiting a planet, that's okay. A planet orbiting a planet, it doesn't work so well :)
Post by: RuiAce on December 03, 2016, 06:12:03 pm
I believe the fact that it is elliptical means that it is more accurate to do this:Kinda reminds me of the formula for binary stars in astrophysics tbh...
That is, the mass of the orbiting body does play a small role. Provided that we are talking about a planet orbiting a star, or a satellite orbiting a planet, that's okay. A planet orbiting a planet, it doesn't work so well :)
Edit: It IS that formula ahaha
Post by: katnisschung on December 06, 2016, 08:43:17 pm
i did question 3 of this past paper
https://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20solhttps://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20sol
unfortunately there are no answers for this q.
my answer...
probably wrong...
if the scale was moved further away from the plane in which the ball moves, this adds an unknown extra
distance to the ball's projectile path. Assuming the ball is thrown with the same force,
the initial horizontal velocity would decrease as the distance is increased. Therefore, as
the initial horizontal velocity decreases, the initial velocity must decrease...
thanks ruiace or jamon or jake or anyone willing to answer another one of my questions :)
Post by: katnisschung on December 06, 2016, 08:44:56 pm
into how much detail should u know about the michelson-morely
experiment... yes we just covered it at school...yes i'm concerned
lmao missed out on 2 weeks+ of lessons becos of hallelujah practice :P
Post by: FallonXay on December 06, 2016, 09:04:08 pm
also quick question
into how much detail should u know about the michelson-morely
experiment... yes we just covered it at school...yes i'm concerned
lmao missed out on 2 weeks+ of lessons becos of hallelujah practice :P
Relatively well. Normally, the most they'd throw at you on this question would be a 4, possibly 5 marker.
• You should be able to draw a labelled diagram of the experiment
• Identify the experimental aim (to measure the relative velocity of the Earth through the aether wind)
• Describe the experiment (splitting of beam at 90 degrees via a half-silvered mirror which were then reflected and combined to produce an interference pattern etc) and how it produced a 'null result' (The interference pattern was the same when the apparatus was rotated)
• Effects/ impact of the Michelson Morley Experiment on scientific views (i.e debates about the validity of the experiment/ doubt on the existence of the aether, use as one of the pieces of evidence supporting Einstein's Theory of Special Relativity)
• and I remember being asked in a practice trial paper once about the properties of the aether (medium through which light travelled through, permeated all matter in the universe, believed to be an absolute frame of reference, undetectable elastic material)
EDIT: Jamon explains the Michelson and Morley experiment/ aether really well in his guide here: Physics: A Complete Guide to the Course! (It's at the top of the Relativity section, you'll probably recognise the diagram)
Post by: RuiAce on December 06, 2016, 09:14:01 pm
hi again!You seem to be answering 3b) so I'll leave 3a) alone for now.
i did question 3 of this past paper
https://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20solhttps://thsconline.github.io/s/?view=6521&id=0ByEFYhkkDQBKeDNxaHFsOG5GQkU&n=Hurlstone%202007%20w.%20sol
unfortunately there are no answers for this q.
my answer...
probably wrong...
if the scale was moved further away from the plane in which the ball moves, this adds an unknown extra
distance to the ball's projectile path. Assuming the ball is thrown with the same force,
the initial horizontal velocity would decrease as the distance is increased. Therefore, as
the initial horizontal velocity decreases, the initial velocity must decrease...
thanks ruiace or jamon or jake or anyone willing to answer another one of my questions :)
Here's my attempt:
I wouldn't say that a fictitious distance was added in. However, if the data logger was built with a scale set for specifically, say, 3m away, then if you move the data logger back there will definitely be problems.
Rather than talking about an unknown extra distance, I would just use common sense here. In the same way that if you stand further away from a television you see things smaller, the trajectory of the ball relative to the data logger becomes smaller. Because of the fact the distance got shrunk (and obviously nothing changed to time), the velocity must supposedly decrease as well. Hence the data logger will register an underestimated velocity (keyword here is underestimated, because we're talking about how quantitative data was incorrectly predicted).
Which does, of course, mean that I agree with your final answer. Just the explanation was a bit weird...
Or I could probably use the inverse square law to argue my point but I'm a bit too scared to do that. The others would be able to give you a more confident answer.
Post by: katnisschung on December 06, 2016, 10:07:10 pm
according to an observer in the ship watching a ball being
dropped in the ship with constant velocity why does it
appear to fall directly down? (does the parabolic path of the ball 'compensate'
for the distance traveled by the ship thus drop directly below the observer?)
also what would happen in the situation that the ship was accelerating?
Post by: RuiAce on December 06, 2016, 10:27:07 pm
I was doing a past paper question and it got me considering this....If the ship was moving at a constant velocity it is its own inertial frame of reference. The ball belongs to this frame of reference. So to the observer in the ship, it will just see the ball drop vertically.
according to an observer in the ship watching a ball being
dropped in the ship with constant velocity why does it
appear to fall directly down? (does the parabolic path of the ball 'compensate'
for the distance traveled by the ship thus drop directly below the observer?)
also what would happen in the situation that the ship was accelerating?
Note that the person in the ship actually thinks everything else around the ship is what's moving. Not that he is the one moving; he thinks he is stationary.
If the ship was accelerating, then we'd have a non-inertial frame of reference, to which THEN we see the parabolic path; not falling straight down.
Post by: FallonXay on December 07, 2016, 10:03:06 pm
If the ship was moving at a constant velocity it is its own inertial frame of reference. The ball belongs to this frame of reference. So to the observer in the ship, it will just see the ball drop vertically.
Note that the person in the ship actually thinks everything else around the ship is what's moving. Not that he is the one moving; he thinks he is stationary.
If the ship was accelerating, then we'd have a non-inertial frame of reference, to which THEN we see the parabolic path; not falling straight down.
Hiya Rui~
Just a quick question, 'cos this reminded me of something some people were talking about in my class earlier this year: In the non-inertial frame of reference, if the ball is dropped, how come it's a parabolic path (not diagonally linear)?
Post by: RuiAce on December 07, 2016, 10:32:58 pm
Hiya Rui~I eventually taught myself how this works by combining both physics and maths together (although mostly still physics). You're pretty capable so I'll let you try to decipher it, but come back if you need further help :) (It's kinda weird to be fair - I visualised it all in my head in a matter of seconds but an explanation takes ages)
Just a quick question, 'cos this reminded me of something some people were talking about in my class earlier this year: In the non-inertial frame of reference, if the ball is dropped, how come it's a parabolic path (not diagonally linear)?
The idea is to use projectile motion principles. Pretend that you are about to throw a ball. Whilst the ball is in your hand, you have full control of its motion. However, at the instant the ball leaves your hand, it becomes a projectile.
The point of this, is that at the instant the ball leaves your hand, that's when you can guarantee that its horizontal velocity is constant. When it was in your hand, you could literally just move the ball back and forth, and that's obviously not a constant horizontal velocity.
Back to relativity.
Pretend we are in an inertial frame of reference first. Suppose you're travelling at some velocity (for visual purposes, let's consider small velocities such as 20ms-1). Additionally, first suppose that the ball is still in your hand. Then it's obviously travelling at 20ms-1 with you. Then suppose you release the ball. The ball is in the same inertial frame of reference as before you left it; it's also travelling at 20ms-1. Hence because both you and the ball are travelling at 20ms-1, you see it drop vertically.
But why is it, that when you drop the ball, it's still "moving at 20ms-1"? (Note that I'm trying to keep away from Einsteinian physics here; I'm kinda using Newtonian physics).
Consider a stationary observer watching you travel at 20ms-1. To him, it appears as though the ball already had an initial velocity of 20ms-1! You were holding the ball at 20ms-1 and that's what he saw, but the instant you let go of the ball you basically released it AT horizontal velocity 20ms-1. So the observer can actually confirm that once the ball is dropped, you are travelling at the same speed as with the ball.
Now for the non-inertial frame of reference.
Once again, you start off by holding the ball still. Except this time you're accelerating. Acceleration is indeed what's going to play a key role here.
Suppose you start accelerating from, idk, at rest. You accelerate from being at rest (at some magnitude) all the way up to 20ms-1, and THEN you drop the ball. This is where the analogy from earlier comes into play.
When you were accelerating, you had full control of the velocity of the ball. So if you were accelerating, the ball accelerated with you. But once you dropped the ball, it isn't with you anymore; you lost control over the ball! The ball was released when it acquired a velocity of 20ms-1, so because it's no longer affected by anything (i.e. your hand went away) it now travels by 20ms-1 by itself.
But whilst the ball is travelling at 20ms-1 by itself now, what are you doing? You're still accelerating! Because you're getting faster than 20ms-1 you start seeing the ball 'lag behind' you now.
The point, is that once you let go of the ball, it formed its own new INERTIAL frame of reference. You remain as a non-inertial frame of reference because you're accelerating, but nothing drives the ball into accelerating anymore so it's now in an inertial frame. This is also what the stationary observer would see - you're getting faster but the ball appears to stay at the same speed.
Post by: FallonXay on December 07, 2016, 10:44:45 pm
The point, is that once you let go of the ball, it formed its own new INERTIAL frame of reference. You remain as a non-inertial frame of reference because you're accelerating, but nothing drives the ball into accelerating anymore so it's now in an inertial frame. This is also what the stationary observer would see - you're getting faster but the ball appears to stay at the same speed.
Haha 'Quick Question' ;) (But in all seriousness, thanks for the detailed answer)
So what I don't understand is that: since you are an observer on the accelerating vehicle and in your frame you are 'stationary', wouldn't you observe the ball the be accelerating away from you (isn't the ball only travelling at a constant speed if the observer were also in the inertial frame/ at travelling at a velocity of 0 m/s)? So wouldn't this observer see a horizontal acceleration?
(Also, what effect does having two different frames of reference have? - i.e the ball in the non-inertial and the observer in the inertial)
EDIT: I mean, the ball in the inertial, and observer in the non-inertial
Post by: RuiAce on December 07, 2016, 10:56:13 pm
Haha 'Quick Question' ;) (But in all seriousness, thanks for the detailed answer)Yeah I giggled at "quick" when I reread that :P
So what I don't understand is that: since you are an observer on the accelerating vehicle and in your frame you are 'stationary', wouldn't you observe the ball the be accelerating away from you (isn't the ball only travelling at a constant speed if the observer were also in the inertial frame/ at travelling at a velocity of 0 m/s)? So wouldn't this observer see a horizontal acceleration?
(Also, what effect does having two different frames of reference have? - i.e the ball in the non-inertial and the observer in the inertial)
Yep. So here's a twist. Normally, if you're the one accelerating you would know that you're accelerating. This is from the fact that we can't "feel" velocity as we can't "feel" momentum. Rather, we can feel acceleration because we can feel force.
Most of the analogies are given where you don't know that you're accelerating. i.e. You think everything around you is accelerating. (Just like how in physics we studied as though everything around you was moving.)
In this scenario, you're right. Somewhere in there I wrote that it appears as though the ball "lags behind" you. This is because indeed, the ball appears to carry a negative acceleration relative to you!
This is a part of why non-inertial frames of reference and general relativity isn't looked at in the course; it becomes too complex to describe. (Jake might know a thing or fifty.) To quantify the weird happening we introduce things known as "fictitious forces".
To a stationary observer, they'll probably know that some force is being exerted on you, and hence you're accelerating. As for you, these fictitious forces are a mere way to justify why the ball DOES appear to have a negative acceleration. The reason why the negative acceleration is as stated in my above post, but the outcome is basically what you determined.
Wikipedia provides this definition:
Quote
A fictitious force, also called a pseudo force, d'Alembert force or inertial force, is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.The basic idea is that because you are actually the one accelerating, you see these weird things that go on. That's really much it! You don't realise you're accelerating, maybe, but you just are so these fictitious forces happen to 'exist'!
(As an aside, another example of a fictitious force is the 'centrifugal force', which only exists to justify Newton's third law of motion with the centripetal force.)
Having two different frames of references besides your own just means you now have three frames of references in total. It just means there's now more things to consider. It's like you're playing an MMORPG and you're the noob being overshadowed by the higher damage of the Lv20 toon, but there's a Lv60 toon behind them dealing even more explosive damage :P
Post by: FallonXay on December 07, 2016, 11:14:23 pm
Yeah I giggled at "quick" when I reread that :P
Yep. So here's a twist. Normally, if you're the one accelerating you would know that you're accelerating. This is from the fact that we can't "feel" velocity as we can't "feel" momentum. Rather, we can feel acceleration because we can feel force.
Most of the analogies are given where you don't know that you're accelerating. i.e. You think everything around you is accelerating. (Just like how in physics we studied as though everything around you was moving.)
In this scenario, you're right. Somewhere in there I wrote that it appears as though the ball "lags behind" you. This is because indeed, the ball appears to carry a negative acceleration relative to you!
This is a part of why non-inertial frames of reference and general relativity isn't looked at in the course; it becomes too complex to describe. (Jake might know a thing or fifty.) To quantify the weird happening we introduce things known as "fictitious forces".
To a stationary observer, they'll probably know that some force is being exerted on you, and hence you're accelerating. As for you, these fictitious forces are a mere way to justify why the ball DOES appear to have a negative acceleration. The reason why the negative acceleration is as stated in my above post, but the outcome is basically what you determined.
Wikipedia provides this definition:The basic idea is that because you are actually the one accelerating, you see these weird things that go on. That's really much it! You don't realise you're accelerating, maybe, but you just are so these fictitious forces happen to 'exist'!
(As an aside, another example of a fictitious force is the 'centrifugal force', which only exists to justify Newton's third law of motion with the centripetal force.)
Having two different frames of references besides your own just means you now have three frames of references in total. It just means there's now more things to consider. It's like you're playing an MMORPG and you're the noob being overshadowed by the higher damage of the Lv20 toon, but there's a Lv60 toon behind them dealing even more explosive damage :P
So the ball only 'appears' to be accelerating - but in reality it's travelling at a constant velocity? Would it be correct/ 'appropriate' to say that the observation of the ball as accelerating is incorrect (to scientifically make this conclusion from the observation)? I'm a little confused: I don't really understand the effect of the fictitious forces in affecting 'reality'~
Post by: RuiAce on December 07, 2016, 11:30:20 pm
So the ball only 'appears' to be accelerating - but in reality it's travelling at a constant velocity? Would it be correct/ 'appropriate' to say that the observation of the ball as accelerating is incorrect (to scientifically make this conclusion from the observation)? I'm a little confused: I don't really understand the effect of the fictitious forces in affecting 'reality'~See this is why it's out of the HSC course. It is complicated, and what the actual correct answer is, well I have no idea.
But basically that's how I analyse the scenario.
Post by: jamonwindeyer on December 07, 2016, 11:36:51 pm
So the ball only 'appears' to be accelerating - but in reality it's travelling at a constant velocity? Would it be correct/ 'appropriate' to say that the observation of the ball as accelerating is incorrect (to scientifically make this conclusion from the observation)? I'm a little confused: I don't really understand the effect of the fictitious forces in affecting 'reality'~
Tagging in ;) yes, it would be appropriate to say that it is incorrect, it is more appropriate to say that it is an inaccuracy due to taking our measurement from a non-inertial frame of reference :) like, it is still a frame of reference, and so it has value, but the observation needs to be interpreted correctly.
The fictitious forces thing is best explained with an example. Imagine being in a plane. Constant speed, no turbulence, all the windows are shut. There is absolutely no way to prove that you are moving in the sky. You are in an inertial frame of reference.
Now, pretend the plane gives a burst of speed, lurches upwards, etc. This will cause things to fly around and for you to feel a force, possibly get knocked off your feet.
At this point, in your magic plane that you can't see out of, you have two options. You can accept that you are in fact flying, and have indeed accelerated. That's the logical choice. If, instead, you want to maintain the fact that you are in an inertial frame of reference and that you aren't accelerating, you have to then invent something else that caused all the commotion. This is the fictitious force, you invent it to avoid accepting the fact that you yourself are accelerating.
It's kind of like believing in the Tooth Fairy when you are young. The logical choice is the easiest, just accept that the tooth fairy doesn't exist. If you want to try and maintain the fact that she does exist, then you need to invent a lot of fiction, do a lot of storytelling, to make your view make sense ;D
Post by: FallonXay on December 07, 2016, 11:42:33 pm
See this is why it's out of the HSC course. It is complicated, and what the actual correct answer is, well I have no idea.
But basically that's how I analyse the scenario.
Ahhh k. Cheers for the explanation/ discussion Rui, was very interesting :)
(Also, I think my brain melted in the process of making sense of everything haha)
The fictitious forces thing is best explained with an example. Imagine being in a plane. Constant speed, no turbulence, all the windows are shut. There is absolutely no way to prove that you are moving in the sky. You are in an inertial frame of reference.
and very nice example, helped with the whole fictitious force thing :P . Thanks Jamon~
P.s if you're interested in where this sparked from there was a similar question except it was about a pendulum hanging on a roof in which was cut (Though I'm not sure that the string would affect anything because there isn't tension anymore when the string is cut?...) - 2010 Q23b
Post by: jamonwindeyer on December 08, 2016, 12:10:13 am
P.s if you're interested in where this sparked from there was a similar question except it was about a pendulum hanging on a roof in which was cut (Though I'm not sure that the string would affect anything because there isn't tension anymore when the string is cut?...) - 2010 Q23b
I remember this question, I had a feeling that this was the catalyst!! The string doesn't affect anything after it is cut, it is just used as a way to indicate the fact that the train is accelerating, without saying that it is ;D
Post by: RuiAce on December 08, 2016, 10:30:33 am
Post by: katnisschung on December 08, 2016, 06:38:34 pm
i know its either a or c
(starts at 0m/s then accelerates downward at a constant rate due to acceleration)
but then why does the velocity go into the negative for when it comes back up?
pls help
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/physics_01.pdf
Post by: jamonwindeyer on December 08, 2016, 06:52:04 pm
could anyone explain why it is (d) for q 15 (multiple choice)
i know its either a or c
(starts at 0m/s then accelerates downward at a constant rate due to acceleration)
but then why does the velocity go into the negative for when it comes back up?
pls help
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/physics_01.pdf
Hey! Great question; it sounds like you pretty much have it!
First of all, the graph in D has straight lines; this makes sense because the acceleration is constant. So any changes to velocity should always be linear (the gradient of those lines would be \(9.8\), to match with acceleration due to gravity).
The reason that the line goes into the negative region is because it is, after the bounce, travelling in the opposite direction to what it did initially. So say the ball is travelling \(30ms^{-1}\) downwards, after the bounce it might be travelling \(28ms^{-1}\) upwards. The difference in direction is what that shift reflects.
That said, it would be a better graph if it was negatively sloped. That is, went into the negative first, then shifted into the positive. But it asks for the best representation, so D will have to do ;) does that help?
Post by: katnisschung on December 08, 2016, 07:01:29 pm
Post by: jamonwindeyer on December 08, 2016, 07:03:23 pm
ahh so they have nominated upwards as negative?
Yeah, for some reason they have! Probably because it started travelling downwards, so that was made the positive direction. There is nothing wrong with it, it's just a little bit of a confusing choice :P
Post by: katnisschung on December 08, 2016, 07:06:12 pm
but why is it a straight line drop in velocity?
shouldn't it be angled like a (taking into account that m= acceleration due to gravity)
Post by: jamonwindeyer on December 08, 2016, 07:08:42 pm
ahhh im probably overthinking it
but why is it a straight line drop in velocity?
shouldn't it be angled like a (taking into account that m= acceleration due to gravity)
If you were looking at position, then it would be curved for sure!! But this is velocity; the formula for velocity is this:
It's actually a straight line! It's a straight line because, yes, the gradient is the acceleration due to gravity. The acceleration due to gravity is constant. A line with a constant gradient is a straight line! :)
Post by: teapancakes08 on December 08, 2016, 11:15:50 pm
"A projectile fired up into the air from the top if a 75m high cliff hits the ground 500m out from the base, considering the projectile being fired at an angle and landing on a surface below that from which it was fired. Find the initial velocity, horizontal velocity, vertical velocity, range, maximum height, time taken to reach maximum height, and time of flight."
From the question I'm given the range, but am confused on where to start with everything else. It'd be easier if I could find with the maximum height though, since everything works out from there, but it doesn't seem likely to find it first. So which value should I try to find first?
Post by: jamonwindeyer on December 09, 2016, 12:22:10 am
How should I approach this question:
"A projectile fired up into the air from the top if a 75m high cliff hits the ground 500m out from the base, considering the projectile being fired at an angle and landing on a surface below that from which it was fired. Find the initial velocity, horizontal velocity, vertical velocity, range, maximum height, time taken to reach maximum height, and time of flight."
From the question I'm given the range, but am confused on where to start with everything else. It'd be easier if I could find with the maximum height though, since everything works out from there, but it doesn't seem likely to find it first. So which value should I try to find first?
Hey!! Are you sure that you've been given all the required information here? Any missing part of the question? I ask because I don't think there is enough information to answer (happy for someone to correct me) :P
Post by: Jakeybaby on December 09, 2016, 12:31:18 am
Hey!! Are you sure that you've been given all the required information here? Any missing part of the question? I ask because I don't think there is enough information to answer (happy for someone to correct me) :PI do agree with you, seems that all equations will be left with 2+ unknown variables. One part seems to be missing from the question imo.
Post by: jamonwindeyer on December 09, 2016, 01:10:00 am
I do agree with you, seems that all equations will be left with 2+ unknown variables. One part seems to be missing from the question imo.
Yeah I think so too! I can get it into an equation with only \(\theta\) and \(V\), that's best I got! :P
Post by: teapancakes08 on December 09, 2016, 08:22:23 am
Yeah I think so too! I can get it into an equation with only \(\theta\) and \(V\), that's best I got! :P
Question 9 on this sheet.
Post by: jamonwindeyer on December 09, 2016, 03:13:13 pm
Question 9 on this sheet.
Yeah I think there might be a piece of information missing! You can tell because the question right below it gives the same information, plus one bit extra. I don't think the information given in Question 9 is enough :P could just be a 'faulty' question? :)
Post by: Aaron12038488 on December 16, 2016, 02:22:36 pm
Is it good to start on the syllabus now for physics
Thanks,
Also, The first dot point which is Describe the energy transformations required in a mobile phone, is this satisfactory.
Spoiler
Sound Waves (voice) converted into electrical signals via microphone. Digitised (binary code) is transmitted as radio waves to base station. Each base station is connected to switching centre by an optic fibre network that carries signal as electrical impulses- produced by radio wave energy interacting with aerial. Switching centres are connected to other switching centres and base stations.
Sound Energy → Electrical Signals → Transmitted as Radio Waves → Light Energy → Sound Energy
Call between mobile and landline at the base station – signal is converted to light and will travel along an optic fibre network to distant to switching centre. Call between centre will be transformed to another switching centre close to a base station near the receiving mobile. The signal is then fed to the base station as an electrical signal and is broadcast as radio waves. In the mobile, radio waves are converted to electrical impulses which is then converted to sound from the speakers.
Call between mobile and landline at the base station – signal is converted to light and will travel along an optic fibre network to distant to switching centre. Call between centre will be transformed to another switching centre close to a base station near the receiving mobile. The signal is then fed to the base station as an electrical signal and is broadcast as radio waves. In the mobile, radio waves are converted to electrical impulses which is then converted to sound from the speakers.
Sound Energy → Electrical Signals → Transmitted as Radio Waves → Light Energy → Sound Energy
Call between mobile and landline at the base station – signal is converted to light and will travel along an optic fibre network to distant to switching centre. Call between centre will be transformed to another switching centre close to a base station near the receiving mobile. The signal is then fed to the base station as an electrical signal and is broadcast as radio waves. In the mobile, radio waves are converted to electrical impulses which is then converted to sound from the speakers.
Call between mobile and landline at the base station – signal is converted to light and will travel along an optic fibre network to distant to switching centre. Call between centre will be transformed to another switching centre close to a base station near the receiving mobile. The signal is then fed to the base station as an electrical signal and is broadcast as radio waves. In the mobile, radio waves are converted to electrical impulses which is then converted to sound from the speakers.
I've started the Physics Prelim Syllabus as my yearly report for science was very dismal. As I'm doing Physics, Chem and Bio, I'm really anxious in how I will perform in Year 11 and 12. Thoughts??
Mod Edit: Merged posts and added spoiler. You can use the 'Modify' button on the top right to add to your post to avoid posting twice in a row ;D
Post by: jamonwindeyer on December 16, 2016, 02:31:40 pm
Personally, I'd just chill out these holidays. You can do a bit of light reading if you like of course, but the Year 11 content doesn't impact your HSC, and it is important to have a break before the hard work begins. I admire your drive, but you can probably reign it in a bit, there is definitely no need to start Year 11 content this early :)
In saying that, seems like you have a great understanding of the first dot point!!
Don't stress about your Year 10 Science results; HSC science is completely different. New skills, new start. If you work really hard then there is no reason for any of that to be reflected in your senior studies! ;D
Post by: bluecookie on December 17, 2016, 02:39:58 pm
Post by: RuiAce on December 17, 2016, 02:43:44 pm
What does the 'position' in a position time graph mean? Is it the distance or displacement or something else?In physics, position is just another word for displacement.
Post by: jamonwindeyer on December 17, 2016, 03:07:34 pm
What does the 'position' in a position time graph mean? Is it the distance or displacement or something else?
Welcome to the forums bluecookie! ;D let me know if you need a hand finding things :)
Post by: bluecookie on December 17, 2016, 03:19:42 pm
In physics, position is just another word for displacement.Thank you!
Welcome to the forums bluecookie! ;D let me know if you need a hand finding things :)Thanks :D
Mod Edit: Merged posts. You can use the 'Insert Quote' button below your typing window to add quotes from multiple users ;D
Post by: bluecookie on December 18, 2016, 07:03:14 pm
Post by: RuiAce on December 18, 2016, 07:15:11 pm
What does the area under a displacement velocity graph represent?Check what you said. This is not in the HSC physics course. All graphs are something-time graphs.
Note, however, the area under a velocity-time graph represents the displacement travelled over that period of time.
Post by: bluecookie on December 18, 2016, 09:16:08 pm
Check what you said. This is not in the HSC physics course. All graphs are something-time graphs.
Note, however, the area under a velocity-time graph represents the displacement travelled over that period of time.
Sorry, I meant displacement time graph.
Post by: Jakeybaby on December 18, 2016, 09:29:26 pm
Sorry, I meant displacement time graph.I can't see that value having any important physical meaning, nor can I see it being tested at HSC level. I imagine it is just what you said, the area under a displacementp-time graph, nothing more, nothing less.
Post by: RuiAce on December 18, 2016, 09:31:19 pm
Sorry, I meant displacement time graph.Precisely as mentioned above. It is of no use whatsoever.
(Mathematically, integrating the displacement tells us nothing useful)
It's just like how the slope of the acceleration-time graph is of no use (in the HSC course that is; in the real world of physics it's called jerk)
Post by: Aaron12038488 on December 22, 2016, 10:40:00 am
Post by: jamonwindeyer on December 22, 2016, 10:46:16 am
I was wondering how you make notes. I want to get in the habit of making notes sooner so I get the general jist and will be fine when exams come up.
Hey Aaron! There are all sorts of good things to do, but the main thing I'd recommend is to make sure they are completely based off the syllabus dot points. That will make sure that you don't miss anything. I also, personally, think it is better to write your notes progressively as you learn the content, not in big chunks ;D
Besides that, this article might be worth a read ;D
Post by: Aaron12038488 on December 22, 2016, 10:49:30 am
Post by: RuiAce on December 22, 2016, 10:55:18 am
Thanks for the rapid reply. Also for Year 11, will all the tests and Pracs be based on the Syllabus Dot Points?EVERYTHING is based on the syllabus dot points.
Post by: jamonwindeyer on December 22, 2016, 10:55:57 am
Thanks for the rapid reply. Also for Year 11, will all the tests and Pracs be based on the Syllabus Dot Points?
For Year 11 and 12, definitely. That's the whole point of a syllabus after all; it dictates what you need to know for the course! Any assessment must link back to the syllabus. You could be asked to do research into some of the points in more detail than is necessary, but that would be a take-home assignment; not a test or a practical :)
Post by: Aaron12038488 on December 22, 2016, 11:03:16 am
Post by: jamonwindeyer on December 22, 2016, 11:07:20 am
Will the school start from the very first topic or would it be random?
For Year 11, I actually started with the very last topic. It does depend on preference. It is more common to start with either the very first topic, or Moving About ;D
Post by: Aaron12038488 on December 22, 2016, 11:20:02 am
Post by: RuiAce on December 22, 2016, 11:25:09 am
I'm looking at the schools prelim booklet, and it says syllabus components covered in Physics. It has P2, P11, P12, P13, etc. What does P2 and the rest mean?The outcomes (whilst not bad to read if you have time) serve no real purpose for you as a student whatsoever. It serves as a reference for teachers in setting assessment tasks to determine what skills a student must have.
Post by: Aaron12038488 on December 25, 2016, 09:59:24 am
How much info should you put for each dot point?
Post by: Happy Physics Land on December 25, 2016, 10:06:35 pm
How do u make notes?
How much info should you put for each dot point?
Hey Aaron!
I can definitely help you there. When making notes, especially for physics, I dont like to include a tonne of information. You need to especially avoid having model answers in your notes that you can memorise, because physics questions are never the same and just like maths, they require an open mind and real understanding of physics concepts. When making notes, try to use a lot of colour, diagrams, tables and flowcharts. You want your notes to be easy to read and interesting to read so that you actually WANT TO read your notes. Try to include diagrams such as the galvanometer, loudspeakers and BCS theory in your notes (best if you hand-draw them and take photos and paste them into your notes). These images will come in very handy in an exam. Below is a sample of my notes for solar cell:
(http://i.imgur.com/ifNh9eP.png)
Physics is all about processes. That's why we use flowchart to explain a lot of things in physics. Instead of memorising written answers, we understand each part of the process of why something happens. This way we can properly answer any question that's thrown at us during exams. Below is an example of my medical physics notes on ultrasound:
(http://i.imgur.com/2JL0Vqc.png)
You definitely don't need to write passages, the most you will ever need to do is brief dot points. Try to make each dotpoint straightforward and simple, avoid combining various ideas in one dot point.
Best of luck!
Best Regards
Happy Physics Land
Post by: Aaron12038488 on December 26, 2016, 10:03:39 am
Post by: jamonwindeyer on December 26, 2016, 11:18:04 am
Do you also make notes for the skills section?
I never did that, they are skills so it's hard to make notes on them when notes are inherently content focused ;D I'm just not sure what you'd write there! You are wayyy better off just doing practice questions and such to get those skills down :)
Post by: and1_98 on December 31, 2016, 01:44:56 pm
Not sure whether a question like this is assessable...? Only based on the first line of the question; 'Using your knowledge of solar eclipses...'
Thanks in advance
Post by: jakesilove on December 31, 2016, 04:17:00 pm
I got 1968km for the Moon's radius. Not sure if this is the answer though (there aren't answers, although Google says it's 1737km).
Not sure whether a question like this is assessable...? Only based on the first line of the question; 'Using your knowledge of solar eclipses...'
Thanks in advance
Interesting question! Personally, I would have had a go at using similar triangles. This method is definitely not assessable, but it's interesting, so let's give it a go.
Using the 'knowledge of solar eclipses', we know that the moon appears to JUST block out the sun. We can draw ourselves a diagram, which would look something like this.
(http://i.imgur.com/N6p431H.png)
Okay, so, we can quickly find the acute angle between the earth and the tangent lines. The 'opposite' side is 700,000km and the 'adjacent' side is 150,000,000km plus the radius (700,000km), giving us 150,700,000km. So,
So, theta is equal to 0.26614 degrees. Now, let's use the 'moon' triangle. The angle is the same, but the 'opposite' and 'adjacent' sides are different.
That's pretty close to the real value! However, DEFINITELY not assessable in Physics (but a good 2U maths question!).
Thinking of a Physics way is a bit trickier. In fact, I'm not 100% sure there is a better way. How did you do it? An answer of 1968km, in my opinion, is probably correct; it's very close to the real value, which would be unlikely if you had done something wrong!
Post by: bluecookie on December 31, 2016, 10:21:26 pm
what are the units for T?
Post by: RuiAce on December 31, 2016, 10:25:24 pm
in r^3/T^2=GM/4(pi)^2seconds.
what are the units for T?
Always assume seconds for anything related to time.
Post by: bluecookie on December 31, 2016, 10:56:01 pm
seconds.
Always assume seconds for anything related to time.
Thank you ^^
Post by: bluecookie on January 01, 2017, 02:36:18 pm
Post by: RuiAce on January 01, 2017, 02:41:37 pm
For the right hand grip rule, does the thumb point in the direction of conventional current flow or actual current flow?Always conventional current.
Post by: bluecookie on January 01, 2017, 03:03:43 pm
Post by: bluecookie on January 01, 2017, 03:10:15 pm
Post by: jakesilove on January 01, 2017, 03:22:19 pm
When two wires are anti-parallel to each other, how does the left hand wire experience a force to the left?
Draw the field lines. Use crosses to signify lines into the page, and dots to signify lines out of the page. For anti-parallel wires, you'll find that you get only dots between the wires (or only crosses). As like charges repel, both wires will repel each other, resulting in the left wire experiencing a force to the left.
Post by: Rathin on January 02, 2017, 12:28:25 pm
Post by: Happy Physics Land on January 02, 2017, 02:00:44 pm
When the coil of wire is perpendicular to the magnetic field and the current is in the same direction as the coil as its a DC circuit it will oppose the motion of torque, why is that? (split ring commutators are used to overcome this problem but I am not 100% on why this problem arises).
Hey Rathin!
Im guessing here that you are talking about how when the rotor coil changes the direction of its rotation after half a revolution (i.e. when it becomes perpendicular to the magnetic field). There's nothing that's opposing the original torque here, the only thing that has changed is the direction of the current through the rotor coil relative to the magnetic field. And according to our right hand palm rule, when the current direction reverses relative to the magnetic field, the force acting upon the coil would change as well. Be aware that the net current direction isn't changing here since a DC voltage is supplied. Current direction is only reversed RELATIVE TO THE MAGNETIC FIELD. Therefore since the direction of forces acting on the sides of the rotor coil changes, the direction of torque on the coil must change as well and thats why it starts rotating in a reverse direction.
What the commutator essentially does is it changes the direction of current flow in the rotor coil once every half a revolution to ensure that the current flow RELATIVE TO THE MAGNETIC FIELD is in a constant direction and therefore the force acting on the sides of the coil can be constant and hence torque/rotation can also be constant.
Best Regards
Happy Physics Land
Post by: shreya_ajoshi on January 02, 2017, 08:04:56 pm
If the distances of these moons form the planet's centre are R,9R and 16R respectively, calculate the ratio of their orbital speeds.
a) If the moons have the same orbital speeds, find the ratio of their orbital radii
b) If the orbital radii of the moons are the same, find the ratio of their orbital speeds
Mod edit: Merged duplicate posts. At times like this, please resort to the modify button instead of posting multiple times.
Post by: RuiAce on January 02, 2017, 08:47:13 pm
Post by: Happy Physics Land on January 02, 2017, 08:51:46 pm
Three moons around planet X have masses M,9M and 16M
If the distances of these moons form the planet's centre are R,9R and 16R respectively, calculate the ratio of their orbital speeds.
a) If the moons have the same orbital speeds, find the ratio of their orbital radii
b) If the orbital radii of the moons are the same, find the ratio of their orbital speeds
Mod edit: Merged duplicate posts. At times like this, please resort to the modify button instead of posting multiple times.
a)
Multiplying everything by 144:
Post by: shreya_ajoshi on January 03, 2017, 12:47:56 pm
Thank you!
However the answer to the first question is different to the answers the book provided.According to them, the answer is 12;4;3
Post by: shreya_ajoshi on January 03, 2017, 12:50:50 pm
Could I please have help with this question. I have difficulty identifying t0 and tv
Star X is 8.0ly from Earth. A spaceship travels at 0.5c to reach the star.
a) Calculate how long the trip takes as measured by an observer on Earth
b) Calculate how long the trip takes as measure by the astronauts in the ship
c) Calculate the distance travelled as measure by the astronauts
d) Calculate the speed of the ship as measured by the astronauts
Post by: Rathin on January 03, 2017, 01:52:02 pm
Hi! :)
Could I please have help with this question. I have difficulty identifying t0 and tv
Star X is 8.0ly from Earth. A spaceship travels at 0.5c to reach the star.
a) Calculate how long the trip takes as measured by an observer on Earth
b) Calculate how long the trip takes as measure by the astronauts in the ship
c) Calculate the distance travelled as measure by the astronauts
d) Calculate the speed of the ship as measured by the astronauts
a) t=d/s
∴t=8/0.5
∴t=16 years
b) using the formula t(v)=(t(o))/sqrt((1-(v^2/c^2)))
sub in the values such that t(o)=16 and v=0.5c
∴t(v)=18.48 years
c)d=s*t
∴d=0.5c*18.48
∴d=2.77x10^9 m
d) This is a trick question as the astronauts cannot measure the speed inside their inertial reference frame unless sourced from an external reference frame.
Post by: RuiAce on January 03, 2017, 07:27:19 pm
Thank you!That's correct if the masses were all M, M, M and not M, 9M, 16M. But I don't think that was what you typed at the time though...
However the answer to the first question is different to the answers the book provided.According to them, the answer is 12;4;3
Post by: shreya_ajoshi on January 03, 2017, 08:12:14 pm
a) t=d/s
∴t=8/0.5
∴t=16 years
b) using the formula t(v)=(t(o))/sqrt((1-(v^2/c^2)))
sub in the values such that t(o)=16 and v=0.5c
∴t(v)=18.48 years
c)d=s*t
∴d=0.5c*18.48
∴d=2.77x10^9 m
d) This is a trick question as the astronauts cannot measure the speed inside their inertial reference frame unless sourced from an external reference frame.
Thank you very much :)
Post by: shreya_ajoshi on January 03, 2017, 08:14:58 pm
That's correct if the masses were all M, M, M and not M, 9M, 16M. But I don't think that was what you typed at the time though...
Oh okay, but the question does say M,9M and 16, so I'm not too sure anymore haha
Post by: Happy Physics Land on January 03, 2017, 08:46:29 pm
a) t=d/s
∴t=8/0.5
∴t=16 years
b) using the formula t(v)=(t(o))/sqrt((1-(v^2/c^2)))
sub in the values such that t(o)=16 and v=0.5c
∴t(v)=18.48 years
c)d=s*t
∴d=0.5c*18.48
∴d=2.77x10^9 m
d) This is a trick question as the astronauts cannot measure the speed inside their inertial reference frame unless sourced from an external reference frame.
Good stuff there rathin my friend
Post by: bluecookie on January 04, 2017, 12:22:43 pm
Draw the field lines. Use crosses to signify lines into the page, and dots to signify lines out of the page. For anti-parallel wires, you'll find that you get only dots between the wires (or only crosses). As like charges repel, both wires will repel each other, resulting in the left wire experiencing a force to the left.My book said nothing about like charges repelling. It just said to use the right hand grip rule for the first wire, and then the right hand palm rule for the second wire. It worked for parallel lines. However when I tried it with anti parallel lines (with the first line pointing up for reference) I got crosses between the wires, and then using the right hand palm rule on the second wire I got a force going to the right. However, seeing as the first wire generates a magnetic field pointing to the right, doesn't it mean the first wire moves to the right also? So I got they're both going right, but the book says they should repel. I don't understand how the first wire is meant to be moving to the left?
Post by: RuiAce on January 04, 2017, 12:26:34 pm
My book said nothing about like charges repelling. It just said to use the right hand grip rule for the first wire, and then the right hand palm rule for the second wire. It worked for parallel lines. However when I tried it with anti parallel lines (with the first line pointing up for reference) I got crosses between the wires, and then using the right hand palm rule on the second wire I got a force going to the right. However, seeing as the first wire generates a magnetic field pointing to the right, doesn't it mean the first wire moves to the right also? So I got they're both going right, but the book says they should repel. I don't understand how the first wire is meant to be moving to the left?Like charges repel is the same thing as saying
Two positive charges repel
Two negative charges repel
(But a positive charge and a negative charge attract)
Just like how with magnets, two north poles repel.
Post by: bluecookie on January 04, 2017, 12:48:55 pm
Like charges repel is the same thing as saying
Two positive charges repel
Two negative charges repel
(But a positive charge and a negative charge attract)
Just like how with magnets, two north poles repel.
Thanks :)
Post by: Aaron12038488 on January 04, 2017, 06:05:54 pm
Post by: FallonXay on January 04, 2017, 06:07:46 pm
To achieve a band 6 for any subject, do you have to get 90/100 in the HSC exam?
After the mark is aligned, yes. (It isn't a 90 raw mark)
and extension subjects should be 45+ for an E4
Post by: Rathin on January 04, 2017, 06:10:12 pm
To achieve a band 6 for any subject, do you have to get 90/100 in the HSC exam?
Its normally around 82 Raw mark which aligns to 90. But give or take one depending on the difficulty.
Post by: RuiAce on January 04, 2017, 06:16:05 pm
To achieve a band 6 for any subject, do you have to get 90/100 in the HSC exam?If we're talking about raw marks, no.
Alignment is a process that the BOSTES uses (you can research it if you want) to address the fact that different courses have harder exams in different years. A board of members called 'judges' are responsible for determining the alignment algorithm that gets applied that year.
Because the algorithm is different, it is never possible to say with certainty how much you need to score raw to get aligned to 90 (or whatever mark in question). As per the raw marks database, however, historically it is approximately 83. Subject to about plus/minus 4 inaccuracy.
Post-alignment, however, you must have achieved at least an average of 90. This average is between your moderated AND aligned internal mark, as well as your aligned HSC exam mark.
For further reading, find Jamon's article
Post by: Aaron12038488 on January 04, 2017, 06:16:57 pm
Post by: FallonXay on January 04, 2017, 06:32:45 pm
So I'm going to start Year 11 in a couple of weeks. What equipment should I take with me. My school hasn't told us what to bring.
The usual school equipment?
• Notebooks
• Diary (personal preference, or my school provided one)
• Laptops (if your school needs one)
• Pencil case (Pens, pencil, highlighter, whiteout, glue, scissors, sharpener, ruler, etc.)
• A calculator
• A geometry set (maybe not needed at the beginning of the year, but nice to have if you're doing mathy/sciency subjects and want to be neat :) )
Post by: Aaron12038488 on January 05, 2017, 01:44:46 pm
Post by: jakesilove on January 05, 2017, 01:59:11 pm
This holiday I have been studying The World Communicates Topic. I'm almost finished it, so should I memorise everything before school starts. This will give me an edge and make me confident.
If it's going to make you feel better about your course work, and make you more confident, then go for it! Whilst not absolutely necessary, you should be studying in whatever way makes you the most comfortable
Post by: jamonwindeyer on January 05, 2017, 02:00:29 pm
This holiday I have been studying The World Communicates Topic. I'm almost finished it, so should I memorise everything before school starts. This will give me an edge and make me confident.
You can if you like! But remember that your Year 11 content isn't assessable at the HSC Level (for World Communicates, you essentially need to know a formula in the HSC, which they give you on the reference sheet anyway), so don't stress about it too much! Year 11 is a time to develop good habits and figure out how you work best - I regret putting as much stress into Year 11 as I did :) just studying the topic is way more than I ever did! :)
Post by: Swagadaktal on January 05, 2017, 02:07:47 pm
This holiday I have been studying The World Communicates Topic. I'm almost finished it, so should I memorise everything before school starts. This will give me an edge and make me confident.TBH I wouldn't even recommend doing this for a year 12 study, letalone a year 11 one!
It's not sustainable to be doing this so early on and it doesn't really provide you with much of an edge, because there is more than enough time during the actual year to learn the content!
I'd strongly recommend taking a more sustainable approach :)
Post by: Aaron12038488 on January 05, 2017, 02:49:37 pm
Post by: Aaron12038488 on January 06, 2017, 03:38:52 pm
How should I go about answering this dot point.
Is this alright, any improvements.
The sun is a producer of all the EM waves sending all bandwidth to Earth.
The earth’s atmosphere has the ability to absorb any incoming energy EM radiation
Ultimately only visible light, radio waves, and microwaves are able to reach the surface of the Earth
Too much exposure to UV radiation can result in cancers and dangerous reactions
Too much exposure to X-rays and gamma radiation would quickly kill us
Post by: Rathin on January 06, 2017, 03:49:43 pm
Identify the electromagnetic wavebands filtered out by the atmosphere, especially UV, X-rays and gamma rays.
How should I go about answering this dot point.
Is this alright, any improvements.
The sun is a producer of all the EM waves sending all bandwidth to Earth.
The earth’s atmosphere has the ability to absorb any incoming energy EM radiation
Ultimately only visible light, radio waves, and microwaves are able to reach the surface of the Earth
Too much exposure to UV radiation can result in cancers and dangerous reactions
Too much exposure to X-rays and gamma radiation would quickly kill us
I dont think you need that much detail as its an identify verb but it is good to know.
But this is also the relevant information you will need to know.
Radio Waves: Not Absorbed
Microwaves: Not Absorbed
Infrared: Partially Absorbed
Visible Light: Not Absorbed
UV Ray: Absorbed
X Ray: Absorbed
Gamma Ray: Absorbed
Post by: jamonwindeyer on January 06, 2017, 04:10:22 pm
Post by: Aaron12038488 on January 06, 2017, 04:20:38 pm
Post by: Rathin on January 06, 2017, 04:24:38 pm
rathin, did you just copy that from KC Notes?
Not sure, It copied directly from my Prelim Notes..I used many sources so It is a possibility.
Also, for this dot point the most I have seen while doing past papers in yr 11 was MC question which asked which of these rays are absorbed. So if you remember that Gamma Ray, X Ray and UV Ray are absorbed fully by the Atmosphere you will be fine.
Post by: jamonwindeyer on January 09, 2017, 01:22:44 pm
Jamon are these decent notes so far?
Looks great to me! Very visual, no huge chunks of information, nicely done Aaron! :)
Post by: Aaron12038488 on January 09, 2017, 01:31:46 pm
Post by: katnisschung on January 09, 2017, 05:13:45 pm
the star ship enterprise Enterprise is stationary in space to make repairs to the warp
drive. Suddenly a Kinglon vessel appears from the right and flashes towards and past the Enterprise at half the speed
of light.
When it is exactly in front of the Enterprise, Captain Kirk fires port and starboard at the same time.
so this is how i interpreted it... (too lazy to upload a pic)
i have Enterprise stationary (on a planet) with Kinglon whizzing past going left...
and then i get lost with where the lasers are going...
Post by: Rathin on January 09, 2017, 05:32:53 pm
im having trouble visualising this... pls help!
the star ship enterprise Enterprise is stationary in space to make repairs to the warp
drive. Suddenly a Kinglon vessel appears from the right and flashes towards and past the Enterprise at half the speed
of light.
When it is exactly in front of the Enterprise, Captain Kirk fires port and starboard at the same time.
so this is how i interpreted it... (too lazy to upload a pic)
i have Enterprise stationary (on a planet) with Kinglon whizzing past going left...
and then i get lost with where the lasers are going...
To the observer in the Kinglon the lasers go straight as they are in their own inertial reference frame (assumed constant motion). To an observer the laser has travelled the opposite direction in which Kinglon is travelling. I am also kinda confused on what the info is with all these weird names and unstructured info but here is my 2 cents..hope it helps.
Post by: Aaron12038488 on January 09, 2017, 05:51:53 pm
Outline how modulation of amplitude or frequency of visible light, microwaves and/or radio waves can be used to transmit information
Post by: RuiAce on January 09, 2017, 05:57:50 pm
for this dot point, do i have to talk about all three or can I just do Radio Waves?It is the same for all three of them.
Outline how modulation of amplitude or frequency of visible light, microwaves and/or radio waves can be used to transmit information
Post by: Rathin on January 09, 2017, 06:04:23 pm
for this dot point, do i have to talk about all three or can I just do Radio Waves?
Outline how modulation of amplitude or frequency of visible light, microwaves and/or radio waves can be used to transmit information
For this normally you either have to describe/explain/discuss AM and FM. So make sure you know how the carrier way has its Amplitude or Frequency modulated for information to be carried and make sure you have Advantages and Disadvantages for both and discuss questions are..no pun intended..very frequent.
Post by: katnisschung on January 09, 2017, 06:37:49 pm
because the speed of light is absolute (3 x 10^8)
and light is an em wave its safe to assume
the speed of ALL EM waves are absolute
Post by: RuiAce on January 09, 2017, 06:41:49 pm
ok so probably a stupid question butYes. The speed of all EMR is constant
because the speed of light is absolute (3 x 10^8)
and light is an em wave its safe to assume
the speed of ALL EM waves are absolute
Of course, when Einstein published his theory of special relativity, it discussed this matter with respect to visible light. But the nature of EMR allows this point to be strengthened into the fact that the speed of all EMR must be c. That being said, you should still refer to the speed of 'light' being constant, when writing 3 mark responses regarding special relativity.
Post by: jamonwindeyer on January 09, 2017, 11:24:18 pm
im having trouble visualising this... pls help!
the star ship enterprise Enterprise is stationary in space to make repairs to the warp
drive. Suddenly a Kinglon vessel appears from the right and flashes towards and past the Enterprise at half the speed
of light.
When it is exactly in front of the Enterprise, Captain Kirk fires port and starboard at the same time.
so this is how i interpreted it... (too lazy to upload a pic)
i have Enterprise stationary (on a planet) with Kinglon whizzing past going left...
and then i get lost with where the lasers are going...
This is majorly confusing; I don't blame you for being lost :P
I'm with Rathin's interpretation. To simplify it, here is (I think) the exact same scenario without all the Star Trek references:
A train travelling at half the speed of light races past a platform, from right to left. The train is the same length as the platform. When the train and platform are directly aligned, fireworks are let off at either end of the platform.
Here, the ships are replaced with a platform and train, and the lasers with fireworks. It is just a weird version of Einstein's thought experiment, I think :)
Post by: Aaron12038488 on January 10, 2017, 09:36:07 am
How do I go about in this dot point.
Post by: jamonwindeyer on January 10, 2017, 10:49:13 am
Discuss problems produced by the limited range of the electromagnetic spectrum available for communication purposes.
How do I go about in this dot point.
Basically your points here will concern the fact that we only have a limited number of channels. Because we can only use a certain range of the EM spectrum safely, that limits how many frequencies we can use to communicate.
Practically, this causes a congestion of frequencies. We need to allocate frequencies to certain purposes and ensure no one overlaps; without these precautions, you'd get interference everywhere. For example, FM radio stations get frequencies somewhere near 80-110 MHz in Australia (with some research you could probably get the actual allocations), these are the only ones they can use.
Really, you can summarise this in your notes as just congestion of frequencies, I didn't have much more than that :)
Post by: Aaron12038488 on January 10, 2017, 10:52:44 am
Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer.
For this dot point, do we have to talk about each one, or just in general
Post by: jamonwindeyer on January 10, 2017, 10:54:55 am
Thanks, one other question,
Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer.
For this dot point, do we have to talk about each one, or just in general
Each one; they are all slightly different (light is especially different to radio/micro waves) :)
Post by: Rathin on January 10, 2017, 10:55:47 am
Discuss problems produced by the limited range of the electromagnetic spectrum available for communication purposes.
How do I go about in this dot point.
The main reason of the limiting the range assigned for communication is that the government is attempting to avoid interference with other radio signals such as the emergency services. This creates problems as it limits the bandwidth allocated for communication from 500 kHz to 300 GHz which then creates a congestion of frequencies.
To minimise these problem scientists are researching the possibility of using the frequencies near Infrared waves to increase the number of users.
The use of cables also increases the number of users as it allows far greater volume of communications as cable networks have no bandwidth limitations and extra cables can be installed.
Post by: Rathin on January 10, 2017, 10:57:40 am
Thanks, one other question,
Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer.
For this dot point, do we have to talk about each one, or just in general
- Reflection of light is used in optic fibres which allows massive amounts of information transfer.
- Radio waves reflects off the ionosphere which allows the information to be transferred long distances.
- Microwaves are used in radars which reflect off objects to determine the distance of that object.
Post by: jamonwindeyer on January 10, 2017, 11:00:40 am
Post by: Aaron12038488 on January 10, 2017, 11:02:33 am
How much maths is involved in Preliminary and HSC Physics?
Post by: RuiAce on January 10, 2017, 11:03:13 am
Just out of curiosity,About the same difficulty as general maths.
How much maths is involved in Preliminary and HSC Physics?
Quantity wise, there's probably a 30%-70% split between maths and theory, or something like that
Post by: jamonwindeyer on January 10, 2017, 11:05:21 am
Just out of curiosity,
How much maths is involved in Preliminary and HSC Physics?
There is probably a bit more in Prelim, because you'll deal with vectors, which involves a fair bit of trigonometry!
As long as you can rearrange equations, do some basic right-angled trigonometry (and perhaps a few other little bits and pieces, a bit of geometry) you'll be totally sweet :)
Post by: jakesilove on January 10, 2017, 11:18:13 am
Just out of curiosity,
How much maths is involved in Preliminary and HSC Physics?
Basically, if you can sub numbers into equations, and do some algabraic rearranging, you'll be fine :) I know plenty of people who weren't very confident with mathematics, but after enough practice could smash out Physics maths questions with ease! Whether you're brilliant at Maths, or shudder at the thought of it, with enough practice you'll be fine :)
Post by: Rathin on January 10, 2017, 11:29:41 am
Post by: jamonwindeyer on January 10, 2017, 11:31:26 am
Surprisingly the knowledge in Vectors in prelim physics helped me in complex vectors!
Yeah ditto!! The new syllabus has 3 dimensional vectors being used, so in a few years it will be even more useful :)
Post by: RuiAce on January 10, 2017, 11:35:33 am
Surprisingly the knowledge in Vectors in prelim physics helped me in complex vectors!In my MX2 class, there were only 3 students including me. 2 of which did physics.
The guy who didn't do physics was the one having difficulty at the start with vectors. :P
Yeah ditto!! The new syllabus has 3 dimensional vectors being used, so in a few years it will be even more useful :)I hope they use the matrix notation though and not the dumb i j k stuff... :P
Post by: claudia.augustine on January 12, 2017, 05:08:20 pm
Post by: RuiAce on January 12, 2017, 05:13:07 pm
How is the gravitational potential energy formula (E= -GMm/r) derived? I get why it's negative, im just confused on why the r is not squared like that in Newton's universal gravitation lawGo back to prelim.
W=Fs. What was that formula again?
Post by: Rathin on January 12, 2017, 05:17:33 pm
How is the gravitational potential energy formula (E= -GMm/r) derived? I get why it's negative, im just confused on why the r is not squared like that in Newton's universal gravitation law
Equate F(G)=GMm/r^2 and F(G)=mg
to get E(p)=GMm/r
BUT We take a reference point at infinity where E(p)=0 so any finite distances will be trivially negative..THUS E(p)=-GMm/r
Post by: RuiAce on January 12, 2017, 05:24:00 pm
Equate F(G)=GMm/r^2 and F(G)=mgNot too sure how that one works... that just gives g=GM/r^2
to get E(p)=-GMm/r
Post by: Rathin on January 12, 2017, 05:26:46 pm
Not too sure how that one works... that just gives g=GM/r^2
Yes that is correct, now we take a reference point at infinity at which E(p)=0 so any finite distances (aka below our reference point) is negative..thus the whole quantity 'Gravitational Potential Energy' is always negative.
Post by: RuiAce on January 12, 2017, 05:28:31 pm
Yes that is correct, now we take a reference point at infinity at which E(p)=0 so any finite distances (aka below our reference point) is negative..thus the whole quantity 'Gravitational Potential Energy' is always negative.And how does g=GM/r^2 magically become Ep=-GmM/r?
Look again. I don't believe you answered the question
Post by: claudia.augustine on January 12, 2017, 05:36:35 pm
W=Fs. What was that formula again?
Ahh so it would be,
w= Fs
= GMm/r^2 x r
= GMm/r
Post by: RuiAce on January 12, 2017, 05:37:14 pm
Ahh so it would be,The basic idea, yep. :)
w= Fs
= GMm/r^2 x r
= GMm/r
Post by: Rathin on January 12, 2017, 05:39:37 pm
And how does g=GM/r^2 magically become Ep=-GmM/r?
Look again. I don't believe you answered the question
Oh damm.. didn't read question properly..I thought OP was asking for why its negative..soz
Post by: Iminschool on January 12, 2017, 05:53:33 pm
Post by: RuiAce on January 12, 2017, 05:57:36 pm
I genuinely don't know how to do this questionThis is related to the motors and generators topic. Have you been taught stuff regarding electromagnetic induction and Lenz's law?
Post by: Iminschool on January 12, 2017, 06:04:49 pm
Post by: RuiAce on January 12, 2017, 06:30:00 pm
Recall that by the principle of electromagnetic induction, a change in magnetic flux is required to induce a force into an object. For this response, I will assume that the iron rod is just sitting there, and not a part of the circuit. This is because the spring is what's submerged in the mercury solution.
At the instant the switch is turned on, conventional current will flow out of the positive terminal and upwards. The current will then flow into the spring. Recall from studies in preliminary that if an electrically conductive substance (wire or something) has a charge flowing through it, it will produce its own magnetic field.
By using the right-hand grip rule, if the fingers point in the direction of the spring (the spring is a COILED wire), the thumb points in the direction of the magnetic field produced. Hence, the magnetic field travels through the spring from top to bottom (so the bottom is a north).
This suddenly newly produced magnetic field invoked a change in flux around the iron rod. As per the principle of electromagnetic induction, the iron rod will produce its own magnetic flux (briefly), and as per Lenz's law this will be to counteract the change in flux imposed by the spring.
This change in flux caused by the iron rod then affects the spring. And thus as per electromagnetic induction, the spring is now affected by a force, which causes it to bounce.
JAMON you need to fix up my response
Post by: jamonwindeyer on January 12, 2017, 07:40:16 pm
I genuinely don't know how to do this question
Here's my explanation, use it with Rui's ;D
We need to know first about electromagnetic induction. A conductor that experiences a changing magnetic flux will have a current induced through it. This current is a manifestation of the conservation of energy (but that isn't necessary to explain here). The current acts to establish a new magnetic field (remember all currents are surrounded by a magnetic field), and according to Lenz's Law, this magnetic field will act in the opposite direction to the one that created it. That is, it opposes the change that created it.
Now when the switch is closed, it is a direct current that flows through the spring. Normally, we don't associate direct currents with changing magnetic fields and electromagnetic induction, since the current flows in a single direction and so establishes a constant magnetic field, not a changing one. However, that's long term - There is still a brief moment where the magnetic field around the spring changes. As the current goes from nothing, to something, that causes a changing magnetic field. This is the tough bit to understand - If it's a little confusing let me know!
We don't really care about direction in this scenario. The spring acts as a solenoid, and as the current through it increases, that creates a changing magnetic field. The iron rod experiences this changing magnetic field and will briefly have currents induced through it (eddy currents). These currents create a new magnetic field, and this interacts with the magnetic field of the spring to exert a force which causes it to bounce. Kind of like a bar magnet pushing on another bar magnet, sort of. This interaction is very quick, it only happens as the current begins flowing through the spring. Once it is flowing properly the force on the spring is gone.
If you want to get super technical, the movement of the bouncing spring may cause the rod to experience another small change in magnetic flux which could cause another force. However, this process wouldn't sustain itself (you'd lose energy to heat in the rod and the bouncing would fade away anyway) :)
Post by: Iminschool on January 12, 2017, 08:57:58 pm
Here's my explanation, use it with Rui's ;D
We need to know first about electromagnetic induction. A conductor that experiences a changing magnetic flux will have a current induced through it. This current is a manifestation of the conservation of energy (but that isn't necessary to explain here). The current acts to establish a new magnetic field (remember all currents are surrounded by a magnetic field), and according to Lenz's Law, this magnetic field will act in the opposite direction to the one that created it. That is, it opposes the change that created it.
Now when the switch is closed, it is a direct current that flows through the spring. Normally, we don't associate direct currents with changing magnetic fields and electromagnetic induction, since the current flows in a single direction and so establishes a constant magnetic field, not a changing one. However, that's long term - There is still a brief moment where the magnetic field around the spring changes. As the current goes from nothing, to something, that causes a changing magnetic field. This is the tough bit to understand - If it's a little confusing let me know!
We don't really care about direction in this scenario. The spring acts as a solenoid, and as the current through it increases, that creates a changing magnetic field. The iron rod experiences this changing magnetic field and will briefly have currents induced through it (eddy currents). These currents create a new magnetic field, and this interacts with the magnetic field of the spring to exert a force which causes it to bounce. Kind of like a bar magnet pushing on another bar magnet, sort of. This interaction is very quick, it only happens as the current begins flowing through the spring. Once it is flowing properly the force on the spring is gone.
If you want to get super technical, the movement of the bouncing spring may cause the rod to experience another small change in magnetic flux which could cause another force. However, this process wouldn't sustain itself (you'd lose energy to heat in the rod and the bouncing would fade away anyway) :)
Wow, you and Rui went in hard on this one and its much appreciated. Mb i forgot to think of the little details in my answer. Your explanation was really good and helped me further understand electromagnetic induction.
Cheers
Post by: jamonwindeyer on January 12, 2017, 10:04:37 pm
Wow, you and Rui went in hard on this one and its much appreciated. Mb i forgot to think of the little details in my answer. Your explanation was really good and helped me further understand electromagnetic induction.
Cheers
Awesome! Happy to help ;D
Post by: kiwiberry on January 14, 2017, 08:40:39 pm
In a DC motor, is the torque on the coil maximum when the plane of the coil is parallel to the magnetic field?
Post by: bsdfjnlkasn on January 14, 2017, 09:16:07 pm
Hey,
In a DC motor, is the torque on the coil maximum when the plane of the coil is parallel to the magnetic field?
Hey kiwiberry,
You're right and here's why:
Since the expression for torque is: τ=nBIA cos θ we must look at the angle between the magnetic field and the plane of the coil to address cos θ. When they are parallel the angle is 0° and we know that cos0=1 essentially making τ=nBIA . This would give the maximum value for torque in a DC motor as any angle >0 will result in a smaller τ value.
Hopefully that helped! :)
Post by: kiwiberry on January 15, 2017, 12:15:16 am
Hey kiwiberry,
You're right and here's why:
Since the expression for torque is: τ=nBIA cos θ we must look at the angle between the magnetic field and the plane of the coil to address cos θ. When they are parallel the angle is 0° and we know that cos0=1 essentially making τ=nBIA . This would give the maximum value for torque in a DC motor as any angle >0 will result in a smaller τ value.
Hopefully that helped! :)
Ah that makes sense, thank you!! :)
Post by: katnisschung on January 19, 2017, 02:39:21 pm
idk it helps me when i derive it...
Im trying to derive the length dilation formula and im stuck
(probably becos my algebra skills are terrible and ask me why i took physics...becos i was a fool)
i get that u need to equate the time dilation equation to get length dilation
i just dont understand how the denominator in t2 came to be positive??
becos forward journey= ct1=L_v+vt1 (note v is supposed to be an underscore for L)
t1=(L_v)/(c-v)
return journey = ct2=L_v+vt2
how did the denominator become c+v??
am i missing sth i don't understand
Post by: jamonwindeyer on January 19, 2017, 03:18:28 pm
ok i suppose i could go about memorising the formulas but
idk it helps me when i derive it...
Im trying to derive the length dilation formula and im stuck
(probably becos my algebra skills are terrible and ask me why i took physics...becos i was a fool)
i get that u need to equate the time dilation equation to get length dilation
i just dont understand how the denominator in t2 came to be positive??
becos forward journey= ct1=L_v+vt1 (note v is supposed to be an underscore for L)
t1=(L_v)/(c-v)
return journey = ct2=L_v+vt2
how did the denominator become c+v??
am i missing sth i don't understand
Hey Katniss! Purely algebraically you are right, but I think you'll find it has something to do with the direction of \(v\) - Since it is moving in the opposite direction on return, \(v\) will take a new sign. At least, that's my guess - Without seeing the full proof I'm not sure, and this is only something I've seen derived once (and a long time ago too, lol).
Just saying, you are way beyond the scope of the course at this point. I'm a massive fan of deriving formulae, but this is definitely not worth stressing over! Good on you for pushing hard to extend yourself. :)
Post by: katnisschung on January 19, 2017, 03:25:11 pm
oh i see now.... i was racking my brains before trying to figure out why it was pos.
yeah i might stick to memorising the relativity formulae ahha
Post by: kiwiberry on January 19, 2017, 03:52:07 pm
ok i suppose i could go about memorising the formulas but
idk it helps me when i derive it...
Im trying to derive the length dilation formula and im stuck
(probably becos my algebra skills are terrible and ask me why i took physics...becos i was a fool)
i get that u need to equate the time dilation equation to get length dilation
i just dont understand how the denominator in t2 came to be positive??
becos forward journey= ct1=L_v+vt1 (note v is supposed to be an underscore for L)
t1=(L_v)/(c-v)
return journey = ct2=L_v+vt2
how did the denominator become c+v??
am i missing sth i don't understand
Just to elaborate :)
The return journey is supposed to be ct2=L_v-vt2. On the way back, as the train is travelling at velocity v, it will have moved a distance of vt2 in the opposite direction that the light is travelling before the light reaches the end. This means that on the return journey, the light travels a shorter distance than the forward journey to reach the end, this shorter distance being L_v-vt2!
Post by: Rathin on January 19, 2017, 05:08:26 pm
Post by: bholenath125 on January 20, 2017, 04:58:01 pm
Is there a site that can explain it better?
Post by: jamonwindeyer on January 20, 2017, 05:26:17 pm
I am having significant difficulty in motors and generators in understanding how the motor rotates. Like F= bniCostheta
Is there a site that can explain it better?
I wrote a little guide on this here which could shed some light!
You should also try looking on Youtube for a video explaining it - A visual picture of how the forces involved produce a rotation can be really beneficial in making it a little clearer ;D try looking up 'HSC How a Motor Works' or something similar ;D
Post by: katnisschung on January 21, 2017, 04:47:14 pm
but i get the quickest responses in this thread :)
what topics r u guys planning to cover in the
physics
2u maths
and biology lecture?
i know schools can tend to cover topics in different order
so just wanting a heads up... also really enjoyed the lectures
u guys ran in sept/oct
Post by: kiwiberry on January 21, 2017, 07:55:47 pm
probably not the most appropriate thing to post here
but i get the quickest responses in this thread :)
what topics r u guys planning to cover in the
physics
2u maths
and biology lecture?
i know schools can tend to cover topics in different order
so just wanting a heads up... also really enjoyed the lectures
u guys ran in sept/oct
i would also like to know about the chem and 3u maths lectures :)
Post by: Rathin on January 21, 2017, 08:27:33 pm
Post by: Aaron12038488 on January 22, 2017, 10:09:04 am
Post by: jamonwindeyer on January 22, 2017, 10:21:48 am
So my parents are deciding to get two tutors for me for Physics. I have started tutoring for 1 of them which is individual tutoring. Is it bad to have two tutors, as they may teach different methods. In the upside is that I have more resources and greater understanding.
Two tutors is two more than I ever had 8)
Like, I personally think its a bit excessive to have two tutors for the same subject. Of course - More understanding, more resources, more time spent on the subject. But like, two tutors is (probably) ludicrously expensive and they are teaching you the exact same stuff. They may have different approaches/methodologies, and that could help in small ways, but the fact is that once you start the second tutoring session, you are going to be playing a session essentially on repeat. You should, if the first tutor did their job, understand things relatively well before the second tutor even says a word.
I don't see anything bad happening because of two tutors mind you. It's just like, why, you know what I mean? You'd be better off spending that money on buying a practice question booklet and doing individual practice questions, to be honest (in my opinion obviously).
PS - When you say your parents decided, do YOU want two tutors?
Post by: Aaron12038488 on January 22, 2017, 11:11:36 am
Post by: jamonwindeyer on January 22, 2017, 11:45:49 am
My second tutor is not individual, it is a group class. My parents believe as my tutor is a Uni student, that he will not be fully committed and also has no experience so as a precaution they will be looking for another tutor. Personally I totally agree with you, I will rather spend on practice books and past papers so I can apply my knowledge rather than learning content again. With school starting in a week, I Personally believe that my current tutor is doing a good job and with school lurking around the corner, I think I will be fine.
Ahh right, cool cool. Well whatever works for you - Nothing bad will happen because of having two tutors. The Physics content is fairly non-interpretive and usually there is only one way to do a problem. Just be aware that two classes/tutors isn't going to give anywhere near the value for money as having just one person :)
Post by: Rathin on January 22, 2017, 12:27:52 pm
Post by: bsdfjnlkasn on January 24, 2017, 09:48:55 am
1. Would a planet rotating quicker or slower increase the gravitational acceleration according to
2. How is an object's GPE and Work related (on Earth)? So for expressions: GPE = mgh and W = Fs
Post by: jakesilove on January 24, 2017, 10:32:01 am
Hey I have a few Space questions that I was hoping to get some help with :)
1. Would a planet rotating quicker or slower increase the gravitational acceleration according to
2. How is an object's GPE and Work related (on Earth)? So for expressions: GPE = mgh and W = Fs
1. Rotation doesn't actually affect the formula, and definitely isn't something relevant to the curriculum. However, as we know that high velocities increase the mass of an object, greater spin will result in higher mass. That increases g!
2. An object doesn't have 'work', but it does have GPE. 'Work' is a force times a distance, so only occurs when work is done ON or BY the object.
Jake
Post by: bsdfjnlkasn on January 24, 2017, 10:47:27 am
2. An object doesn't have 'work', but it does have GPE. 'Work' is a force times a distance, so only occurs when work is done ON or BY the object.
Jake
Sorry I should have been clearer with my second question.
What I meant was: What makes work when done on an object (across a horizontal plane) comparable to GPE which acts vertically? Is there no difference? Is it something I just have to accept - that they're essentially the same just GPE accounts for gravity because it exists and therefore will affect the force?
Also, could you please check my notes for the equation GPE = mgh?
o This equation is only really accurate when we use it to calculate the GPE of an object on Earth as g is taken as a constant and hence doesn’t account for g decreasing as altitude increases
• It assumes that any object which has a GPE=0 on Earth can be taken as a reference point for all GPE calculations in Space
• To rectify this, we need a universal definition of GPE that takes into account the equation for Gravitational acceleration
And some more questions: 8)
When trying to understand why the expression for potential energy is negative:
we take the theoretical case that an object is infinitely away from a gravitational field and so has a GPE=0. So when it is pushed towards this field, it begins losing Ep (as kinetic E is gained) which because it was initially 0 means a negative Ep value. Would the rate at which the Ep decreases resemble an exponential graph? I'm just thinking this because the object obviously has to accelerate (does it ever hit terminal velocity?) which stops there from being a linear relationship between the transfer of Ep to kinetic E.
Sorry i'm not being really technical here and could be completely wrong but I just thought this might be an easy question to answer (my textbook doesn't say anything about this so the question could also be completely irrelevant lol)
Thanks Jake!
Post by: Shadowxo on January 24, 2017, 12:19:09 pm
Hey I have a few Space questions that I was hoping to get some help with :)
1. Would a planet rotating quicker or slower increase the gravitational acceleration according to
2. How is an object's GPE and Work related (on Earth)? So for expressions: GPE = mgh and W = Fs
Hi,
Not sure if you need to know for HSC but
1. The centripetal (rotational force) = mv2/r, so as v increases, the centripetal force increases
When a planet rotates, the centripetal force = gravitational force
So if the velocity is increased, the centripetal force increases and therefore so must the gravitational force, and as Fg = GMm/r2, r decreases.
Basically, it rotates faster when it's closer to the centre of the earth, and if it's closer then g is more.
Extra:
mv2/r = GMm/r2
v2=GM/r
GM (constant for a particular planet) = v2r
So as velocity increases, r decreases
2. Work is like a transfer of energy.
Work = Fd = ma*d.
gpe = mgh
You can do work on an object vertically to give it more gpe, as the change in work = change in gpe (as a of object= g)
So you can increase an object's gpe by doing work on it (vertically). It's like a transfer of energy, if done vertically the work is converted into gpe
Horizontally, the work is converted into ke (gpe unaffected horizontally)
Don't know if this will help but hope so :)
With your notes :
- Yes when g = 9.8 but g=GM/r2 so can be different. To accurately find gpe you'd need a graph showing the change in g for long distances where this value would change
- gpe isn't really calculated in space, just know that it's higher further away. gpe is mostly used for changes in energy, so change in (∆) gpe = mg∆h
Not sure about that last question, don't recall learning it :P
Post by: davidss on January 24, 2017, 09:34:17 pm
can someone help with these multiple choice questions. Thanks :)
Post by: kiwiberry on January 24, 2017, 10:30:34 pm
Hi,
can someone help with these multiple choice questions. Thanks :)
The last one can be done using Kepler's Law
Remember that M refers to the mass of the central body (Saturn), not the mass of the moons orbiting. So firstly we need to find M by subbing in the given data for Titan (and using the orbital radius for r as this already includes the radius of the planet)
Now we just use this to find T for Rhea!
So the answer is B :)
Post by: RuiAce on January 24, 2017, 11:21:22 pm
Hi,Q10 and Q9 both involve considering the principle of electromagnetic induction.
can someone help with these multiple choice questions. Thanks :)
In Q10, the graph for coil X should be obvious. Once it's turned on, current starts flowing through it. Once it's turned off, it will no longer be. This limits your answer to A and C.
Recall that by the principle of electromagnetic induction, a current is induced into a conductor if there is a CHANGE in the magnetic flux around it. At what point does the change in flux occur?
At the instant the switch is turned on, coil X produces a magnetic field (you learnt this in prelim, and the direction is determined by the right-hand grip rule). This means that all in a sudden, there was a CHANGE in magnetic flux. Because the magnetic field was not there prior to the switch being turned on, and once the switch is turned on it will stay there.
So because ONLY at THAT instant, there was a change in flux, you can already see why A must be the correct answer. I'll leave you to figure out the rest of the puzzle that guarantees it.
(I don't know how to properly explain Q9 anymore. Will let someone take over.)
The last one can be done using Kepler's LawNice LaTeX
Remember that M refers to the mass of the central body (Saturn), not the mass of the moons orbiting. So firstly we need to find M by subbing in the given data for Titan (and using the orbital radius for r as this already includes the radius of the planet)
Now we just use this to find T for Rhea!
So the answer is B :)
Post by: davidss on January 25, 2017, 06:52:55 am
Post by: bsdfjnlkasn on January 25, 2017, 10:25:07 pm
I know lectures are underway and that it's a busy time so I won't mind if my questions aren't answered for a few days - they're not too urgent. (Plus, you guys deserve a break! :P)
So regarding the outcome: "Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket" I'm a little confused about some of the details that relate to Earth specifically.
My textbook first gives the definition for angular speed as
"Angular speed: How fast the angle of a line that joins the object with the centre is changing" and this is really confusing to me as I don't understand how this angle can change if an object travelling in a circle sustains tangential velocity. Doesn't this mean that the angle held at any instant by the circular-moving object and the centre is 90o ?
2. What is an object's linear speed? If it increases, will it increase the object's angular speed also?
3. How do larger linear speed reduce the effects of gravity?
4. Is Earth's elliptical shape a result of it's rotational axis?
5. What provides Earth's centripetal force which allows it to orbit? Is it the gravitational fields of other, larger celestial bodies?
Now for some questions that are separate from the quoted outcome:
1. How does the linear orbit velocity acting on a satellite cause it to move away from the Earth?
Thank you :)
Post by: Spencerr on January 26, 2017, 12:21:27 am
So regarding the outcome: "Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket" I'm a little confused about some of the details that relate to Earth specifically.
Thank you :)
In this question which is generally worth 3/4 marks. There are two key things you have to mention.
1) Launching from the Earth's equator in an easterly direction to utilise the Earth's rotational velocity.
2) Using the Earth's orbital velocity around the Sun
A rocket launched from Earth would be launched from the equator in an easterly direction in order to add the Earth's rotational velocity to it's own velocity. This adding of velocities is quite similar to the idea of launching an aeroplane from a moving platform. Let's say the aeroplane is launched at 5 m/s towards the right. If the platform is still and not moving, the aeroplane would only travel at 5 m/s towards the right. However if the platform was also moving towards the right at 10m/s. The aeroplane would be able to add on the platform's velocity to its own velocity and instead travel at 5 m/s. Intuitively, you can imagine a similar thing happening when javelin throwers throw their spears. The point of their run up is to increase the javelin's speed, which will allow it to travel further through the air. In the case of Earth, Earth rotates at a certain high speed around it's axis and this speed is a MAXIMUM at the equator. So if a rocket was launched at the equator, it would gain Earth's rotational velocity around its axis for FREE (without the additional use of costly fuel). This is very helpful as it can help the rocket reach the necessary ESCAPE VELOCITY and enter a stable parking orbit around the Earth before the next event occurs.
The Earth orbits around the sun with a certain velocity (that is it's tangential/linear velocity). Once the rocket enters into a parking orbit around the Earth, it can be boosted out of the orbit and into Space. However, there is a relatively tight time frame involved as the Earth's tangential velocity must align with the trajectory of the rocket. (You can imagine the tangential velocity of Earth's orbit around the Sun constantly changing as the Earth moves around the Sun). Similar situation with the moving platform here- once the rocket is boosted out of orbit, it can add to it's own velocity relative to the sun, the orbital velocity of the Earth without the use of additional fuel. This is really helpful as it reduces the amount of fuel needed for the rocket to attain such a high velocity and it increases the payload the rocket can travel, allowing interplanetary travel possible!
Hope this has helped :) I've included more information than what is needed for a 3/4 marker though!
Post by: Mayalily on January 26, 2017, 11:08:29 am
The first part of the question requires you to find the Ep of a satellite of M=500kg, at 10,000km altitude.
i've subbed this into -G m1m2/r
hence:
-6.67*10-11* [500*(6.0*1024)]/(6371000+10,000,000) - Where 6371000 is the radius of the earth in metres and 10,000,000 is the altitude in metres.
But I get -1.222*1010J, rather than the answer; -3.119*1010J
Am I doing something wrong that i've missed or is the wrong answer written down?
Post by: wyzard on January 26, 2017, 12:01:13 pm
I have a question which i am getting close to the answe with, but not quite, and I have no idea why and I'm quite confused and frustrated.
The first part of the question requires you to find the Ep of a satellite of M=500kg, at 10,000km altitude.
i've subbed this into -G m1m2/r
hence:
6.67*10-11* [500*(6.0*1024)]/(6371000+10,000,000) - Where 6371000 is the radius of the earth in metres and 10,000,000 is the altitude in metres.
But I get 1.222*1010J, rather than the answer; -3.119*1010J
Am I doing something wrong that i've missed or is the wrong answer written down?
You forgot the negative sign.
The only possible other place where there might be an error is that is the question referring to Earth or some other planet, since it is unspecified?
I've crunched the numbers and got the same answer as you did, so very likely the answer they'd given is wrong. Might want to verify with others in this forum though :P
Post by: Rathin on January 26, 2017, 12:18:57 pm
I have a question which i am getting close to the answe with, but not quite, and I have no idea why and I'm quite confused and frustrated.
The first part of the question requires you to find the Ep of a satellite of M=500kg, at 10,000km altitude.
i've subbed this into -G m1m2/r
hence:
6.67*10-11* [500*(6.0*1024)]/(6371000+10,000,000) - Where 6371000 is the radius of the earth in metres and 10,000,000 is the altitude in metres.
But I get 1.222*1010J, rather than the answer; -3.119*1010J
Am I doing something wrong that i've missed or is the wrong answer written down?
Also got -1.22x10^10 J
Post by: bsdfjnlkasn on January 26, 2017, 12:29:14 pm
Hi,
Not sure if you need to know for HSC but
1. The centripetal (rotational force) = mv2/r, so as v increases, the centripetal force increases
When a planet rotates, the centripetal force = gravitational force
So if the velocity is increased, the centripetal force increases and therefore so must the gravitational force, and as Fg = GMm/r2, r decreases.
Basically, it rotates faster when it's closer to the centre of the earth, and if it's closer then g is more.
Hey shadowxo,
Spoiler
Thanks for the detailed answer! We definitely learn about the centripetal force and how it is provided by the gravitational attraction force in space.
I'm a bit confused by how you connected centripetal acceleration with gravitational acceleration (g) in the last line. Is it just because the formula for g has the same variable r (or d because they represent the same distance) in it and because r decreases in the attraction force, it must also decrease for the gravitational acceleration? I'm just not really seeing the jump between the gravitational force and acceleration. Also, thank you for providing the 'extra' bit an explanation of work and Ep - they made perfect sense.
I'm a bit confused by how you connected centripetal acceleration with gravitational acceleration (g) in the last line. Is it just because the formula for g has the same variable r (or d because they represent the same distance) in it and because r decreases in the attraction force, it must also decrease for the gravitational acceleration? I'm just not really seeing the jump between the gravitational force and acceleration. Also, thank you for providing the 'extra' bit an explanation of work and Ep - they made perfect sense.
For anyone else who wants to help me out with my questions - I still have a few left :)
1. I get really confused by angular velocity, my textbook first gives the definition for angular speed as
"Angular speed: How fast the angle of a line that joins the object with the centre is changing" and this is really confusing to me as I don't understand how this angle can change if an object travelling in a circle sustains tangential velocity. Doesn't this mean that the angle held at any instant by the circular-moving object and the centre is 90°?
2. What is an object's linear speed? If it increases, will it increase the object's angular speed also?
3. How do larger linear speed reduce the effects of gravity?
4. Is Earth's elliptical shape a result of it's rotational axis?
5. What provides Earth's centripetal force which allows it to orbit? Is it the gravitational fields of other, larger celestial bodies?
6. How does the linear orbit velocity acting on a satellite cause it to move away from the Earth?
Thank you :)
Post by: de on January 26, 2017, 01:44:38 pm
For anyone else who wants to help me out with my questions - I still have a few left :)
1. I get really confused by angular velocity, my textbook first gives the definition for angular speed as
"Angular speed: How fast the angle of a line that joins the object with the centre is changing" and this is really confusing to me as I don't understand how this angle can change if an object travelling in a circle sustains tangential velocity. Doesn't this mean that the angle held at any instant by the circular-moving object and the centre is 90°?
2. What is an object's linear speed? If it increases, will it increase the object's angular speed also?
3. How do larger linear speed reduce the effects of gravity?
4. Is Earth's elliptical shape a result of it's rotational axis?
5. What provides Earth's centripetal force which allows it to orbit? Is it the gravitational fields of other, larger celestial bodies?
6. How does the linear orbit velocity acting on a satellite cause it to move away from the Earth?
Thank you :)
1. Angular velocity is easy to think about to start with as an average speed. So the average angular speed is defined as
2. If you mean by linear speed the tangential velocity, then yep (assuming the point stays the same distance from the centre of the circle).
If you mean the speed with which the centre is travelling. That is independent of the speed of a orbiting point.
3. Not quite sure what you mean here or what gravity you're talking about or what you mean by linear speed.
4. Again no idea. Surely the shape of the earth isn't a result of the fact you can draw a line through it? I might be being dense...
5. Yes other planets, the sun, etc, their fields act on the earth providing a radial acceleration and thus various orbits occur.
6. Linear velocity doesn't really "act" on a satellite. But, yes, if an external force acts on a satellite tangentially to its orbit then the velocity component in that direction increases and since v^2/r=a and the gravitational field is constant (a) r must also increase.
Post by: Rathin on January 26, 2017, 02:52:30 pm
Hey shadowxo,SpoilerThanks for the detailed answer! We definitely learn about the centripetal force and how it is provided by the gravitational attraction force in space.
I'm a bit confused by how you connected centripetal acceleration with gravitational acceleration (g) in the last line. Is it just because the formula for g has the same variable r (or d because they represent the same distance) in it and because r decreases in the attraction force, it must also decrease for the gravitational acceleration? I'm just not really seeing the jump between the gravitational force and acceleration. Also, thank you for providing the 'extra' bit an explanation of work and Ep - they made perfect sense.
For anyone else who wants to help me out with my questions - I still have a few left :)
1. I get really confused by angular velocity, my textbook first gives the definition for angular speed as
"Angular speed: How fast the angle of a line that joins the object with the centre is changing" and this is really confusing to me as I don't understand how this angle can change if an object travelling in a circle sustains tangential velocity. Doesn't this mean that the angle held at any instant by the circular-moving object and the centre is 90°?
2. What is an object's linear speed? If it increases, will it increase the object's angular speed also?
3. How do larger linear speed reduce the effects of gravity?
4. Is Earth's elliptical shape a result of it's rotational axis?
5. What provides Earth's centripetal force which allows it to orbit? Is it the gravitational fields of other, larger celestial bodies?
6. How does the linear orbit velocity acting on a satellite cause it to move away from the Earth?
Thank you :)
Q4: When the Earth rotates around its axis it creates a apparent centrifugal effect which is greatest at the equator and less at the poles. This cohesive force causes the earth to 'bulge' at the equator due to the apparent value of g decrease and 'flatten' at the poles due to a increase in the value of g.
Q5: Gravitational Force is like a frictional force that provides the centripetal force. E.g When a car is going around a circle a centripetal force is provided by the friction on the road, similarly Centripetal force is provided by the Gravitational force thus we get the Fc=Fg (this is how the escape velocity formula is derived)
EDIT: Fixed my silly mistake pointed out by Jamon
Post by: Shadowxo on January 26, 2017, 03:20:58 pm
Hey shadowxo,SpoilerThanks for the detailed answer! We definitely learn about the centripetal force and how it is provided by the gravitational attraction force in space.
I'm a bit confused by how you connected centripetal acceleration with gravitational acceleration (g) in the last line. Is it just because the formula for g has the same variable r (or d because they represent the same distance) in it and because r decreases in the attraction force, it must also decrease for the gravitational acceleration? I'm just not really seeing the jump between the gravitational force and acceleration. Also, thank you for providing the 'extra' bit an explanation of work and Ep - they made perfect sense.
For anyone else who wants to help me out with my questions - I still have a few left :)
1. I get really confused by angular velocity, my textbook first gives the definition for angular speed as
"Angular speed: How fast the angle of a line that joins the object with the centre is changing" and this is really confusing to me as I don't understand how this angle can change if an object travelling in a circle sustains tangential velocity. Doesn't this mean that the angle held at any instant by the circular-moving object and the centre is 90°?
2. What is an object's linear speed? If it increases, will it increase the object's angular speed also?
3. How do larger linear speed reduce the effects of gravity?
4. Is Earth's elliptical shape a result of it's rotational axis?
5. What provides Earth's centripetal force which allows it to orbit? Is it the gravitational fields of other, larger celestial bodies?
6. How does the linear orbit velocity acting on a satellite cause it to move away from the Earth?
Thank you :)
For an object to be rotating around a planet, it has to be at a certain speed to maintain that rotation. That speed is faster as you get closer to the surface of the earth, so faster means closer to the earth. As you get closer to the earth, g increases. So a faster an object travels, the closer it is to the centre of the earth and therefore the greater gravitational force acting on it.
I think your other questions have been answered by these great posters :D
Post by: jamonwindeyer on January 26, 2017, 03:22:41 pm
Q4: When the Earth rotates around its axis it creates a apparent centrifugal effect which is greatest at the equator and less at the poles. This cohesive force causes the earth to 'bulge' at the equator due to the apparent value of g increased and 'flatten' at the poles due to a decrease in the value of g.
Wait, isn't it \(g\) decreasing at the equator and compared to the poles? An apparent increase in gravitational acceleration would make things seem heavier, which would suggest things flattening compressing, so the poles should have the apparent increase right?
(Could have misinterpreted what you said, apologies in advance if I have)
Post by: Rathin on January 26, 2017, 03:40:43 pm
Wait, isn't it \(g\) decreasing at the equator and compared to the poles? An apparent increase in gravitational acceleration would make things seem heavier, which would suggest things flattening compressing, so the poles should have the apparent increase right?
(Could have misinterpreted what you said, apologies in advance if I have)
Omg I am so sorry, that's what I meant..accidentally wrote it the other way around!! The value of g is increased at the poles thus then being flattened and decreased at equator causing them to bulge.
Post by: jamonwindeyer on January 26, 2017, 03:54:32 pm
Omg I am so sorry, that's what I meant..accidentally wrote it the other way around!! The value of g is increased at the poles thus then being flattened and decreased at equator causing them to bulge.
Don't be sorry! All good - I know you know your shit so I literally said my post aloud before I posted it. I'm like, "I'm about to contradict Rathin, this is deep water." ;)
Post by: Aaron12038488 on January 26, 2017, 06:57:37 pm
Post by: bsdfjnlkasn on January 26, 2017, 08:18:35 pm
Thank you for these comprehensive answers, they've really cemented my understanding of the topics in question :) :) . I'll be sure to ask more soon ;D
Post by: Rathin on January 26, 2017, 08:49:15 pm
Is Dot Point Preliminary Physics good?
Its okay..but some questions are really useless. For practice of questions I would suggest past papers>dot point books but they are not bad for passive practice.
Post by: kiwiberry on January 26, 2017, 09:07:35 pm
Is Dot Point Preliminary Physics good?
Hey! Some of my friends who used the dot point books said that they were helpful to review their learning after class. However, I know that others (including me) ended up being too lazy to do the questions regularly and never got around to using it. My phys teacher also said that the answers in the back are too brief and definitely not band 6 quality. I guess it's okay as a starting point and for revision, but personally I managed to survive on past papers haha :) I'm biased though so keep that in mind, you may find it useful!
Post by: bsdfjnlkasn on January 26, 2017, 10:04:14 pm
Q5: Gravitational Force is like a frictional force that provides the centripetal force. E.g When a car is going around a circle a centripetal force is provided by the friction on the road, similarly Centripetal force is provided by the Gravitational force thus we get the Fc=Fg (this is how the escape velocity formula is derived)
Hey Rathin,
When you say gravitational force in Q5 do you mean gravitational attraction force because I have this in my notes: "The centripetal force which supports circular motion in space is provided by the force of gravitational attraction." Sorry if this is a silly clarification.
Post by: Rathin on January 26, 2017, 10:06:35 pm
Hey Rathin,
When you say gravitational force in Q5 do you mean gravitational attraction force because I have this in my notes: "The centripetal force which supports circular motion in space is provided by the force of gravitational attraction." Sorry if this is a silly clarification.
Yup exactly the same thing :)
Post by: bsdfjnlkasn on January 26, 2017, 10:31:03 pm
In the Space topic, do we use the formula v= ωr often in practical applications? Is angular speed often tested?
2. Could a definition for orbital velocity be: the speed at which an object orbits another
3. Is an object's linear speed it's instantaneous speed at a particular point? As all objects describing a circle are actually travelling tangentially to the centre of the circle, can we take the proposed definition to be true? (Textbook just name drops 'linear speed' in the context of centripetal force and circular motion, which to me doesn't make much sense)
Post by: jamonwindeyer on January 26, 2017, 10:42:12 pm
Hello and I'm back :)
In the Space topic, do we use the formula v= ωr often in practical applications? Is angular speed often tested?
2. Could a definition for orbital velocity be: the speed at which an object orbits another
3. Is an object's linear speed it's instantaneous speed at a particular point? As all objects describing a circle are actually travelling tangentially to the centre of the circle, can we take the proposed definition to be true? (Textbook just name drops 'linear speed' in the context of centripetal force and circular motion, which to me doesn't make much sense)
Welcome back!
Angular speed is never tested (I didn't know what \(\omega\) was until university), so no stresses there. That formula is never used.
I'd say that's a good definition! I normally prefer to define it mathematically, \(v=\frac{2\pi R}{T}\), but that is a solid worded definition.
Annnnnd yep, pretty much agree with you there. I never use the term 'linear speed' because it is confusing in that regard. In my opinion, just say instantaneous velocity, which in an orbit is always tangential to the orbit ;D
Post by: Aaron12038488 on January 27, 2017, 02:18:10 pm
Post by: bsdfjnlkasn on January 27, 2017, 03:25:52 pm
I'm going to begin Yr 11, and my first test is a Prac Test. Is there any chance it would be the 'The World Communicates Topic'.
Hey Aaron,
The test will probably be on the topic you're studying in term 1. The only prac exam I had was in my prelim exams, the main focus being on the Electrical Energy unit.
Post by: bsdfjnlkasn on January 27, 2017, 03:30:13 pm
Welcome back!
Angular speed is never tested (I didn't know what \(\omega\) was until university), so no stresses there. That formula is never used.
I'd say that's a good definition! I normally prefer to define it mathematically, \(v=\frac{2\pi R}{T}\), but that is a solid worded definition.
Annnnnd yep, pretty much agree with you there. I never use the term 'linear speed' because it is confusing in that regard. In my opinion, just say instantaneous velocity, which in an orbit is always tangential to the orbit ;D
Hey Jamon!
Great job on the maths lecture it was really well paced and thorough :)
Would it be safe to say that we don't need to know the definition for angular velocity then? Because I still don't really have a strong grasp on what it is.
Or, does it come up in longer response questions?
What does the R in your formula for orbital velocity stand for?
Also, do we need to know the specifics (as in the method) of the Michelson-Morley experiment or just the result of it?
Thank you for the rest of the answers, though! Will add them to my notes now 8)
Post by: bsdfjnlkasn on January 27, 2017, 07:52:40 pm
If the train is moving and as a result the observer is as well, how is the distance that the light has to travel to the observer still the same and only the time different? It's probably got something to do with the different frame of reference but I'm not really seeing how it connects to the outside world (where the fireworks are). But I do get why the time taken is longer for the tail of the train, I just don't understand why this doesn't apply to the distance as well.
Also in the equations relevant to this scenario, how does vt arise (in the numerator?)
Post by: kiwiberry on January 28, 2017, 01:33:07 am
I was wondering if someone could explain to me the last few lines on the following screenshot. I attached the entire explanation for context, but my confusion arises from the last bracketed statement "(since d is still equal to d')."
If the train is moving and as a result the observer is as well, how is the distance that the light has to travel to the observer still the same and only the time different? It's probably got something to do with the different frame of reference but I'm not really seeing how it connects to the outside world (where the fireworks are). But I do get why the time taken is longer for the tail of the train, I just don't understand why this doesn't apply to the distance as well.
Also in the equations relevant to this scenario, how does vt arise (in the numerator?)
I think the time for the light from the rear to the observer is supposed to be \(t'=\frac{d'+vt}{c}\), not sure though
\(vt\) is the distance that the train moves in the time that it takes for the light to reach the observer (distance = speed x time, so distance=vt since the train is travelling at velocity v). So the light from the front will travel \(vt\) less than it would at rest, ie \(d-vt\)
Post by: jabuibui on January 28, 2017, 02:52:58 am
So if they ask you to calculate the g-force the astronaut experiences during launch off on Mars ( g.a = 3.71 m/s^2), do you also change the bottom g from 9.8 to 3.71?
Post by: jakesilove on January 28, 2017, 09:24:53 am
Just wondering with the G-force equation (a + g/ g), do you change both the 'g' values according to the planet's gravitational acceleration?
So if they ask you to calculate the g-force the astronaut experiences during launch off on Mars ( g.a = 3.71 m/s^2), do you also change the bottom g from 9.8 to 3.71?
Really great question! I got asked the same thing in my lecture the other day. I'm fairly sure that you do have to change one of the 'g's to the acceleration on that planet, but I think it's the g on top of the equation. ie.
Think about it this way. Say acceleration on a given planet was 1m/s instead of 9.8m/s. Presumably, you feel heaps 'lighter'. Thus, if you were travelling UPWARDS at 8.8m/s, you would definitely feel 'heavier' than normal, but you would probably only feel as heavy as you feel on earth. ie
However, this will never come up in the HSC!
Post by: bsdfjnlkasn on January 28, 2017, 10:55:54 am
I think the time for the light from the rear to the observer is supposed to be \(t'=\frac{d'+vt}{c}\), not sure though
\(vt\) is the distance that the train moves in the time that it takes for the light to reach the observer (distance = speed x time, so distance=vt since the train is travelling at velocity v). So the light from the front will travel \(vt\) less than it would at rest, ie \(d-vt\)
Hey that makes much more sense! I was just a bit hesitant to question the textbook because obviously you wouldn't expect for things to be written incorrectly. I'll make sure to clarify this with my teacher when we cover it in class.
Thanks for your help :)
Post by: bsdfjnlkasn on January 28, 2017, 12:19:26 pm
I'm just a bit confused by time dilation, specifically how it can be 'observed' on Earth like in the example of looking at a rocket travelling at c.
My textbook explanation is this "time dilation can be summarise as a 'moving clock appears to run slower'"
I find it a bit vague especially since all the previous explanations were about observing a light on a train by two people, one on the train and one outside on Earth. Also, just to quickly clarify with this example; time dilation is best observed by considering distances and how their apparent length varies with different frames of reference. Is it right to draw a connection between the two and hence say that if the distance appeared to increase then time dilation was observed (as t = d/c in both instances)?
Now back to the textbook definition, Is it just an accepted fact that when looking at something closely, time goes slower? I just don't understand how this definition helps with an understanding of time dilation (like at all :P). So when, I try and apply this definition (key word being try haha) to the twin paradox I don't understand how the Earth twin is observing time dilation by seeing the 'spacecraft ticking more slowly than time on Earth.' Does the rocket appear to be moving slower than it actually is? If that's the case then I really don't understand time dilation at all.
IDK I guess i'm just looking for a clear explanation of time dilation so that I can better 'imagine' it /understand it practically through thought experiments.
Post by: Whitey on January 28, 2017, 12:50:28 pm
I have had a look for the answer to my question on the forum, but was unsuccessful. I do apologise if my question has already been answered and I have not been able to locate it.
I was looking through my space notes to see if they are up to date for the start of school when I noticed an inconsistency with my notes and the syllabus. I have written in my notes for the Law of Gravitation "F = G(mM/r^2)" when the syllabus says "F=G(mM/d^2)". I would just like to know why there is a differences between the too. I'm not sure if I am completely missing something.
I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!
Post by: Iminschool on January 28, 2017, 12:58:20 pm
Hello!
I have had a look for the answer to my question on the forum, but was unsuccessful. I do apologise if my question has already been answered and I have not been able to locate it.
I was looking through my space notes to see if they are up to date for the start of school when I noticed an inconsistency with my notes and the syllabus. I have written in my notes for the Law of Gravitation "F = G(mM/r^2)" when the syllabus says "F=G(mM/d^2)". I would just like to know why there is a differences between the too. I'm not sure if I am completely missing something.
I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!
r ----> Radius of planetary body
d ----> Distance from the centre of a planetary body
Therefore they're the same thing
Post by: Iminschool on January 28, 2017, 12:59:26 pm
r ----> Radius of planetary body
d ----> Distance from the centre of a planetary body
Therefore they're the same thing
P.S 'd' may also include altitude if your point of reference isn't on the surface
Post by: jamonwindeyer on January 28, 2017, 01:35:25 pm
Hello!
...
I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!
Welcome to the forums! Let us know if you need a hand finding stuff ;D I'll let Jake know his slides are being put to good use!! :)
Post by: Whitey on January 28, 2017, 02:38:36 pm
r ----> Radius of planetary body
d ----> Distance from the centre of a planetary body
Therefore they're the same thing
Thanks mate! This clarified the differences I had. I will now adjust my notes accordingly.
Welcome to the forums! Let us know if you need a hand finding stuff ;D I'll let Jake know his slides are being put to good use!! :)
Thank you! Finding stuff shouldn't be too much of a hassle. The forums are neat so everything is easy to find, good job on your accessibility!
Post by: Rathin on January 29, 2017, 11:59:41 am
Every millennium Super Thuc and Super Brian fly to planet X to have a rap battle, which is 30 light years from their home planets. Planet Beast and Planet Guan respectively. But Thuc, having perfected his speed, travels at 0.8c whilst Brian only does 0.6c. How much older (or younger) will Thuc be then Brian.
This is probably the hardest HSC physics can get in terms of relativity calculations. Good Luck :)
Post by: Whitey on January 29, 2017, 12:28:57 pm
Every millennium Super Thuc and Super Brian fly to planet X to have a rap battle, which is 30 light years from their home planets. Planet Beast and Planet Guan respectively. But Thuc, having perfected his speed, travels at 0.8c whilst Brian only does 0.6c. How much older (or younger) will Thuc be then Brian.
I'm no maths wizard and I haven't done much practice over the holidays, so I'm just having a crack. I'm thinking you have to use a combination of length and time contraction because light years is a distance and their travel speed is velocity? Or am I completely off track?
Post by: katnisschung on January 29, 2017, 04:14:28 pm
for this mass dilation question... its pretty simple but
i'm getting the wrong percentage
calculate the percentage increase in the mass of an Apollo rocket
as it escapes from Earth's gravity at 11.2 kp/s (yeahs i converted to m/s)
Post by: Whitey on January 29, 2017, 04:27:12 pm
can someone pls run me through the working out
for this mass dilation question... its pretty simple but
i'm getting the wrong percentage
calculate the percentage increase in the mass of an Apollo rocket
as it escapes from Earth's gravity at 11.2 kp/s (yeahs i converted to m/s)
I'm not expert, but I'll try to help. Is there an initial mass, because when I try to plug the numbers into the formula I have two unknown m values. Also just another thought of mine is that do you have to subtract 9.8 as gravity acts against the acceleration of the rocket?
Post by: jakesilove on January 29, 2017, 04:35:46 pm
can someone pls run me through the working out
for this mass dilation question... its pretty simple but
i'm getting the wrong percentage
calculate the percentage increase in the mass of an Apollo rocket
as it escapes from Earth's gravity at 11.2 kp/s (yeahs i converted to m/s)
I assume you mean kilometres per second? Well, we know that the mass dilation formula looks like
Where Mv is the mass of the fast-moving object, and Mo is the mass of the object at rest. Plugging our info into the equation (after converting to m/s) we get
If you solve this, you show that the change is mass is basically zero. Like, you get something like
Clearly, mass has increased, but by next to nothing. Did you type out the question correctly?
Jake
Post by: katnisschung on January 29, 2017, 04:39:47 pm
m=1kg and then figured the percentage out as 1.0 x 10^-7%
but the answers says it 1.0x10^-8%
yeah i typed it correctly sighs our teacher gave us the old version of surfing
Post by: jakesilove on January 29, 2017, 04:41:59 pm
no there isn't an initial mass... i hypothetically let
m=1kg and then figured the percentage out as 1.0 x 10^-7%
but the answers says it 1.0x10^-8%
yeah i typed it correctly sighs our teacher gave us the old version of surfing
Interesting! Potentially the answers are wrong, but honestly you're so close that you clearly understand the material well enough :)
Also interesting that my answer is between yours and the solutions given in the book, but closer to yours :)
Post by: katnisschung on January 29, 2017, 05:23:03 pm
calculate the speed that would cause a length contraction of 15%...
Post by: RuiAce on January 29, 2017, 05:27:19 pm
another question...
calculate the speed that would cause a length contraction of 15%...
Post by: beau77bro on January 29, 2017, 09:04:45 pm
Post by: RuiAce on January 29, 2017, 09:19:54 pm
question: "calculate the altitude of a 500kg satellite which has a gravitational potential energy of 1.5x10^9J." im so bad at potential energy someone help ahhaha
You could probably use mgh, but I never use that formula. Find it too unreliable.
Post by: jamonwindeyer on January 29, 2017, 09:28:26 pm
You could probably use mgh, but I never use that formula. Find it too unreliable.
I normally go with the \(E=mgh\) formula for altitudes under 1000km ;D in this case, since the altitude is unknown, I'd say you made the best choice :)
Post by: RuiAce on January 29, 2017, 09:30:17 pm
I normally go with the \(E=mgh\) formula for altitudes under 1000km ;D in this case, since the altitude is unknown, I'd say you made the best choice :)Aha that's good to hear :D
I very rarely go to mgh though... don't trust its accuracy one bit
Post by: beau77bro on January 29, 2017, 09:34:07 pm
I normally go with the \(E=mgh\) formula for altitudes under 1000km ;D in this case, since the altitude is unknown, I'd say you made the best choice :)
i think E = -Gm1m2/r won out i finally got the right answer... mostly because i miss calculated hahaha thankyou guys
Post by: jamonwindeyer on January 29, 2017, 09:39:25 pm
i think E = -Gm1m2/r won out i finally got the right answer... mostly because i miss calculated hahaha thankyou guys
Great to hear!!
I very rarely go to mgh though... don't trust its accuracy one bit
I remember I once did a test and found that for altitudes in the hundreds of kilometres, it only makes a difference from the second decimal point or something. Maybe I should do that again :P
Post by: Caitlyn_ on January 30, 2017, 06:12:15 pm
a) The spacecraft's length and width at this speed
b) The spacecraft's mass at this speed
c) The time that elapses on the spacecraft if 1 hour passes on earth whilst the craft is travelling at this relativistic speed.
Not sure how to set out and structure this particular question.
Post by: jamonwindeyer on January 30, 2017, 06:25:58 pm
A 4000kg spaceship is 40.00m long and 8.00m wide when stationary. If the spacecraft now travels at 0.99c in the direction of the length of the craft, calculate, from the frame of reference of an observer on Earth:
a) The spacecraft's length and width at this speed
b) The spacecraft's mass at this speed
c) The time that elapses on the spacecraft if 1 hour passes on earth whilst the craft is travelling at this relativistic speed.
Not sure how to set out and structure this particular question.
Hi Caitlyn! Welcome to the forums! For this question, it will actually be helpful to do a separate calculation of:
It appears in every formula, so we should just do it once to get it done!
So, we can now use this value for other questions. Take your length contraction question for example. We are given the rest length of the spacecraft as 40 metres, that's \(l_0\). We pop that into the formula for length contraction on your reference sheet:
Note that for this question it also wants the width - The width is unaffected! So it is still 8 metres wide; this is because length contraction only occurs in the direction of travel. So it only contracts length-ways.
The other questions are just substitution questions just like above! Take the value we found above, and the value of \(t_0\) or \(m_0\) from the question, then use it to find the relativistic value \(t_v\) or \(m_v\). Does that help? ;D
Post by: Whitey on January 30, 2017, 06:44:58 pm
A 4000kg spaceship is 40.00m long and 8.00m wide when stationary. If the spacecraft now travels at 0.99c in the direction of the length of the craft, calculate, from the frame of reference of an observer on Earth:I tired these myself for practice and I could do a and b without trouble but c gave me some strife.
c) The time that elapses on the spacecraft if 1 hour passes on earth whilst the craft is travelling at this relativistic speed.
I originally got
Post by: bsdfjnlkasn on January 30, 2017, 10:11:09 pm
I tired these myself for practice and I could do a and b without trouble but c gave me some strife.
I originally gotbut then I realised that time shortens on the really really fast object so I rearranged so I got
I'm not sure which is right as I don't think I wholly grasp the concept of time dilation yet. Can someone help please?
Hey i'm hoping to help out a bit here - I only recently learnt these formulas but think this would be a good way to apply my knowledge :)
So we know that the spacecraft is moving relative to the Earth which means to = 1 (as this is the time that has passed on Earth)
Now on to the calculation:
Subbing in known values:
Evaluating:
tv = 7 hours and 5 minutes
We can also check that our answer is correct as the spacecraft (where v = 0.99c) will experience time dilation. This means our answer needs to be longer than 1 hour. By considering the words which come up in our formulae (contraction and dilation) we can distinguish what the two mean with simple definitions. Contraction means becoming smaller, and dilation means becoming larger.
I hope that this made sense, please excuse my LaTex I have no clue what i'm doing lol
Post by: jamonwindeyer on January 30, 2017, 10:30:01 pm
Hey i'm hoping to help out a bit here - I only recently learnt these formulas but think this would be a good way to apply my knowledge :)
I hope that this made sense, please excuse my LaTex I have no clue what i'm doing lol
Just verifying you got it spot on... Well done! ;D
Rui wrote a cool guide on LaTex, if you are interested. Otherwise I learned from Google if you are keen! Really useful thing to know for Uni ;D
Post by: RuiAce on January 30, 2017, 11:36:02 pm
Just verifying you got it spot on... Well done! ;DOh gees. Thanks for reminding me I have to update this
Rui wrote a cool guide on LaTex, if you are interested. Otherwise I learned from Google if you are keen! Really useful thing to know for Uni ;D
Post by: Whitey on January 31, 2017, 06:31:39 am
We can also check that our answer is correct as the spacecraft (where v = 0.99c) will experience time dilation. This means our answer needs to be longer than 1 hour. By considering the words which come up in our formulae (contraction and dilation) we can distinguish what the two mean with simple definitions. Contraction means becoming smaller, and dilation means becoming larger.
Thank you so much!! I get the calculations and where to put tv and to on the formula but my understanding of the concept was not proper, even though we have done it in class.
That paragraph at the end differing between contraction and dilation is pure gold for me! I'm pretty sure that I understand them now!
Post by: Caitlyn_ on January 31, 2017, 11:17:38 am
Hi Caitlyn! Welcome to the forums! For this question, it will actually be helpful to do a separate calculation of:
It appears in every formula, so we should just do it once to get it done!
So, we can now use this value for other questions. Take your length contraction question for example. We are given the rest length of the spacecraft as 40 metres, that's \(l_0\). We pop that into the formula for length contraction on your reference sheet:
Note that for this question it also wants the width - The width is unaffected! So it is still 8 metres wide; this is because length contraction only occurs in the direction of travel. So it only contracts length-ways.
The other questions are just substitution questions just like above! Take the value we found above, and the value of \(t_0\) or \(m_0\) from the question, then use it to find the relativistic value \(t_v\) or \(m_v\). Does that help? ;D
Thank you so much. I've only just learnt this topic and still wasn't confident on it. So what you have said has helped out a lot!
Post by: jamonwindeyer on January 31, 2017, 10:33:54 pm
Thank you so much. I've only just learnt this topic and still wasn't confident on it. So what you have said has helped out a lot!
Really glad to hear it! Let us know if you have any further questions ;D
Post by: bsdfjnlkasn on February 01, 2017, 08:44:02 am
Thank you so much!! I get the calculations and where to put tv and to on the formula but my understanding of the concept was not proper, even though we have done it in class.
That paragraph at the end differing between contraction and dilation is pure gold for me! I'm pretty sure that I understand them now!
Yay, I'm so glad to have helped!! 8)
Post by: beau77bro on February 04, 2017, 11:38:10 am
thankyou
Post by: jamonwindeyer on February 04, 2017, 11:45:09 am
so my half yearly is coming up and its only focusing on mod 1 and 2, and i have already smashed these out notes wise and everything. im finding i have very limited resources in terms of questions and exam questions. currently im doing surfing just as like a breezy kinda review set of questions, and dot point multiple choice questions because i find multiple choice can be quite tricky and the questions are quite good and also include extension sections that allow me to test extensive knowledge over a give part of the course. but i have no options in terms of actual hsc/exam type long response or harder questions? suggestions??? i can only find a couple of half yearly papers and im not sure that will last very long/ provide a wide enough range of questions to prepare for the exam. suggestions??? honestly links to more half yearly papers would be amazing but i might end up trying some trial papers. but id much rather follow the advice of someone who knows what they are dping hahahahah.
thankyou
Hey! Sounds like you are nice and prepared early on, nice job! As a first option, this site has a collection of Physics papers, including Half Yearly Papers and Trials. Great first stop for practice questions, if you haven't seen it already!
If you are willing to spend a bit of money, you can buy this. It's a set of topic tests for all the core topics in Physics, with detailed solutions to BOSTES style questions handwritten by a Band 6 student. As a disclaimer, that student was me and I wrote the tests, so I'm definitely biased! That said, I do think that it is useful for precisely this purpose ;D
Beyond that, your textbook questions, combined with a bit of extra practice from some combo of those two sources above (and anything else you have access to), should definitely be enough prep for your half yearlies ;D
Post by: beau77bro on February 04, 2017, 11:55:03 am
Hey! Sounds like you are nice and prepared early on, nice job! As a first option, this site has a collection of Physics papers, including Half Yearly Papers and Trials. Great first stop for practice questions, if you haven't seen it already!
If you are willing to spend a bit of money, you can buy this. It's a set of topic tests for all the core topics in Physics, with detailed solutions to BOSTES style questions handwritten by a Band 6 student. As a disclaimer, that student was me and I wrote the tests, so I'm definitely biased! That said, I do think that it is useful for precisely this purpose ;D
Beyond that, your textbook questions, combined with a bit of extra practice from some combo of those two sources above (and anything else you have access to), should definitely be enough prep for your half yearlies ;D
OMG WHAT THE TOPIC TESTS ARE READY!!!!!!!????? thats perfect. yassssssss omg thankyou sooo much im buying them now, they are soo good for chem. thankyouuuu, and where are the past papers hahaha sorry?
THANLYOU FOR TELLING ME THIS
how long will it take to deliver?
Post by: jakesilove on February 04, 2017, 12:04:21 pm
how long will it take to deliver?
Generally, the product will be delivered to you within the week :)
Post by: beau77bro on February 04, 2017, 12:05:26 pm
Post by: jamonwindeyer on February 04, 2017, 12:34:39 pm
... thankyouuuu, and where are the past papers hahaha sorry?
Google THSC, it should be the first option! HUGE bank of past papers and resources ;D
Post by: beau77bro on February 04, 2017, 01:42:23 pm
Post by: jamonwindeyer on February 04, 2017, 02:15:14 pm
oh yea, theres only a couple of half yearlies? should i just do trials if i finish them?
There's 10 half yearlies, that in conjunction with textbook questions should be heaps! Like, do Trials if you need to, but 10 half yearly practice papers with a few dozen textbook questions is definitely a heap of practice :) if you do every paper, mark it, consolidate then go back and re-answer things you got wrong - That's a solid 3-4 hours per paper, so you've got a solid 30-40 hours of study there ;D
Post by: bluecookie on February 04, 2017, 05:10:43 pm
When you combine these two phenomena together the current that is induced in the conductor produces its own magnetic field that will interact with the external magnetic field that produced the current in the first place!"
^^ This is what it says in my textbook. I'm a bit confused, does the current produced change direction in regards to the initial current? (is it independent of the initial current is what I'm trying to say)
And doesnt that mean the current/magnetic field are changing all the time? Cause if the current produces its own magnetic field that then interacts with the external magnetic field...that means a new current is produced. But then the production of that new current changes the initial magnetic field, which then changes the interaction with the external magnetic field, so does that mean there's a constant flow of new currents and magnetic fields when it comes to current carrying conductors?
Post by: jakesilove on February 04, 2017, 06:13:52 pm
"A current-carrying conductor will produce its own magnetic field that can interact with an external magnetic field producing a force (the torque effect). The relative motion between an external magnetic field and a conductor will produce a current within that conductor (electromagnetic force).
When you combine these two phenomena together the current that is induced in the conductor produces its own magnetic field that will interact with the external magnetic field that produced the current in the first place!"
^^ This is what it says in my textbook. I'm a bit confused, does the current produced change direction in regards to the initial current? (is it independent of the initial current is what I'm trying to say)
And doesnt that mean the current/magnetic field are changing all the time? Cause if the current produces its own magnetic field that then interacts with the external magnetic field...that means a new current is produced. But then the production of that new current changes the initial magnetic field, which then changes the interaction with the external magnetic field, so does that mean there's a constant flow of new currents and magnetic fields when it comes to current carrying conductors?
Honestly, you've nailed it. If you're confused, that's because it's confusing, but you're 100% correct. The current produces ANOTHER field, which produces ANOTHER current, which opposes the direction of the original current (Ie. it is dependant on the initial current; the induced current will run opposed to the supplied current!). And yes, things are constantly changing and shifting, however it all eventually settles down into a sort of equilibrium. I have to say, though, that you seem to understand the principle of back-emf really well! It's bloody confusing, it took quite a few goes for me to understand what was happening. Keep revising!
Post by: bluecookie on February 05, 2017, 12:12:30 am
Honestly, you've nailed it. If you're confused, that's because it's confusing, but you're 100% correct. The current produces ANOTHER field, which produces ANOTHER current, which opposes the direction of the original current (Ie. it is dependant on the initial current; the induced current will run opposed to the supplied current!). And yes, things are constantly changing and shifting, however it all eventually settles down into a sort of equilibrium. I have to say, though, that you seem to understand the principle of back-emf really well! It's bloody confusing, it took quite a few goes for me to understand what was happening. Keep revising!Thank you :)
Post by: bluecookie on February 05, 2017, 02:04:22 am
When they say the electron moves is excited and moves from one energy level to another, does the electron move from the first energy level to the second and then back? Or does it move from the first to the last energy elvel and then back? Or is it random movement in either direction?
Also, when they say one second is defined by the duration of [inset number I can't be bothered to type haha] periods of radiation corresponding to the transition between two hyperfine levels, what is one period defined by? Is it the one singular jump between energy levels? Or is one period defined by the jump from the inner energy level to the outer? (if it orbits in that way)
Sorry, just really confused :P
Post by: jamonwindeyer on February 05, 2017, 10:05:40 am
What are hyperfine and fine levels in the atomic clock caesium-133?
When they say the electron moves is excited and moves from one energy level to another, does the electron move from the first energy level to the second and then back? Or does it move from the first to the last energy elvel and then back? Or is it random movement in either direction?
Also, when they say one second is defined by the duration of [inset number I can't be bothered to type haha] periods of radiation corresponding to the transition between two hyperfine levels, what is one period defined by? Is it the one singular jump between energy levels? Or is one period defined by the jump from the inner energy level to the outer? (if it orbits in that way)
Sorry, just really confused :P
Hey hey! While someone a little more rehearsed (like Jake) might be able to help you here, just know that you definitely don't need to know how an atomic clock works in the HSC ;D is this as background knowledge for the time dilation section in Space? You will never be expected to explain how an atomic clock works ;D
Edit: Woops unless this is from an Option in which case ignore me and hopefully Jake can help you out! ;D
Post by: bluecookie on February 05, 2017, 11:55:13 am
Hey hey! While someone a little more rehearsed (like Jake) might be able to help you here, just know that you definitely don't need to know how an atomic clock works in the HSC ;D is this as background knowledge for the time dilation section in Space? You will never be expected to explain how an atomic clock works ;D
Edit: Woops unless this is from an Option in which case ignore me and hopefully Jake can help you out! ;D
Ah sorry, my bad. It was extension work the teacher gave us. ^^
Post by: jakesilove on February 05, 2017, 12:40:01 pm
What are hyperfine and fine levels in the atomic clock caesium-133?
When they say the electron moves is excited and moves from one energy level to another, does the electron move from the first energy level to the second and then back? Or does it move from the first to the last energy elvel and then back? Or is it random movement in either direction?
Also, when they say one second is defined by the duration of [inset number I can't be bothered to type haha] periods of radiation corresponding to the transition between two hyperfine levels, what is one period defined by? Is it the one singular jump between energy levels? Or is one period defined by the jump from the inner energy level to the outer? (if it orbits in that way)
Sorry, just really confused :P
Hey! I'm only going to answer your second question, because the other two are DEFINITELY way outside the curriculum. The second, however, can form a part of the Q2Q option, and is also not too difficult to explain.
You can imagine that light, of a certain energy/wavelength, is incident on an electron in the first energy level. The electron will absorb that energy if and only if it is the exact energy required to pump it up a certain number of discrete energy levels. So, say the difference in energy between n=1 and n=2 was 10kJ, and n=2 and n=3 was 15kJ. Then, a photon with 10kJ of light could cause an n=1 electron to go to n=2 (and back), but not an n=2 to n=3. Similarly, a photon with 25kJ of energy could cause an n=1 electron to go to n=3 (and back).
Note that there isn't a 'last' energy level.
Hope that explanation makes sense!
Post by: bluecookie on February 05, 2017, 04:29:23 pm
Hey! I'm only going to answer your second question, because the other two are DEFINITELY way outside the curriculum. The second, however, can form a part of the Q2Q option, and is also not too difficult to explain.
You can imagine that light, of a certain energy/wavelength, is incident on an electron in the first energy level. The electron will absorb that energy if and only if it is the exact energy required to pump it up a certain number of discrete energy levels. So, say the difference in energy between n=1 and n=2 was 10kJ, and n=2 and n=3 was 15kJ. Then, a photon with 10kJ of light could cause an n=1 electron to go to n=2 (and back), but not an n=2 to n=3. Similarly, a photon with 25kJ of energy could cause an n=1 electron to go to n=3 (and back).
Note that there isn't a 'last' energy level.
Hope that explanation makes sense!
Thank you!
Post by: imda.beast on February 05, 2017, 04:50:56 pm
Post by: jakesilove on February 05, 2017, 04:57:10 pm
Hi, I want to know how should I answer "Identify a rocket scientist and assess two of his major contributions to rocketry" for Von Braun, since most of von Brauns work is practical based rather than theory based, I find it difficult to assess his contribution. Thnx jake
Hey, and welcome to the forums! Doesn't matter whether his contribution is practical or theory, as long as you list two of them. For instance, if you cited the development of the Saturn V rockets, and various shuttle engineering developmenets, that would be plenty, as long as you assess the contributions as well (eg. Saturn V took us to the moon!).
Post by: Aaron12038488 on February 05, 2017, 06:43:24 pm
Thanks
Post by: Rathin on February 05, 2017, 06:52:40 pm
I know I've asked this question before. I'm having trouble with this dot point: Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer. What do I write for Light and Microwaves.
Thanks
- Reflection of light is used in optic fibres which allows massive amounts of information transfer.
- Radio waves reflects off the ionosphere which allows the information to be transferred long distances.
- Microwaves are used in radars which reflect off objects to determine the distance of that object.
Post by: kiwiberry on February 05, 2017, 06:55:40 pm
I know I've asked this question before. I'm having trouble with this dot point: Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer. What do I write for Light and Microwaves.
Thanks
Some more for light:
- Forms virtual images of objects placed in front of plane mirror, both real and virtual from curved mirrors
- Telescopes use parabolic concave mirrors to reflect light from stars
- Torches and driving lights have parabolic reflectors – adjust light source for flood or spot beams
Post by: bluecookie on February 10, 2017, 10:24:37 am
V(a rel b) = V(a rel C) – V(b rel c)
Post by: jamonwindeyer on February 10, 2017, 10:54:02 am
I don't really get the relative velocity formula...
V(a rel b) = V(a rel C) – V(b rel c)
Hey! What the formula actually does mechanically is a little tricky to explain. Essentially, it lets you measure the relative velocity between two frames of references. Best done with an example I think.
Say that you're sitting down, watching two people run past you in the same direction. You measure the first runner as 9 kilometres an hour, and the second at 7 kilometres an hour. For the formula, that's \(v_{ac}\) and \(v_{bc}\); the two velocities relative to a third reference frame - You! So in the formula, C represents the frame of reference of the measurer, and A and B represents two moving bodies.
The formula lets you calculate the velocity of A relative to B, meaning, the second runner is watching the first runner. How fast does he measure? The answer is, \(v_{ab}=v_{ac}-v_{bc}=9-7=2\) kilometres an hour. This makes sense, the first runner is moving 2 kilometres faster than the second runner, so from the frame of reference of the second runner, they'd say the first runner is moving away at 2 kilometres per hour.
The trick here is the frames of reference, and recognising that we don't always have to measure velocities with respect to the ground. We can measure them with respect to any inertial frame of reference. In this case, we choose the second runner; the first runner is moving 2 kilometres faster than that, so the relative velocity is 2 kilometres an hour.
The formula should only be used once you've got an intuitive understanding of how it works, otherwise it leads to trouble ;D I hope this helps!!
Post by: Mayalily on February 10, 2017, 07:09:40 pm
Post by: jamonwindeyer on February 10, 2017, 08:01:28 pm
Hey, I'm doing my HSC this year, and in relation to the option topic, my teacher wants to do medical physics, which I'm really uninterested in. He said it would be fine if I just studied for the option I want to do (Quanta to Quarks) but I'm worried that this may make it significantly harder to get the marks that I want in it.
Hey Mayalily! I think its fair enough that you want to study another option (although I did Medical Physics and it is awesome ;) )
There are no options that are significantly harder than others directly; the difficulty will come from needing to teach everything to yourself. If you are self motivated and driven, why not! Quanta is a popular option so there are a HEAP of resources for it; you won't have trouble in that regard. You just need to ask yourself; can you self teach an entire Option? Will you do the work to find the resources? Not doing the work with your class is definitely a disadvantage but if it means you can study something you are more interested in, I'm all for it!
Of course we will help you every step of the way too ;D if you were super keen you could even get a tutor to help you with that Option maybe? :)
Post by: bsdfjnlkasn on February 10, 2017, 08:42:32 pm
Post by: jamonwindeyer on February 10, 2017, 08:48:08 pm
Hey is gravitational field strength the same as the gravitational acceleration?
Hey! It definitely is, we define gravitational field strength as 'force experienced per unit mass,' so \(\text{Strength}=\frac{F}{m}\), which is also gravitational acceleration! ;D
Post by: bsdfjnlkasn on February 10, 2017, 08:50:39 pm
Hey! It definitely is, we define gravitational field strength as 'force experienced per unit mass,' so \(\text{Strength}=\frac{F}{m}\), which is also gravitational acceleration! ;D
Thanks Jamon, one more: whigh formula would you use to find the weight of an object on Earth's surface if you're given a value for the Earth's radius?
Post by: jamonwindeyer on February 10, 2017, 08:55:10 pm
Thanks Jamon, one more: whigh formula would you use to find the weight of an object on Earth's surface if you're given a value for the Earth's radius?
No worries! If you are given the radius of the earth, chances are it will be the full derivation using \(F=\frac{GMm}{d^2}\), rather than \(W=mg\), because you will probably be dealing with an object significantly higher than the surface of the earth. Remember, the value of \(g\) is only valid at or near the earth's surface, too far above and you need the full version ;D
Post by: bsdfjnlkasn on February 10, 2017, 09:03:00 pm
No worries! If you are given the radius of the earth, chances are it will be the full derivation using \(F=\frac{GMm}{d^2}\), rather than \(W=mg\), because you will probably be dealing with an object significantly higher than the surface of the earth. Remember, the value of \(g\) is only valid at or near the earth's surface, too far above and you need the full version ;D
Soo that F expression is actually equivalent to weight on Earth?
Post by: jamonwindeyer on February 10, 2017, 09:07:08 pm
Soo that F expression is actually equivalent to weight on Earth?
If you substitute the earths radius and mass \(M\) into that formula, with the weight of the object as \(m\), then yep! You'll get the weight force on the object ;D
Post by: bsdfjnlkasn on February 10, 2017, 09:08:30 pm
If you substitute the earths radius and mass \(M\) into that formula, with the weight of the object as \(m\), then yep! You'll get the weight force on the object ;D
That's sick! Thank you for all your help :) :)
Post by: jamonwindeyer on February 10, 2017, 09:09:38 pm
That's sick! Thank you for all your help :) :)
Gotta love it when stuff fits together ;) no worries at all! ;D
Post by: bsdfjnlkasn on February 11, 2017, 10:00:06 pm
If a 10kg mass were to be released from 1000km above the surface of Earth, initially being stationary, determine its velocity just before it hits the surface of the Earth.
Post by: kiwiberry on February 11, 2017, 11:14:30 pm
Hey, could I get some help with the following:
If a 10kg mass were to be released from 1000km above the surface of Earth, initially being stationary, determine its velocity just before it hits the surface of the Earth.
I'm not sure, but I think you can use \(a=\frac{GM}{r^{2}}\) to find acceleration, and then sub into \(v^{2}=u^{2}+2as\) to find v!
Post by: jamonwindeyer on February 11, 2017, 11:55:19 pm
I'm not sure, but I think you can use \(a=\frac{GM}{r^{2}}\) to find acceleration, and then sub into \(v^{2}=u^{2}+2as\) to find v!
I think that's the general idea! But, the acceleration at 1000km above the surface is different to the acceleration at the surface. Indeed, it changes slightly the whole way down. To do it super accurately, you'd actually need a more complete formula that took that into account those changes (I think there'd be Calculus involved) :P
In this case, we can use the reasonable estimate that the acceleration is just a uniform \(g=9.8ms^{-2}\) the whole way through the drop. At an altitude of 1000km, that's a reasonable estimate. Then you just pop that value for acceleration into \(v^2=u^2+2as\) ;D
As a side note, we didn't get earths radius, so that's a clue we can't use \(a=\frac{GM}{r^2}\), even though it is really tempting! ;D
Post by: beau77bro on February 12, 2017, 06:33:42 pm
Post by: jakesilove on February 12, 2017, 08:57:24 pm
Why does special relativity apply to inertial frames of reference? What does acceleration do to affect the theory? Speed of light is constant I assume still holds? But does it ahhahaha? Please feel free to drop bombs on me but I don't think I could hack like a Uni level paper explaining hahaha basics I'm dumb lmao
Hey Beau! Don't stress too much about accelerating frames etc; it's beyond the curriculum, and beyond Special relativity. Essentially, the whole idea of special relativity stems from two propositions; that the speed of light is constant, and that the laws of physics hold true in all inertial frames of reference. From there, using trains as analogies etc etc, you can show that time must dilate etc when inertial frames move in relation to each other. So, why must the frames be inertial (ie. non-accelerating)? Well, that's basically part of the definition of the principle of relativity!
Bit of a circular answer, but hope that makes sense.
Post by: RuiAce on February 12, 2017, 09:03:07 pm
But yeah, not much to add onto Jake's answer. "Acceleration warps space-time" but you don't know what space-time even is in the HSC course.
Post by: beau77bro on February 12, 2017, 09:23:17 pm
I suppose you could say that special relativity was built on inertial frames of reference. Of course, something must apply to what it was built on.
But yeah, not much to add onto Jake's answer. "Acceleration warps space-time" but you don't know what space-time even is in the HSC course.
"Warps space time" hahaha ok that's all I needa know, thankyou both of you. Cheers Jake and Rui. Absolute gods
Post by: beau77bro on February 12, 2017, 10:10:53 pm
Post by: jakesilove on February 13, 2017, 11:05:22 am
Stroboscopic experiments--> we didn't really do them and I keep seeing questions on it and I have no idea how to do them? Are we supposed to be able to do them? I feel my class doesn't do the required pracs, opinions? Thankyouuuu
Stroboscopic measurement is definitely not in the syllabus. Where are you finding the questions you're talking about? Don't worry about it at all, it isn't relevant!
Post by: beau77bro on February 13, 2017, 06:52:12 pm
Stroboscopic measurement is definitely not in the syllabus. Where are you finding the questions you're talking about? Don't worry about it at all, it isn't relevant!
really? omg praise the lord. its in surfing, dot point and like this prac book. also a physics topic test i ordered hasnt arrived?
Post by: jakesilove on February 13, 2017, 07:04:34 pm
really? omg praise the lord. its in surfing, dot point and like this prac book. also a physics topic test i ordered hasnt arrived?
Hey Beau! Very weird; would you like to private message me your name and when you ordered the topic tests, and I'll look into that for you?
Post by: Iminschool on February 14, 2017, 08:04:59 pm
Post by: jamonwindeyer on February 16, 2017, 12:32:00 am
For this question, the answer is B however i wanted to know how the current was reversed for it not to be A, is it because of the initial direction of current?
Hey! Sorry for the late reply - I really don't even think that question is 'answerable,' because it doesn't define the direction of current properly. Like, the flow of current can only take a positive/negative value if you DEFINE it that way, by saying, "Okay, if it flows this way, it is positive." Usually we set this definition with voltages or whatever, but we don't get that here. So, I don't think it is appropriate in that regard :P
The only thing I can think of is, using the mathematical version of Faraday's Law:
That negative symbol represents Lenz's Law - Perhaps they think it most appropriate to put the current negative to start with, since the induced current is opposing the motion of the generator? That seems a bit of a stretch though, in my opinion :)
Post by: RuiAce on February 16, 2017, 12:33:45 am
Hey! Sorry for the late reply - I really don't even think that question is 'answerable,' because it doesn't define the direction of current properly. Like, the flow of current can only take a positive/negative value if you DEFINE it that way, by saying, "Okay, if it flows this way, it is positive." Usually we set this definition with voltages or whatever, but we don't get that here. So, I don't think it is appropriate in that regard :PYep that was what I thought - Lenz's law. But I wasn't certain so I didn't want to post.
The only thing I can think of is, using the mathematical version of Faraday's Law:
That negative symbol represents Lenz's Law - Perhaps they think it most appropriate to put the current negative to start with, since the induced current is opposing the motion of the generator? That seems a bit of a stretch though, in my opinion :)
Post by: jamonwindeyer on February 16, 2017, 12:48:45 am
Yep that was what I thought - Lenz's law. But I wasn't certain so I didn't want to post.
Yeah neither am I... Don't stress about it too much Iminschool, HSC would never be that vague ;D
Post by: Iminschool on February 16, 2017, 05:30:53 am
Yeah neither am I... Don't stress about it too much Iminschool, HSC would never be that vague ;D
Haha, awesome
Post by: Kle123 on February 16, 2017, 09:05:49 pm
Post by: RuiAce on February 16, 2017, 09:43:04 pm
I am referring to question 10 of a past HY paper of my school. I thought the answer was D, however the marking guideline says C. It may be wrong though because question 4 was a complete dud (as seen in marking guideline). Could someone explain to me why the answer is C.Where's the question?
Post by: Kle123 on February 16, 2017, 10:25:12 pm
Post by: RuiAce on February 16, 2017, 10:57:42 pm
whoops should have previewed my post.I actually don't see how it could possibly be C...
Post by: Kle123 on February 16, 2017, 11:03:41 pm
I actually don't see how it could possibly be C...wow, my school is so dodgy. Thanks RuiAce for helping and responding so fast to me today.
Post by: jamonwindeyer on February 16, 2017, 11:10:45 pm
I actually don't see how it could possibly be C...
Just as extra assurance, definitely agree with Rui - And I agree with your answer of D too Kle123 ;D
Post by: Bubbly_bluey on February 16, 2017, 11:48:13 pm
I feel like this response is no enough for the 3 marks... :/ ??? ???
btw thanks jamon and elyse (if you're here) for your positive feedback for english, literally made my night :)
Post by: jamonwindeyer on February 17, 2017, 12:00:04 am
Hey guys! i need a little help with this past hsc question. From what i hopefully understand so far, BC and the current of the wire are running in parallel in the same direction so they would attract. AD runs in the opposite direction as the wire so they would repel. I'm not so sure what happens to AB and CD: do they not experience a force? And so what would happen to the overall square wire because it can't rotate unless there is a magnetic field...
I feel like this response is no enough for the 3 marks... :/ ??? ???
btw thanks jamon and elyse (if you're here) for your positive feedback for english, literally made my night :)
Hey Bubbly! You are welcome ;D let me see if I can help here, a few points:
- You're correct on all counts; AB and CD definitely do not experience the force! This is because the currents are perpendicular; you don't get a force in that circumstance :)
- You have identified all the major points, except you should specify that the attractive force due to BC is larger than the repulsive force due to AD. Therefore, the net force on the loop is attractive. There is no rotation, the loop just moves towards the straight wire (or, as a better answer, just say the loop experiences a net force upwards! ;D
Post by: katnisschung on February 17, 2017, 05:07:03 pm
so for motors and generators, when calculating the force on a current
carrying conductor within a magnetic field, is the angle measured
from the line parallel to that of the magnetic field or from an
'imaginary' line perpendicular to the magnetic field?
thanks... i hope this makes sense
Post by: Kle123 on February 17, 2017, 06:20:09 pm
Hi there,
so for motors and generators, when calculating the force on a current
carrying conductor within a magnetic field, is the angle measured
from the line parallel to that of the magnetic field or from an
'imaginary' line perpendicular to the magnetic field?
thanks... i hope this makes sense
you would use the angle between the current carrying wire and the magnetic field. So the answer to your question is parallel (as all magnetic field lines i've seen in force questions i imagine are from areas really close to a magnet's pole or from uniform magnetic plates). Also, It wouldn't matter if you use the obtuse or acute angle as sine would give the same answer
Post by: Kle123 on February 17, 2017, 06:30:16 pm
Post by: jamonwindeyer on February 18, 2017, 01:00:49 am
In this question I had some idea of how to answer it, however in marking it afterwards i gave myself a zero as i didnt talk about kepler's law of periods. I still can't relate kepler's law into my answer. Please help ataranote's leaders :). Thanks alot!
Hey Kle! So this question first requires a knowledge of WHAT the required manoeuvre is - What was your previous answer (even roughly)? Just want to see whether you've got that down before I explain how Kepler's can be linked to it - Want to make sure I give you the right answer for where you are at ;D
Ps - Thanks for your awesome answer above! ;D
Post by: Kle123 on February 18, 2017, 04:27:22 pm
I wrote in my answer, assuming that the altitude is high enough that that there is negligible orbital decay, to most efficiently dock the craft, only the thrust is to be applied in the direction of the space station for only a minimal time, so that a small velocity is required which eventually moves the spacecraft to the station. As approaching the station, the thrust is to be forced in the opposite direction for the exact same amount of time so that the ship would decelerate back to being stationary relative to the station. As the craft is now hovering on top of the station, the thrust is to be forced upwards relative to the space station floor for a minimal amount of time so that it can slowly move into docking mode.
I didn't know whether the 5000m is important and also don't know how docking looks like.
Post by: jamonwindeyer on February 18, 2017, 05:18:21 pm
Thanks Jamon,
I wrote in my answer, assuming that the altitude is high enough that that there is negligible orbital decay, to most efficiently dock the craft, only the thrust is to be applied in the direction of the space station for only a minimal time, so that a small velocity is required which eventually moves the spacecraft to the station. As approaching the station, the thrust is to be forced in the opposite direction for the exact same amount of time so that the ship would decelerate back to being stationary relative to the station. As the craft is now hovering on top of the station, the thrust is to be forced upwards relative to the space station floor for a minimal amount of time so that it can slowly move into docking mode.
I didn't know whether the 5000m is important and also don't know how docking looks like.
Great! Solid answer - I love that you notice that orbital decay could play a role, I like that. One issue with your answer - If you speed up, will you be in the same orbit as the Space Station? No! Speeding up would mean you break away from the earth's orbit, so your answer doesn't quite work. Instead, what we need to do is apply forces such that we increase/decrease our orbital radius. To start, we apply a force to decrease our orbital radius. According to Kepler's 3rd Law, that will also reduce our period, and thus, increase our orbital velocity! We start moving faster, kind of like we are moving to overtake the station on the inside. At the precisely correct moment, we apply another force to return to our initial orbit and dock with the station, having used our reduction in radius to speed ourselves up :)
Does that make sense? It's a little tricky to comprehend, but if you run it through your head a few times hopefully something clicks!! Once you understand the manoeuvre, it becomes a little easier to tie Kepler's Law into it (it is the reason it works) ;D
Post by: Kle123 on February 18, 2017, 06:28:08 pm
Great! Solid answer - I love that you notice that orbital decay could play a role, I like that. One issue with your answer - If you speed up, will you be in the same orbit as the Space Station? No! Speeding up would mean you break away from the earth's orbit, so your answer doesn't quite work. Instead, what we need to do is apply forces such that we increase/decrease our orbital radius.
OMG that makes so much sense thanks!
One thing though... how would we decrease our orbital radius? Do we provide thrust in front of direction of motion to slow down? But wouldn't that cause us to follow an elliptical orbit/or slowly fall into the Earth's atmosphere / or does slowing down automatically drop us into a lower orbit?
So to reduce orbital radius do we slow down through providing forward thrust then accelerate to the required speed a lower orbit (which is what i'm thinking)?
Post by: bsdfjnlkasn on February 18, 2017, 10:43:22 pm
Define the term ‘g force’, and discuss the relevance of g force during the early stage of rocket launching
I find the definition for 'g force' in my textbook really unsatisfactory so if someone could please explain the concept of these forces to me- that would be greatly appreciated!!
Post by: Kle123 on February 19, 2017, 12:26:42 am
Hey there, I'm a tad confused with what to include my response to the following:
Define the term ‘g force’, and discuss the relevance of g force during the early stage of rocket launching
I find the definition for 'g force' in my textbook really unsatisfactory so if someone could please explain the concept of these forces to me- that would be greatly appreciated!!
The CONCEPT-
G force is a person’s apparent weight expressed a multiple of their true weight on earth. G-force is a scale used to measure different forces that are experienced at different rates of acceleration (when in contact with an accelerating body).
As a rocket accelerates people will feel different forces in comparison to another person. This variation is due to the different mass of different people which consequently correlates to different weight force(W=mg) and the force a body experiences due to Newton's second law.
If this raw scale of force due to acceleration was used to describe the effects on every person, it would be highly complicated. Thus g-force scale is used instead to simplify and communicate the same relative forces on different masses, as it only takes into account acceleration in which everyone experiences the same.
_______________________________________________________
During early stage of rocket launching, where rockets acceleration is greatest, g-force is significant due to the effects such as unconsciousness due to blood draining away(Inertia) from certain areas of the body and death if extreme g-forces are sustained. Consequently techniques have to be utilised such that astronauts aren't seriously harmed
Post by: jamonwindeyer on February 19, 2017, 01:04:06 am
OMG that makes so much sense thanks!
One thing though... how would we decrease our orbital radius? Do we provide thrust in front of direction of motion to slow down? But wouldn't that cause us to follow an elliptical orbit/or slowly fall into the Earth's atmosphere / or does slowing down automatically drop us into a lower orbit?
So to reduce orbital radius do we slow down through providing forward thrust then accelerate to the required speed a lower orbit (which is what i'm thinking)?
Hmm, yep I think from memory the best way to answer it is indeed your way! Provide an opposing thrust forwards, this reduces the orbital radius. Maintain this orbit to catch up to the space station, then get back into the initial orbit to dock ;D
Post by: f_tan on February 19, 2017, 12:20:53 pm
1. Two significant problems that will affect a manned spaceflight to mars are:
- the changes in gravitation energy
- protecting the space vehicle from high-speed electrically charged particles from the sun.
Use your understanding of physics to analyse these problems (8 marks)
2. In 1970 NASA launched Apollo 13, their third mission planned to land humans on the Moon. Half-way to the Moon a huge explosion crippled the spacecraft. The only way home for the astronauts was to fly around the back of the Moon and then fire the rocket engine to take the craft out of lunar orbit and put it into an Earth-bound trajectory
At the completion of the rocket engine burn, mission leader Jim Lovell was heard to say, 'We just put Isaac Newton in the driver's seat.'
Given that the spacecraft returned safely to Earth, justify Jim Lovell's statement (4 marks)
Any help is appreciated, thank you!
Post by: jamonwindeyer on February 19, 2017, 03:55:36 pm
Hi, not sure how to answer these past HSC questions:
1. Two significant problems that will affect a manned spaceflight to mars are:
- the changes in gravitation energy
- protecting the space vehicle from high-speed electrically charged particles from the sun.
Use your understanding of physics to analyse these problems (8 marks)
2. In 1970 NASA launched Apollo 13, their third mission planned to land humans on the Moon. Half-way to the Moon a huge explosion crippled the spacecraft. The only way home for the astronauts was to fly around the back of the Moon and then fire the rocket engine to take the craft out of lunar orbit and put it into an Earth-bound trajectory
At the completion of the rocket engine burn, mission leader Jim Lovell was heard to say, 'We just put Isaac Newton in the driver's seat.'
Given that the spacecraft returned safely to Earth, justify Jim Lovell's statement (4 marks)
Any help is appreciated, thank you!
Hey f_tan!
That first question you should answer in two parts. The first bit, you are talking about changes in gravitational potential energy. You could (and should) interpret this two ways - The challenge in obtaining that energy, and the challenge in then getting rid of it safely to land on Mars. Your answer to this part of the question should discuss GPE (perhaps include the formula), and discuss the methods by which we can obtain and shed this energy as required. It will likely require a discussion of orbits, as well as the rotational and orbital motion of the earth (think - How do we use the earths rotation to make it easier to gain GPE?) :)
The second part involving high speed electrically charged particles, you are discussing insulate shielding, much the same as you would discuss in a re-entry into the atmosphere question ;D
The second question is, basically, asking you to explain the Physics principles involved in the described manoeuvre. Essentially what happened is that the spacecraft was crippled while in orbit around the moon. The spacecraft maintained that orbit until a suitable time, where the engines were used to escape that orbit and enter an orbit around the earth (not so much an orbit, but a trajectory, as it had them gradually descend to the surface). Your job in answering the question is to explain how Newton's Laws play a role in that scenario. Newton's law of Universal Gravitation will need mentioning, since that is what governs the nature of orbits, and the 2nd Law (\(F=ma\)) is related to the acceleration of the rocket during the engine burn ;D
I hope that helps :)
Post by: f_tan on February 19, 2017, 04:41:32 pm
Hey f_tan!
That first question you should answer in two parts. The first bit, you are talking about changes in gravitational potential energy. You could (and should) interpret this two ways - The challenge in obtaining that energy, and the challenge in then getting rid of it safely to land on Mars. Your answer to this part of the question should discuss GPE (perhaps include the formula), and discuss the methods by which we can obtain and shed this energy as required. It will likely require a discussion of orbits, as well as the rotational and orbital motion of the earth (think - How do we use the earths rotation to make it easier to gain GPE?) :)
The second part involving high speed electrically charged particles, you are discussing insulate shielding, much the same as you would discuss in a re-entry into the atmosphere question ;D
The second question is, basically, asking you to explain the Physics principles involved in the described manoeuvre. Essentially what happened is that the spacecraft was crippled while in orbit around the moon. The spacecraft maintained that orbit until a suitable time, where the engines were used to escape that orbit and enter an orbit around the earth (not so much an orbit, but a trajectory, as it had them gradually descend to the surface). Your job in answering the question is to explain how Newton's Laws play a role in that scenario. Newton's law of Universal Gravitation will need mentioning, since that is what governs the nature of orbits, and the 2nd Law (\(F=ma\)) is related to the acceleration of the rocket during the engine burn ;D
I hope that helps :)
Thank you so much!!
Post by: kiwiberry on February 20, 2017, 10:20:00 pm
Post by: RuiAce on February 20, 2017, 10:23:16 pm
Hey, when back emf cancels out supply emf in a motor, why does it move at a constant speed if there's no net force?Think about what no net force means.
By Newton's second law of motion, we know that \(\Sigma F=ma\). Since all of our objects never have 0 mass, if net force = 0 then acceleration = 0.
If something is not accelerating, then its velocity does not change. It either remains at rest, or remains at the same velocity it was always at..
Worth mentioning: In fact, Newton's first law of motion (law of inertia) also states that an object tends to stay in its current state of motion (whatever velocity it was in) unless acted upon by an external force.
Post by: jamonwindeyer on February 20, 2017, 10:32:38 pm
Hey, when back emf cancels out supply emf in a motor, why does it move at a constant speed if there's no net force?
Just to add to Rui's answer - This is for ideal motors - Meaning no friction! If there was friction, it would slow down and gradually stop due to frictional forces ;D
Post by: RuiAce on February 20, 2017, 10:33:45 pm
Just to add to Rui's answer - This is for ideal motors - Meaning no friction! If there was friction, it would slow down and gradually stop due to frictional forces ;DHey Jamon, just gonna digress for a bit. Imagine if someone successfully made the frictionless surface 8)
Post by: jamonwindeyer on February 20, 2017, 10:35:34 pm
Hey Jamon, just gonna digress for a bit. Imagine if someone successfully made the frictionless surface 8)
No one could stop them ;)
Post by: kiwiberry on February 20, 2017, 10:41:14 pm
Think about what no net force means.
By Newton's second law of motion, we know that \(\Sigma F=ma\). Since all of our objects never have 0 mass, if net force = 0 then acceleration = 0.
If something is not accelerating, then its velocity does not change. It either remains at rest, or remains at the same velocity it was always at..
Worth mentioning: In fact, Newton's first law of motion (law of inertia) also states that an object tends to stay in its current state of motion (whatever velocity it was in) unless acted upon by an external force.
Just to add to Rui's answer - This is for ideal motors - Meaning no friction! If there was friction, it would slow down and gradually stop due to frictional forces ;DOhhh I see! Thanks guys :)
Post by: strawberriesarekewl on February 22, 2017, 10:01:31 am
Post by: jamonwindeyer on February 22, 2017, 10:31:02 am
I know this sounds like a very silly question but at school our teachers for the sciences are making us do past HSC questions in class and I know for a fact most HSC students leave HSC papers till the HSC and do past trial papers in order to prep for trials. Is it a good idea to do a couple of HSC questions during non-HSC periods? I have this feeling that the teachers will take some past trial questions for our school exams...
Hey strawberries! Welcome to the forums :)
So I see where you are coming from, but doing HSC questions now can only be a good thing. Practice is always a good thing. There are heaps of HSC exams to practice with now, so those combined with Trials means you won't have a shortage come September. It doesn't really matter what exams you do in the lead up to the HSC - Obviously HSC exams are a little better but everything is going to be almost equally useful ;D
Plus, doing HSC questions now, you won't remember the questions in September - You can always redo them :)
Post by: strawberriesarekewl on February 22, 2017, 10:47:25 am
Hey strawberries! Welcome to the forums :)
So I see where you are coming from, but doing HSC questions now can only be a good thing. Practice is always a good thing. There are heaps of HSC exams to practice with now, so those combined with Trials means you won't have a shortage come September. It doesn't really matter what exams you do in the lead up to the HSC - Obviously HSC exams are a little better but everything is going to be almost equally useful ;D
Plus, doing HSC questions now, you won't remember the questions in September - You can always redo them :)
Thanks for the solid advice
A quite "cliche" question but why is GPE negative? I know it has to do with the reference point being infinity or something and that when an object is being brought closer to the earth it has "negative work done" but it still kind of confuses me how GPE is negative
Post by: jamonwindeyer on February 22, 2017, 11:16:51 am
Thanks for the solid advice
A quite "cliche" question but why is GPE negative? I know it has to do with the reference point being infinity or something and that when an object is being brought closer to the earth it has "negative work done" but it still kind of confuses me how GPE is negative
Basically, it's a matter of definition! We choose GPE to be zero at an infinite distance from the centre of the field - That choice is convenient and makes the most sense for large scale applications.
We need GPE to decrease as we move closer to the centre of the field - As something loses 'altitude' it loses GPE. So, if we have chosen GPE to be zero at an infinite distance from the field, then every distance closer than that should be less. The only way to be less than zero is to be negative :)
That's the best explanation of WHY it happens. You can also talk about work. When we do work on an object, we give it energy. Bringing an object closer to the earth requires negative work (which makes sense, gravity pulls things closer), so the energy it acquires will be negative ;D
Note that you'll never be required to give a detailed explanation of why GPE is negative - You just need to know it is the case :)
Post by: kiwiberry on February 22, 2017, 03:40:52 pm
So I know to derive it, you sub V=IR into P=VI, but why does power turn into power loss when you do so?
Also, why can't you sub in I=V/R instead to get Ploss=V2/R? I know that this formula doesn't work because stepping up the voltages significantly reduces power loss, so it wouln't make sense to have Ploss proportional to V2. I just don't understand why subbing in the same formula in a different way doesn't work. Could someone please clarify why this is the case? :)
Post by: f_tan on February 22, 2017, 09:15:04 pm
1. The power of the sun is approximately 3.86 x 10^26 W. Calculate the mass conversion that must take place if all of this power is the result of the conversion of mass into energy. -Is power just the same thing as energy?
2. A cyclotron can be used to accelerate protons to relativistic velocities as they travel repeatedly around a circular, evacuated tube. Explain why the ultimate speed of the protons is limited and account for the fact that the kinetic energy of such a proton has no theoretical limit. - I thought the first bit might be because of mass dilation, but i'm not sure about kinetic energy?
Thank you!
Post by: strawberriesarekewl on February 22, 2017, 09:57:37 pm
Post by: jamonwindeyer on February 22, 2017, 11:03:40 pm
Hey, I'm a bit confused about the power loss formula.
So I know to derive it, you sub V=IR into P=VI, but why does power turn into power loss when you do so?
Also, why can't you sub in I=V/R instead to get Ploss=V2/R? I know that this formula doesn't work because stepping up the voltages significantly reduces power loss, so it wouln't make sense to have Ploss proportional to V2. I just don't understand why subbing in the same formula in a different way doesn't work. Could someone please clarify why this is the case? :)
Step into my office ;)
Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.
The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need :)
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference ;D
Post by: jamonwindeyer on February 22, 2017, 11:08:10 pm
Having trouble with these questions:
1. The power of the sun is approximately 3.86 x 10^26 W. Calculate the mass conversion that must take place if all of this power is the result of the conversion of mass into energy. -Is power just the same thing as energy?
2. A cyclotron can be used to accelerate protons to relativistic velocities as they travel repeatedly around a circular, evacuated tube. Explain why the ultimate speed of the protons is limited and account for the fact that the kinetic energy of such a proton has no theoretical limit. - I thought the first bit might be because of mass dilation, but i'm not sure about kinetic energy?
Thank you!
Hey!
1. Almost - Power is a rate of change of energy! 1 Watt = 1 Joule per second. So your answer will be a rate of change of mass, \(kg/s\) ;D
2. Both ideas are linked to mass dilation. Say we keep pushing the protons around the ring, meaning we keep giving them extra energy. The speed will approach but never reach the speed of light - Mass dilation is the reason for that! So if speed isn't increasing, where does that energy go? It goes to mass, that's the idea of mass dilation in the first place. Unlike the speed, the mass can actually increase as much as it likes, approaching infinity! Theoretically, if we keep pushing the proton, it will keep getting heavier. Since kinetic energy is defined as \(\frac{1}{2}mv^2\), if mass keeps increasing without a limit, so does kinetic energy ;D
Does that help? :)
Post by: jamonwindeyer on February 22, 2017, 11:10:31 pm
In HSC physics what are the hardest aspects in the following Modules, Space, motors and generators, ideas to implementation and quanta to quarks (option topic)
Hmm, if I had to pick one thing based on personal opinion (didn't do Quanta though):
Space: Special Relativity explanations - The calculations not so much
Motors and Generators: Three phase induction motors
Ideas to Implementation: Deflection plates in CRT's
;D
Post by: RuiAce on February 22, 2017, 11:10:43 pm
Step into my office ;)Flabbergasted. What the hell.
Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.
The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need :)
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference ;D
Life actually makes sense!!!
Post by: f_tan on February 22, 2017, 11:14:51 pm
Hey!
1. Almost - Power is a rate of change of energy! 1 Watt = 1 Joule per second. So your answer will be a rate of change of mass, \(kg/s\) ;D
2. Both ideas are linked to mass dilation. Say we keep pushing the protons around the ring, meaning we keep giving them extra energy. The speed will approach but never reach the speed of light - Mass dilation is the reason for that! So if speed isn't increasing, where does that energy go? It goes to mass, that's the idea of mass dilation in the first place. Unlike the speed, the mass can actually increase as much as it likes, approaching infinity! Theoretically, if we keep pushing the proton, it will keep getting heavier. Since kinetic energy is defined as \(\frac{1}{2}mv^2\), if mass keeps increasing without a limit, so does kinetic energy ;D
Does that help? :)
Yes! Thanks so much!
Post by: kiwiberry on February 22, 2017, 11:18:13 pm
Step into my office ;)
Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.
The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need :)
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference ;D
Oh my god that makes so much sense now!!! Thanks Jamon, that was a really good explanation :D
Post by: jamonwindeyer on February 22, 2017, 11:19:19 pm
Flabbergasted. What the hell.
Life actually makes sense!!!
Yes! Thanks so much!
Oh my god that makes so much sense now!!! Thanks Jamon, that was a really good explanation :D
WOO! Knowledge is power ;) most welcome guys so glad I could be helpful! ;D
Post by: strawberriesarekewl on February 22, 2017, 11:19:43 pm
(Also anyone else willing to put some input so then i can get a somewhat better idea)
Post by: jamonwindeyer on February 22, 2017, 11:34:40 pm
What other stuff would you of had considered difficult other than those aspects jamon?
(Also anyone else willing to put some input so then i can get a somewhat better idea)
I'd say the whole course is fairly difficult - Those are standouts for me personally, but like, difficult questions can be asked on practically any dot point ;D
Post by: jakesilove on February 23, 2017, 03:17:28 pm
What other stuff would you of had considered difficult other than those aspects jamon?
(Also anyone else willing to put some input so then i can get a somewhat better idea)
I second everything Jamon has said.
Post by: strawberriesarekewl on February 23, 2017, 03:48:35 pm
what is the hardest thing in quanta to quarks in your opinion
Post by: bsdfjnlkasn on February 23, 2017, 03:57:40 pm
Step into my office ;)
Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.
The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need :)
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference ;D
Hey Jamon,
This definitely is a gap in my knowledge from Year 11 so I would really appreciate some additional examples (even though some really bright people have already moved on ;) ). I just never understood the reason or method for calculating potential drops across resistors, so any help would be greatly appreciated ;D ;D
Post by: Sukakadonkadonk on February 23, 2017, 04:40:08 pm
Could someone please explain to me what an electromotive force actually is? I keep seeing it used in strange ways and would appreciate an accurate definition.
Thanks!!
Post by: jamonwindeyer on February 23, 2017, 07:31:59 pm
Hey Jamon,
This definitely is a gap in my knowledge from Year 11 so I would really appreciate some additional examples (even though some really bright people have already moved on ;) ). I just never understood the reason or method for calculating potential drops across resistors, so any help would be greatly appreciated ;D ;D
Sure!! I think a worked example in a video will be best - I'll put something together tonight and upload it ;D
Post by: jamonwindeyer on February 23, 2017, 07:42:11 pm
Hey guys, :)
Could someone please explain to me what an electromotive force actually is? I keep seeing it used in strange ways and would appreciate an accurate definition.
Thanks!!
Hey! So basically, it's a voltage. It's an abstract concept; force does not mean Newtons, but instead energy per unit charge. So, an electromotive force is something that gives things with charge, energy. Just like gravity gives things with mass, energy! ;D
This could be a voltage generated by a battery, or induction, or an electric field. But it's a voltage, an electric field, and it causes things with charge to gain energy! ;D
It is genuinely difficult to explain this in a semi-informal way, without the proper formal definitions of it (which wikipedia has and does a decent job with if you like), but does that help at all? :)
Post by: jakesilove on February 23, 2017, 07:43:42 pm
Hey! So basically, it's a voltage. It's an abstract concept; force does not mean Newtons, but instead energy per unit charge. So, an electromotive force is something that gives things with charge, energy. Just like gravity gives things with mass, energy! ;D
This could be a voltage generated by a battery, or induction, or an electric field. But it's a voltage, an electric field, and it causes things with charge to gain energy! ;D
It is genuinely difficult to explain this in a semi-informal way, without the proper formal definitions of it (which wikipedia has and does a decent job with if you like), but does that help at all? :)
Just to add to that; without a formal understanding of the underlying mathematics, it's really hard to explain the difference between EMF and voltage. In fact, I'm supposed to understand the underlying mathematics and I'm still bloody confused. So, for the sake of the HSC, think of it as the push on an electron
Post by: strawberriesarekewl on February 23, 2017, 07:59:56 pm
How is the physics poster going? Also how do I study effectively by looking at a poster? (serious question because I am genuinely curious to know)
I like how the chemistry one was super summarised and what not but I think those posters just have the bare bones
Post by: jamonwindeyer on February 23, 2017, 08:11:12 pm
Hey jake/jamon
How is the physics poster going? Also how do I study effectively by looking at a poster? (serious question because I am genuinely curious to know)
I like how the chemistry one was super summarised and what not but I think those posters just have the bare bones
A poster is always going to be bare bones content - To put all the minor details of the whole course on a single piece of paper would cover a wall, aha :) The poster isn't meant to be the same as the set of notes - The notes cover the intricacies, the poster is a (super effective) tool for memorising the basic concepts in a visual and succinct way. You can't teach yourself the course from the poster, like you could the Notes :)
Having the posters in places you view often (above your study space, for example), gives you the content in an easy to digest format. I used to have posters above my bed for main concepts and formulas - I'd just study them while I was moving around my room, just to jog my memory for free while I wouldn't normally be studying :)
Post by: Sukakadonkadonk on February 23, 2017, 08:37:55 pm
Hey! So basically, it's a voltage. It's an abstract concept; force does not mean Newtons, but instead energy per unit charge. So, an electromotive force is something that gives things with charge, energy. Just like gravity gives things with mass, energy! ;D
This could be a voltage generated by a battery, or induction, or an electric field. But it's a voltage, an electric field, and it causes things with charge to gain energy! ;D
It is genuinely difficult to explain this in a semi-informal way, without the proper formal definitions of it (which wikipedia has and does a decent job with if you like), but does that help at all? :)
Yep, thanks Jake, Jamon,
Definitely a better description than some other resources I've used.
So if during an exam, they ask a question on EMF's would the first few of your sentences be a good way of answering it?
Post by: Bubbly_bluey on February 23, 2017, 09:01:01 pm
Another question: I know 3 hand grip rules
1) right hand palm slap rule
2) right hand grip rule
3) Left hand palm rule
I keep getting confused when to use which rule to use.
Post by: jamonwindeyer on February 23, 2017, 09:28:24 pm
Yep, thanks Jake, Jamon,
Definitely a better description than some other resources I've used.
So if during an exam, they ask a question on EMF's would the first few of your sentences be a good way of answering it?
As a backwards way of answering your question, I highly doubt they'd ask it ;)
But if I got:
Question: What is an electromotive force?
Answer: A potential difference/electric field which increases the energy per unit charge of objects within said field.
Or, something similar to that :) but yep, highly doubt they'd ask it, just too finicky imo! :)
Post by: jamonwindeyer on February 23, 2017, 09:35:37 pm
hello! i'm having a great deal of trouble trying to figure out how "induced currents in rings and loops" work. I did a worksheet in class (photo) with answers but i obviously don't know whats going on. How do you determine the direction of the induced current of rings when they enter into a denser or less dense magnetic flux?
Another question: I know 3 hand grip rules
1) right hand palm slap rule
2) right hand grip rule
3) Left hand palm rule
I keep getting confused when to use which rule to use.
Hey! This is a really tricky thing, I remember doing a really similar task and having heaps of trouble. Thankfully, it doesn't get asked much in the HSC! :)
So in the scenarios shown on your sheet, you need to ask yourself one question: What extra magnetic field lines are being added to your loop? Is it additional field lines INTO the page, or OUT OF the page. The current will flow to create magnetic field lines in the loop in the opposite direction to what is being introduced - That is Lenz's Law.
Let's do Loop B as an example. It is moving into a less dense magnetic field, so, removing lines into the page. We can analogise this to mean adding lines out of the page - It means the same thing! So, the current will act to introduce lines INTO the page, to do the opposite to what the movement is doing.
Here, we turn to the right hand grip rule. We need to introduce lines INTO the page, meaning the North Pole should face INTO the page as well (remember, magnetic field lines flow towards north inside a loop/coil). So, our thumb faces into the page, which means our fingers wrap clockwise - There's your current direction :)
So there is two steps: Figure out the change that is being introduced; the current will do the opposite. Then, use the right hand grip rule (as shown above) to yield the current direction.
Oh, and on those rules:
Right Hand Grip: Direction of current in a loop/coil/solenoid
Right Hand Slap: Direction of current in any other scenario
Left Hand Slap: Don't use it unless you are confident, but it is the direction of electron flow in any other scenario. Think eLEFTron ;)
Hope this helps!
Post by: kiwiberry on February 23, 2017, 09:45:26 pm
Hey! This is a really tricky thing, I remember doing a really similar task and having heaps of trouble. Thankfully, it doesn't get asked much in the HSC! :)
So in the scenarios shown on your sheet, you need to ask yourself one question: What extra magnetic field lines are being added to your loop? Is it additional field lines INTO the page, or OUT OF the page. The current will flow to create magnetic field lines in the loop in the opposite direction to what is being introduced - That is Lenz's Law.
Let's do Loop B as an example. It is moving into a less dense magnetic field, so, removing lines into the page. We can analogise this to mean adding lines out of the page - It means the same thing! So, the current will act to introduce lines INTO the page, to do the opposite to what the movement is doing.
Here, we turn to the right hand grip rule. We need to introduce lines INTO the page, meaning the North Pole should face INTO the page as well (remember, magnetic field lines flow towards north inside a loop/coil). So, our thumb faces into the page, which means our fingers wrap clockwise - There's your current direction :)
So there is two steps: Figure out the change that is being introduced; the current will do the opposite. Then, use the right hand grip rule (as shown above) to yield the current direction.
Oh, and on those rules:
Right Hand Grip: Direction of current in a loop/coil/solenoid
Right Hand Slap: Direction of current in any other scenario
Left Hand Slap: Don't use it unless you are confident, but it is the direction of electron flow in any other scenario. Think eLEFTron ;)
Hope this helps!
Just to add, the right hand palm/slap rule is also used to find the direction of the force experienced by a current carrying conductor in a magnetic field - ie the motor effect :)
Post by: jamonwindeyer on February 23, 2017, 09:46:14 pm
Just to add, the right hand palm/slap rule is also used to find the direction of the force experienced by a current carrying conductor in a magnetic field - ie the motor effect :)
Ahh yep yep, good catch kiwiberry, tah! ;D
Post by: strawberriesarekewl on February 23, 2017, 10:22:53 pm
Post by: jamonwindeyer on February 23, 2017, 10:52:48 pm
I know this goes back to prelim but how do we determine the direction of certain things in the space and motors and generators module
You might have to be a bit more specific than "certain things," aha :P
Post by: jamonwindeyer on February 24, 2017, 11:02:19 am
Post by: Bubbly_bluey on February 24, 2017, 12:12:33 pm
Thank you :)
Post by: jakesilove on February 24, 2017, 12:18:12 pm
Hi we just did "Lenz's Law on Moving Conductors (Sheets) and i am really confused as to how a metal plate with slits has a slower stopping motion than a full plate when swung between a magnetic field. I understand in a full plate that lenz's laws apply to try to oppose the entry and attract it but i don't know how the slit plates work. There is also something about Eddy currents which I also don't understand.
Thank you :)
Hey!
If you draw a sheet of metal, and then the eddy currents, you can imagine them like great big circles. Lots and lots of circles of electrons are moving around the sheet, and that movement of electrons creates the opposing magnetic field.
However, if a sheet of metal has slits in it, then the eddy currents themselves must get smaller. Draw a sheet with slits in it, and try to draw some circles to represent the eddy currents. They just can't be as big; they don't have enough space! So, the eddy currents are literally 'weaker', and so create a weaker magnetic field, which opposes the movement less.
It's all about the physical space that the eddy currents (ie. electrons) have to travel. Does that make sense?
Post by: Bubbly_bluey on February 24, 2017, 07:10:55 pm
It's all about the physical space that the eddy currents (ie. electrons) have to travel. Does that make sense?
Yes thank you. but quick question: are eddy currents generated to oppose the metal plates entering the magnetic field like induced current?
Post by: jamonwindeyer on February 24, 2017, 07:35:00 pm
Yes thank you. but quick question: are eddy currents generated to oppose the metal plates entering the magnetic field like induced current?
You got it ;D
Post by: bsdfjnlkasn on February 25, 2017, 07:14:56 pm
Any explanations/diagrams would be greatly appreciated :) :)
Post by: Maraos on February 25, 2017, 09:19:51 pm
I don't know if this has already being asked before but my question is how do you get ready for a year 12 Physics skills examination?
I have a 'skills' examination this Friday (3rd of March) on Ideas to implementation and I'm not really sure how I should spend my time
My exam notification is also very vague (probably on purpose) it states;
Work covered:
All work during practical lessons. See also the above skills outcomes from the syllabus
Format:
The exam will have 5 minutes reading time and 45 minutes working time. The exam will consist of two parts; an information sheet and question section totaling 35 marks.
Any advice or help will be greatly appreciated!
Thanks!
Post by: Maraos on February 25, 2017, 09:27:53 pm
Hey sorry if this is a really broad/silly question, but why don't satellites fall back to Earth?
Any explanations/diagrams would be greatly appreciated :) :)
Satellites don't fall back to earth due to their extremely high velocities. This is the same reason why the moon doesn't crash into the earth, because it is in an orbit around another celestial body. Even though the gravity of Earth has an influence on the satellite and is pulling it towards the center the satellite has enough velocity to maintain an orbit. Think about shooting a cannon on-top of a very big mountain, it will travel some distance but then eventually fall. However, with enough velocity the cannon ball will continually circle the planet . So in a sense a satellite is actually falling to earth, it just has enough velocity to continually miss the atmosphere.
Post by: jamonwindeyer on February 26, 2017, 12:16:39 am
Hi,
I don't know if this has already being asked before but my question is how do you get ready for a year 12 Physics skills examination?
I have a 'skills' examination this Friday (3rd of March) on Ideas to implementation and I'm not really sure how I should spend my time
My exam notification is also very vague (probably on purpose) it states;
Work covered:
All work during practical lessons. See also the above skills outcomes from the syllabus
Format:
The exam will have 5 minutes reading time and 45 minutes working time. The exam will consist of two parts; an information sheet and question section totaling 35 marks.
Any advice or help will be greatly appreciated!
Thanks!
Hey! Thanks heaps for your awesome answer to the above question - Couldn't have answered it better myself ;D
So for your skills task, I recommend you jump to Page 38 of the syllabus and have a bit of a read - That's the stuff that will be assessed (it could be the stuff on your notification, and probably is all there, but just in case). It's a bit of a mish-mash of stuff, but think things like:
- Identifying variables
- Proposing experiment designs/improvements
- Reliability, accuracy, validity
- Graphing (especially lines of best fit)
- Using equipment properly
- Research skills (is a source reliable or not and why?)
These are always tough to prepare for, but I'd spend my time going over the notes for all the practical tasks you've done for I2I so far - Understand how they work and why they work in quite a bit of detail. Read over any practical reports you've written in the last 12 months to re-familiarise with key concepts - Reliability, accuracy, variables, etc. Besides that, unless you were given a sample task (which if not you may be able to find here, not sure) then there isn't much else you can do!!
I hope that helps, at least a little - Good luck! ;D
Post by: Maraos on February 26, 2017, 12:25:11 am
Hey! Thanks heaps for your awesome answer to the above question - Couldn't have answered it better myself ;D
So for your skills task, I recommend you jump to Page 38 of the syllabus and have a bit of a read - That's the stuff that will be assessed (it could be the stuff on your notification, and probably is all there, but just in case). It's a bit of a mish-mash of stuff, but think things like:
- Identifying variables
- Proposing experiment designs/improvements
- Reliability, accuracy, validity
- Graphing (especially lines of best fit)
- Using equipment properly
- Research skills (is a source reliable or not and why?)
These are always tough to prepare for, but I'd spend my time going over the notes for all the practical tasks you've done for I2I so far - Understand how they work and why they work in quite a bit of detail. Read over any practical reports you've written in the last 12 months to re-familiarise with key concepts - Reliability, accuracy, variables, etc. Besides that, unless you were given a sample task (which if not you may be able to find here, not sure) then there isn't much else you can do!!
I hope that helps, at least a little - Good luck! ;D
Thanks for the advice!
The notification does list some syllabus outcomes so I will go back and have a look at them.
After that i guess ill just have to hope for the best ;D ;D
Post by: jamonwindeyer on February 26, 2017, 12:28:31 am
Thanks for the advice!
The notification does list some syllabus outcomes so I will go back and have a look at them.
After that i guess ill just have to hope for the best ;D ;D
Unfortunately that's about all you can do! Very hard to prep for skills tests without samples - Not much more you can practice with besides drawing graphs of random data sets ::)
Post by: beau77bro on February 26, 2017, 09:24:02 pm
(http://uploads.tapatalk-cdn.com/20170226/1080de26f23c66275e23e2b53d6634ff.jpg)
Post by: kiwiberry on February 26, 2017, 10:25:04 pm
Someone please help I don't know how to even approach this/ think about it
(http://uploads.tapatalk-cdn.com/20170226/1080de26f23c66275e23e2b53d6634ff.jpg)
So this question is addressing the relativity of simultaneity!
Captain Y is in the same frame of reference as the lasers, so will see them hit at the same time. This rules out C and D
Captain X is moving with respect to the lasers, and so is in a different frame of reference. Ok, the question is a bit unclear because technically Captain X would see laser 1 hit before laser 2 if the lasers were fired after Captain X had flown past, because he'll be sitting closer to laser 1 no matter what. But, assuming Captain X was in the middle of the two lasers when they were fired (like in the train thought experiments), because the spaceship is moving at 0.5c left, Captain X will move closer to laser 1 and further away from laser 2 before they strike. Therefore, due to the constancy of the speed of light, the light from laser 1 will travel a shorter distance and reach Captain X before laser 2. So the answer should be A :)
Post by: beau77bro on February 27, 2017, 08:52:44 am
Post by: jamonwindeyer on February 27, 2017, 09:45:40 am
Wow thankyou kiwi. But wait how is laser 1 gonna hit first. He's moving away from it but the ship isn't? Like the lasers are perpendicular to his velocity, isn't it different to the thought experiment? I'm just a tad confused but I don't get how his motion factors into the lights distance travelled when they aren't in the same plane.
The lasers will definitely strike the ship at the same time with respect to Captain Y. But here's the issue - The Captain of X will only see the lasers hit the ship once the light from the point of impact reaches him. The scenario is literally the same as the thought experiment - Instead of letting off fireworks at either end, you have laser explosions! It's acknowledging that the Captain of X doesn't see the laser strikes as they happen, but once the light from those explosions reaches him, that is the tricky bit.
The other reason the lasers don't hit at the same time? It's not an option - In questions like this, if you aren't totally sure of the correct answer, sometimes you just have to go with the most sensible answer! It will often get you out of trouble ;D
Post by: winstondarmawan on March 02, 2017, 06:19:16 pm
I know this is not a physics specific question, but I just wanted to know the best possible way/structure to answer the following types of questions:
- Assess the validity of this statement.
- Assess the validity/reliability/accuracy of this experiment.
Thanks in advance!
Post by: inu99 on March 02, 2017, 06:45:53 pm
A) Determine the size of the torque when the coil is in the horizontal position
B) How would this torque affect the motion of the coil
c) Determine the size of the torque when the coil has moved through an angle of 30 degrees
d) Determine the size of the torque when the coil has reached th vertical position
Post by: jakesilove on March 02, 2017, 06:46:30 pm
Hey!
I know this is not a physics specific question, but I just wanted to know the best possible way/structure to answer the following types of questions:
- Assess the validity of this statement.
- Assess the validity/reliability/accuracy of this experiment.
Thanks in advance!
Hey! The best answer for ANY assess question is to weigh up to potential advantages against potential disadvantages. In this case, generally start by stating why something may be valid (ie. testing what is trying to be tested), and then go into a bit of depth as to where validity may fall down. Things to think about when it comes to validity/accuracy/reliability:
Validity
- Is the experiment testing what you are trying to test?
- Are there uncontrolled variables that may affect the data?
- Is there another explanation/factor that will affect your measurements?
Accuracy
- How good is your measuring equipment?
- How close is your values to the 'real' value?
- Are there any errors in your experiment (human error etc.)
Reliability
- If you repeated the experiment, would you get the same data?
- How many times did you repeat the experiment?
- Did you take any averages? Plot any graphs?
Hope this helped!
Post by: jakesilove on March 02, 2017, 06:54:27 pm
The armature of a DC motor consists of 40 turns of circular coils with a diameter of 20cm. A voltage of 6.0v is supplied to the coil. If the coil has a total resistance of 0.50 ohms and the magnetic field has a strenght of 0.20T:
A) Determine the size of the torque when the coil is in the horizontal position
B) How would this torque affect the motion of the coil
c) Determine the size of the torque when the coil has moved through an angle of 30 degrees
d) Determine the size of the torque when the coil has reached th vertical position
A)
The relevant formula is
We have to first current first, using
Now,
B)
The torque affects the motion of the coil by exerting a turning force. This turning force causes the coil to spin (either clockwise or anti-clockwise) around the central axis of the coil.
C)
D)
When the coil is vertical, the angle will be 90 degrees. Plugging this into the formula, the torque is zero!
Post by: Maraos on March 02, 2017, 08:17:29 pm
How exactly would you improve the validity of an experiment?
I understand the validity is whether or not your experimental design is adequate in solving the aim and whether or not all the controlled variables were kept and constant and the independent variable was changed.
But how could you make improvements to increase the validity of an experiment? .
Thanks!
Post by: Maraos on March 02, 2017, 11:21:52 pm
Need an answer quick (Physics Prac test tomorrow)After doing some of my own research and looking at past papers I think I've got it (can someone please confirm if my thinking is correct??)
How exactly would you improve the validity of an experiment?
I understand the validity is whether or not your experimental design is adequate in solving the aim and whether or not all the controlled variables were kept and constant and the independent variable was changed.
But how could you make improvements to increase the validity of an experiment? .
Thanks!
Improving validity is 'making sure' that the controlled variables are 100% controlled. And removing or adding/improving unreliable equipment/methods to ensure that the experimental design is adequate in achieving the aim.
So essentially i think that it is making adjustments to your experimental design in-order to ensure that controlled variables are kept at a constant and that the experimental design is actually solving your intended aim.....
Is this right? ;D ;D I dunno....
Post by: kiwiberry on March 02, 2017, 11:37:05 pm
After doing some of my own research and looking at past papers I think I've got it (can someone please confirm if my thinking is correct??)Hey, I think you're right! I can't think of anything other than controlling variables haha
Improving validity is 'making sure' that the controlled variables are 100% controlled. And removing or adding/improving unreliable equipment/methods to ensure that the experimental design is adequate in achieving the aim.
So essentially i think that it is making adjustments to your experimental design in-order to ensure that controlled variables are kept at a constant and that the experimental design is actually solving your intended aim.....
Is this right? ;D ;D I dunno....
Post by: jamonwindeyer on March 02, 2017, 11:58:17 pm
Post by: bsdfjnlkasn on March 03, 2017, 05:07:28 pm
This one is sort of open to anyone who has been required to do a practical task for Space. If you guys could offer me the types of experiments you did and the style of questions (will probably need to graph something) and whether there were any specific things that were integral to the experiment. I know this question is super vague but I just want to get a broader understanding of the range of experiments I could be asked to consider and what skills will help me excel in them.
I know this has probably been asked a million times but what would you recommend to be a general skill set that will cover the outcomes tested in these exams? I want to be super prepared for this exam and hopefully can get some good tips (as practice for the most part is really limited - any resources therefore would be greatly appreciated ;D)
Thanks again and I really hope it doesn't seem like I'm cheating. I really need to improve on my physics rank and settle in nicely before trials :-\ so if you could help a fellow physics student out that would be lovely :)
Post by: Bubbly_bluey on March 03, 2017, 06:44:28 pm
I have two questions:
1) What is a clear explanation of a back emf? My current "understanding" is that when an induced current is produced to create the spinning motion of the ammature, the law of conservation applies because it can not spin indefinitely faster. Does it act like a resistant to the normal flow of the current? I'm not what exactly it is.
2)Does Lenz's law only apply to DC and AC generators and not motors?
Thanks:)
Post by: jamonwindeyer on March 03, 2017, 08:17:32 pm
Hey there another prac related question :)
This one is sort of open to anyone who has been required to do a practical task for Space. If you guys could offer me the types of experiments you did and the style of questions (will probably need to graph something) and whether there were any specific things that were integral to the experiment. I know this question is super vague but I just want to get a broader understanding of the range of experiments I could be asked to consider and what skills will help me excel in them.
I know this has probably been asked a million times but what would you recommend to be a general skill set that will cover the outcomes tested in these exams? I want to be super prepared for this exam and hopefully can get some good tips (as practice for the most part is really limited - any resources therefore would be greatly appreciated ;D)
Thanks again and I really hope it doesn't seem like I'm cheating. I really need to improve on my physics rank and settle in nicely before trials :-\ so if you could help a fellow physics student out that would be lovely :)
Hey! It's definitely not cheating to ask for some friendly advice mate! ;D
As far as I know, the big experiments for space will either involve a pendulum or analysis of projectiles. So be sure to revise your kinematic formulae, and the pendulum formula, \(T=\sqrt{\frac{2\pi}{L}}\) - There's a pretty good chance you'll need one of them :)
In terms of practice, I'd revise:
- Graphing/Lines of Best Fit
- Variables (Independent/Dependent/Controlled)
- Variability/Accuracy/Reliability
- Sources of Error
You might be able to find some practice experiment tasks here ;D
Besides that, some general advice - Do the experiment twice if you have time (or take lots of measurements), it increases reliability. Ensure you graph your data with the independent variable on the horizontal axis and the dependent variable on the vertical axis, and use pencil (trust me, just use pencil). Human error is always a source of error ;) and just take it slow and be careful with your setup!
Post by: jamonwindeyer on March 03, 2017, 08:24:35 pm
hello all!
I have two questions:
1) What is a clear explanation of a back emf? My current "understanding" is that when an induced current is produced to create the spinning motion of the ammature, the law of conservation applies because it can not spin indefinitely faster. Does it act like a resistant to the normal flow of the current? I'm not what exactly it is.
2)Does Lenz's law only apply to DC and AC generators and not motors?
Thanks:)
Hey!!
1. Back EMF is the name given to the induced emf in a motor that opposes the supply emf. It is formed because the coil experiences a changing magnetic flux as it spins, thus inducing an emf by Faraday's Law. It must flow to oppose the supply current that created it, to adhere to the Conservation of Energy and Lenz's Law.
^ That would be my succinct explanation. I think you are on the right track, you need to work on understanding that back emf is another, opposing voltage/current. That is, the current in your coil is given by:
I actually wrote a guide on induction last year, which contains a bit of elaboration on Lenz's Law and back emf! It might be of some help ;D
2. Lenz's Law is a universal concept, governing the direction of all induced emf's ;D
Post by: kiwiberry on March 03, 2017, 08:37:01 pm
Hey there another prac related question :)Just to add to Jamon's answer :)
This one is sort of open to anyone who has been required to do a practical task for Space. If you guys could offer me the types of experiments you did and the style of questions (will probably need to graph something) and whether there were any specific things that were integral to the experiment. I know this question is super vague but I just want to get a broader understanding of the range of experiments I could be asked to consider and what skills will help me excel in them.
I know this has probably been asked a million times but what would you recommend to be a general skill set that will cover the outcomes tested in these exams? I want to be super prepared for this exam and hopefully can get some good tips (as practice for the most part is really limited - any resources therefore would be greatly appreciated ;D)
Thanks again and I really hope it doesn't seem like I'm cheating. I really need to improve on my physics rank and settle in nicely before trials :-\ so if you could help a fellow physics student out that would be lovely :)
I feel like the pendulum prac will most likely be tested because it's easy to set up and there's quite a lot you could ask about it in terms of calculations and stuff on validity/accuracy/reliability (but don't ignore the others!). A few things to consider with the pendulum prac off the top of my head:
- Timing 10 oscillations to increase accuracy - too many oscillations and the pendulum will be affected by air resistance, but too little swings will lead to a larger margin of error
- Only timing after the pendulum has reached a constant swing to increase accuracy as well
- Experiment repeated with different lengths of string multiple times to increase reliability
Some other random thoughts:
- There's a high chance that you'll get some sort of question which asks you to graph something, find the gradient of the graph, and then use that value to calculate something. Eg for the pendulum, you could be asked to graph L vs T2 and then use this to find g using \(T=2\pi \sqrt{\frac{L}{g}}\)
- Line of best fit - only draw from the x-coordinate of your first point to the x-coordinate of your last point. It doesn't have to start and end at your first and last points - just make sure there's an even number of points on either side of it. Bring a 30cm ruler, trust me, it makes life so much easier when drawing these haha
- Calculating gradient - don't use points that you've already plotted! Pick something on your line of best fit that's in between, and show which points you pick on your graph. Omg and remember that gradients need units too!!! I've lost marks so many times for this :'(
The teachers at my school are really picky though so you might not even have to follow all these rules
I haven't done a prac exam on Space so keep that in mind, but hope this helps a bit! :)
Post by: jamonwindeyer on March 03, 2017, 08:46:58 pm
OK but i cant make a connection with Lenz's labelling applied on motors. Like isn't a motor powered by the motor effect with no induced emf?
A motor is definitely 'powered' by the motor effect - It's the primary principle that allows it to work ;D but you still get induced emf in motors, that is back emf, and what governs the direction of back emf in motors? That's Lenz's Law ;D
Post by: Bubbly_bluey on March 03, 2017, 08:55:09 pm
A motor is definitely 'powered' by the motor effect - It's the primary principle that allows it to work ;D but you still get induced emf in motors, that is back emf, and what governs the direction of back emf in motors? That's Lenz's Law ;D
I'm sorry last question for today promise XD
So is the induced emf generated when the ammature cuts the magnet flux, and the system does not like that so it will create back emf to counter that?
Post by: jamonwindeyer on March 03, 2017, 09:01:19 pm
I'm sorry last question for today promise XD
So is the induced emf generated when the ammature cuts the magnet flux, and the system does not like that so it will create back emf to counter that?
Ahaha no worries! Yep that is pretty much it - The changing magnetix flux experienced by the armature is caused by the rotation, and as you say, it "doesn't like it" (Physics hates change) and an opposing emf (back emf) is induced to counteract that change ;D
Post by: bsdfjnlkasn on March 04, 2017, 08:36:54 am
Sorry I wasn't sure of how to quote both of your responses but I really wanted to thank you for helping me out! I am quite convinced we'll have a prac on analysing projectiles but i've taken down the extra pendulum information (which was super detailed and helpful btw :D!) just in case. We only just started learning about projectiles so what do you think we could be asked to do in relation to them - probably proving their parabolic shape if subject to constant acceleration? But then how could they incorporate graphs and gradients? Also what would you do in an exam if your results were really far off the theoretical but you don't have time to change them - I guess it gives you more to talk about in the accuracy/reliability/validity sections but surely they'll have to deduct marks if it's just outright inaccurate - really hoping this doesn't happen :-[
Also, when it comes down to asking questions on all three (accuracy/reliability/validity), would it be wise to address them sometimes together? Or are they so distinct that there is little room to join them and show how decreasing accuracy comes at the expense of decreasing validity for example. Does that make sense? I just feel like i'd run out of things to say if I had to focus on one specific thing. Are there any stand out things I should mention for each one? A google search isn't particularly helpful in this so as always, any help would be greatly appreciated!!
Post by: helpsme on March 04, 2017, 12:12:49 pm
2.Why does the air cool as it rises?
3.Why is so little heat transferred from the Earth's surface to its atmosphere by conduction?
Post by: jamonwindeyer on March 04, 2017, 12:58:38 pm
Hey Jamon and kiwiberry!
Sorry I wasn't sure of how to quote both of your responses but I really wanted to thank you for helping me out! I am quite convinced we'll have a prac on analysing projectiles but i've taken down the extra pendulum information (which was super detailed and helpful btw :D!) just in case. We only just started learning about projectiles so what do you think we could be asked to do in relation to them - probably proving their parabolic shape if subject to constant acceleration? But then how could they incorporate graphs and gradients? Also what would you do in an exam if your results were really far off the theoretical but you don't have time to change them - I guess it gives you more to talk about in the accuracy/reliability/validity sections but surely they'll have to deduct marks if it's just outright inaccurate - really hoping this doesn't happen :-[
Also, when it comes down to asking questions on all three (accuracy/reliability/validity), would it be wise to address them sometimes together? Or are they so distinct that there is little room to join them and show how decreasing accuracy comes at the expense of decreasing validity for example. Does that make sense? I just feel like i'd run out of things to say if I had to focus on one specific thing. Are there any stand out things I should mention for each one? A google search isn't particularly helpful in this so as always, any help would be greatly appreciated!!
Projectile questions will usually involve either something to do with their parabolic flight path, or you could even use projectile analysis to determine a value for acceleration due to gravity. They'll all be fairly similar though: Launch a projectile horizontally from a given height, measure the range (usually) ;D
If you get a significantly (>10% error) inaccurate result, use your discussion to extrapolate as to the reasons why. Believe it or not, if you perform a valid experiment, take results correctly, analyse them correctly and discuss why they could be inaccurate, you should still perform quite well! The majority of the marks will be on your skills, there shouldn't be a heap assigned to your results actually being accurate :) you could of course redo the experiment if you have time, but that's a push ;D
You can definitely discuss those three things together, though you MUST address each of them in some way. Rui wrote a great guide on how they might interrelate ;D
Post by: Shadowxo on March 04, 2017, 04:41:29 pm
Hey Jamon and kiwiberry!
Sorry I wasn't sure of how to quote both of your responses but I really wanted to thank you for helping me out! I am quite convinced we'll have a prac on analysing projectiles but i've taken down the extra pendulum information (which was super detailed and helpful btw :D!) just in case. We only just started learning about projectiles so what do you think we could be asked to do in relation to them - probably proving their parabolic shape if subject to constant acceleration? But then how could they incorporate graphs and gradients? Also what would you do in an exam if your results were really far off the theoretical but you don't have time to change them - I guess it gives you more to talk about in the accuracy/reliability/validity sections but surely they'll have to deduct marks if it's just outright inaccurate - really hoping this doesn't happen :-[
Also, when it comes down to asking questions on all three (accuracy/reliability/validity), would it be wise to address them sometimes together? Or are they so distinct that there is little room to join them and show how decreasing accuracy comes at the expense of decreasing validity for example. Does that make sense? I just feel like i'd run out of things to say if I had to focus on one specific thing. Are there any stand out things I should mention for each one? A google search isn't particularly helpful in this so as always, any help would be greatly appreciated!!
Just adding to jamonwindeyer (from a VCE course perspective)
Often you'll have to resolve the speed into it's different components, and use this to find a variable. Some common ones are how long it's in the air, how far it travels horizontally, the speed it hits the ground at, etc. If you can resolve the speed into it's components and use the basic physics equations it should be fairly straightforward. They may also ask questions such as "when tested, the projectile didn't travel as far as calculated, explain why" - air resistance, and other similar questions. Overall I found projectiles straightforward so you should be fine. Good luck :)
Post by: jamonwindeyer on March 04, 2017, 05:00:05 pm
Just adding to jamonwindeyer (from a VCE course perspective)
Often you'll have to resolve the speed into it's different components, and use this to find a variable. Some common ones are how long it's in the air, how far it travels horizontally, the speed it hits the ground at, etc. If you can resolve the speed into it's components and use the basic physics equations it should be fairly straightforward. They may also ask questions such as "when tested, the projectile didn't travel as far as calculated, explain why" - air resistance, and other similar questions. Overall I found projectiles straightforward so you should be fine. Good luck :)
Just giving this the HSC Tick of Approval: All definitely directly relevant ;D
Post by: f_tan on March 04, 2017, 08:56:41 pm
(http://i.imgur.com/xELoSAk.png)
Thanks!
Post by: lsong on March 04, 2017, 11:29:07 pm
I have 2 questions that I think are pretty general, but I'm not sure what to talk about.
1) Contrast the operation of a galvanometer with that of a simple generator (2)
For this question, I feel like two points need to be made. One of those could be that one requires a current to function while the other creates the current. For the second point would it be okay to contrast the structure?
2) Manufacturers state that induction cook tops can only be used with saucepans made from iron and purchasers are advised to use a refrigerator magnet to test their saucepans.
Since currents can be induced in metals other than iron, discuss the reasons for designing induction cook tops that require iron based saucepans. (3)
For this question, I know that iron is conductive but has high resistance which means the eddy currents collide with the lattice structure of the iron. I also know that iron is ferromagnetic, but I'm not sure how this could be related.
Any help would be great thanks! :)
Post by: jamonwindeyer on March 04, 2017, 11:34:06 pm
Is the first hypothesis/experiment about the aether model, or is it Einstein's thought experiments for his Theory of Special Relativity?
(http://i.imgur.com/xELoSAk.png)
Thanks!
Hey! I'd say the Aether (and Michelson and Morley's inability to detect it) was the observation that raised the problem, and Einstein's Special Theory of Relativity was the new hypothesis. The initial experiments (thought experiments) provided some validation for the model, which is why it was adopted so quickly, but it was the real world experiments conducted in the late 20th/early 21st century which really hammered it home :)
Post by: jamonwindeyer on March 04, 2017, 11:52:24 pm
Hi,
I have 2 questions that I think are pretty general, but I'm not sure what to talk about.
1) Contrast the operation of a galvanometer with that of a simple generator (2)
For this question, I feel like two points need to be made. One of those could be that one requires a current to function while the other creates the current. For the second point would it be okay to contrast the structure?
2) Manufacturers state that induction cook tops can only be used with saucepans made from iron and purchasers are advised to use a refrigerator magnet to test their saucepans.
Since currents can be induced in metals other than iron, discuss the reasons for designing induction cook tops that require iron based saucepans. (3)
For this question, I know that iron is conductive but has high resistance which means the eddy currents collide with the lattice structure of the iron. I also know that iron is ferromagnetic, but I'm not sure how this could be related.
Any help would be great thanks! :)
Hey hey! ;D
For the first one, I like the way you think! Definitely a peculiar question, I'd say your planned answer would work really well ;D
For the second one, you've also got it. The main idea is that iron is ferromagnetic (the resistance is important too but not as much I don't think). As a ferromagnetic substance, iron is very susceptible to magnetisation, which (basically) also makes it very susceptible to eddy current flow. Explaining the specifics of ferromagnetism properly is beyond the syllabus - It is a quantum mechanical thing - But it should make intuitive sense. Ferromagnetic substance means easily magnetised, meaning easily affected by magnetic field. Therefore, the eddy currents will form more easily and thus dissipate more heat ;D
So my answer would focus on iron being ferromagnetic, and thus, that the effects of induction cooktops are more pronounced. This would involve a brief mention of how an induction cooktop works in the first place ;D
Post by: lsong on March 05, 2017, 09:09:58 am
Post by: bsdfjnlkasn on March 06, 2017, 11:24:59 pm
Post by: Syndicate on March 06, 2017, 11:31:15 pm
Hey I've been asked to "calculate the distance a satellite must be from Earth to be in geostationary orbit" how do even go about solving this? Any help would be greatly appreciated :)
T = 24 hours (a geostationary satellite is one which orbits the earth whilst remaining over a certain location on earth at all times.
Convert it to seconds, and use the formula: r^3 = (T^2 x GM)/4pi^2
r = the radius of the satellite from the centre of the earth. Therefore, to calculate the satellite's distance from the Earth's surface, you will need to subtract it by the radius of the Earth.
Post by: Bubbly_bluey on March 07, 2017, 06:27:28 pm
Initially I chose B because the galvanometer would deflect more if the current in the secondary increase by the formula. Isecondary/ Iprimary= n primary/ n secondary. And so a decrease in n secondary??
Post by: Iminschool on March 07, 2017, 07:20:10 pm
Hi! I'm having a trouble understanding the answer to this question.i agree with your explanation, i reckon it is B too as it is a step-down transformer. Thus, as V decreases, I increases
Initially I chose B because the galvanometer would deflect more if the current in the secondary increase by the formula. Isecondary/ Iprimary= n primary/ n secondary. And so a decrease in n secondary??
Post by: jamonwindeyer on March 07, 2017, 07:42:12 pm
Hi! I'm having a trouble understanding the answer to this question.
Initially I chose B because the galvanometer would deflect more if the current in the secondary increase by the formula. Isecondary/ Iprimary= n primary/ n secondary. And so a decrease in n secondary??
Hey Bubbly!!
So this is probably going to be the only time I'll ever say this - Don't worry about this question. It's really stupid, the given answer (A) can only be obtained through some really backwards thinking that I'm honestly not too sure about, because it makes assumptions about the questions I don't like. Some think the Galvanometer is actually measuring voltage, some think it is to do with the resistance introduced by adding more coils (the transformer is ideal, so don't buy that) - The one thing I believe about the question is that the voltage source is DC not AC, as indicated by the lines. But that shouldn't affect the answer, and of course, BOSTES hasn't given a solution either :P
So, don't worry about it. It's either a really convoluted/poorly written question, an incorrect answer by BOSTES, or perhaps the Illuminati. But your reasoning is solid, it makes the most sense, and it shows you understand the content! ;D
Post by: Bubbly_bluey on March 07, 2017, 10:19:06 pm
Hey Bubbly!!
So this is probably going to be the only time I'll ever say this - Don't worry about this question. It's really stupid, the given answer (A) can only be obtained through some really backwards thinking that I'm honestly not too sure about, because it makes assumptions about the questions I don't like. Some think the Galvanometer is actually measuring voltage, some think it is to do with the resistance introduced by adding more coils (the transformer is ideal, so don't buy that) - The one thing I believe about the question is that the voltage source is DC not AC, as indicated by the lines. But that shouldn't affect the answer, and of course, BOSTES hasn't given a solution either :P
So, don't worry about it. It's either a really convoluted/poorly written question, an incorrect answer by BOSTES, or perhaps the Illuminati. But your reasoning is solid, it makes the most sense, and it shows you understand the content! ;D
Oh Yes i see what you mean!!! But I thought a galvanometer measures only current like an ammeter. If it was measuring voltage wouldn't it be a volt meter?
Post by: jamonwindeyer on March 07, 2017, 10:20:17 pm
Oh Yes i see what you mean!!! But I thought a galvanometer measures only current like an ammeter. If it was measuring voltage wouldn't it be a volt meter?
I totally agree, hence why I hate those explanations ;D
Post by: Sukakadonkadonk on March 08, 2017, 05:37:41 pm
Is there any difference between a metal being magnetic and a metal being Ferromagnetic?
Thanks.
Post by: jakesilove on March 08, 2017, 08:41:04 pm
Hey guys,
Is there any difference between a metal being magnetic and a metal being Ferromagnetic?
Thanks.
Hey! Generally, Ferromagnetism is just a term used to describe metals that exhibit strong, permanent magnetic properties (most notable, Iron). In the HSC syllabus, though, you can basically use the two terms interchangeably as far as I know :)
Jake
Post by: kiwiberry on March 10, 2017, 09:11:32 pm
Post by: RuiAce on March 10, 2017, 09:15:29 pm
This is from HSC 2013, why is the answer D?A gets ruled out because the magnetic field is radial.
Now, note that the x-axis represents the speed of the motor. The speed of the motor will (because of electromagnetic induction and Lenz's law and blah) cause back EMF to appear in the motor. The back EMF is pretty much reducing the value of I in \(\tau = nBIA\cos \theta\). Hence, as the motor speeds up, the greater back EMF produced will work against the effect of the torque, thus bringing it back down.
Poor explanation because I'm shit at physics now.
Post by: kiwiberry on March 10, 2017, 10:19:46 pm
A gets ruled out because the magnetic field is radial.
Now, note that the x-axis represents the speed of the motor. The speed of the motor will (because of electromagnetic induction and Lenz's law and blah) cause back EMF to appear in the motor. The back EMF is pretty much reducing the value of I in \(\tau = nBIA\cos \theta\). Hence, as the motor speeds up, the greater back EMF produced will work against the effect of the torque, thus bringing it back down.
Poor explanation because I'm shit at physics now.
Thanks Rui!! :)
Post by: Aaron12038488 on March 12, 2017, 06:10:43 pm
Post by: jamonwindeyer on March 12, 2017, 07:44:29 pm
Just wondering what I should aim for (marks overall) at the end of Yr 11 and 12. I want to get >95.
100 ☺
Seriously, if you are really aiming high, you should just be aiming to get the absolute highest mark you can. Why restrict yourself to getting a mark of X, just aim for 100!
Realistically you'd want to aim for a Band 6, but you shouldn't think that way ☺
Post by: Iminschool on March 13, 2017, 04:34:06 pm
Q. The function of a split-ring commutator in a motor is to:
a) Prevent back EMF from building up
b) Keeps the rotor spinning by converting AC to DC
c) Keeps the rotor spinning by converting DC to AC
d) None of the above
Thanks,
Post by: Alexicology on March 13, 2017, 04:47:21 pm
Simple question here but my teacher marked it weirdly last year
Q. The function of a split-ring commutator in a motor is to:
a) Prevent back EMF from building up
b) Keeps the rotor spinning by converting AC to DC
c) Keeps the rotor spinning by converting DC to AC
d) None of the above
Thanks,
The Answer is c) because a DC motor uses DC voltage and a split ring commutator is needed to reverse the direction of the current so that the force on each side of the coil is constant. therefore AC current.
Post by: jamonwindeyer on March 13, 2017, 04:56:47 pm
Simple question here but my teacher marked it weirdly last year
Q. The function of a split-ring commutator in a motor is to:
a) Prevent back EMF from building up
b) Keeps the rotor spinning by converting AC to DC
c) Keeps the rotor spinning by converting DC to AC
d) None of the above
Thanks,
It is definitely not A or B because they don't make sense - But I don't really like the wording of C. It needs to reverse the direction of current every half turn, but that isn't really converting DC to AC. It is very inaccurate wording, and that makes me think it is D...
The Answer is c) because a DC motor uses DC voltage and a split ring commutator is needed to reverse the direction of the current so that the force on each side of the coil is constant. therefore AC current.
Undoubtedly this is correct too and makes sense, but I think they are trying to trick you with wording! But I wouldn't be surprised if the answer was C either ;D
Post by: Alexicology on March 13, 2017, 06:14:40 pm
It is definitely not A or B because they don't make sense - But I don't really like the wording of C. It needs to reverse the direction of current every half turn, but that isn't really converting DC to AC. It is very inaccurate wording, and that makes me think it is D...
Undoubtedly this is correct too and makes sense, but I think they are trying to trick you with wording! But I wouldn't be surprised if the answer was C either ;D
I kind of think it is D too because the wording of the answer is weird, its function is reversing the current every half turn so thank you for correcting me :)
Post by: jamonwindeyer on March 13, 2017, 06:47:41 pm
I kind of think it is D too because the wording of the answer is weird, its function is reversing the current every half turn so thank you for correcting me :)
Oh definitely not a correction, your reasoning is correct too! Like, from the point of view of the coil it might as well be AC ;) just tricky questions being tricky :P
Post by: Alexicology on March 13, 2017, 06:50:34 pm
Oh definitely not a correction, your reasoning is correct too! Like, from the point of view of the coil it might as well be AC ;) just tricky questions being tricky :P
Oh ok
Post by: Shafidontcry on March 14, 2017, 10:15:49 am
Post by: jakesilove on March 14, 2017, 10:21:10 am
What do you guys think?
Hey! I would suspect that the graph has to be continuous; the conductor will speed up, slow down and stop with continuous movement, and thus the change is voltage would have to be continuous. Thus, I suspect it is C?
Post by: jamonwindeyer on March 14, 2017, 11:08:05 am
What do you guys think?
Remember that the size of your induced EMF (voltage) is proportional to the RATE OF CHANGE of magnetic field. So, the size of your voltage is proportional to the speed at which that conductor is moved through the field.
We don't know whether the movement is at a constant speed or if there is acceleration, but we do know that there is movement to the left, then the right, then the left. So we should expect a positive voltage (based on the graphs), then a negative voltage, then a positive one again.
That rules out B and C. A doesn't seem right to me either, because it has a constant acceleration in the positive direction besides a quick sudden change in the negative direction in the middle there. Like, A would imply the rod accelerates to the left at a constant rate, then immediately starts moving in the opposite direction instantaneously, slowing down and stopping, then accelerating back left. It just doesn't seem sensible, though I think it might be possible.
I'm going with D. It shows periods of sustained speed and acceleration, which makes more sense for the scenario at hand!
Edit: Welcome to the forums btw!
Post by: Bubbly_bluey on March 14, 2017, 02:23:23 pm
Thank you :D
Post by: dux99.95 on March 14, 2017, 06:03:00 pm
My question is, if you had to quickly study for SPACE and Motors & generators, how would you do it?
Like, whats most important in each e.g focus on pracs more in SPACE, focus on calculations more in Motors & generators (these are just examples, idk if you should actually do that).
Obviously I should study EVERYTHING but half yearlies are soon - so what should I do?
Post by: jamonwindeyer on March 14, 2017, 07:15:46 pm
Hi guys! i honestly don't know where to start in this question ???
Thank you :D
Hey Bubbly!
So for this one, we need to know the right hand grip rule for finding the direction of magnetic field around a current carrying wire. It's not hard, just give yourself a thumbs up with your right hand. If the thumb points in the direction of current, your fingers wrap around the wire in the direction of the magnetic field!
If you apply this analysis to this question, you'll find that two wires cause a field into the page, and one causes a field out of the page. So the strength of the field in the middle will be \(B\) - The strength of the field from a single wire, because the other two cancel out :) the direction, by the same logic, will be into the page - The answer is A ;D
Post by: jamonwindeyer on March 14, 2017, 07:21:11 pm
Hey
My question is, if you had to quickly study for SPACE and Motors & generators, how would you do it?
Like, whats most important in each e.g focus on pracs more in SPACE, focus on calculations more in Motors & generators (these are just examples, idk if you should actually do that).
Obviously I should study EVERYTHING but half yearlies are soon - so what should I do?
Hey Saloni! If I had to prioritise my content, it would be being able to explain stuff. So, being able to see a dot point and go, "Yep, I can explain Back EMF/Time Dilation/Conditions of Reentry/etc etc." You need to understand your content. The calculations aren't as important to practice (if you must choose) because you get your formula sheet, and that is half the battle. The practicals are important, but ultimately won't be worth as many marks as understanding the theory behind your syllabus dot points! :)
So I'd be dot pointing the more explanatory dot points and focusing on those, or even just brainstorming around the syllabus dot points to see which ones you know and which you don't! Then, just study your notes and do practice questions to get back on track ;D
I hope that helps! Be sure to come to us if you have any questions about the content before your exam, best of luck! :)
Post by: dux99.95 on March 15, 2017, 10:07:47 pm
Post by: jamonwindeyer on March 15, 2017, 11:28:57 pm
Yep that definitely helped! Other question (not physics related tho), how do I see the direct reply to my question? I click on the notification thingo and it leads me to this entire post. Then I have to click on the last page and sift to find my answer - Is it supposed to be like that?
Sure! When you view the notification (I don't use them myself so memory is fuzzy), you should see page numbers below the thread title? If so, you can just click the most recent page and your response will almost always be there!!
Right now there is no way to jump straight to the person who quoted you - Trust me, that's something on my wish list! ;D
Post by: Bubbly_bluey on March 16, 2017, 09:45:13 pm
Post by: kiwiberry on March 17, 2017, 08:09:29 pm
A compass placed on a table points north. In which direction should a conductor, placed directly above the compass, carry current in order to reverse the direction of the compass needle?
Thanks :)
Post by: jamonwindeyer on March 17, 2017, 09:00:58 pm
Hey guys! I find this question get asked a lot in exams but i still can't get my head around the concept of momentum during rocket launches. I know that it is conserved by newton's third law but my question is: does the momentum for the rocket and gas change during different stages of the launch? and if so how does it change? (please dumb down the explain wherever you can :) )
Hey Bubbly!
So basically, the total momentum of the rocket and fuel system is constant (and remains equal). The momentum of the rocket itself changes, as does the change in the momentum of the fuel, but they will always exactly cancel each other out!
The explanation of exactly what happens to a rockets velocity and mass (and thus momentum) throughout launch is something I've covered here - Give it a read a few times and hopefully it makes sense! Otherwise definitely send any follow up questions my way!
Post by: jamonwindeyer on March 17, 2017, 09:13:38 pm
Hey! I'm having trouble understanding this question:
A compass placed on a table points north. In which direction should a conductor, placed directly above the compass, carry current in order to reverse the direction of the compass needle?
Thanks :)
Hey Kiwi!
So we need to use the right hand grip rule here, but backwards. We need a magnetic field running south (compasses align with magnetic field lines). So, we need a magnetic field running down the compass. Hold up your right hand in a thumbs up, then align your fingers to point downwards on your imaginary compass. That should have your thumb pointing to the left. So, we need a current flowing to the left (WEST), assuming we have a conductor above the compass running from East to West ;D
Aha, this is hard to explain with text, does this make sense kiwi? ;D
Post by: kiwiberry on March 17, 2017, 09:21:32 pm
Hey Kiwi!
So we need to use the right hand grip rule here, but backwards. We need a magnetic field running south (compasses align with magnetic field lines). So, we need a magnetic field running down the compass. Hold up your right hand in a thumbs up, then align your fingers to point downwards on your imaginary compass. That should have your thumb pointing to the left. So, we need a current flowing to the left (WEST), assuming we have a conductor above the compass running from East to West ;D
Aha, this is hard to explain with text, does this make sense kiwi? ;D
Yes that does!! Thanks Jamon :D
Post by: Bubbly_bluey on March 18, 2017, 05:31:20 pm
Post by: liamwindeyer on March 18, 2017, 05:37:56 pm
Hey! just a quick little question. When calculating gravitational potential energy, does the radius include the distance between the two centres of the planets or only the radius of one mass + the aplitude?
It would be the distance between the centres of the planets, like Newton's Universal Law of Gravitation formula :)
Edit - Should've added that the GPE formula and Newton's formula are practically the same concept; used for different scenarios. GPE is usually used for when a smaller satellite is orbiting a planet, whilst Newton's is usually used for the effect of gravity between planets (or two similarly sized objects)
Post by: smile123 on March 18, 2017, 07:48:47 pm
Post by: jakesilove on March 18, 2017, 08:03:12 pm
PLEASE HELP :)
Hey! Let's use trigonometry to find the horizontal distance the pilot will need to fly before they crash. We know that the angle between the ground and the rising slope is 4.3 degrees. The height (therefore, the opposite side) will have a height of 35m. Draw yourself a diagram if this isn't clear, but we can use trigonometry to show that
Great! So, the pilot will be level with (and therefore hit) the ground in 465.49m. The pilot is travelling
Therefore
Poor pilot
Post by: smile123 on March 18, 2017, 08:18:09 pm
Post by: kiwiberry on March 18, 2017, 08:40:39 pm
Post by: Abi21 on March 18, 2017, 09:21:08 pm
I am in year 10 and am currently choosing my VCE subjects. I have one spot left and I'm trying to choose between physics?chemistry. I noticed that you've done both subjects and i was wondering what are some of the pros of Physics and why should i choose it instead of chem. Sorry if this is a little irrelevant. :D
Post by: TooLazy on March 18, 2017, 09:26:32 pm
Hey Jake,
I am in year 10 and am currently choosing my VCE subjects. I have one spot left and I'm trying to choose between physics?chemistry. I noticed that you've done both subjects and i was wondering what are some of the pros of Physics and why should i choose it instead of chem. Sorry if this is a little irrelevant. :D
I find physics more interesting and easier, but chem for me is straight cancer
Post by: Bubbly_bluey on March 18, 2017, 11:36:04 pm
when an object travels at 0.9c,its mass apparently gets converted into energy but what is that energy? does it get released somewhere?
Thanks for putting up with me ;D
Post by: jakesilove on March 19, 2017, 11:24:39 am
Why is it harder to turn a hand-powered generator attached to an electrical device than one connected to nothing? Does it have something to do with back emf?
Hey! When the device is connected to an electrical device, electricity is being removed from the system (ie. current is being drawn away to power that device). Thus, the person turning the generator has to add extra energy (kinetic to electrical) to make up for this net loss of energy in the system. So, the generator is harder to turn! Additionally, there may be eddy currents produced in the electrical device, which will also draw energy away.
Jake
Post by: jakesilove on March 19, 2017, 11:26:57 am
Hey Jake,
I am in year 10 and am currently choosing my VCE subjects. I have one spot left and I'm trying to choose between physics?chemistry. I noticed that you've done both subjects and i was wondering what are some of the pros of Physics and why should i choose it instead of chem. Sorry if this is a little irrelevant. :D
Hey!
I personally prefer Physics. The concepts are broad and fundamental, and it contains a lot more mathematical operations. As a maths-head myself, the Maths surrounding Chemistry was a bit trivial. At the end of the day, it's all about what excites you more. Look at the curriculums briefly (I'm in NSW, so don't know what the VCE curriculum looks like) and see whether there is anything that interests you. Do you care more about space, or chemical reactions? Electricity, or processes that allow the human body to function? Think about what you've learned up until now, and whether you've enjoyed Chemistry more than Physics or vice versa.
Let me know if you have any more questions!
Post by: jakesilove on March 19, 2017, 11:28:26 am
hello! i am having trouble understanding the equivalence between mass and energy, more specifically einstein's famous: e= mc2.
when an object travels at 0.9c,its mass apparently gets converted into energy but what is that energy? does it get released somewhere?
Thanks for putting up with me ;D
Hey! Really good question; my answer is that this goes way beyond the curriculum. In simple terms, when an object speeds up, it JUST gains energy. This energy comes in the form of kinetic energy. However, since E=mc^2, energy and mass are COMPLETELY equivalent. So, an increase in energy IS an increase in mass!
ie. Object gets faster. Energy increases. Mass therefore also increases. Badabing, badaboom.
Post by: kiwiberry on March 19, 2017, 02:21:51 pm
Hey! When the device is connected to an electrical device, electricity is being removed from the system (ie. current is being drawn away to power that device). Thus, the person turning the generator has to add extra energy (kinetic to electrical) to make up for this net loss of energy in the system. So, the generator is harder to turn! Additionally, there may be eddy currents produced in the electrical device, which will also draw energy away.
Jake
Thanks Jake :)
More questions haha:
Would the answer to the first one be D? And I have no idea what the second question means, but the answer was 3<1<2
Post by: jakesilove on March 19, 2017, 03:03:14 pm
Thanks Jake :)
More questions haha:
Would the answer to the first one be D? And I have no idea what the second question means, but the answer was 3<1<2
I would say so! The more you speed up the motor (ie. accelerate), the greater the back EMF will cancel out the supplied current. So, I'd go with D :)
Post by: jamonwindeyer on March 19, 2017, 07:04:04 pm
Thanks Jake :)
More questions haha:
Would the answer to the first one be D? And I have no idea what the second question means, but the answer was 3<1<2
That second question isn't examinable - Where's it from ;D
The actual reason it is 3>1>2 is because voltage is a change in electric field strength. Electric fields are stronger when their field lines are closer! So in 3, the potential difference is positive because the field is stronger at Point B. Then it is zero for 1 because the field is the same in both. Then it is negative for 2 because the field is stronger at point A.
Post by: jakesilove on March 19, 2017, 07:20:36 pm
That second question isn't examinable - Where's it from ;D
The actual reason it is 3>1>2 is because voltage is a change in electric field strength. Electric fields are stronger when their field lines are closer! So in 3, the potential difference is positive because the field is stronger at Point B. Then it is zero for 1 because the field is the same in both. Then it is negative for 2 because the field is stronger at point A.
I'm fairly sure that 'fields are stronger when field lines are closer together' is within the curriculum; I definitely learnt it in year 12, and used this principle to draw diagrams etc. From there, I definitely think this question is not examinable, however can probably be solved intuitively.
Post by: f_tan on March 19, 2017, 07:22:56 pm
(http://i.imgur.com/RQ23HOu.png)
Post by: kiwiberry on March 19, 2017, 07:31:21 pm
That second question isn't examinable - Where's it from ;D
The actual reason it is 3>1>2 is because voltage is a change in electric field strength. Electric fields are stronger when their field lines are closer! So in 3, the potential difference is positive because the field is stronger at Point B. Then it is zero for 1 because the field is the same in both. Then it is negative for 2 because the field is stronger at point A.
It's from one of my school's past papers! :/ Good to know that it shouldn't be examinable, it freaked me out for a bit haha. Thanks for the explanantion though :)
Post by: f_tan on March 19, 2017, 07:54:53 pm
Thank you!
(http://i.imgur.com/vvVAVle.png)
Post by: jamonwindeyer on March 19, 2017, 08:54:46 pm
I'm fairly sure that 'fields are stronger when field lines are closer together' is within the curriculum; I definitely learnt it in year 12, and used this principle to draw diagrams etc. From there, I definitely think this question is not examinable, however can probably be solved intuitively.
Yep, that bit is indirectly examinable because you learn about flux density as field strength for magnetic fields, and you touch it again in Ideas to Implementation - But linking electric field to voltage in this way is a real stretch. Like, a switched on Year 12 student could answer it. But I doubt they'd ever expect anyone to do it - If the syllabus was the American/Mexican border this is just beyond Trump's impending wall imo :P
Post by: jamonwindeyer on March 19, 2017, 08:56:35 pm
Why is there no induced current for the wire on the right?
(http://i.imgur.com/RQ23HOu.png)
Hey! You've got a DC current that sets up a constant magnetic field - You need a changing magnetic field for an induced current ;D if it was an AC current that would be a different story!
Post by: f_tan on March 19, 2017, 09:00:13 pm
Hey! You've got a DC current that sets up a constant magnetic field - You need a changing magnetic field for an induced current ;D if it was an AC current that would be a different story!
But if current increases, wouldn't that cause the magnetic field to increase too and therefore change?
Post by: jamonwindeyer on March 19, 2017, 09:04:59 pm
But if current increases, wouldn't that cause the magnetic field to increase too and therefore change?
Hmmm, yeah sorry you're totally right... At the very instant it increases the magnetic field would increase and that would induce a brief EMF - But perhaps the questions means a sustained EMF or something?
Is this from a practice paper where the answer suggests there is no induced current or? :)
Post by: f_tan on March 19, 2017, 09:18:52 pm
Is this from a practice paper where the answer suggests there is no induced current or? :)
This is what the answer says, but I dont really get it:
(http://i.imgur.com/vtJwcjK.png)
Post by: jamonwindeyer on March 19, 2017, 09:32:59 pm
How do I do this question?
Thank you!
(http://i.imgur.com/vvVAVle.png)
Right, so this one is relying on the mathematical version of Faraday's Law:
Now to my knowledge, this isn't directly examinable either (lots of weird questions tonight, or maybe I've just forgotten the whole syllabus ;)) - But not too difficult if you have learned that formula!
As a side note, this formula doesn't apply exactly this way exactly each time - But since it is a basic coil situation I'm fairly sure we just do this, and I see no other way anyhow! ;D
So the change in time is 0.2 seconds - What's the change in magnetic flux? Well we relate flux to flux density (field strength) and area with the formula \(\phi=BA\).
Initially, the area of the circular loop is is \(A_i=\pi r^2=0.04\pi\) (using SI units). The final area will be half of this value, \(0.02\pi\), so the change in area is \(0.02\pi\). So \(\Delta\phi=BA=2\times0.02\pi=0.04\pi\).
We use the formula to obtain:
I'm loosely confident this is the approach - Does it match any solutions you might have to this question?
Irrespective of this, the graph we can do. Remember that \(A=\pi r^2\), so decreasing radius at a constant rate will actually result in an accelerating change to area! Think of it this way, we know the area at 20cm is \(0.04\pi\). At 15cm, it will be \(0.0225\pi\). At 10cm, it will be \(0.01\pi\). The decrease in radius from 20cm to 15cm has a slightly larger effect on area than the decrease from 15cm to 10cm. Hence, the rate of decrease of area is highest initially, slowly decreasing.
Thus, the graph we sketch should have a peak right at the start, then a slow decrease, as the rate of change of area slowly decreases (and thus, so will the induced emf) ;D
Post by: jamonwindeyer on March 19, 2017, 09:38:16 pm
This is what the answer says, but I dont really get it:
(http://i.imgur.com/vtJwcjK.png)
Hmmm, this is interesting! I think when it says 'link' it is referring to flux linkage, like you'd get in a transformer with the iron core. But like, those don't make the emf, they significantly increase the size but it isn't a make or break.
The other explanation could be that there is no induced current, because there is no direction the current could flow to oppose the change that created it. That is, no current could flow in the conductor to introduce a field out of the page at that point - Like, the magnetic field due to that induced current wouldn't do anything to oppose the other field. So therefore, why should it be induced? :)
Post by: bsdfjnlkasn on March 19, 2017, 09:41:16 pm
I got the gradient = 0.73 which should give me the time I need to sub into one of the projectile equations. I'm just not sure which one to use and how direction plays into this because i'm putting -9.8 for g but then that gives me a negative distance (delta y) so i'm really confused???
Any help would be great!
Post by: f_tan on March 19, 2017, 09:42:46 pm
I'm loosely confident this is the approach - Does it match any solutions you might have to this question?
Thanks so much! I couldn't find any solutions for the question.
Post by: bsdfjnlkasn on March 19, 2017, 09:48:13 pm
Sorry if this is really stupid but how would you count the dots in a question like this? Would you count the first one on the right or not because it's when the initial velocity is 0 i.e. when t=0 - or am I completely overthinking this - please help :) :)
EDIT: I've attached the relevant questions and my answers - if someone could please help me out with c and check the rest I'd really appreciate it!!(http://uploads.tapatalk-cdn.com/20170319/e25245e7c76b3d44d35d2909bc42ea9e.jpg)
Post by: bsdfjnlkasn on March 19, 2017, 10:09:03 pm
I'm also stuck with how to interpret part d)
(This is a practice test by the way im not making you guys do an exam for me
- there just weren't any answers provided online)(http://uploads.tapatalk-cdn.com/20170319/484f9b2a7a82ee4e11147116f9b66dd5.jpg)
Post by: jamonwindeyer on March 19, 2017, 10:11:31 pm
Hey there how would I find the height of the platform of this question?
I got the gradient = 0.73 which should give me the time I need to sub into one of the projectile equations. I'm just not sure which one to use and how direction plays into this because i'm putting -9.8 for g but then that gives me a negative distance (delta y) so i'm really confused???
Any help would be great!
You did the right thing! You would have gotten about -2.5 metres? That represents the fact that, after the time of flight, you are 2.5 metres BELOW your launch point. It's because the projectile is lower than when you started :)
Height of the platform is about 2.5 metres ;D
Hey me again,
Sorry if this is really stupid but how would you count the dots in a question like this? Would you count the first one on the right or not because it's when the initial velocity is 0 i.e. when t=0 - or am I completely overthinking this - please help :) :)
EDIT: I've attached the relevant questions and my answers - if someone could please help me out with c and check the rest I'd really appreciate it!!(http://uploads.tapatalk-cdn.com/20170319/e25245e7c76b3d44d35d2909bc42ea9e.jpg)
You are over thinking a bit I think ;) counting an extra dot will only make a slight difference to your estimates and thus your calculations, so I don't think it would matter!
For C - You need to identify that the horizontal velocity is constant. You can tell because the horizontal distance between each of your dots is relatively constant ;D
The others look good on a quick glance - Your first one it is 0.67m below the HIGHEST point, not the initial point (you've redefined the origin to be the peak point by using the formula in that way). So just 0.67 metres should be the answer. For D, you might need to include the horizontal component of velocity as well using pythagoras?
Post by: jamonwindeyer on March 19, 2017, 10:17:30 pm
Hey is there anything I could add to my answers for projectile motion here (Q5)
I'm also stuck with how to interpret part d)
(This is a practice test by the way im not making you guys do an exam for me
(http://uploads.tapatalk-cdn.com/20170319/484f9b2a7a82ee4e11147116f9b66dd5.jpg)
For A, give a brief explanation why (horizontal motion has no effect on vertical)
I think you've misinterpreted B, C and D - The vehicle is accelerating, the frame of reference is not inertial. So the person outside the vehicle would see a parabolic motion - But the person inside the vehicle would see the mass accelerating horizontally (because the car is accelerating away from the falling mass).
If the person in the vehicle couldn't tell they were moving, they'd need to say, "Well, some magical force must have accelerated that mass." This reflects the fact if we are in a non-inertial reference frame, we need to introduce 'fictional' forces to account for observations ;D
Post by: bsdfjnlkasn on March 19, 2017, 10:34:33 pm
For D, you might need to include the horizontal component of velocity as well using pythagoras?
Hey Jamon!
Thanks so much for the prompt replies ;D I was just wondering if you could offer a bit more guidance for part d) as I just don't see where we can get enough information to do a vector sum (and then apply pythagoras). We're not given any angles so can't resolve the horizontal/vertical components and so I'm genuinely lost with how to approach this question.
Thanks again for everything!
Post by: jamonwindeyer on March 19, 2017, 10:37:29 pm
Hey Jamon!
Thanks so much for the prompt replies ;D I was just wondering if you could offer a bit more guidance for part d) as I just don't see where we can get enough information to do a vector sum (and then apply pythagoras). We're not given any angles so can't resolve the horizontal/vertical components and so I'm genuinely lost with how to approach this question.
Thanks again for everything!
All good! Hmm, you are right - I don't see how we can find the horizontal component either. Unless you somehow use the distance in B as a measure, but that would mean the diagram has to be to scale and they don't specify that...
It must just mean vertical velocity as you did initially? Unless anyone else has any ideas ;D
Post by: bsdfjnlkasn on March 19, 2017, 10:49:52 pm
All good! Hmm, you are right - I don't see how we can find the horizontal component either. Unless you somehow use the distance in B as a measure, but that would mean the diagram has to be to scale and they don't specify that...
It must just mean vertical velocity as you did initially? Unless anyone else has any ideas ;D
Ok phew :D
Here's a similar question - could you again check that I've approached everything correctly? Does the last question also then mean I should find the vertical velocity?(http://uploads.tapatalk-cdn.com/20170319/9839a5a28e882ed1845d456e53d5e6a8.jpg)
Post by: TooLazy on March 19, 2017, 10:58:02 pm
Hey me again,
Sorry if this is really stupid but how would you count the dots in a question like this? Would you count the first one on the right or not because it's when the initial velocity is 0 i.e. when t=0 - or am I completely overthinking this - please help :) :)
EDIT: I've attached the relevant questions and my answers - if someone could please help me out with c and check the rest I'd really appreciate it!!(http://uploads.tapatalk-cdn.com/20170319/e25245e7c76b3d44d35d2909bc42ea9e.jpg)
Horizontal velocity is constant, acceleration = 0.
Only force acting on the projectile is gravity (neglecting air resistance)
If you know the horizontal distance and time, you can use the formula x=ut to get the horizontal velocity
Post by: TooLazy on March 19, 2017, 10:59:37 pm
Ok phew :D
Here's a similar question - could you again check that I've approached everything correctly? Does the last question also then mean I should find the vertical velocity?(http://uploads.tapatalk-cdn.com/20170319/9839a5a28e882ed1845d456e53d5e6a8.jpg)
For the overall velocity in part d,
you need to first calculate both components of the final velocity and add them together using a vector diagram.
This will enable you to find the final velocity
Post by: khitnay on March 21, 2017, 01:46:41 pm
My assignment requires me to build either a motor, a generator or a transformer with complex parts such as commutators and a complex design.
I have chosen to build a three phase (to make the design complex) AC generator with stationary coils and a rotor with magnets on it unlike the statonary magnetic field and rotating coil as learnt in the curriculum. However, I'm not sure whether such a model would require slip ring commutators or even how to start building it.
All the models of alternators I've seen on youtube are self excited with a dc generator connected. Is it possible to build an alternator that is not self excited? If so, how should I go about building it?
Post by: RuiAce on March 21, 2017, 02:05:29 pm
Hi! Anyone who could help me with this physics assignment would be deeply appreciatedSince the three-phase motor is an example of an induction motor, if you had a commutator it wouldn't be for the motor itself; it would be for the magnet rotor. Which may make the design a bit unnecessarily complicated.
My assignment requires me to build either a motor, a generator or a transformer with complex parts such as commutators and a complex design.
I have chosen to build a three phase (to make the design complex) AC generator with stationary coils and a rotor with magnets on it unlike the statonary magnetic field and rotating coil as learnt in the curriculum. However, I'm not sure whether such a model would require slip ring commutators or even how to start building it.
All the models of alternators I've seen on youtube are self excited with a dc generator connected. Is it possible to build an alternator that is not self excited? If so, how should I go about building it?
This chose is pretty advanced. (Back when I had this assignment I just built an ordinary DC motor.) You have to somehow ensure that the wires are going to be running in sync. I haven't examined the validity of these links but here is a video and an article you can read about it. (Taken from a quick Google search.)
https://www.youtube.com/watch?v=DAYz3v8HwEE
https://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=0ahUKEwiAzoGVzebSAhWGkJQKHddvDRYQFggsMAM&url=https%3A%2F%2Fwww.electrical4u.com%2Fconstruction-of-three-phase-induction-motor%2F&usg=AFQjCNH4ZsiQclUPg6gWP-Ald3gXaBCstg&sig2=VEuh92PB7wn5HcQxExYEkg
Horizontal velocity is constant, acceleration = 0.Thank you for the answer but I have a request. To help ease the students' understanding, please write down formulae as they are given on the data sheet
Only force acting on the projectile is gravity (neglecting air resistance)
If you know the horizontal distance and time, you can use the formula x=ut to get the horizontal velocity
Post by: arunasva on March 23, 2017, 10:29:07 pm
Post by: jamonwindeyer on March 24, 2017, 12:57:37 am
https://drive.google.com/file/d/0B3wlCPQFQN1tNkxUYXRXNGZSYU0/view help with this if anyone can. Please T.T
Welcome to the forums!! ;D remember that G-force is closely related to weight force. The G-force experienced by the astronauts will be equal to:
This formula acknowledges the effects of gravity normally, being added to any additional acceleration \(a\) of the rocket. We notice that the only thing that will differ between the astronauts is their mass - Everything else is constant. So, the ratio of the G-forces is equal to the ratio of their masses! For this reason, I'm fairly certain the answer should be B ;D
Post by: arunasva on March 24, 2017, 04:13:11 am
Post by: Syndicate on March 24, 2017, 08:25:24 am
I have an exam in 8 hrs (HSC OUTSIDE AUSTRALIA) I am so done for dont know how to use light years. Please help.
Hi arunasva,
9 light year is just a unit of time. For the time dilation formula, it doesn't matter what units are used for the time as long as your proper time, and dilated time has the same units.
For question a, t0 = 9 light year, v = 0.75c, t =? (9 is the proper time, as it is the actual time).
t = t0Y
t = 9 x 1.512
t = 13.61 light years
b) Use the speed = distance/ time formula. Why? Because the time calculated by the astronaut is the proper time in this case.
distance = 9 lights years
Speed = 0.75 c
Time = 9/0.75
time = 12 light years
Post by: arunasva on March 24, 2017, 09:37:00 pm
Post by: jamonwindeyer on March 24, 2017, 09:49:17 pm
Yup there was a light years question. I messed it up. Messed up physics pretty darn bad. Sigh im stuck only 5 units and physics is so fucking hard dont even get projectile motion look at the question and sit there not knowing what to do. I should've dropped Physics in Year 11 but I got 90 % and parents were like nah we can't let you drop that. Ill Fail again in the trials. Oh well.
Don't give up! If you go through the year thinking you'll fail, you'll fail. If you go through the year willing to go hard and improve, you never know what can happen! Never too late to turn things around my friend! You're always welcome to pop any questions here and we'll help you any way we can :)
Post by: arunasva on March 25, 2017, 04:26:23 am
Projectile Motion and Special Relativity (Math stuff the theory is alright) I tried practising everything but couldnt do anything without looking at the answers.
Post by: jamonwindeyer on March 25, 2017, 10:22:55 am
What do you have to do if you do not get a concept at all ? I have 2
Projectile Motion and Special Relativity (Math stuff the theory is alright) I tried practising everything but couldnt do anything without looking at the answers.
Getting used to the Math questions is just practice - It can take a bit, but eventually you will start noticing patterns and getting better!! ;D
A few things that might help:
- Guide to Math Questions
- Guide to Rocket Launches
- Guide to Relativity
Post by: bluecookie on March 27, 2017, 11:13:05 pm
vp/vs=is/ip (if u let the power of the primary equal the power of the secondary que law of conservation and sub p=iv and crossmultiply)
They say if you increase vp by factor, is decreases by same factor and vice vercea, but it's just not making sense to me atm. If you multiply vp by a factor of k, you must multiply RHS by factor of k so they can cancel out and the equation becomes the same. So you multiply k to the numerator and isn't is increased by a factor of k as well? Not decreased?
Post by: bluecookie on March 27, 2017, 11:19:36 pm
Post by: arunasva on March 27, 2017, 11:51:20 pm
On similar note. Does np>ns apply for step up or step down? (and vice vercea for ns>np)
When you increase the Voltage of the primary Current of the Primary decreases. Vp/Vs = Is/Ip. Yeah if you multiply both numerators by K you will increase the value of the Voltage of the Primary and Current Current Of The Secondary. Therefore current of the primary(the denominator Ip) decreases because the numerator Is gets bigger than Ip. So you increased the voltage for the Primary coil and the Current in that coil (Ip) decreased while the current in the other coil Increased (Is)
Post by: kiwiberry on March 28, 2017, 08:07:49 am
On similar note. Does np>ns apply for step up or step down? (and vice vercea for ns>np)
np>ns for step down because emf is proportional to the number of coils, so a smaller number will induce a smaller emf in the secondary coil :) and vice versa for step up
Post by: jamonwindeyer on March 28, 2017, 08:32:40 am
Awesome answers above, cheers guys
Post by: lsong on March 28, 2017, 02:55:15 pm
Any ideas would be great thanks!
Post by: jamonwindeyer on March 28, 2017, 03:17:54 pm
Hi, one of the dot points in the Motors and Gen syllabus asks to identify how transmission lines are insulated from supporting structures. I can only find one out of the three (which is the number that my teacher asked for). Is there three possible examples for this?
Any ideas would be great thanks!
For the exam you'd only need to know one, the only example I ever learned was ceramic insulating discs :P but hopefully someone can lend a hand with more ideas! :)
Post by: lsong on March 28, 2017, 05:19:21 pm
For the exam you'd only need to know one, the only example I ever learned was ceramic insulating discs :P but hopefully someone can lend a hand with more ideas! :)
Ah ok. Yeah in tutoring I only learnt about the one, might have to ask my teacher.
Thank you though! ;D
Post by: Sukakadonkadonk on March 28, 2017, 08:41:15 pm
Could someone please explain this question for me please? Especially the potential energy part (how they did it in the solutions).
"An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet."
Thanks!
Post by: jamonwindeyer on March 28, 2017, 08:54:06 pm
Hey guys,
Could someone please explain this question for me please? Especially the potential energy part (how they did it in the solutions).
"An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.
Calculate how much more work needs to be done to move the object to a stationary point 80 000 km from the centre of the planet."
Thanks!
Hey hey! I can't see the solution, could you upload it if you haven't already? That way I can explain the approach they have taken for you ;D
Post by: Sukakadonkadonk on March 28, 2017, 09:10:18 pm
Hey hey! I can't see the solution, could you upload it if you haven't already? That way I can explain the approach they have taken for you ;D
Ok, here is the solution they gave.
Is there an easier way to do it?
Post by: jamonwindeyer on March 28, 2017, 10:00:10 pm
Ok, here is the solution they gave.
Is there an easier way to do it?
Cool! They've done it the best/easiest way, let me explain! ;D
We know that a certain amount of work, 1MJ, is needed to move the object from 10,000km to 20,000km. This energy goes directly to a change in GPE. So, the 4 lines at the top are saying, "We know that the difference in GPE at the two heights is equal to work done. Let's use that." In this case, we use it to find the grouped value of the constants \(GMm\).
So now we have \(GMm\) - And we need to find the energy required to move from 20,000km to 80,000km. This is now just formula work - The last 3 lines involve calculating the difference in GPE at those two radii ;D
(Note - They do factor out the \(GMm\), don't let that confuse you, it is still just GPE! :)
Does that help give it some context?
Post by: Sukakadonkadonk on March 29, 2017, 07:30:04 am
Cool! They've done it the best/easiest way, let me explain! ;D
We know that a certain amount of work, 1MJ, is needed to move the object from 10,000km to 20,000km. This energy goes directly to a change in GPE. So, the 4 lines at the top are saying, "We know that the difference in GPE at the two heights is equal to work done. Let's use that." In this case, we use it to find the grouped value of the constants \(GMm\).
So now we have \(GMm\) - And we need to find the energy required to move from 20,000km to 80,000km. This is now just formula work - The last 3 lines involve calculating the difference in GPE at those two radii ;D
(Note - They do factor out the \(GMm\), don't let that confuse you, it is still just GPE! :)
Does that help give it some context?
Yes, thank you very much :)
Post by: Kle123 on March 31, 2017, 09:14:53 am
Post by: jakesilove on March 31, 2017, 09:20:52 am
The answer is A. Could someone please explain?
Hey! We want to force the path of the particle DOWN. Remember that electric fields point in the direction that a positive charge will move. So, by adding an electric field denoted in A, we push the positive particle back towards the target. Does that make sense?
Post by: katnisschung on March 31, 2017, 02:34:39 pm
revising the michelson morely experiment and i was wanting to confirm their result was considered a "null result for two reasons"
1. there were no discernible changes to the interference pattern when the interferometer was rotated
2. the interference pattern would result in two waves in phase (considering the two beams of light are split and travel equidistantly, if the aether existed they probably presumed this..becos one would be travelling with the aether...thus reach the detector faster and result in an interference pattern where the waves were out of phase )
but would result in two waves in phase with the same reasoning--> travelling equidistantly.
idk nothing really said the second reason explicitly..maybe i'm overthinking it. thoughts?
Post by: jamonwindeyer on March 31, 2017, 02:47:34 pm
bonjour! :)
revising the michelson morely experiment and i was wanting to confirm their result was considered a "null result for two reasons"
1. there were no discernible changes to the interference pattern when the interferometer was rotated
2. the interference pattern would result in two waves in phase (considering the two beams of light are split and travel equidistantly, if the aether existed they probably presumed this..becos one would be travelling with the aether...thus reach the detector faster and result in an interference pattern where the waves were out of phase )
but would result in two waves in phase with the same reasoning--> travelling equidistantly.
idk nothing really said the second reason explicitly..maybe i'm overthinking it. thoughts?
Hey Katniss! The first reason is the reason, your second one isn't necessarily right, though it is correct in principle. It would be next to impossible to make two separate beams of light travel perfectly equidistantly, like, a speck of dust would kill that! So there will almost definitely be an interference pattern - It's only if it changes in the experiment that we could have had evidence to support the Aether model ;D
Post by: Kle123 on March 31, 2017, 11:12:33 pm
Hey! We want to force the path of the particle DOWN. Remember that electric fields point in the direction that a positive charge will move. So, by adding an electric field denoted in A, we push the positive particle back towards the target. Does that make sense?
yes thankyou jake
Post by: Bubbly_bluey on March 31, 2017, 11:28:16 pm
In MnG,there a graph on the relationship between the magnetic flux vs time as the ammature turns.
Assuming that the initial postion is parallel to the field, the graph shows one 360 revolution of the ammature. So my question is: why is this relationship a sin curve? like 3/4 of the revolution the ammature is experiencing maximum flux (b/c lots of field lines are going through it) but the graph shows a minimum. basically imasking why isn't it shape like camel bumps.
Thanks :)
Post by: jamonwindeyer on April 01, 2017, 12:22:30 am
hey guys!
In MnG,there a graph on the relationship between the magnetic flux vs time as the ammature turns.
Assuming that the initial postion is parallel to the field, the graph shows one 360 revolution of the ammature. So my question is: why is this relationship a sin curve? like 3/4 of the revolution the ammature is experiencing maximum flux (b/c lots of field lines are going through it) but the graph shows a minimum. basically imasking why isn't it shape like camel bumps.
Thanks :)
Hey Bubbly! I'm having a little trouble picturing your scenario, but I think I'm with you. The reason it doesn't look like camel bumps, for starters, comes from the fact that magnetic flux is a vector - It has a direction. So, the magnetic flux through the coil has to be negative just as much as positive because the armature is rotating through a full 360 degrees. The reason why it is sinusoidal is, pretty much, the same reason that the formula for torque has a cosine term, \(\tau=BAIn\cos{\theta}\) - It is just a consequence of the math and the interactions involved, you don't need to explain it in depth! ;D
I can try and explain it a little better if I could see the diagrams specifically, if it helps ;D
Post by: Zainbow on April 02, 2017, 11:50:34 am
Just a small question: If the question gives us data in, say, MJ, do we answer in joules or MJ as well? (SI or whatever they give us?)
Post by: jamonwindeyer on April 02, 2017, 11:53:24 am
Hey!
Just a small question: If the question gives us data in, say, MJ, do we answer in joules or MJ as well? (SI or whatever they give us?)
Hey Zainbow! My policy has always been that using just joules with scientific notation is always acceptable provided they don't specify, but you can definitely use MJ or GJ or whatever they use if that is your preference - You'd get paid for the answer regardless ;D
Post by: Zainbow on April 02, 2017, 11:57:38 am
Hey Zainbow! My policy has always been that using just joules with scientific notation is always acceptable provided they don't specify, but you can definitely use MJ or GJ or whatever they use if that is your preference - You'd get paid for the answer regardless ;D
Thanks Jamon :)
Post by: Aaron12038488 on April 02, 2017, 01:54:33 pm
Post by: pikachu975 on April 02, 2017, 03:53:58 pm
I have my practical tomoz, which is the World Communicates Topic. Has anyone done a prac on this topic and if so any advice will be very appreciative.
I can't remember since it was last year but I think we did one with light boxes and refraction of light waves through glass. Should be easy enough I can't remember though if that's even the right topic it's under...
- I do the prac in the second half so I can see how others do it and make sure I do it right.
Post by: jamonwindeyer on April 02, 2017, 04:35:47 pm
I can't remember since it was last year but I think we did one with light boxes and refraction of light waves through glass. Should be easy enough I can't remember though if that's even the right topic it's under...
- I do the prac in the second half so I can see how others do it and make sure I do it right.
That's the one
I did it too, I had to determine the refractive index of glass, some variation on the light box setup is the most common experiment that students get given in this topic! Similar to pikachu, I remember I did a little trial run of the experiment with quick calculation to check that I get the expected answer, before you get stuck into doing it properly it is nice to know you are on the right track
Post by: winstondarmawan on April 02, 2017, 04:37:00 pm
Thanks in advance.
Post by: Aaron12038488 on April 02, 2017, 04:56:16 pm
Post by: jamonwindeyer on April 02, 2017, 04:58:06 pm
How would you do the question in the attachment below?
Thanks in advance.
This one is nasty - I remember this, it is a copy of a question from a 2004 paper, with some dates changed!
So the idea is this. Consider Kepler's Law:
Mars has a larger orbital radius than the Earth, so according to this equation, it must also have a larger orbital period. We can't calculate exactly how much, especially given that the diagram is not to scale (or at least specified to be), but we know that it is larger. What does this mean? Well, in the 10 months time span you have been given, earth will complete about 5/6 of a revolution. Mars will complete less than this - I'd say about 1/2 of a revolution might be what you want.
We also know that they are going anticlockwise. Why? Well, that's just what we need for this scenario to make sense. Putting all that together, I'd get this:
(http://i.imgur.com/xpbvh4b.png)
Not 100% confident on this, but this would be my best crack (also excuse the shitty diagram lol) ;D
Post by: jamonwindeyer on April 02, 2017, 04:59:24 pm
Thanks, also is this prac done individually or in groups. I'm so nervous as it is worth 20%. :-\ :-X ::) ???
Probably individually, but it depends on your school! :) don't be nervous! It is only Year 11 after all, the results this year don't count in the long run - Just use this as a gauge to see how you'd go in a prac in Year 12 :)
Post by: bluecookie on April 02, 2017, 10:55:37 pm
When you increase the Voltage of the primary Current of the Primary decreases. Vp/Vs = Is/Ip. Yeah if you multiply both numerators by K you will increase the value of the Voltage of the Primary and Current Current Of The Secondary. Therefore current of the primary(the denominator Ip) decreases because the numerator Is gets bigger than Ip. So you increased the voltage for the Primary coil and the Current in that coil (Ip) decreased while the current in the other coil Increased (Is)
Oh, so it's inversely proportional with the primary voltage/current, not the primary voltage/secondary current. Ah, I see :) Thanks for clearing up that misunderstanding for me! (:
Post by: bluecookie on April 02, 2017, 10:56:05 pm
np>ns for step down because emf is proportional to the number of coils, so a smaller number will induce a smaller emf in the secondary coil :) and vice versa for step up
Thanks :)
Post by: winstondarmawan on April 03, 2017, 02:35:39 pm
This one is nasty - I remember this, it is a copy of a question from a 2004 paper, with some dates changed!
So the idea is this. Consider Kepler's Law:
Mars has a larger orbital radius than the Earth, so according to this equation, it must also have a larger orbital period. We can't calculate exactly how much, especially given that the diagram is not to scale (or at least specified to be), but we know that it is larger. What does this mean? Well, in the 10 months time span you have been given, earth will complete about 5/6 of a revolution. Mars will complete less than this - I'd say about 1/2 of a revolution might be what you want.
We also know that they are going anticlockwise. Why? Well, that's just what we need for this scenario to make sense. Putting all that together, I'd get this:
(http://i.imgur.com/xpbvh4b.png)
Not 100% confident on this, but this would be my best crack (also excuse the shitty diagram lol) ;D
So say if I assumed the diagram was to scale like and calculated the approximate location of Mars that way (it turned out to be 1/7th of the rotation) would that be okay if I didn't state any calculations?
Post by: jamonwindeyer on April 03, 2017, 03:31:51 pm
So say if I assumed the diagram was to scale like and calculated the approximate location of Mars that way (it turned out to be 1/7th of the rotation) would that be okay if I didn't state any calculations?
Hmm, I would say that assuming the diagram is to scale is probably not a correct thing to do. But then, given how little info you are given, it is a logical assumption to make. You'd probably get paid anyway as long as the rest of your reasoning was okay - Definitely don't assume diagrams are to scale in HSC/Trial exams though :)
Post by: Aaron12038488 on April 03, 2017, 04:50:46 pm
Need urgent advice to improve on pracs and hopefully doesnt occur in year 12.
Post by: winstondarmawan on April 03, 2017, 09:43:25 pm
Thanks in advance.
Post by: ProfLayton2000 on April 04, 2017, 12:52:09 pm
Post by: jamonwindeyer on April 04, 2017, 01:54:01 pm
I completely flunked physics, the prac which is light going through a glass perspex. My CRO didnt work which wasted like 5-10 min. I didnt know how to answer the questions and its worth 20%. I'm so bad at practicals, however when it comes to tests, I always ace it.
Need urgent advice to improve on pracs and hopefully doesnt occur in year 12.
Damn! That's bad luck Aaron, fingers crossed your mark comes out a little better than you anticipate! Once you get your feedback, just try and recognise, "Where did my marks go?" Did you understand the theory? Was your process correct? Do you need to work on time management?
Learn from mistakes my friend - That is what Year 11 is for!
Post by: jamonwindeyer on April 04, 2017, 01:57:07 pm
Would appreciate help with the question in the attachments.
Thanks in advance.
Hey Winston! Actually doing this experiment, the graph would be two peaks, one negative one positive. Whichever peak you put second would be larger.
To explain the shape: As the magnet falls it exposes the tube to a changing magnetic field, inducing a current. This reaches a peak before minimising as the magnet reaches the centre of the tube. Then, as it falls out, the changing field is now in the opposite direction, so you get a peak in the other direction. It is moving faster than it did in the first half, so the peak is larger.
Hope that helps!!
Post by: jamonwindeyer on April 04, 2017, 02:18:40 pm
Hello, why doesn't an AC induction motor have any back emf?
Hey ProfLayton! AC induction motors don't have back emf because we don't feed them a current - We actually RELY on induced currents to make it spin!! Back EMF is normally considered as an induced current flow due to our supply current - No supply means no back emf
Post by: Aaron12038488 on April 04, 2017, 05:22:44 pm
Post by: jamonwindeyer on April 04, 2017, 06:40:31 pm
Out of curiosity, what is an open-investigation. I saw my assessment booklet, and it stated that.
Pretty much just an investigation/research task without a definite answer - Pretty much any take home task you receive for Physics will fall under this category ;D
Post by: katnisschung on April 04, 2017, 07:28:53 pm
i got r to be 1.00x10^17 and then i can't exactly sub
this into the escape velocity formula cos it isn't the radius ..
Post by: Bubbly_bluey on April 05, 2017, 02:37:07 pm
Thanks
Post by: jamonwindeyer on April 05, 2017, 02:56:39 pm
Hey guys! I've never see a question likethis before. How do I even start? Does the ball have both and initial horizontal and vertical velocity because the lift? ???
Thanks
Hey hey! This was a doozy last year. Think of it this way - When you are in a lift coming to the top, or a rollercoaster slowing down near the top about to drop, do you know how you feel like you are lifting off the ground? Or out of the chair? That's a slight upwards acceleration due to the motion of your reference frame, and that is what the ball experiences!
This slight upwards acceleration acts against gravity, and increases the flight time (the ball reduces height more slowly). The answer is therefore C ;D
This is a tough one to explain - Does this make sense? :)
Post by: teapancakes08 on April 05, 2017, 05:04:56 pm
....
I'm not really sure how to start this off honestly...
Post by: jamonwindeyer on April 05, 2017, 09:05:18 pm
"Two massless parallel current carrying wires are supported by strings as shown in the diagram. If each of the wires is 1 m long and carries a current of 50 mA, and the supporting strings are 85 cm long, calculate the force acting on each of the wires when the angle between the strings is 60º."
....
I'm not really sure how to start this off honestly...
Hey! So this is the formula we need:
So, we know the wires are 1 metre long (\(L=1\)), we know the currents are 50mA (\(I_1=I_2=0.05\), and \(k\) is a constant. Only thing missing is the distance between them! We find that by drawing an imaginary line between the two wires and using the cosine rule:
Note that we could also get this result by noticing that the triangle is equilateral (since it is isosceles and the two base angles need to add to 120, they must be 60 degrees each), but this is the way you'd do it in general. So, with \(d=85\text{cm}=0.85\text{m}\) we can plug all our values into that formula above to get our answer ;D
Hopefully that makes sense! ;D
Post by: teapancakes08 on April 05, 2017, 10:01:18 pm
Hey! So this is the formula we need:
So, we know the wires are 1 metre long (\(L=1\)), we know the currents are 50mA (\(I_1=I_2=0.05\), and \(k\) is a constant. Only thing missing is the distance between them! We find that by drawing an imaginary line between the two wires and using the cosine rule:
Note that we could also get this result by noticing that the triangle is equilateral (since it is isosceles and the two base angles need to add to 120, they must be 60 degrees each), but this is the way you'd do it in general. So, with \(d=85\text{cm}=0.85\text{m}\) we can plug all our values into that formula above to get our answer ;D
Hopefully that makes sense! ;D
I see ^^ (guess this is where have a maths background comes in handy :P)
I was really confused because the answers in the textbook threw me off...
Post by: jamonwindeyer on April 05, 2017, 10:05:10 pm
I see ^^ (guess this is where have a maths background comes in handy :P)
I was really confused because the answers in the textbook threw me off...
I feel like those answers are possibly incorrect! Because their method would make sense, but that angle at the top isn't 12.5 degrees. They've basically cut that top angle in half - So it should be 30 degrees!
Post by: teapancakes08 on April 05, 2017, 10:11:22 pm
I feel like those answers are possibly incorrect! Because their method would make sense, but that angle at the top isn't 12.5 degrees. They've basically cut that top angle in half - So it should be 30 degrees!
That's what I thought too! But then since it was the textbook and my physics skills aren't very impressive I just stared at it for god knows how long until I decided to put it on ATARNotes...
Post by: jamonwindeyer on April 05, 2017, 10:13:50 pm
That's what I thought too! But then since it was the textbook and my physics skills aren't very impressive I just stared at it for god knows how long until I decided to put it on ATARNotes...
Ahaha don't worry, I was always afraid to contradict my textbook too!! Always good to check with us if you just don't think it seems quite right - No textbook is immune from errors ;D
Post by: Bubbly_bluey on April 07, 2017, 06:01:22 pm
A and B)was about the astoriod having a stable orbit around the sun
c) The force of the astoriod is negliable
D) Theres and equal opposite reaction
Thanks heaps! ;D So sorry if the question is very vague (brain fart)
Post by: jamonwindeyer on April 07, 2017, 06:39:11 pm
Hey! There was a question that goes along the lines of: An astronaut lands and stands on an astoroid that is orbiting the Sun between (I forgot the planets) Mars and Jupitur. Why does the astronaut not fall off?? or why does the astronaut feel weightless? -Something along those lines. It was a multiple choice:
A and B)was about the astoriod having a stable orbit around the sun
c) The force of the astoriod is negliable
D) Theres and equal opposite reaction
Thanks heaps! ;D So sorry if the question is very vague (brain fart)
Hey Bubbly! Sorry mate, but I can't really decipher it with the info you've given - Might just have to wait until you get the paper back. How do you think you went? :)
Post by: Bubbly_bluey on April 07, 2017, 08:43:08 pm
Hey Bubbly! Sorry mate, but I can't really decipher it with the info you've given - Might just have to wait until you get the paper back. How do you think you went? :)haha its ok ill try get back to you when I get the question. As for the exam, a couple of questions from the exam were past HSC and I happened to have studied briefly the exact same questions! How crazy was that XD. But seriously wished I had paid more attention to the solutions to word my responses better ::)
Post by: jamonwindeyer on April 07, 2017, 08:59:13 pm
haha its ok ill try get back to you when I get the question. As for the exam, a couple of questions from the exam were past HSC and I happened to have studied briefly the exact same questions! How crazy was that XD. But seriously wished I had paid more attention to the solutions to word my responses better ::)
Ahaha gotta love it! A few people have had that - Definitely lucky! I had an exam in university last year that was literally one half the Exam for 2014, one half the Exam for 2015... Crazy
well done!! Enjoy your break friend, you've earned it!
Post by: pikachu975 on April 09, 2017, 10:59:00 pm
a little clueless as to how they got v..
i got r to be 1.00x10^17 and then i can't exactly sub
this into the escape velocity formula cos it isn't the radius ..
Pretty sure it's asking for escape velocity FROM 1x10^17 m away from Earth, so just sub that as r and you're done!
Question:
If they tell you the torque acting on the WHOLE coil, would the torque acting on ONE side be half of the original torque?
Post by: jamonwindeyer on April 10, 2017, 12:08:51 am
Pretty sure it's asking for escape velocity FROM 1x10^17 m away from Earth, so just sub that as r and you're done!
Question:
If they tell you the torque acting on the WHOLE coil, would the torque acting on ONE side be half of the original torque?
Totally missed that question, sorry Katniss! Think you are almost definitely right pikachu too, cheers heaps ;D
Hmm, that's an interesting question. Torque is a rotational force, I don't think it would quite make sense to define it on one side of the coil only. Because if you separate one half, it's more just a linear force on one side if you catch my drift? Have never really thought about it before!
If you loosen the definition a bit though, I think it would be! The torque on a coil is generated by the forces acting on either side, and those forces are equal in magnitude. So yeah, I think it stands to reason that the torque would be half ;D
Post by: jakesilove on April 10, 2017, 09:00:51 am
Totally missed that question, sorry Katniss! Think you are almost definitely right pikachu too, cheers heaps ;D
Hmm, that's an interesting question. Torque is a rotational force, I don't think it would quite make sense to define it on one side of the coil only. Because if you separate one half, it's more just a linear force on one side if you catch my drift? Have never really thought about it before!
If you loosen the definition a bit though, I think it would be! The torque on a coil is generated by the forces acting on either side, and those forces are equal in magnitude. So yeah, I think it stands to reason that the torque would be half ;D
Torque doesn't act on an 'area' of the coil; it acts on the whole coil (or rather, the plane of the coil). That's why, when you try to figure out the angle relevant for the calculation of torque, you use the PLANE instead of a particular wire. If a question EVER asked you for the torque, even if it's like 'on wire XY', you still need to take the entire coil into account (as the magnitude is the same everywhere, you just do the normal calculation). So, I wouldn't focus too much on the physical representation of Torque, and just sub the hell out of the formula they've given you :)
Post by: beau77bro on April 14, 2017, 07:23:41 pm
Post by: jamonwindeyer on April 14, 2017, 09:38:04 pm
can someone please explain what is meant by this graph of black body radiation, i thought id just started to understand blackbody radiation, but i dont get what the axises mean in the graph. very much appreciated if someone could explain the features of it thanks.
Hey! Sure thing, I'll try and make it super simple ;D
So what we are looking at here is radiation curves for black bodies (which is, for the most part, the same thing as the thermal radiation given off by everyday objects, even you and I). Black bodies emit radiation at all frequencies - Meaning you and I (assuming we are black bodies, which we aren't, but you know, dramatic effect) are emitting X-rays right now. The issue is how much X-ray radiation is being emitted.
That's what these curves show. The horizontal axis is wavelength, the vertical is intensity, which is just (speaking basically) how much of each wavelength we get. So if we look at the red line, there is a HEAP of energy at the 0.5 micrometer wavelength, but not much at all at the 2 micrometer wavelength. This means that it emits a lot of 0.5 micrometer radiation, not a lot of 2 micrometer radiation.
Every black body has what is called a characteristic frequency/wavelength. This is just the wavelength/frequency where a black body emits the MOST radiation.
The different lines represent bodies of different temperatures - Indeed, for ideal black bodies, the only thing affecting the characteristic wavelength is the temperature. Objects with higher temperatures have lower characteristic wavelengths (and higher characteristic frequencies, by the \(c=f\lambda\) formula) :) :)
Let me know if I can clarify any of this for you! :) I tried to keep it as simple as I could but definitely happy to help pick any bits of it apart that you'd like ;D
Post by: winstondarmawan on April 15, 2017, 12:06:05 am
Thanks in advance.
Post by: jamonwindeyer on April 15, 2017, 12:29:22 am
Hello! Another question I would like help with.
Thanks in advance.
Hey! Great question this one. So the first step is ALWAYS to resolve the velocity into its vertical and horizontal components. The launch velocity is known, \(V\), so:
You could also get decimal approximations to these values, or leave them in trig form - The exact values I've done you might only know if you do 2U ;D So now let's examine what we know. For the horizontal axis, we know that:
So that's given us a formula linking velocity to time! Now in the vertical direction:
This is going to get messy - Replace \(V_y\) and \(t\) with the expressions involving \(V\) we've found, and we want a vertical displacement of -34 (34 metres BELOW our launch point):
Now if you solve this for \(V\), I think you get about \(21.42\) metres per second! :) does this working out make sense? It's a tough one! ;D
Post by: itssona on April 15, 2017, 01:39:44 pm
Like negative is decceleration? or is it accelerartion in a negative DIRECTION
Post by: RuiAce on April 15, 2017, 01:44:46 pm
hey this is a silly question but whats the difference between pos and negative acceleration??'Deceleration' isn't a thing in physics. You're either accelerating in the positive or the negative direction. Deceleration is just a colloquial word for the act of slowing down, but we don't consider this in physics. We always consider it in that we are accelerating, just the other way around.
Like negative is decceleration? or is it accelerartion in a negative DIRECTION
For an example, take the positive direction to be right, and negative to be left. You are accelerating in the positive direction if you are accelerating towards your right. You are accelerating in the negative direction if you are accelerating towards your left.
For another example, take the positive direction to be up and negative to be down. If you drop a ball, it will accelerate downwards due to gravity. So gravity is an example of negative acceleration in this scenario.
Note that the concept of acceleration only makes sense provided you can define a positive direction to begin with.
Post by: beau77bro on April 15, 2017, 02:13:06 pm
Hey! Sure thing, I'll try and make it super simple ;D
So what we are looking at here is radiation curves for black bodies (which is, for the most part, the same thing as the thermal radiation given off by everyday objects, even you and I). Black bodies emit radiation at all frequencies - Meaning you and I (assuming we are black bodies, which we aren't, but you know, dramatic effect) are emitting X-rays right now. The issue is how much X-ray radiation is being emitted.
That's what these curves show. The horizontal axis is wavelength, the vertical is intensity, which is just (speaking basically) how much of each wavelength we get. So if we look at the red line, there is a HEAP of energy at the 0.5 micrometer wavelength, but not much at all at the 2 micrometer wavelength. This means that it emits a lot of 0.5 micrometer radiation, not a lot of 2 micrometer radiation.
Every black body has what is called a characteristic frequency/wavelength. This is just the wavelength/frequency where a black body emits the MOST radiation.
The different lines represent bodies of different temperatures - Indeed, for ideal black bodies, the only thing affecting the characteristic wavelength is the temperature. Objects with higher temperatures have lower characteristic wavelengths (and higher characteristic frequencies, by the \(c=f\lambda\) formula) :) :)
Let me know if I can clarify any of this for you! :) I tried to keep it as simple as I could but definitely happy to help pick any bits of it apart that you'd like ;D
thanks so much jamon, cleared it up alot for me. the detail and extra info really helped as well. i think i get it alot more thankyou
Post by: itssona on April 15, 2017, 02:58:19 pm
'Deceleration' isn't a thing in physics. You're either accelerating in the positive or the negative direction. Deceleration is just a colloquial word for the act of slowing down, but we don't consider this in physics. We always consider it in that we are accelerating, just the other way around.
For an example, take the positive direction to be right, and negative to be left. You are accelerating in the positive direction if you are accelerating towards your right. You are accelerating in the negative direction if you are accelerating towards your left.
For another example, take the positive direction to be up and negative to be down. If you drop a ball, it will accelerate downwards due to gravity. So gravity is an example of negative acceleration in this scenario.
Note that the concept of acceleration only makes sense provided you can define a positive direction to begin with.
Ohh! Okay that makes sense :O
so is decceleration more like "decreasing rate of acceleration"
http://webs.mn.catholic.edu.au/physics/emery/assets/prelim60.gif
in a veloctity time graph, positive gradient is positive accleration but how would we know by just knowing velocity? how do they just conclude that negative gradient is negative acceleration when negative acceleration is based on direction
Anyways thank you sooo much for clearing that bit up!! :D :D
Post by: jamonwindeyer on April 15, 2017, 03:51:57 pm
Ohh! Okay that makes sense :O
so is decceleration more like "decreasing rate of acceleration"
http://webs.mn.catholic.edu.au/physics/emery/assets/prelim60.gif
in a veloctity time graph, positive gradient is positive accleration but how would we know by just knowing velocity? how do they just conclude that negative gradient is negative acceleration when negative acceleration is based on direction
Anyways thank you sooo much for clearing that bit up!! :D :D
I'd call de-acceleration, decreasing magnitude of acceleration. Meaning, whatever direction you are accelerating in, deceleration would mean reducing how much you are accelerating in that direction. That's how I'd interpret it at least ;D
We define velocity and acceleration directions to be the same. Positive velocity and positive acceleration occur in the same direction. It doesn't make sense to define it any other way ;D so, since we have a positive gradient, it just makes sense to define acceleration as positive. I suppose the answer is, why would you choose to define acceleration as negative when your rate of change of velocity (gradient) is positive? That would be seriously counterintuitive, which is why we never do it ;D
Post by: itssona on April 15, 2017, 04:22:43 pm
I'd call de-acceleration, decreasing magnitude of acceleration. Meaning, whatever direction you are accelerating in, deceleration would mean reducing how much you are accelerating in that direction. That's how I'd interpret it at least ;D
We define velocity and acceleration directions to be the same. Positive velocity and positive acceleration occur in the same direction. It doesn't make sense to define it any other way ;D so, since we have a positive gradient, it just makes sense to define acceleration as positive. I suppose the answer is, why would you choose to define acceleration as negative when your rate of change of velocity (gradient) is positive? That would be seriously counterintuitive, which is why we never do it ;D
Ohhhhh I finally get it :O Now I see how silly my thinking was LOL
Thank you a tonne Jamon! :D
Post by: smile123 on April 15, 2017, 06:26:21 pm
Post by: beau77bro on April 15, 2017, 07:33:17 pm
Post by: jakesilove on April 15, 2017, 07:40:05 pm
HELP
Hey!
We need to drive a total of 270km. The first 100km is fine, and as you've travelled at 100km/h, this distance will take 1 hour to travel. So, at this point it is 9:00am. Then, you slow down to 42km/h for 43km. This will take
So, this takes us through to 10:01am. We need to get to the interview by 11:15am, and have travelled 143 km. So, we have 270-143=127km to travel in 1 hour and fourteen minutes! We can convert this to 1.23333 hours. To find the minimum speed required to make this distance, we simply divide the distance by the time,
Post by: jakesilove on April 15, 2017, 07:43:02 pm
not entirely sure what this means, mostly the 'overlapping in a continuous fashion' portion any help would be appreciated.
Definitely a clumsy sentence, but to be honest, this entire area of the Physics curriculum is clumsy. All you need to know is that we define two bands when it comes to conduction; the valence band, and the conduction band. Electrons need to be able to jump from the valence band to the conduction band in order for conduction to occur. Thus, the bands are touches in Metals, closeish to each other in semi-conductors, and far away in insulators. In reality, there is crazy quantum mechanical stuff going on here, and so the description you've posted is just an attempt to avoid all of that (which is totally fair enough).
The below is all that is worth having in your head when you think about something like this
(https://www.learner.org/courses/physics/visual/img_lrg/bands.jpg)
(Ignore the Fermi level)
Post by: beau77bro on April 15, 2017, 08:09:48 pm
Post by: jakesilove on April 15, 2017, 08:19:49 pm
hahah oh ok, sweet thankyou. so we dont need to know it? and i would honestly love to hear the crazy quantum stuff if u can be bothered to type it. quantum mechanics seems super interesting, i was the guy that kept asking u questions at the lectures btw. thanks for all the help and advice. you guys have an awesome site going
Unfortunately, it would be genuinely impossible to go into any real depth about the band structure of semiconductors etc. Suffice it to say that the approximation described above is plenty for the HSC, and for any real life application. Beyond that, it's a lot of mathematical formalism and electrodynamics. Stick within the curriculum for now, and satiate your hunger for knowledge on YouTube or at Uni :)
Post by: beau77bro on April 16, 2017, 12:05:21 am
Post by: Dragomistress on April 16, 2017, 11:40:52 am
Post by: RuiAce on April 16, 2017, 11:44:09 am
For all equations in physics, do they all use following units: Newtons, metres, seconds, kilograms, degrees?For example you forgot all the units in electricity e.g. Amps, Volts, Ohms.
It may be worth mentioning though, that seconds, metres and kilograms are S.I. base units and yes they will always be used. Degrees just happen to pop up, and 1 N is just 1 kg m s-2. So they definitely appear as well.
Post by: katnisschung on April 16, 2017, 01:23:46 pm
i assume that its going to experience different forces due to the motor effect
and its going to be continually deflected upwards and downwards (how do i determine the direction of the current through the copper wire)
here's the question
two solenoid coils are suspended as shown in the diagrams below (diagram one is exactly the same as the second just doesn't have the copper wire) BOTH ARE FREE TO SWING FROM SIDE TO SIDE
In both the investigations the magnet is moved towards the coils
Explain what effect the motion of the magnet has in both cases.
my answer is pretty long so here's it summed up (yesh i did explain it but i can't upload it cos of the file size limit)
in figure 1 north pole introduced into the solenoid, emf induced according to electromagnetic induction. According to Lenz's law this emf will give rise to a current which creates a magnetic field that opposes this increasing magnetic field, thus creates a magnetic north
(when magnetic north is moved away south is created on B point of solenoid by similar logic)
similar phenomenon occurs in figure 2
but whats up with that copper wire?
Post by: jamonwindeyer on April 16, 2017, 02:55:00 pm
guys what happens to the copper wire in the second diagram?
i assume that its going to experience different forces due to the motor effect
and its going to be continually deflected upwards and downwards (how do i determine the direction of the current through the copper wire)
...
but whats up with that copper wire?
Hey Katniss! You've definitely got the main ideas down here - That copper wire in the second solenoid is there to complete the circuit. In the first solenoid, the wire is wrapped around and that is great, but one end isn't connected to the other end! So there is no real closed path for current to flow through. So the induced currents and other effects you discuss actually won't occur for the first solenoid, at least not to any significant level. They only occur in the second solenoid, when that copper wire joins the two sides of the wire together, completing a path for current flow! ;D
Post by: katnisschung on April 16, 2017, 03:15:25 pm
Hey Katniss! You've definitely got the main ideas down here - That copper wire in the second solenoid is there to complete the circuit. In the first solenoid, the wire is wrapped around and that is great, but one end isn't connected to the other end! So there is no real closed path for current to flow through. So the induced currents and other effects you discuss actually won't occur for the first solenoid, at least not to any significant level. They only occur in the second solenoid, when that copper wire joins the two sides of the wire together, completing a path for current flow! ;D
ahaha i see yeah i think i was probably overthinking it a little. Thanks Jamon!
Post by: jamonwindeyer on April 16, 2017, 03:38:18 pm
hahah oh ok, sweet thankyou. so we dont need to know it? and i would honestly love to hear the crazy quantum stuff if u can be bothered to type it. quantum mechanics seems super interesting, i was the guy that kept asking u questions at the lectures btw. thanks for all the help and advice. you guys have an awesome site going
Here are my notes on Quantum Mechanics from 1st Year Higher Physics - They don't explain stuff super nicely, they were just for my revision, but they might spark some further interest or give you some rough understandings that you can improve with research! ;D or whatever! Just clicked that I had them :)
Post by: bluecookie on April 16, 2017, 03:42:35 pm
If an object is dropped from a point, would it still be moving at the same velocity as gravity (9.8ms) when it hits the ground, or would it have accelerated over time?
Similarly, if a brick and a feather are dropped from the same height, would they both hit the ground at the same time?
Post by: jamonwindeyer on April 16, 2017, 03:46:25 pm
Umm, I'm just confused about a concept that I maybe should have gotten years ago when we did Moving About in yr 11 (but I didn't oops), but uh, does acceleration apply to gravity?
If an object is dropped from a point, I know it moves at the velocity of gravity, but does it accelerate over time?
Also, if you drop a brick and a feather, would they both hit the ground at the same time? Why/why not?
Hey! So gravity is acceleration - Gravity applies a force which generates an acceleration of \(9.8ms^{-2}\) directed downwards, to every object on the earth's surface ;D
If we neglect air resistance, yes, a brick and a feather would hit the ground at the same time because they experience the same gravitational acceleration! They checked this on the moon where there is no air (or close to it, I think the surface of the moon is pretty much a vacuum) and it worked. The only reason that bricks hit first normally is because feathers are affected far more by air resistance and other external forces ;D
Post by: bluecookie on April 16, 2017, 03:52:40 pm
Hey! So gravity is acceleration - Gravity applies a force which generates an acceleration of \(9.8ms^{-2}\) directed downwards, to every object on the earth's surface ;D
If we neglect air resistance, yes, a brick and a feather would hit the ground at the same time because they experience the same gravitational acceleration! They checked this on the moon where there is no air (or close to it, I think the surface of the moon is pretty much a vacuum) and it worked. The only reason that bricks hit first normally is because feathers are affected far more by air resistance and other external forces ;D
So the velocity of the object will always be 9.8? And the acceleration is always zero?
Thanks :D
Post by: jamonwindeyer on April 16, 2017, 03:55:17 pm
So the velocity of the object will always be 9.8? And the acceleration is always zero?
Thanks :D
Acceleration will always be \(9.8\text{ms}^{-2}\) - The velocity there is no set rule unfortunately, depends on if the object is being impacted on by something besides gravity ;D
Post by: bluecookie on April 16, 2017, 03:57:52 pm
Acceleration will always be \(9.8\text{ms}^{-2}\) - The velocity there is no set rule unfortunately, depends on if the object is being impacted on by something besides gravity ;D
Thanks :D
Post by: itssona on April 16, 2017, 04:45:59 pm
http://buphy.bu.edu/~duffy/PY106/14e.GIF
like flux is BAcostheta and I understand the way A is the area and costheta is the angle made with the normal but i dont get Faraday's formula :/
thank you
Post by: beau77bro on April 16, 2017, 07:55:47 pm
Here are my notes on Quantum Mechanics from 1st Year Higher Physics - They don't explain stuff super nicely, they were just for my revision, but they might spark some further interest or give you some rough understandings that you can improve with research! ;D or whatever! Just clicked that I had them :)
hahaha thanks jamon. i got abit(very) lost past the photoelectric effect, but i will give understanding it another go later ahahah thankyo
Post by: smile123 on April 16, 2017, 08:31:46 pm
I figured out a) is 18.62m/s
but I can't find b) and c)
Post by: jakesilove on April 16, 2017, 10:10:41 pm
HELP
I figured out a) is 18.62m/s
but I can't find b) and c)
For b) we use
The final velocity is zero, as the rock is stationary for a moment. Thus,
Now, it takes 9.00-1.30=7.70s for the rock to go from the maximum height to the ground. Thus
However, this includes the height ABOVE the building. Subtracting this, 290.5-17.69=272.8m is our answer :)
Post by: jamonwindeyer on April 16, 2017, 11:51:40 pm
Hey so I was reading about EMF and magnetic flux and recapping stuff and the concepts are pretty clear but I'm having trouble understanding Faraday's law's formula
http://buphy.bu.edu/~duffy/PY106/14e.GIF
like flux is BAcostheta and I understand the way A is the area and costheta is the angle made with the normal but i dont get Faraday's formula :/
thank you
Hey! So that version of the formula isn't going to be useful in the HSC - Since you'll never have to deal with Faraday's law numerically! I'll give you another version - But I will explain this one first.
\(\Delta\left(BA\cos{\theta}\right)\) - This is the change in magnetic flux. Note that the \(\Delta\) symbol just means "difference" as always. The reason it is \(BA\) is from the formula \(\Phi=BA\) - You get magnetic flux by multiplying magnetic flux density by area - Which makes intuitive sense ;D The \(\cos\phi\) term just takes care of the angle of the field, basically :)
\(N\) is the number of turns in the coil you are talking about, and then the denominator represents change in time. So in words:
The induced EMF in a solenoid is equal to the rate of change of magnetic flux in that solenoid, multiplied by the number of turns in the solenoid.
Again though, you'll never use this formula algebraically in the HSC. Not assessable - Faraday's Law in this algebraic form isn't even in the syllabus ;D so I instead tell my students to use this simpler version just for the utility:
In words: The magnitude of an induced emf is proportional to the rate of change of magnetic flux. This captures exactly the same information as above, but just in a way less technical way that is far easier to convey in exams.
Oh woops, forgot something. Notice both your formula and my formula have negative signs - We can actually mathematically prove that this negative exists in the formula, but for us, it represents Lenz's Law ;D
Post by: itssona on April 17, 2017, 12:15:08 am
Hey! So that version of the formula isn't going to be useful in the HSC - Since you'll never have to deal with Faraday's law numerically! I'll give you another version - But I will explain this one first.
\(\Delta\left(BA\cos{\theta}\right)\) - This is the change in magnetic flux. Note that the \(\Delta\) symbol just means "difference" as always. The reason it is \(BA\) is from the formula \(\Phi=BA\) - You get magnetic flux by multiplying magnetic flux density by area - Which makes intuitive sense ;D The \(\cos\phi\) term just takes care of the angle of the field, basically :)
\(N\) is the number of turns in the coil you are talking about, and then the denominator represents change in time. So in words:
The induced EMF in a solenoid is equal to the rate of change of magnetic flux in that solenoid, multiplied by the number of turns in the solenoid.
Again though, you'll never use this formula algebraically in the HSC. Not assessable - Faraday's Law in this algebraic form isn't even in the syllabus ;D so I instead tell my students to use this simpler version just for the utility:
In words: The magnitude of an induced emf is proportional to the rate of change of magnetic flux. This captures exactly the same information as above, but just in a way less technical way that is far easier to convey in exams.
Oh woops, forgot something. Notice both your formula and my formula have negative signs - We can actually mathematically prove that this negative exists in the formula, but for us, it represents Lenz's Law ;D
magnetic field density is like magnetic field strength right? so B is always magnetic density?
Phew, so glad it's not tested! Although it is so fkn interesting! Thank you for explaining it, I understand the rest of what you said so i feel so proud rn :D
The simplified formula you showed is awesome omfg, thank you Jamon!! Wait, in the HSC can I still use the formula in an extended mark question just to show off and get marks?
Lenz's law from what I read, apparently opposes motion of the current? So the negative is because it's going in the opposing direction?
Thank you for the amazing answers, much appreciated omg :)
Post by: jamonwindeyer on April 17, 2017, 09:27:05 am
magnetic field density is like magnetic field strength right? so B is always magnetic density?
Phew, so glad it's not tested! Although it is so fkn interesting! Thank you for explaining it, I understand the rest of what you said so i feel so proud rn :D
The simplified formula you showed is awesome omfg, thank you Jamon!! Wait, in the HSC can I still use the formula in an extended mark question just to show off and get marks?
Lenz's law from what I read, apparently opposes motion of the current? So the negative is because it's going in the opposing direction?
Thank you for the amazing answers, much appreciated omg :)
Everything you propose up there is correct - B is flux density which is field strength, and yep, that's the reason the negative is there
You can use the formula in those sorts of questions! But they'd be equally impressed by your version or my version, so pick your poison
Post by: itssona on April 17, 2017, 09:53:20 am
Everything you propose up there is correct - B is flux density which is field strength, and yep, that's the reason the negative is there
You can use the formula in those sorts of questions! But they'd be equally impressed by your version or my version, so pick your poison
Yaaaaayy!! :D
Ooh nice, I shall do that next year in the HSC lol :D
Post by: crazycodpro on April 17, 2017, 11:33:37 am
Post by: wyzard on April 17, 2017, 11:57:37 am
:)
For a), use the formula
Post by: beau77bro on April 17, 2017, 12:28:39 pm
sorry for the billion questions, but appreciate any responses to as many as u can ahhahaha
Post by: jakesilove on April 17, 2017, 12:42:11 pm
can someone please state an exact definition of in and out of phase. i understand that constructive mean they add together to produce a greater amplitude resultant wave and destructive is the reverse. but phase gets me... is 'in phase' when the crests both line up? like is that the ONLY time its in phase? or is just when they are in a phase that makes constructive interference? then out of phase means '180 difference in phase' or like the troughs line up with crests? again is it that single situation? then what about antiphase? is that when they are completely out of phase, or is that when they are out of phase completely and have the same amplitude???? and what does in a constant phase mean on its own omg?
sorry for the billion questions, but appreciate any responses to as many as u can ahhahaha
Yep, you're right on the money! In phase means that the two waves are exactly identical; ie they will entirely constructively interfere. Out of phase just means that they are NOT perfectly aligned; some areas will interfere constructively, but many will interfere destructively. Anti-phase is when they are perfectly unaligned; ie a creset meets a trough and cancels it out.
Post by: beau77bro on April 17, 2017, 01:52:22 pm
Yep, you're right on the money! In phase means that the two waves are exactly identical; ie they will entirely constructively interfere. Out of phase just means that they are NOT perfectly aligned; some areas will interfere constructively, but many will interfere destructively. Anti-phase is when they are perfectly unaligned; ie a creset meets a trough and cancels it out.
Yep, you're right on the money! In phase means that the two waves are exactly identical; ie they will entirely constructively interfere. Out of phase just means that they are NOT perfectly aligned; some areas will interfere constructively, but many will interfere destructively. Anti-phase is when they are perfectly unaligned; ie a creset meets a trough and cancels it out.
ohhh ok thankyou jake, you really cleared it up. thankyouuuu
Post by: beau77bro on April 17, 2017, 02:02:24 pm
also when is the next lecture from you guys?
Post by: beau77bro on April 17, 2017, 02:20:38 pm
Post by: jakesilove on April 17, 2017, 02:26:49 pm
why is it important that x-rays have wavelengths of similar distance to the distance between atoms in the crystal lattice in the x-ray diffraction experiments by the braggs?
also when is the next lecture from you guys?
To be honest, when it comes to that dotpoint I wouldn't bother knowing WHY, just that it is an important factor. Basically, the waves need to be able to 'get between' the layers, and thus create a diffraction pattern. So, wavelengths of similar size sort of allow that.
Keep an eye out for future lecture dates; we'll be doing some mid-year Trial prep lectures :)
No, you don't need to know the first paragraph of information.
Post by: beau77bro on April 17, 2017, 03:17:52 pm
Post by: beau77bro on April 17, 2017, 07:27:01 pm
Post by: jakesilove on April 17, 2017, 07:28:45 pm
ok so question, i have a couple of sources saying x-rays have wavelengths the size of the atoms, then others saying they are the size of the distance between atoms in the lattice? which is it? could it be both since x-rays are portion of the spectrum?
Both, because the distance between atoms in a lattice is often pretty close to the size of the atom itself (in atomic units, so like even if the distance between atoms is a couple of times bigger than the atom itself, it's still goddamned small).
Post by: bsdfjnlkasn on April 17, 2017, 08:23:33 pm
For b) we use
The final velocity is zero, as the rock is stationary for a moment. Thus,
Now, it takes 9.00-1.30=7.70s for the rock to go from the maximum height to the ground. Thus
However, this includes the height ABOVE the building. Subtracting this, 290.5-17.69=272.8m is our answer :)
Hey Jake,
SO sorry if i'm wrong here, but shouldn't the time for part c) be 9 - 1.9 = 7.1 seconds? Maybe i'm missing some info somewhere so just ignore me if i'm wrong haha. Thank you!!
Post by: jakesilove on April 17, 2017, 08:36:05 pm
Hey Jake,
SO sorry if i'm wrong here, but shouldn't the time for part c) be 9 - 1.9 = 7.1 seconds? Maybe i'm missing some info somewhere so just ignore me if i'm wrong haha. Thank you!!
Yep you're probably right, it's possible that I misread the question (whatever the real time is was stated in the question).
Post by: itssona on April 23, 2017, 05:59:42 pm
so I have this velcoty time graph and my teacher gave notes for it but I am skeptical about it (maybe its just me not being good lol)
but so in the segment C, he said : deaccelerating, stopping, and then accelerating to the left
isnt that wrong?
also, for segment D, he said uniform decceleration????
and for G he said its accelerating while for H he said its deccelerating, but isnt it the OTHER way round?
Post by: itssona on April 23, 2017, 06:06:41 pm
Post by: ellipse on April 23, 2017, 06:25:52 pm
Heey, could someone help me with this:
so I have this velcoty time graph and my teacher gave notes for it but I am skeptical about it (maybe its just me not being good lol)
but so in the segment C, he said : deaccelerating, stopping, and then accelerating to the left
isnt that wrong?
also, for segment D, he said uniform decceleration????
and for G he said its accelerating while for H he said its deccelerating, but isnt it the OTHER way round?
He is correct.
In segment C, the object initially has a high velocity, but it slows down (as seen by the magnitude of velocity deceasing). When the velocity is 0, it is not moving, thus it is stopping. Soon after that, although the velocity is getting negative, the magntitude still increases. This means that it is accelearting to the left (if the right is the positive direction)
In segment D, the velicity changes from -3 to 0. Again, if you look at the magntidue, it decreases, thus it is decelarating.
Similarly if you look at the magntiude of the velocity, it increases at G (goes from 0 to 3 in the left) thus it is accelerating (non-uniformly) and at H the magnitude decreases, thus it decelerates.
Hope that helps :)
Post by: itssona on April 23, 2017, 06:33:24 pm
He is correct.
In segment C, the object initially has a high velocity, but it slows down (as seen by the magnitude of velocity deceasing). When the velocity is 0, it is not moving, thus it is stopping. Soon after that, although the velocity is getting negative, the magntitude still increases. This means that it is accelearting to the left (if the right is the positive direction)
In segment D, the velicity changes from -3 to 0. Again, if you look at the magntidue, it decreases, thus it is decelarating.
Similarly if you look at the magntiude of the velocity, it increases at G (goes from 0 to 3 in the left) thus it is accelerating (non-uniformly) and at H the magnitude decreases, thus it decelerates.
Hope that helps :)
OMG thank you!! :D
I guess I was confused but now that you explained it in terms of magnitude, it makes sense! Thankss :)
Post by: ellipse on April 23, 2017, 06:40:16 pm
OMG thank you!! :D
I guess I was confused but now that you explained it in terms of magnitude, it makes sense! Thankss :)
No worries! Happy to help :)
Post by: itssona on April 23, 2017, 07:05:56 pm
all I can see is that its not in the centre :/ but im missing something since its 2 marks
Post by: bsdfjnlkasn on April 23, 2017, 08:04:49 pm
hey also could someone please help with this:
all I can see is that its not in the centre :/ but im missing something since its 2 marks
Hey!
So we must remember that validity is concerned with how the controlled variables/setup in an experiment is kept the same. The question only asks for a reason why the experiment is invalid so we'll need to explain for the second mark. Because it's asking why it is invalid, we must look at what isn't actually being sufficiently controlled
You could probably consider how the eventual direction of the bottom spring will be influenced by additional forces in the springs themselves therefore reducing the force's magnitude and (at least initially) changing the direction of the springs as they contract and stretch (this is all assuming that the springs are subject to gravity i.e. the set up is perpendicular to the ground). I'm not sure what impact the whole thing not being in the centre will be as we can see that the strings are of different lengths - this was most probably intentional as it will show how different forces (in this case tension in the string) acting in different directions will affect the bottom spring which will show the net force. OR we could just consider how the forces are not acting perpendicular to one another meaning a vector sum can't even be performed in the first place haha that would probably be the simplest answer.
Hopefully that helped a little bit, if someone could confirm/give a better answer that would be much appreciated :)
Post by: itssona on April 23, 2017, 08:24:44 pm
Hey!ohh spring force! I didnt consider that at all, thank you so much! :D !!!!
So we must remember that validity is concerned with how the controlled variables/setup in an experiment is kept the same. The question only asks for a reason why the experiment is invalid so we'll need to explain for the second mark. Because it's asking why it is invalid, we must look at what isn't actually being sufficiently controlled
You could probably consider how the eventual direction of the bottom spring will be influenced by additional forces in the springs themselves therefore reducing the force's magnitude and (at least initially) changing the direction of the springs as they contract and stretch (this is all assuming that the springs are subject to gravity i.e. the set up is perpendicular to the ground). I'm not sure what impact the whole thing not being in the centre will be as we can see that the strings are of different lengths - this was most probably intentional as it will show how different forces (in this case tension in the string) acting in different directions will affect the bottom spring which will show the net force. OR we could just consider how the forces are not acting perpendicular to one another meaning a vector sum can't even be performed in the first place haha that would probably be the simplest answer.
Hopefully that helped a little bit, if someone could confirm/give a better answer that would be much appreciated :)
Post by: itssona on April 23, 2017, 09:11:00 pm
a.Calculate how long it would take a boat that can travel at 6m/s in still water to cover a distance of 180m down river and back 180m. The water current is 4m/s.
b.The boat is then driven slightly angled against the current so that it travels exactly across the river, whose width is 120m.
Calculate how long it would take the boat to cover the river and back.
I know relative velocity but i never learnt stream questions :/
Post by: jamonwindeyer on April 23, 2017, 09:43:20 pm
can someone please help with this:
a.Calculate how long it would take a boat that can travel at 6m/s in still water to cover a distance of 180m down river and back 180m. The water current is 4m/s.
b.The boat is then driven slightly angled against the current so that it travels exactly across the river, whose width is 120m.
Calculate how long it would take the boat to cover the river and back.
I know relative velocity but i never learnt stream questions :/
Hey! So the first one, you just think of it as a constant \(4\) metres per second of velocity, downstream, being added to however fast the boat can travel. So, if it is travelling downstream, it can travel at \(6+4=10\text{ms}^{-1}\). If upstream it is going against the current, so \(6-4=2\text{ms}^{-1}\). So to travel 180m downstream then 180m upstream (in whatever order), we just use the idea that time is distance divided by speed:
The next bit is trickier - The boat driver angles themselves so that they travel directly across the river. What this means is, they make their velocity such a direction that the component of velocity parallel to the stream cancels with the flow of the river, leaving only the horizontal. You'll draw a right angled velocity triangle, with \(4\text{ms}^{-1}\) as the longer arm and \(6\text{ms}^{-1}\) as the hypotenuse (how fast the boat actually travels). Using pythag:
So to cover the 120 metre river:
This might be a little hard to picture without the diagram in front of you - Try to tackle it again having read this and see if it can help you guide your way through :)
Post by: itssona on April 23, 2017, 10:04:06 pm
Hey! So the first one, you just think of it as a constant \(4\) metres per second of velocity, downstream, being added to however fast the boat can travel. So, if it is travelling downstream, it can travel at \(6+4=10\text{ms}^{-1}\). If upstream it is going against the current, so \(6-4=2\text{ms}^{-1}\). So to travel 180m downstream then 180m upstream (in whatever order), we just use the idea that time is distance divided by speed:the second part took me a while but I read what you said and re-attempted it, and I got it now! Thank you so much Jamon :) !!
The next bit is trickier - The boat driver angles themselves so that they travel directly across the river. What this means is, they make their velocity such a direction that the component of velocity parallel to the stream cancels with the flow of the river, leaving only the horizontal. You'll draw a right angled velocity triangle, with \(4\text{ms}^{-1}\) as the longer arm and \(6\text{ms}^{-1}\) as the hypotenuse (how fast the boat actually travels). Using pythag:
So to cover the 120 metre river:
This might be a little hard to picture without the diagram in front of you - Try to tackle it again having read this and see if it can help you guide your way through :)
Post by: beau77bro on April 24, 2017, 01:42:31 pm
But how would this apply to say an electric motor? the lack of resistance would change what exactly?
also does anyone know any good bcs explanations? a youtube video or explanation on line would be great thankyou
Mod Edit: Post merge :)
Post by: jamonwindeyer on April 24, 2017, 04:47:43 pm
I'm just trying to work a bit more on my understanding. would i be correct in saying that superconductivity is viable for the transmission of electricity as the action of dropping the resistance to zero removes any power loss as per P = I2R. but would prevent a conductor from doing work, as without resistance there is no potential difference, which is a measure of the work done between two points. so a superconducting light would be pointless, as resistance creates the heat which produces a given light.
Exactly correct on all counts ;D
Quote
But how would this apply to say an electric motor? the lack of resistance would change what exactly?
Hmm, that's an interesting question! I believe that using a superconductor for the coil wouldn't actually achieve the desired effect, because superconductors exclude magnetic flux. This means that you don't get forces on current carrying wires in a magnetic field - Because the current doesn't interact with the magnetic field. I could be wrong there - Perhaps the Meisner effect would manifest in such a way as to create the force by some other means. I'm honestly not sure!
What we can (and do as of fairly recently I believe) use superconductors for in motors is to generate the magnetic fields. Superconductors, having zero resistance, are an extremely efficient way to generate powerful magnetic fields - Because they can carry HUGE amounts of current ;D and that is with, again according to the formula, no power loss!! ;D
Quote
also does anyone know any good bcs explanations? a youtube video or explanation on line would be great thankyou
I wrote a guide on Superconductors here - The BCS theory is really tough to understand just from reading though. Not sure of any good video explanations - I reckon some of the Notes in the free notes section would have good diagrammatical ones! Otherwise, a handy Youtube search will set you straight I am sure ;D
Edit: Rathin made some recommendations here ;D
Post by: smile123 on April 26, 2017, 03:19:10 pm
Post by: jamonwindeyer on April 26, 2017, 03:20:24 pm
HELP PLEASE!!! :)
Hey smile123! This definitely isn't a question you need to know how to do in HSC Physics - Where did you find it? ;D
Post by: Rathin on April 26, 2017, 09:11:37 pm
Post by: jamonwindeyer on April 26, 2017, 09:22:15 pm
How to do Part C?
Hey Rathin! Can't quite step through the Math right this second, but I reckon you'd be okay with a breakdown:
- The mass being placed puts a force on the coil due to its gravitational acceleration. Call this \(F_m\), and \(F_m=mg\).
- This force is applied tangentially to the rotation of the coil, so produces a torque. \(\tau=F_md=mgd\) where \(d\) is the 'radius' of the coil, the distance from the centre, 0.4 metres.
- You need to produce an equal and opposite torque to balance the torque caused by the mass, and thus hold the mass in place. Torque due to a motor is \(\tau=BAIn\) (the coil is parallel to the field so we don't need the cosine term, \(\cos{90}=1\)). So you are solving the following equation for \(I\):
You have everything you need in terms of values there I think! Hope this helps ;D
Post by: teapancakes08 on April 28, 2017, 10:16:23 pm
Negatively charged latex spheres are introduced between two charged plates and are held stationary by the electric field. Each sphere has a mass of 2.4 × 10^−12 kg and the strength of the field required to counter their weight is 4.9 × 10^7 NC^−1. Sketch this arrangement, identifying the positive and the negative plate, and determine the charge on the spheres.
I drew the diagram (I think...), but I don't know which formula to start with given the information... :-\
Post by: pikachu975 on April 29, 2017, 02:15:41 am
To be honest, I have no idea how to approach this question:
Negatively charged latex spheres are introduced between two charged plates and are held stationary by the electric field. Each sphere has a mass of 2.4 × 10^−12 kg and the strength of the field required to counter their weight is 4.9 × 10^7 NC^−1. Sketch this arrangement, identifying the positive and the negative plate, and determine the charge on the spheres.
I drew the diagram (I think...), but I don't know which formula to start with given the information... :-\
So basically you draw horizontal parallel plates first and a sphere in between it (as it is falling down). Now, you consider gravitational force because it is falling down, F = mg. The electric plates HOLD IT STATIONARY as given, so it must be attracted to the top plate. Since the negatively charged spheres will attract to the positive plate, the positive plate is up, because it would counter the downward force due to gravity.
F = mg
F = 2.4x10^-12 x 9.8
F = 2.352 x 10^-11 N
Now to counter this downward force, the electric field exerts the same force upwards. We can now just sub into E = F/q to find the sphere's charge.
E = F/q
q = F/E
q = 2.352x10^-11 / 4.9x10^7
q = 4.8 x 10^-19
Therefore the top plate is positive, bottom plate is negative, and the sphere's charge is -4.8x10^-19 (negative because it's given).
Hope this helped!
Post by: arunasva on April 29, 2017, 04:15:42 pm
Post by: jamonwindeyer on April 29, 2017, 04:40:56 pm
How are triodes exactly used for amplification ?
Hey! This is a really common question, and the way I always answer is, don't worry. It is a really complicated piece of Physics, I only learned it fairly recently in Electrical Engineering and it's still something I struggle with, have to rewatch videos or rescan notes a lot of the time.
In the HSC, they'll never assess you beyond knowing that it can be used for amplification. They'll never ask how :)
This is the best video I've ever seen that explains how a transistor works as a SWITCH (not an amplifier), and it covers a lot of HSC relevant material too!
Post by: itssona on April 30, 2017, 10:31:23 am
so when a car accelerates, we push on the accelerator, and the driving force increases, and because the driving force is more than the resistive forces, there is a net force to accelerate it?
alsooo, when a car slows down- we push the brakes, the friciton from the brakes slows the wheels? im confused because (its hard to word it) but like, what about driving force? does driving force become zero when braking? and does driving force basically control the speed of the wheels? :/
sorry for writing in such a messy way, any help is appreciated :)
Post by: gilliesb18 on May 01, 2017, 10:00:49 am
Thanks in advance!!!
Sam is on his push bike riding at 4 m/s. He comes across a huge hill, with a gradient of 50 degrees, and realises his breaks don’t work. He goes speeding down the hill in a panic for and minute and two seconds, until he comes face to face with a brick wall. If he, plus his bike has a mass of 70 kg, and there was a friction force of 113N, then what was his velocity before he hit the wall?
Post by: jamonwindeyer on May 01, 2017, 11:19:18 am
Hello- Can someone please help me with this one??? I've had someone explain it to me already :-[ (not on this forum) but for some reason i forgot how to do it!!!
Thanks in advance!!!
Sam is on his push bike riding at 4 m/s. He comes across a huge hill, with a gradient of 50 degrees, and realises his breaks don’t work. He goes speeding down the hill in a panic for and minute and two seconds, until he comes face to face with a brick wall. If he, plus his bike has a mass of 70 kg, and there was a friction force of 113N, then what was his velocity before he hit the wall?
Hey gillies! Welcome to the forums! ;D
This is a really tough question for HSC level! But here we go ;D
So we are considering forces along the direction of the hill. So, we need to see what component of gravity will act along the hill (picture a vector at 53 degrees to the horizontal, then another perpendicular to that forming a right angled triangle with gravity as the hypotenuse. We'll call this \(F_h\):
Now, we have a frictional force acting against this, so the total force acting down the hill:
Now this force acts for 1 minute and two seconds, starting at a velocity of 4 metres per second. We know the acceleration will be given by:
So we can use the velocity formula:
Now I may have stuffed up some maths above (I'm procrastinating in a lecture, lol) but I get about 370 metres per second, exceeding the speed of sound. This doesn't quite seem sensible, but the question seems a little 'humorous' so maybe! If it is the answer, Sam is... Not in a good state ;)
Does this make sense?
Post by: Maraos on May 01, 2017, 05:13:26 pm
I've got a question regarding one of the physics syllabus dot points (3.2.1)
What would be the best example to this dot point: "Plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current."
Could you speak about the experiment that involves moving a magnet in and out of a solenoid that is hooked up to a galvanometer. Because aren't you technically producing an alternating current?
Post by: jakesilove on May 01, 2017, 06:04:43 pm
Hello! :)
I've got a question regarding one of the physics syllabus dot points (3.2.1)
What would be the best example to this dot point: "Plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current."
Could you speak about the experiment that involves moving a magnet in and out of a solenoid that is hooked up to a galvanometer. Because aren't you technically producing an alternating current?
Yep, you're 100% right; that's the perfect practical task to use! The Galvanometer clearly shows that the current changes directions, so you've produced AC current :)
Post by: Maraos on May 01, 2017, 06:53:06 pm
Yep, you're 100% right; that's the perfect practical task to use! The Galvanometer clearly shows that the current changes directions, so you've produced AC current :)Ah okay thanks for the clarification! :)
Post by: gilliesb18 on May 02, 2017, 10:08:40 am
I've been working it out, but is there any way that you could upload a triangle diagram of it??? Is that asking waaay too much??
Thanks heaps....
Post by: jamonwindeyer on May 02, 2017, 10:11:47 am
Thanks Jamon, that definitely helps....
I've been working it out, but is there any way that you could upload a triangle diagram of it??? Is that asking waaay too much??
Thanks heaps....
Definitely can do that!! I'll draw one up soon when I get a chance! ;D
Post by: itssona on May 02, 2017, 10:16:51 am
what happens to the 2m/s though? I dont get how the 2 even gets involved argh
please answer within 3 hours since my exam is then aha, thank you guys :)
Post by: jamonwindeyer on May 02, 2017, 10:44:45 am
heeey for this question (question 10), would we do mass (72) x 9.8 and that is weight force going down, so that means the force going up is also this?
what happens to the 2m/s though? I dont get how the 2 even gets involved argh
please answer within 3 hours since my exam is then aha, thank you guys :)
Hey! You are right on the money - If we have a constant velocity that means no net force, so the force upwards needs to balance the force downwards (which is why we use an acceleration of 9.8) - The velocity actually doesn't come into the question! Velocity is constant - We only need to apply enough force to balance the weight force, the velocity is a red herring
Post by: itssona on May 02, 2017, 10:52:12 am
Hey! You are right on the money - If we have a constant velocity that means no net force, so the force upwards needs to balance the force downwards (which is why we use an acceleration of 9.8) - The velocity actually doesn't come into the question! Velocity is constant - We only need to apply enough force to balance the weight force, the velocity is a red herringahhh thank you Jamon! Ugh I'll have to make it a point to be cautious of such red herrings! XD
Post by: Rathin on May 02, 2017, 11:06:57 am
Post by: jamonwindeyer on May 02, 2017, 01:15:29 pm
Thanks Jamon, that definitely helps....
I've been working it out, but is there any way that you could upload a triangle diagram of it??? Is that asking waaay too much??
Thanks heaps....
(http://i.imgur.com/husFVLP.jpg)
Post by: jamonwindeyer on May 02, 2017, 01:24:53 pm
Help with ii) please
Sure! So we know that the induced emf is:
Now \(\Phi=BA\), so the change in emf is completely dependent on change in area (since time and field strength are not changing). But area is given by \(A=\pi r^2\), so, if the radius is decreasing, you don't have a constant rate of decrease in area. Now, the rate of change in area is actually linear.
You know Calculus, so we can actually (optionally) consider it this way:
So, the area decreases more slowly as \(r\) decreases. You start with a higher induced EMF, then linearly decrease to a lower value (immediately dropping to zero after the time has elapsed) :) this is a tough one, does it make sense? :)
Post by: Bubbly_bluey on May 02, 2017, 02:02:21 pm
Post by: jakesilove on May 02, 2017, 02:45:34 pm
Hello! Just curious but why does the fluorescence emitted by cathode rays green?
No particular reason here; the film itself would be designed such that, when the lattice is excited (ie. by having electrons hit it), it glows green! It could probably be set up to glow blue if you wanted to.
Post by: gilliesb18 on May 02, 2017, 05:42:23 pm
Post by: Sukakadonkadonk on May 02, 2017, 07:16:19 pm
Does anyone know what the difference is between 1 rayl and 1 kg m^-2 s^-1
in terms of acoustic impedance? Can they be used interchangeably?
or is 1 rayl just equal to 10^6 1 kg m^-2 s^-1
Thanks.
Post by: Aaron12038488 on May 02, 2017, 07:47:42 pm
Post by: cxmplete on May 02, 2017, 07:51:34 pm
Does anyone know how to write a 6 mark response for this hsc question, because I'm pretty stuck rn?
(and no I don't have the success one book)
Post by: Sukakadonkadonk on May 02, 2017, 07:52:49 pm
is nanometers x 10-6, whats 10*-9?
Millimetres is x10^-3
Micrometres is x10^-6
Nanometres is x10^-9
Post by: scienceislife on May 02, 2017, 08:43:59 pm
Post by: jamonwindeyer on May 02, 2017, 10:56:32 pm
Hey Guys,
Does anyone know what the difference is between 1 rayl and 1 kg m^-2 s^-1
in terms of acoustic impedance? Can they be used interchangeably?
or is 1 rayl just equal to 10^6 1 kg m^-2 s^-1
Thanks.
Hey! They are interchangeable - The rayl is just a more convenient name for that unit of measure, \(kgs^{-1}m^{-2}\) - Kind of like how the Newton is interchangeable with \(kgms^{-2}\) (but of course we always use Newton, and just the same, we usually use Rayl, purely because it is easier) ;D
Post by: jamonwindeyer on May 02, 2017, 11:03:16 pm
Hi,
Does anyone know how to write a 6 mark response for this hsc question, because I'm pretty stuck rn?
(and no I don't have the success one book)
Hey! So the sorts of things to cover here would be:
- The nature of p-type and n-type semiconductors, specifically the majority charge carriers for each. Just briefly, enough to explain how the diode works.
- What happens when you bring the PN junction together, specifically the formation of the depletion layer. Most of the response will probably be here.
- The photoelectric effect, and what happens when light shines on the P-type semiconductor
- The net result of the current flow from the above step
There's a lot to understand here, be sure to let us know if there's anything you don't quite understand :)
Post by: jamonwindeyer on May 02, 2017, 11:06:17 pm
How was a standing wave produced by Hertz in his experiment?
Hey! Basically, he directed the radio waves at a reflecting screen. When the waves reflected, they superimposed with the waves being sent to the screen. You don't really need to know the specifics of how, but basically this causes the waves to sort of, partially 'cancel' each other out, and you get the formation of a standing wave as a result! :)
Post by: gilliesb18 on May 03, 2017, 05:16:57 pm
Consider the object shown in the diagram. Several forces act on this object which is at rest on a smooth, horizontal surface. Calculate the net force on the object.
BTW if the left force is too hard to read, its 26 N
Post by: jamonwindeyer on May 03, 2017, 05:28:31 pm
Ak so another question on forces....
Consider the object shown in the diagram. Several forces act on this object which is at rest on a smooth, horizontal surface. Calculate the net force on the object.
BTW if the left force is too hard to read, its 26 N
Hey gillies! So the total force on this object will be \(26-6=20N\) to the right, and \(15N\) upwards. We use those two values to form a right angled triangle, with 20 on the horizontal and 15 on the vertical. The magnitude of the force will be the hypotenuse:
The direction of the force can be found by using simple right angled trigonometry:
This will cause the object to accelerate in that direction ;D does this make sense? :)
Post by: gilliesb18 on May 04, 2017, 10:16:53 am
a) How come loud noises from bass loud speakers can cause windows to rattle??
b) Solar hot water systems heat water
c) waves at the beach can knock you over
d) Some public telephones are powered by photovoltaic cells (whatever they are...)
Thanks!!!!
BG
Oh and thanks heaps for the last reply Jamon!!! It helps heaps!!!
Mod Edit: Post merge :)
Post by: jamonwindeyer on May 04, 2017, 10:49:30 am
Can some one help me on a few wave questions??
a) How come loud noises from bass loud speakers can cause windows to rattle??
b) Solar hot water systems heat water
c) waves at the beach can knock you over
d) Some public telephones are powered by photovoltaic cells (whatever they are...)
Thanks!!!!
BG
Hey! So let's see:
a) I'm reasonably certain that what is happening could be for two reasons. First, human hearing gets less sensitive at lower frequencies. To hear it, you need a more intense sound wave and thus more air pressure. So it is possible that you are getting vibrations purely due to that greater air pressure. More likely though is that the frequency happens to correspond to the resonant frequency of the object.
All objects have a resonant frequency, which is a frequency they respond to more readily than others. Picture a park swing - You know how you settle into the rhythm? That's the resonant frequency of that swing. You can try and swing faster or slower, but it never works quite as well! Indeed, if you expose an object to sounds of its resonant frequency, then it will vibrate far more strongly at that frequency (note, objects always vibrate and rattle when exposed to sound, but we never perceive it unless it is strong).
This idea of resonant frequency is something that engineers have to take very seriously, because all objects do it. If you've not seen it, this video shows what ignoring resonant frequency can do to what you build.
b) Yes, they do! The workings of a solar panel specifically are reasonably complex, but oncoming light waves are converted into electrical energy, which is then converted into heat to radiate into the water!
c) Yes, they can! They carry energy/momentum, its a mass of water moving with a velocity, so there is kinetic energy and momentum there. Get hit, get bowled over if the force is large enough :)
d) Photovoltaic Cells = Solar Cells ;D
Post by: bsdfjnlkasn on May 04, 2017, 06:54:34 pm
With the I2I outcome which focuses on whether cathode rays were charged particles or electromagnetic waves, who are the main profiles we need to know to accurately detail the development of the theory? So far I've got Hertz and Thomson but elsewhere I've read about Varley, Stoney, Goldstein and Helmholtz. Did these four make notable contributions that would be worth discussing in a question if one were to come up? The outcome doesn't exactly detail the historical development so it would be nice to have some confirmation as to where to head with the info for this outcome.
Any help would be greatly appreciated!
Post by: jakesilove on May 04, 2017, 07:17:49 pm
Hey there,
With the I2I outcome which focuses on whether cathode rays were charged particles or electromagnetic waves, who are the main profiles we need to know to accurately detail the development of the theory? So far I've got Hertz and Thomson but elsewhere I've read about Varley, Stoney, Goldstein and Helmholtz. Did these four make notable contributions that would be worth discussing in a question if one were to come up? The outcome doesn't exactly detail the historical development so it would be nice to have some confirmation as to where to head with the info for this outcome.
Any help would be greatly appreciated!
Hey! You don't actually need to know the personalities of the story; just the experiments here. So, know a few experiments that were used to 'show' that electrons were particles, a few experiments that were used to 'show' that electrons were waves, and the definitive experiments that prove it one way or the other!
Post by: jamonwindeyer on May 04, 2017, 07:22:10 pm
Hey there,
With the I2I outcome which focuses on whether cathode rays were charged particles or electromagnetic waves, who are the main profiles we need to know to accurately detail the development of the theory? So far I've got Hertz and Thomson but elsewhere I've read about Varley, Stoney, Goldstein and Helmholtz. Did these four make notable contributions that would be worth discussing in a question if one were to come up? The outcome doesn't exactly detail the historical development so it would be nice to have some confirmation as to where to head with the info for this outcome.
Any help would be greatly appreciated!
Hey! As Jake said, you'll never need the people that conducted the research into these properties, you only really need the properties themselves. But just adding that Hertz and Thomson are good to know because they link to other dot points, I wouldn't worry about remembering the rest either ;D
Post by: bsdfjnlkasn on May 04, 2017, 07:34:26 pm
I'm just looking at the formula F = qvBsina and am still unsure of what the designated angle is, perhaps a diagram might help but I understand if it'd be too hard to sketch so a written explanation is cool 8). Also, my textbook says that if a charged particle enters a uniform magnetic field at 90°, it will move in the arc of a circle. Is there a way to see if the charge enters at 90° or will we always be told this? Also, I don't entirely understand why this would happen, even if it's a minor detail, I would like to know why as it'll probably stick better that way.
Thank you!!
Post by: jamonwindeyer on May 04, 2017, 07:51:57 pm
Awesome, thanks for the help!
I'm just looking at the formula F = qvBsina and am still unsure of what the designated angle is, perhaps a diagram might help but I understand if it'd be too hard to sketch so a written explanation is cool 8). Also, my textbook says that if a charged particle enters a uniform magnetic field at 90°, it will move in the arc of a circle. Is there a way to see if the charge enters at 90° or will we always be told this? Also, I don't entirely understand why this would happen, even if it's a minor detail, I would like to know why as it'll probably stick better that way.
Thank you!!
Hey! So the diagram of the angle is attached below, it is the angle between the field lines and the direction of movement ;D
(http://i.imgur.com/dgFAiC6.png)
You'll always be able to see that angle, if you aren't told it. It is very often 90 degrees ;D usually it is the field going into or out of the page - If that happens, if the particle is moving left/right/up/down, or in any direction around the page, the velocity is always perpendicular to the field :)
The reason the particle moves in a circle if it enters at 90 degrees, is because the force on the object stays perpendicular to its motion. That's just like orbits when gravitational force is perpendicular to orbital velocity. What happens in that circumstance? It is uniform circular motion! ;D
Post by: Sukakadonkadonk on May 05, 2017, 10:00:40 pm
Does anyone know what the voltage time graph looks like for an AC generator which is turned twice as fast?
Thanks.
Post by: jamonwindeyer on May 05, 2017, 10:04:45 pm
Hi,
Does anyone know what the voltage time graph looks like for an AC generator which is turned twice as fast?
Thanks.
Hey! You'll get a similar looking (co)sine wave, but the period will be halved (because it is spun twice as quickly), and the amplitude will double, because the rate of change of magnetic flux is also doubled :)
Post by: scienceislife on May 06, 2017, 07:49:48 am
Post by: beau77bro on May 06, 2017, 02:07:13 pm
Exactly correct on all counts ;D
Hmm, that's an interesting question! I believe that using a superconductor for the coil wouldn't actually achieve the desired effect, because superconductors exclude magnetic flux. This means that you don't get forces on current carrying wires in a magnetic field - Because the current doesn't interact with the magnetic field. I could be wrong there - Perhaps the Meisner effect would manifest in such a way as to create the force by some other means. I'm honestly not sure!
What we can (and do as of fairly recently I believe) use superconductors for in motors is to generate the magnetic fields. Superconductors, having zero resistance, are an extremely efficient way to generate powerful magnetic fields - Because they can carry HUGE amounts of current ;D and that is with, again according to the formula, no power loss!! ;D
I wrote a guide on Superconductors here - The BCS theory is really tough to understand just from reading though. Not sure of any good video explanations - I reckon some of the Notes in the free notes section would have good diagrammatical ones! Otherwise, a handy Youtube search will set you straight I am sure ;D
Edit: Rathin made some recommendations here ;D
OMG Im so sorry i missed this, thankyou sooo much jamon awesome explanation and im gonna look at the link right now thankyou soooo much
Post by: bsdfjnlkasn on May 06, 2017, 03:28:56 pm
I'm just trying to think of an explanation (since I can't find one) as to why electrons in a discharge tube accelerate/decelerate according to an increasing/decreasing electric field around them. Is this because of a concept similar to electric flux? Or should I just accept that this is how striations are explained in the cathode ray experiment.
Any help would be super appreciated, thank you!!
Post by: jamonwindeyer on May 06, 2017, 04:08:28 pm
Hey sorry if this is really silly, but will a moving charge in a changing electric field experience a force?
I'm just trying to think of an explanation (since I can't find one) as to why electrons in a discharge tube accelerate/decelerate according to an increasing/decreasing electric field around them. Is this because of a concept similar to electric flux? Or should I just accept that this is how striations are explained in the cathode ray experiment.
Any help would be super appreciated, thank you!!
Hey! Never a silly question ;D
Moving charges experience a force due to an electric field just the same as they would if stationary. This also goes for if the field is changing. So, according to \(F=Eq\), if the field doubles in strength, the force doubles in strength. It is a direct relationship, it's different to something like induction where the change itself (or the rate of change) is what causes the forces involved. Here, it is just the electric field interacting with charge (be it stationary or moving). If field gets bigger, force gets bigger ;D
Exactly what scenario in the cathode ray section is confusing you? Is it the deflection plates/coils? :)
Post by: jamonwindeyer on May 06, 2017, 04:11:38 pm
What were some of the errors present in Hertz's experiment and how did he reduce them?
Hey! I don't think I'm aware of any errors in the experiment besides inaccuracy of measuring equipment and basic human error, both of which you resolve just by being careful ;D
That question (if it is a question from an exam) could be indirectly asking about the calculation of the velocity of the radio waves. To be most accurate, Hertz set up a reflected standing wave and measured the distance between subsequent nodes/anti-nodes (essentially, measured the wavelength of the wave). He could use this, combined with the known frequency of his transmitter, to determine velocity with the wave formula \(v=\lambda f\), and this is much more accurate than measuring with the tools that he had access to :)
Would love for others to chime in if they have more
Post by: beau77bro on May 06, 2017, 04:55:35 pm
We just had this question in the half yearly. What is the answer and why Idgi. Thankyou very muchhhh
Post by: bsdfjnlkasn on May 06, 2017, 05:51:39 pm
(http://uploads.tapatalk-cdn.com/20170506/2527de8fce985dbf7d3768d95d51915c.jpg)
We just had this question in the half yearly. What is the answer and why Idgi. Thankyou very muchhhh
I think the answer might be A) as it's the only one which obeys the law of conservation of energy. Considering that you can't create nor destroy energy, you know that that as one form increases another must decrease to ensure that the total energy in a system stays the same. So let's apply this to the problem. When the rocket is sitting on the launch pad, it's KE = 0 as it's not moving. We're told that the rocket rises with constant thrust so we we're looking for linear relationships. The faster the rocket goes, the more kinetic energy it has meaning that the potential energy (gravitational in this case) will have to decrease as the total energy in the system has to remain constant. You can show this with the first graph by superimposing the two lines. You will end up with a horizontal line at the height of KE @ t=0 i.e. a constant amount of energy. This obeys the law of conservation of energy unlike any of the others which means a) is our answer.
Hopefully this helped! :)
Post by: bsdfjnlkasn on May 06, 2017, 07:27:04 pm
Describe what happens when electrons hit the screen of a CT. What then happens to the electrons?
I can't seem to find any information on this in my textbook so was wondering if I could get an explanation. My first answer was that the electrons would hit the fluorescent screen and as a result of the collision emit light (energy transfer: Ek to heat and light). This also sort of explains why increasing the frequency of admitted electrons corresponds to a brighter on-screen display.
And now, with following the electrons after they've hit the screen, do they just you know disappear? I suspect the energy used up in the collision is the same amount as their kinetic energy and I might be missing a link here, but are the electrons effectively reduced to this light energy? I just don't know where they'd go haha, like collect at the bottom of the fluorescent screen? Idk it just seems silly thinking about it that way.
Any input would really help me clarify this problem and if someone could advise me how important this detail is, that'd be a good way to put things into perspective as I don't really see this as being relevant to I2I but :P
Thank you!! ;D
Post by: bsdfjnlkasn on May 06, 2017, 07:43:32 pm
Hey! Never a silly question ;D
Moving charges experience a force due to an electric field just the same as they would if stationary. This also goes for if the field is changing. So, according to \(F=Eq\), if the field doubles in strength, the force doubles in strength. It is a direct relationship, it's different to something like induction where the change itself (or the rate of change) is what causes the forces involved. Here, it is just the electric field interacting with charge (be it stationary or moving). If field gets bigger, force gets bigger ;D
Exactly what scenario in the cathode ray section is confusing you? Is it the deflection plates/coils? :)
Hey Jamon, thank you! That makes things so much simpler than M&G :D
We recently covered the discharge tubes prac and had to observe how the different gas pressures influenced the ray produced. One of the features we looked at was striations which appear as pressure is reduced - this made me wonder if we had to know specific explanations for why these occurred. Looking online, I found that they were due to changing electric field strengths (which I now understand to be the result of varying charge concentrations around the discharge tube), but if this isn't relevant to the syllabus then don't worry about my side comment from the previous post haha.
But as you mentioned deflection plates/coils, will we need to know how changing the voltage on the X and Y plates will influence the degree of deflection? I haven't heard of the use of coils in CRTs yet, so if you're free to, could you please offer some resources that do a good job of explaining how coils are involved (or if you have time, possible write something short up - whatever suits, either is awesome :) ).
Thanks again for all your help so far! :) :)
Post by: beau77bro on May 06, 2017, 11:02:51 pm
I think the answer might be A) as it's the only one which obeys the law of conservation of energy. Considering that you can't create nor destroy energy, you know that that as one form increases another must decrease to ensure that the total energy in a system stays the same. So let's apply this to the problem. When the rocket is sitting on the launch pad, it's KE = 0 as it's not moving. We're told that the rocket rises with constant thrust so we we're looking for linear relationships. The faster the rocket goes, the more kinetic energy it has meaning that the potential energy (gravitational in this case) will have to decrease as the total energy in the system has to remain constant. You can show this with the first graph by superimposing the two lines. You will end up with a horizontal line at the height of KE @ t=0 i.e. a constant amount of energy. This obeys the law of conservation of energy unlike any of the others which means a) is our answer.
Hopefully this helped! :)
hmmmm the answer is D in this book, but thankyou bsdfjnlkasn i get where you coming from. ok so i had this in my half yearly, and the answer was B, but in this textbook the answer is D, so which is it and why? -- there is a lot of dispute in my class and im arguing for D, because it escapes so it's Ep is 0 eventually and Ep is always negative, but i really need a solid answer explaining the features any help appreciated.
Post by: Zainbow on May 07, 2017, 12:16:45 am
hmmmm the answer is D in this book, but thankyou bsdfjnlkasn i get where you coming from. ok so i had this in my half yearly, and the answer was B, but in this textbook the answer is D, so which is it and why? -- there is a lot of dispute in my class and im arguing for D, because it escapes so it's Ep is 0 eventually and Ep is always negative, but i really need a solid answer explaining the features any help appreciated.
I'll try to explain this.
Ok so, we know that potential energy increases the further you move away from the planet, and, as you said, it increases to zero. This would make options A and B incorrect, because both their Ep graphs are decreasing. This leaves options C and D, and we have to decide whether the relationship is linear or parabolic. Given the kinetic energy formula, Ek=1/2 mv^2 where v is squared, and knowing that velocity increases due to the rocket accelerating, the graph must be increasing parabolically. Hence, the answer must be D.
I don't know if this makes sense (or if it's right), but I hope it helps.
Post by: jamonwindeyer on May 07, 2017, 12:50:55 am
hmmmm the answer is D in this book, but thankyou bsdfjnlkasn i get where you coming from. ok so i had this in my half yearly, and the answer was B, but in this textbook the answer is D, so which is it and why? -- there is a lot of dispute in my class and im arguing for D, because it escapes so it's Ep is 0 eventually and Ep is always negative, but i really need a solid answer explaining the features any help appreciated.
I'll try to explain this.
Ok so, we know that potential energy increases the further you move away from the planet, and, as you said, it increases to zero. This would make options A and B incorrect, because both their Ep graphs are decreasing. This leaves options C and D, and we have to decide whether the relationship is linear or parabolic. Given the kinetic energy formula, Ek=1/2 mv^2 where v is squared, and knowing that velocity increases due to the rocket accelerating, the graph must be increasing parabolically. Hence, the answer must be D.
I don't know if this makes sense (or if it's right), but I hope it helps.
Throwing my answer behind Zainbow here, the answer is definitely D ;D further, if a rocket has a constant thrust, the acceleration will increase over time as the mass of the rocket decreases (due to lost fuel) - This means velocity will be increasing at an increasing rate, which does more to explain the accelerated increase shown in the final graphs ;D
Beyond that, \(E_p\) must be negative and is increasing as time passes - That's the big thing to spot ;D
I think the answer might be A) as it's the only one which obeys the law of conservation of energy. Considering that you can't create nor destroy energy, you know that that as one form increases another must decrease to ensure that the total energy in a system stays the same. So let's apply this to the problem. When the rocket is sitting on the launch pad, it's KE = 0 as it's not moving. We're told that the rocket rises with constant thrust so we we're looking for linear relationships. The faster the rocket goes, the more kinetic energy it has meaning that the potential energy (gravitational in this case) will have to decrease as the total energy in the system has to remain constant. You can show this with the first graph by superimposing the two lines. You will end up with a horizontal line at the height of KE @ t=0 i.e. a constant amount of energy. This obeys the law of conservation of energy unlike any of the others which means a) is our answer.
Hopefully this helped! :)
I love your way of thinking here, but only one problem - We inject energy into the system from the fuel. So the Conservation of Energy need not apply, at least not without considering the potential chemical energy in the fuel of the rocket ;D
Post by: jamonwindeyer on May 07, 2017, 01:09:27 am
We recently covered the discharge tubes prac and had to observe how the different gas pressures influenced the ray produced. One of the features we looked at was striations which appear as pressure is reduced - this made me wonder if we had to know specific explanations for why these occurred. Looking online, I found that they were due to changing electric field strengths (which I now understand to be the result of varying charge concentrations around the discharge tube), but if this isn't relevant to the syllabus then don't worry about my side comment from the previous post haha.
Definitely totally unnecessary to understand at the HSC level ;D I know Sydney University has a document going into specifics, and perhaps that is what you used, but it is definitely unnecessary to consider it to this level of depth in the HSC (interesting read though, if you haven't already) ;D
For the HSC, you need only know that the different striation patterns are caused by differing air pressures, as it changes the amount of particles in the tube for the electrons to collide with. Speaking roughly, lower air pressures produce less prominent striations for exactly this reason - Less obstructions, and so, less collisions. The colour and nature of the striations can also depend on the gas inside the tube. Understanding these points, and perhaps (even this isn't really necessary but good to know if you can) being able to identify different parts of the patterns (dark spaces and the like), is all you need :)
Quote
But as you mentioned deflection plates/coils, will we need to know how changing the voltage on the X and Y plates will influence the degree of deflection? I haven't heard of the use of coils in CRTs yet, so if you're free to, could you please offer some resources that do a good job of explaining how coils are involved (or if you have time, possible write something short up - whatever suits, either is awesome :) ).
Yes you are, but qualitatively, not quantitatively. Basically, just the knowledge that plates and/or coils are used to deflect an electron beam in a Cathode Ray Television/Oscilloscope is what is required. Exactly how the beam is deflected depends on application:
For a CRO, we need the horizontal axis to represent time, and the vertical to represent some input signal. So, the horizontal deflection plates are set up with a time varying sawtooth wave, which sweeps the beam from the left side of the screen to the right, then immediately jumping back to the starting point to repeat. The vertical deflection plates are connected to an input voltage, so the amount of vertical deflection is completely dependent on your input. The net effect?
(https://i.makeagif.com/media/5-29-2016/jtSBFk.gif)
For a television, we need to sweep the electron beam across the screen. So both vertical/horizontal plates are time varying, with the periods adjusted such that the beam will touch every point on the screen, by zig zagging left and right as it moves up and down.
Hopefully this helps! Let me know if you need anything clarified :)
Post by: katnisschung on May 07, 2017, 03:49:11 pm
in coils in a solenoid increases the magnetic field strength with current kept constant?
thanks :)
Post by: jamonwindeyer on May 07, 2017, 05:15:12 pm
doing some revision questions and i came to wonder what the theory is for how the increase
in coils in a solenoid increases the magnetic field strength with current kept constant?
thanks :)
Hey! So if you have a current carrying wire, it is surrounded by a magnetic field. More coils means more wire, more wire means more field! It's that simple ;D
Post by: beau77bro on May 07, 2017, 05:21:17 pm
Post by: jamonwindeyer on May 07, 2017, 05:25:12 pm
hey so i have an assessment on processing skills and im not entirely sure on what that means. i wanna prepare as best as i can and i get that it's to do with analysing data but thats about it, my teacher said we dont need to memorise experiments or anything but what should i be doing to prepare and what exactly are processing skills other than graphs and stuff.
Hey! So I'd interpret processing skills to be things like:
- Knowing what data means when it is presented to you (EG - identifying independent/dependent variable, etc)
- Being able to analyse data (calculate averages, graph, identify relationships, calculate values based on theory) and draw conclusions from that data
- Discussing significance of the data, exploring sources of error, and other similar things
Preparing for this stuff is really tough - Brush up on practical tasks you've done thus far, particularly focusing on the data/analysis sections over the experimental method itself. Maybe ask your teacher if there is a sample task from a previous year you could look at? :)
Post by: Aaron12038488 on May 07, 2017, 08:14:36 pm
Post by: Sukakadonkadonk on May 07, 2017, 08:24:36 pm
So I just wanted to ask:
If we were asked to graph some information from a given table which does not include data involving 0, do we still extrapolate the line graph to the (0,0) point? Because some of use got marked down for not extending the line. But then a bunch of us complained that it went against what we have been taught in the past so teachers gave us the mark back.
Now I am just confused at what to do in the future for these types of questions.
Another question:
When we are talking about piezoelectric crystals in a transducer, are they usually powered by an AC supply? Or an alternating voltage? I am receiving conflicting answers.
Thank you!! ;D
Post by: armtistic on May 07, 2017, 09:34:13 pm
I've been trying to teach myself Ideas to Implementation because my teacher did a shoddy job and I'm up to the set of dotpoints about black bodies. I can see why the classical ideas about the BB radiation curve didn't make sense but what I don't get is why and how Planck's suggestion that energy occurs in quanta solves this issue. Like what about the fact that energy occurs in packets rather than continuously explains why the graph peaks and falls?
Do we even have to know this?
Also how does the photoelectric effect occur? I've read some sources which say the EMR which strikes an atom causes it to oscillate and if it's charged then the movement of this charged atom releases EMR. I've read other sources which say photons in the EMR strike an electron in the atom and cause it to jump a band, and then this electron falls and releases EMR in doing so.
Post by: jamonwindeyer on May 07, 2017, 10:30:26 pm
a open-ended investigation is always a take home assignment?
Correct ;D
Post by: jamonwindeyer on May 07, 2017, 10:32:16 pm
Hi Guys,
So I just wanted to ask:
If we were asked to graph some information from a given table which does not include data involving 0, do we still extrapolate the line graph to the (0,0) point? Because some of use got marked down for not extending the line. But then a bunch of us complained that it went against what we have been taught in the past so teachers gave us the mark back.
Now I am just confused at what to do in the future for these types of questions.
Hey! So I was never sure either, I never did to be safe. However, if you look at the sample answer to Q21 in the 2015 Physics Exam, the sample answer extrapolated the line through the origin. So, my answer would be, extrapolate the line if you KNOW it should and will pass through the origin. If there is doubt, don't extrapolate ;D
Quote
Another question:
When we are talking about piezoelectric crystals in a transducer, are they usually powered by an AC supply? Or an alternating voltage? I am receiving conflicting answers.
AC supply and alternating voltage are exactly the same thing ;D
Post by: jamonwindeyer on May 07, 2017, 10:43:06 pm
Hey guys,
I've been trying to teach myself Ideas to Implementation because my teacher did a shoddy job and I'm up to the set of dotpoints about black bodies. I can see why the classical ideas about the BB radiation curve didn't make sense but what I don't get is why and how Planck's suggestion that energy occurs in quanta solves this issue. Like what about the fact that energy occurs in packets rather than continuously explains why the graph peaks and falls?
Do we even have to know this?
Also how does the photoelectric effect occur? I've read some sources which say the EMR which strikes an atom causes it to oscillate and if it's charged then the movement of this charged atom releases EMR. I've read other sources which say photons in the EMR strike an electron in the atom and cause it to jump a band, and then this electron falls and releases EMR in doing so.
Hey! Just before I answer, shout out to a whole bunch of short guides I wrote on the course last year. They might act as good little summaries for your self teaching ;D
So a few things for the Black Body Curve. The law that predicted the theoretical shape was called Rayleigh Jeans Law, but you don't need to know that law or WHY it predicted that specific curve. Just know that it did, and why it was an issue (the issue being even beyond not matching observation, that energy can't approach infinity for high frequencies, that makes no sense).
Putting energy in packets, with the energy per packet related to frequency, solves the ultraviolet catastrophe. Think of it like this. A BB releases a quanta due to some change inside the BB. An electron might fall down a band and release the lost energy as EMR, for example. So, the frequency of the emitted photon is directly related to the energy change that occurred in the BB, by \(E=hf\). Now, for a super high frequency photon of EMR, we need a huge energy change all in one go. This is rare. This explains the shape of the curve - At high frequencies, you need a huge energy change in the BB, and these just aren't as common as the smaller energy changes that characterise the big peak in the middle of the curve. The peak of the graph purely represents the frequency corresponding to the most common energy change in a BB of that temperature - This is called the characteristic wavelength. Basically, we get more intensity in the middle, because it is far more likely that an emitted photon falls in that range. More photons, more intensity - And that's where the graph comes from ;D
This is a tough concept - Happy to explain again if you need!
Photoelectric is a little simpler than your sources make it sound (at least for the HSC level) ;D A photon of light strikes an electron and gives its energy to that electron. If the extra energy is enough to break the electrons bonds with the atom, it escapes, with a kinetic energy equal to whatever is leftover from the energy of the photon. We can express this:
Here, \(\Phi\) is the work function of the specific metal, the amount of energy required to free an electron in that metal ;D
Post by: Bubbly_bluey on May 08, 2017, 01:46:29 pm
When a charged particle enters a magnet field, will it deflect in a circular path or a projectile (or is this only for electric field deflections)?
Thank you.
Post by: jamonwindeyer on May 08, 2017, 01:51:30 pm
Hi!
When a charged particle enters a magnet field, will it deflect in a circular path or a projectile (or is this only for electric field deflections)?
Thank you.
Hey! Charged particle in a magnetic field is always circular motion, because the force is perpendicular to the direction of travel at all times ;D
(In electric fields, yes, the force on the particle usually turns it into something similar to a projectile scenario, because the force is in the same direction no matter the motion of the particle) :)
Post by: itssona on May 09, 2017, 09:10:11 pm
So since voltage is an electric potential between two nodes - a battery has voltage because of the difference in electric potential between anode and cathode? ?
Alsoo how would you explain electric potential? Is it kinda like an electric force?
Thanks :)
Post by: jamonwindeyer on May 09, 2017, 09:19:10 pm
Heey am I right by saying this:
So since voltage is an electric potential between two nodes - a battery has voltage because of the difference in electric potential between anode and cathode? ?
Alsoo how would you explain electric potential? Is it kinda like an electric force?
Thanks :)
Hey! Yep, that is definitely correct about the battery ;D
There are proper formal, mathematical definitions of electric potential that you can explore at the tertiary level, but the one I normally go with for HSC students is this: Electric potential is analogous to the electric potential energy of a charged particle at some point in an electric field. Compare it to gravity - Gravitational potential is the energy possessed by an object due to its position in a gravitational field. Electric potential is the same thing, but instead of gravity, it's an electric field ;D
Post by: itssona on May 09, 2017, 09:48:18 pm
Hey! Yep, that is definitely correct about the battery ;D
There are proper formal, mathematical definitions of electric potential that you can explore at the tertiary level, but the one I normally go with for HSC students is this: Electric potential is analogous to the electric potential energy of a charged particle at some point in an electric field. Compare it to gravity - Gravitational potential is the energy possessed by an object due to its position in a gravitational field. Electric potential is the same thing, but instead of gravity, it's an electric field ;D
Omg!! That's the best analogy ever THANK YOU JAMON
Post by: Bubbly_bluey on May 09, 2017, 10:00:25 pm
Thanks :D
Post by: jamonwindeyer on May 09, 2017, 10:17:11 pm
Hi! :) I am having troubles doing the HSC Question in 2016 Q 30 b). (Sorry I can't post the question b/c for some reason it says the screenshot is too big to post. >:( idk). I wrote about how the circuit was incomplete so the mass would fall faster but Idk the other reason for difference in behaviour.
Thanks :D
Hey! So you've pretty much got it. When the circuit is complete, the generator is powering the lightbulb. As the coil spins in the magnetic field, electromagnetic induction causes an opposing torque. This is what is resisting the fall of the mass, because the coil can't spin as quickly (thus limiting the speed the rope can unwind).
When the switch is open, no generator. No opposing torque. No restriction, the mass falls freely under gravitational forces :)
So your answer would probably just explain these things (probably referencing Lenz's Law as well) ;D
Post by: beau77bro on May 11, 2017, 07:15:16 pm
Post by: jamonwindeyer on May 11, 2017, 08:26:48 pm
help the answer is 1.03 x 10^-2 Nm and i have no idea how
Hey! Formula for torque is:
Now the minimum torque is just when \(\theta=90\), the max is when \(\theta=0\). So we just ignore the cosine term:
Now \(N=1\), \(B=0.5\), \(I=1.5\). All we need is the area, which we can get from the radius:
Put it all together to find maximum torque (minimum should be zero):
Hmm, I can't match it either mate! Can anyone spot what we're missing? :)
Post by: arunasva on May 13, 2017, 09:00:48 pm
Also in BCS conduction, how are phonons produced ? Is it that the two positive lattices repel and then get attracted to the electron so it vibrates and hence sound energy ?
Mod Edit: Post merge :)
Post by: arunasva on May 13, 2017, 09:29:59 pm
Hey! This is a really common question, and the way I always answer is, don't worry. It is a really complicated piece of Physics, I only learned it fairly recently in Electrical Engineering and it's still something I struggle with, have to rewatch videos or rescan notes a lot of the time.Thanks
In the HSC, they'll never assess you beyond knowing that it can be used for amplification. They'll never ask how :)
This is the best video I've ever seen that explains how a transistor works as a SWITCH (not an amplifier), and it covers a lot of HSC relevant material too!
Post by: bsdfjnlkasn on May 13, 2017, 09:49:00 pm
We can detect light when our eyes received 2 x 10-17 J of energy. How many photons of light, of wavelength 550nm, is this?
Any help would be appreciated! (I'm sure it's easy, I just don't know where to start)
Thank you!
Post by: jakesilove on May 13, 2017, 10:48:51 pm
Im kinda confused help !!!! The answer's apparently B ?
Hey! This questions asks us to understand the difference between mass and force.
Mass is an absolute(ish) quantity, that will be the same no matter where you are. Weight force, on the other hand, is the mass of an object times the acceleration due to gravity.
So, on Mars, the weight force would be
On Earth, the weight force would be
However, the weight itself doesn't change. Just the weight FORCE!
In BCS conduction, how are phonons produced ? Is it that the two positive lattices repel and then get attracted to the electron so it vibrates and hence sound energy ?
Sort of! However, this is definitely way beyond the curriculum. All you need to know is that a phonon is produced due to vibration energy; nothing more!
Post by: jakesilove on May 13, 2017, 10:56:40 pm
Hey can I please get some help with the following (because I'm being expected to answer these questions before being taught ???)
We can detect light when our eyes received 2 x 10-17 J of energy. How many photons of light, of wavelength 550nm, is this?
Any help would be appreciated! (I'm sure it's easy, I just don't know where to start)
Thank you!
Hey! On the formula sheet, we are given the relation
Where E is the energy, h is Planck's Constant, and f is frequency. First, we need to convert wavelength into frequency.
Now, we can plug the frequency into our Energy equation
Finally, if we want 2*10^(-17)J,
Subject to mathematical error, but that's the method!
Post by: itssona on May 14, 2017, 11:42:30 am
alsooo for Alternating current, would you say that this is a good definition:
AC (Alternating Current) terminals periodically change their polarities. (I.e first terminal starts by being the positive one, then the other.) The AC terminals provide an alternating voltage. If a circuit is connected to the AC terminal, a current will be produced in which the charge carriers move backwards and forwards periodically.
is that basically what AC is?
thank you :)
Post by: Syndicate on May 14, 2017, 12:33:37 pm
why does electric potential energy for a positive charge decrease when its in the direction of an electric field??
alsooo for Alternating current, would you say that this is a good definition:
AC (Alternating Current) terminals periodically change their polarities. (I.e first terminal starts by being the positive one, then the other.) The AC terminals provide an alternating voltage. If a circuit is connected to the AC terminal, a current will be produced in which the charge carriers move backwards and forwards periodically.
is that basically what AC is?
thank you :)
I can answer question 1. So basically, the positive charge is being attracted towards the negative charge, since it's going in the direction of the electric field (positive to negative), it loses Potential Energy, as it moves with no repulsion (instead it gains kinetic energy). On the other hand, if the positive charge was forced to go against a repulsive force, it would gain P. E (think of it as a compressed spring).
Post by: bsdfjnlkasn on May 14, 2017, 02:52:35 pm
Can I get some help with the following:
A light beam has 270 photons passing a given point in one second. If this beam of light is rated as 2.23 x 10-16W, calculate the frequency of this light beam.
Wouldn't it just be 270Hz? I don't understand what the light rating means and if there's a formula we can use considering no wavelength was given ...
Could someone please explain the connection between photons and energy. I also don't see how the number of photons can be relevant (as it has come in previous questions). Like this has been really poorly explained and i'm really confused - like I don't even know how to ask my question. Hopefully someone can help
Thank you!!
Post by: jamonwindeyer on May 14, 2017, 05:33:45 pm
alsooo for Alternating current, would you say that this is a good definition:
AC (Alternating Current) terminals periodically change their polarities. (I.e first terminal starts by being the positive one, then the other.) The AC terminals provide an alternating voltage. If a circuit is connected to the AC terminal, a current will be produced in which the charge carriers move backwards and forwards periodically.
is that basically what AC is?
thank you :)
Yep you've got it! It's just reversing the polarity of voltage (and thus direction of current) at a regular interval (with the number of times it switches back and forth per second being the frequency of the source/signal) ;D
Post by: jamonwindeyer on May 14, 2017, 05:46:24 pm
Hey!
Can I get some help with the following:
A light beam has 270 photons passing a given point in one second. If this beam of light is rated as 2.23 x 10-16W, calculate the frequency of this light beam.
Wouldn't it just be 270Hz? I don't understand what the light rating means and if there's a formula we can use considering no wavelength was given ...
Could someone please explain the connection between photons and energy. I also don't see how the number of photons can be relevant (as it has come in previous questions). Like this has been really poorly explained and i'm really confused - like I don't even know how to ask my question. Hopefully someone can help
Thank you!!
Hey! Don't worry, this trips a lot of people.
When considering EM waves in terms of photons, you can't think of frequency in the usual "the number of waves passing by per second" sense, because it gets confusing. Instead, think of it more abstractly as purely a property of the photons.
Photons are 'packets' of electromagnetic waves, each containing a set amount of energy. That energy is equal to \(E=hf\) - Each photon has an energy proportional to the frequency! The frequency affects energy per photon.
Analysing your question should help clear up the confusion. We have 270 photons passing per second, and the laser is rated for \(2.23\times10^{-16}W\). Now remember, 1 Watt is equivalent to 1 Joule per second. This means that those 270 photons coming past per second, must be delivering a total of \(2.23\times10^{-16}\) joules of energy. To get the energy per photon, divide by 270:
Then, to find the frequency, we use the energy formula for a photon, since we know that each photon must have an energy equal to that we found just then:
So the frequency has nothing to do with photons passing per second - It is instead determined based on the energy per photon required to deliver the specified amount of energy per second (and thus power) :) does this make sense?
I wrote this brief guide on all this Quantum Physics stuff you do for the HSC!
Post by: itssona on May 15, 2017, 11:09:59 am
I can answer question 1. So basically, the positive charge is being attracted towards the negative charge, since it's going in the direction of the electric field (positive to negative), it loses Potential Energy, as it moves with no repulsion (instead it gains kinetic energy). On the other hand, if the positive charge was forced to go against a repulsive force, it would gain P. E (think of it as a compressed spring).
omg thank youuuu!!!
Post by: Bubbly_bluey on May 15, 2017, 12:32:09 pm
Also the graph that goes with it, intensity VS wavelength. I'm not completely understanding what it all means.
Basically the whole concept of black body radiation is confusing.:/
Thanks ;D
Post by: jamonwindeyer on May 15, 2017, 01:15:06 pm
Hi! I'm having trouble visualising what Black body radiation is. (like what it physically is). Are they just normal objects that can absorb all radiations?
Also the graph that goes with it, intensity VS wavelength. I'm not completely understanding what it all means.
Basically the whole concept of black body radiation is confusing.:/
Thanks ;D
Hey! I actually did a big explanation of the black body curves not long ago, you can find that here!
As for what a black body actually IS, it's a perfect absorber and emitter of radiation - Nothing bounces off, if anything hits it, it is absorbed into the body. Note that a perfect black body is a theoretical construct, all objects have some amount of incoming radiation that is reflected ;D
Post by: crazycodpro on May 15, 2017, 05:04:12 pm
Post by: jamonwindeyer on May 15, 2017, 08:58:01 pm
please help :)
Hey! This is a weird one, but force is just mass times acceleration due to gravity:
I think the dimensions are perhaps just red herrings? ;D
Post by: gilliesb18 on May 17, 2017, 12:29:24 pm
Also can someone give me an example of one dimensional, two dimensional, and three dimensional waves?
Thanks in advance.
B...
Post by: jamonwindeyer on May 17, 2017, 12:50:01 pm
Can someone help me understand just the start of the waves topic??? How do i differentiate between louder sound and quieter on a graph?
Also can someone give me an example of one dimensional, two dimensional, and three dimensional waves?
Thanks in advance.
B...
Hey! Louder sound means more energy in the wave, which corresponds to greater peaks in a transverse representation of your sound - See below!
(http://www.yamahaproaudio.com/global/en/Images/training_support_better_sound_p1_p01_4_l_en.jpg)
As for 1/2/3 dimensional waves, hopefully these examples suit!
1 Dimensional Sending a pulse along a spring/slinky (single dimension of movement, along the slinky)
2 Dimensional: Earthquake waves along the earth's surface (two dimensions, up/down and left/right)
3 Dimensional: Sound waves moving in three-dimensional space
Post by: gilliesb18 on May 18, 2017, 10:21:19 am
Hey I have a really really strange question but just recently at school we have been having this discussion about the Bermuda Triangle and someone said something about a Physics law being the source the problems. Is this right?? Or could it just be another theory?
Just thought i'd ask on here cause there's heaps of smarties hanging out on here!!!!
Post by: jakesilove on May 18, 2017, 10:33:16 am
Ok thanks!!! That's great....As far as I can tell, there's nothing Physics related going on there! Just a lot of confirmation bias, in a high-use shipping zone.
Hey I have a really really strange question but just recently at school we have been having this discussion about the Bermuda Triangle and someone said something about a Physics law being the source the problems. Is this right?? Or could it just be another theory?
Just thought i'd ask on here cause there's heaps of smarties hanging out on here!!!!
Post by: gilliesb18 on May 18, 2017, 10:38:19 am
Post by: jamonwindeyer on May 18, 2017, 12:19:02 pm
Ok sure!!! Thanks for confirming that!
In reality, Jake is the cause of the Bermuda Triangle drama. He lives there and shoot Physics lasers at boats that get too close ;)
Post by: gilliesb18 on May 18, 2017, 07:05:32 pm
Post by: Jyrgal on May 20, 2017, 08:24:01 pm
another question: from earth's perspective, if a rocket accelerates from 0.7c to 0.8c and next to it another rocket accelerates from 0.8c to 0.9c, while both rockets have same thrust, would the observer from earth see the rockets accelerate for the same amount of time?
thanks :)
Post by: jamonwindeyer on May 21, 2017, 11:42:05 am
hey! my friend was just asking this to me the other day, if spaceship #1 travelling at 0.8c passes spaceship #2 travelling at 0.9c, what would spaceship #1 see spaceship#2 going at?
another question: from earth's perspective, if a rocket accelerates from 0.7c to 0.8c and next to it another rocket accelerates from 0.8c to 0.9c, while both rockets have same thrust, would the observer from earth see the rockets accelerate for the same amount of time?
thanks :)
Hey! You are asking some hardcore questions there, they both feature relativistic relative velocities. It's the basic idea of relative velocities (throw a ball at 10 kilometres an hour out of a car going 20 kilometres an hour, to a stationary observer those are added to give 30), but you need to add in the effects of time dilation and length contraction :) here's some info on it, but it goes way beyond what you need for the HSC, and way beyond what I've touched in my studies (Jake has done it, he might be able to help!)
For your second question, again I'm honestly not sure, but I think the faster rocket would be seen to accelerate for longer due to increased effects of time dilation? That would be my (very uncertain) guess - These are really cool questions to be asking!! ;D
Post by: jakesilove on May 21, 2017, 02:18:31 pm
hey! my friend was just asking this to me the other day, if spaceship #1 travelling at 0.8c passes spaceship #2 travelling at 0.9c, what would spaceship #1 see spaceship#2 going at?
another question: from earth's perspective, if a rocket accelerates from 0.7c to 0.8c and next to it another rocket accelerates from 0.8c to 0.9c, while both rockets have same thrust, would the observer from earth see the rockets accelerate for the same amount of time?
thanks :)
All of this comes down to Lorentz transformations, which are NOT in the syllabus for some reason. For instance, I think that first question is totally easy, just as straight forward as the usual relativity question, IF YOU WERE GIVEN THE FORMULA! However, you're not, so for now I would just focus on what is actually assessable.
Re your second question; no idea, Relativity doesn't really like acceleration, the maths gets too tough.
Post by: winstondarmawan on May 21, 2017, 04:23:18 pm
Please sort it out!
Thanks in advance.
https://scontent-sjc2-1.xx.fbcdn.net/v/t34.0-12/18601262_1973433546221520_1725643649_n.png?oh=b1bbc6a1c159ab8b0cb10f13786a43d0&oe=59230D4C
Post by: jamonwindeyer on May 21, 2017, 04:28:33 pm
Hello! My friend and I have been arguing about the answer to this question.
Please sort it out!
Thanks in advance.
https://scontent-sjc2-1.xx.fbcdn.net/v/t34.0-12/18601262_1973433546221520_1725643649_n.png?oh=b1bbc6a1c159ab8b0cb10f13786a43d0&oe=59230D4C
Ahaha hey! I think towards 4 at the bottom, towards 1 at the top. If you consider the bottom, we have a N pole moving towards the bottom - So we want to push the N away (Lenz's Law). So, by the right hand grip rule, we want to induce a counter clockwise current - That generates a north pole equivalent field directed towards the magnet to oppose its motion. Then the top is just the opposite - We want a north pole directed downwards to try and bring the S pole back up. Therefore, clockwise current :)
This was just a quick run through - But does it make sense? (who won the argument? Ahaha)
Post by: winstondarmawan on May 21, 2017, 04:35:19 pm
Ahaha hey! I think towards 4 at the bottom, towards 1 at the top. If you consider the bottom, we have a N pole moving towards the bottom - So we want to push the N away (Lenz's Law). So, by the right hand grip rule, we want to induce a counter clockwise current - That generates a north pole equivalent field directed towards the magnet to oppose its motion. Then the top is just the opposite - We want a north pole directed downwards to try and bring the S pole back up. Therefore, clockwise current :)Yes thank you! I did
This was just a quick run through - But does it make sense? (who won the argument? Ahaha)
Post by: jamonwindeyer on May 21, 2017, 04:49:41 pm
Yes thank you! I did
Nicely done ;)
Post by: Jyrgal on May 21, 2017, 05:50:35 pm
All of this comes down to Lorentz transformations, which are NOT in the syllabus for some reason. For instance, I think that first question is totally easy, just as straight forward as the usual relativity question, IF YOU WERE GIVEN THE FORMULA! However, you're not, so for now I would just focus on what is actually assessable.
Re your second question; no idea, Relativity doesn't really like acceleration, the maths gets too tough.
Oh haha, i think the questions r from a sheet my friend got from his teacher... maybe not hsc related, no wonder it had me confused the whole day :'(
Post by: scienceislife on May 25, 2017, 07:20:52 am
Post by: jakesilove on May 25, 2017, 09:17:58 am
Do you find that general physics text books or reliable websites are better for research assessment tasks?
Use both! Markers will look kindly on you if you use a variety of sources, including online and physical sources
Post by: bsdfjnlkasn on May 26, 2017, 05:30:13 pm
Post by: jamonwindeyer on May 26, 2017, 06:02:08 pm
Hey just chiming in with the research assessment question, where can I find good journal articles for semiconductors and transistors (3rd dotpoint set in I2I) as I don't know where to look and what to search. Any advice or links would be super appreciated :) - our bibliography has to assess the accuracy, validity and reliability of the sources and ideally they would all fulfil these requirements :)
Hey! Are you looking for sources that demonstrate how they are used? Any electrical engineering textbooks for uni would suit you - Does it have to be a journal article or? :)
Post by: bsdfjnlkasn on May 26, 2017, 08:19:43 pm
Hey! Are you looking for sources that demonstrate how they are used? Any electrical engineering textbooks for uni would suit you - Does it have to be a journal article or? :)
Mostly just explanations/general info about the dot points in that section of I2I :)
We are required to have a variety of source forms, so a journal article would definitely help (considering I've only used textbooks so far) - if you've got any additional suggestions I'd love to hear them! :)
Post by: arunasva on May 26, 2017, 08:38:33 pm
I have to explain this in a video while conducting the Meissner Effect prac. I am thinking for gravitational field I have to say F=qE is > than -(F=mg) ?
Post by: jamonwindeyer on May 26, 2017, 08:53:18 pm
Mostly just explanations/general info about the dot points in that section of I2I :)
We are required to have a variety of source forms, so a journal article would definitely help (considering I've only used textbooks so far) - if you've got any additional suggestions I'd love to hear them! :)
Hmm - Unfortunately nothing springs to mind, a lot of the journal articles you'd find on that topic would be quite academically rigorous, written as a Thesis perhaps. Hopefully someone else might know one!
(Have you checked any article databases? Normally your library can give you access to some (the names all elude me now but they were super useful back when I did my HSC)
Post by: katnisschung on May 27, 2017, 09:50:57 am
it covers 7 dotpoints (4.4,4.5, 4.6, 4.7 and from the investigations column 4.1.3, 4.1.4, 4.1.5)
I haven't been transcribing any notes this term becos i've been quite busy with work and assessments
will prior knowledge on the previous focus areas of ideas to implementation be necessary to understand these
dotpoints (such as the bcs theory) the test will only be on these dotpoints however I have limited time so looking for the
most efficient manner to do this task (either transcribe everything i've done till this point or just do the research for these dotpoints and transcribe after the exam)
thanks :)
Post by: julies on May 27, 2017, 10:15:34 am
I would just do a lot of questions on research dot points! :)
Post by: pikachu975 on May 27, 2017, 10:52:40 am
I found that prior knowledge of i2i dot points was not that helpful, except maybe the idea of quantised energy for phonons. The most relevant things were probably from motors and generators, i.e. resistance, Lenz's and Faraday's Law (for the Meissner effect and maglev trains)
I would just do a lot of questions on research dot points! :)
Lenz's law doesn't govern how the Meissner effect works despite people saying it does, it works by quantum mechanics. You don't have to say how it works you just need to say what it is.
Post by: jamonwindeyer on May 27, 2017, 12:05:54 pm
I found that prior knowledge of i2i dot points was not that helpful, except maybe the idea of quantised energy for phonons. The most relevant things were probably from motors and generators, i.e. resistance, Lenz's and Faraday's Law (for the Meissner effect and maglev trains)
I would just do a lot of questions on research dot points! :)
Lenz's law doesn't govern how the Meissner effect works despite people saying it does, it works by quantum mechanics. You don't have to say how it works you just need to say what it is.
This is very true - The principle is similar but distinct. I think this paragraph excerpt from the explanation on Wikipedia explains it well (as much as using Wikipedia is discouraged, the explanations it gives for scientific principles are usually quite a nice way to introduce a concept ;D)
Any perfect conductor will prevent any change to magnetic flux passing through its surface due to ordinary electromagnetic induction at zero resistance. The Meissner effect is distinct from this: when an ordinary conductor is cooled so that it makes the transition to a superconducting state in the presence of a constant applied magnetic field, the magnetic flux is expelled during the transition. This effect cannot be explained by infinite conductivity alone. Its explanation is more complex and was first given in the London equations by the brothers Fritz and Heinz London. It should thus be noted that the placement and subsequent levitation of a magnet above an already superconducting material does not demonstrate the Meissner effect, while an initially stationary magnet later being repelled by a superconductor as it is cooled through its critical temperature does.
Placing a magnet above an already superconductive material does not demonstrate the Meissner Effect. A magnet beginning to hover as it becomes superconductive does demonstrate the effect. This is very easy to confuse (and not important at this level of study either, saying that a magnet placed above a superconductor hovers in place due to the Meissner Effect has always done fine from what I've seen, just interesting!)
The big point for you guys is to understand that the Meissner effect is separate and distinct from Lenz's Law/Induction. As pikachu said, you don't need to explain Meissner, just say that it (exclusion of magnetic flux from a superconductor below critical temperature) occurs ;D
Another good explanation here covering some of those more finicky (and not so relevant) points ;D
To Katniss, just research the relevant dot points and make some palm cards to study from - Worry about the rest later ;D
Post by: bsdfjnlkasn on May 27, 2017, 07:44:20 pm
We've just gone through p and n-type semiconductors in class and I have read a few textbooks in addition to consolidate my study. BUT i'm a bit confused as to which explanation to trust; is it one which explains doping in terms of the band structure of the semi-conductor or how the atoms interact in a lattice as a result? In one book it refers to 'dopant' and acceptor levels from which conduction occurs as this is where the electrons jump to from the valence shell (where they leave 'positive' holes). However, most explanations just discuss the complete atom arrangement where electrons from neighbouring Si/Ge atoms try and fill the positive whole gap formed by the doped element. This then constitutes a moving charge and so, current.
Obviously both explanations make sense, but since there's a focus on band theory in the initial content leading explaining valence and conduction bands, should I stick with the band structure explanation? Which one is expected in the HSC?
Please let me know if this question doesn't make any sense!
Any help would be greatly appreciated :D
Post by: pikachu975 on May 27, 2017, 09:11:05 pm
Hey there,
We've just gone through p and n-type semiconductors in class and I have read a few textbooks in addition to consolidate my study. BUT i'm a bit confused as to which explanation to trust; is it one which explains doping in terms of the band structure of the semi-conductor or how the atoms interact in a lattice as a result? In one book it refers to 'dopant' and acceptor levels from which conduction occurs as this is where the electrons jump to from the valence shell (where they leave 'positive' holes). However, most explanations just discuss the complete atom arrangement where electrons from neighbouring Si/Ge atoms try and fill the positive whole gap formed by the doped element. This then constitutes a moving charge and so, current.
Obviously both explanations make sense, but since there's a focus on band theory in the initial content leading explaining valence and conduction bands, should I stick with the band structure explanation? Which one is expected in the HSC?
Please let me know if this question doesn't make any sense!
Any help would be greatly appreciated :D
We went through acceptor levels and donator levels (Depending on group 3 or group 5 dopant) so I guess that
Post by: jamonwindeyer on May 27, 2017, 10:28:41 pm
Post by: beau77bro on May 28, 2017, 02:38:49 pm
thankyou
Post by: katnisschung on May 28, 2017, 02:47:44 pm
i'm trying to self teach myself.... my current understanding is that its a unit of vibrational energy
whereas others are saying its a particle like an electron...confused?
thanks
Post by: jakesilove on May 28, 2017, 04:08:21 pm
can someone please explain this and or conduction bands? like i can't quite visualise it, its just a means to represent how much energy is needed for an electron to flow basically right? then how does this diagram even work?
thankyou
Hey! This is one of those things you just need to memorise; be able to draw the diagram, and describe the bands (ie. When electrons are in the conduction band, they can conduct!). In my HSC, the way I thought about it was that the additional bands (donor/acceptor due to doping) allowed electrons to 'jump' more easily from the Valence to the Conduction band. In essence, the band structure is a diagrammatically way of representing this complex idea; however, the diagram is actually deeply routed in Quantum Physics concepts far beyond the curriculum.
Post by: jakesilove on May 28, 2017, 04:10:31 pm
what the heck is a phonon?
i'm trying to self teach myself.... my current understanding is that its a unit of vibrational energy
whereas others are saying its a particle like an electron...confused?
thanks
Hey! In the HSC, you really don't need to know anything about a phonon. In reality, you just need to be able to say something like 'A phonon (vibrational energy) is produced and transmitted from one cooper pair to another' etc. A phonon is exactly that; a unit of vibrational energy. However, recall that waves/particles are very much the same thing; it's all a scale, not a one-or-the-other type situation. Thus, we say that phonons (like photons) come in a 'packet' of energy, similar in form to an electron (in that there is a defined position etc.). However, it is really just energy. All this is beyond the curriculum!
Post by: beau77bro on May 29, 2017, 11:25:41 am
Hey! This is one of those things you just need to memorise; be able to draw the diagram, and describe the bands (ie. When electrons are in the conduction band, they can conduct!). In my HSC, the way I thought about it was that the additional bands (donor/acceptor due to doping) allowed electrons to 'jump' more easily from the Valence to the Conduction band. In essence, the band structure is a diagrammatically way of representing this complex idea; however, the diagram is actually deeply routed in Quantum Physics concepts far beyond the curriculum.
omg thankyou jake it just clicked, so basically the levels represent the effect of introduced charge carriers? so for the first one, it's showing that the n-type doped conductor has electrons added that can move more easily into the conduction band - so the donor level just means the extra electrons can move easier into the conduction band.
and the acceptor band means the electrons can move into a hole, they do not require as much energy to move into the conduction band but can instead move to adjacent holes?
is that kinda the jist of it? they have to jump lower energy due to the addition of charge carriers?
Post by: beau77bro on May 29, 2017, 11:33:32 am
Post by: jakesilove on May 29, 2017, 12:33:29 pm
omg thankyou jake it just clicked, so basically the levels represent the effect of introduced charge carriers? so for the first one, it's showing that the n-type doped conductor has electrons added that can move more easily into the conduction band - so the donor level just means the extra electrons can move easier into the conduction band.
and the acceptor band means the electrons can move into a hole, they do not require as much energy to move into the conduction band but can instead move to adjacent holes?
is that kinda the jist of it? they have to jump lower energy due to the addition of charge carriers?
You're 100% correct, looks like you have a solid grasp!
also, i was wondering about ur research into quantum tunneling and how that effects processing chips when they start getting super small. we have a research task on transistors and electronics development and effects on society, and one of the parts is about current research and unanswered questions relating to transistors. so i was just wondering if you could give me some insight into the very very basic concepts of your research and some of the issues. i recognise that by making the devices smaller in the integrated circuit, you can fit more making them more powerful processors, and that by making them smaller you reduce energy loss and time between devices, making it faster as well as cheaper to produce as the size decreases. so basically the aim is to make them as small as possible.
You're totally right re making them smaller. However, I think the gist of your question is something along the lines of 'well what does quantum tunnelling have to do with it?'
Since it's way beyond the curriculum, but not too difficult to understand, I'll give you a brief overview. Let's look at an electron, in a classical sense.
We look at an 'infinite square well'.
(http://www.felderbooks.com/papers/images/psi/squarewell.gif)
Why? Well, we imagine that the electrons are 'bound' within a certain location (eg. a part of the transistor), and are allowed to bumble around in that region. A transistor only works if we can control where the electrons are (eg. are they in a p-type semiconductor? An n-type?). Basically, we need to know and be able to affect their movement!
So, in classic physics, we put our electron in an area of the transistor (let's call this our electron well). There are 'walls' stopping the electron from escaping. The 'walls' need to be stronger than the electron, so the electron can't jump over it (I'm talking about energy here, but you can also think of it as physical walls that are taller than an electron can jump). So, we can say if the walls are high enough, our electron stays in the right place, and our transistor works.
However, we make our transistor smaller and smaller. All of a sudden, we need to control EVERY electron, not just 'most' electrons. When we get too small, crazy quantum shit starts to happen.
Our electron is still in this well. However, instead of being 'trapped' by the walls, they have a certain probability of just going straight through! This is insane; it's like me saying that I'll throw a pingpong ball at a wall, and it passes straight through. Or, rather, I try to bounce the ping pong ball OVER the wall, but it's maximum height is half of the height of the wall, and it STILL goes through!!
Now, this messes with our transistor. All of a sudden, we have electrons flowing where we don't want them to flow. This is less of an 'unanswered question' as it is an 'unsolved problem'; we know what's happening, we just don't know the best way to stop it! Dammit, Quantum Physics!
Post by: Bubbly_bluey on May 29, 2017, 12:39:21 pm
Thanks :)
Post by: jakesilove on May 29, 2017, 12:48:38 pm
Hi! How has Planck's theory of Black Body radiation added to existing theories about electromagnetic radiation?
Thanks :)
Hey! Essentially, it comes down to the fact that energy can be emitted/absorbed as discrete packets, rather than as waves. Planck solved the ultraviolet catastrophe by proposing that light (and thus energy) could come in phonons. This completely revolutionised our understanding of electromagnetic radiation, and a few years later Einstein would come and formalise the theory (think all-or-nothing principle, etc.). Additionally, Planck introduced the formula E=hf, from which we could accurately determine the energy/frequency of electromagnetic radiation. However, his calculation of the constant, h, was pretty far off (this would be refined later).
This is a broad overview of the topic area; basically, look at how Planck solved the Black Body problem. If you have any more specific questions, or want any clarification, let me know!
Post by: beau77bro on May 29, 2017, 12:55:50 pm
You're 100% correct, looks like you have a solid grasp!
You're totally right re making them smaller. However, I think the gist of your question is something along the lines of 'well what does quantum tunnelling have to do with it?'
Since it's way beyond the curriculum, but not too difficult to understand, I'll give you a brief overview. Let's look at an electron, in a classical sense.
We look at an 'infinite square well'.
(http://www.felderbooks.com/papers/images/psi/squarewell.gif)
Why? Well, we imagine that the electrons are 'bound' within a certain location (eg. a part of the transistor), and are allowed to bumble around in that region. A transistor only works if we can control where the electrons are (eg. are they in a p-type semiconductor? An n-type?). Basically, we need to know and be able to affect their movement!
So, in classic physics, we put our electron in an area of the transistor (let's call this our electron well). There are 'walls' stopping the electron from escaping. The 'walls' need to be stronger than the electron, so the electron can't jump over it (I'm talking about energy here, but you can also think of it as physical walls that are taller than an electron can jump). So, we can say if the walls are high enough, our electron stays in the right place, and our transistor works.
However, we make our transistor smaller and smaller. All of a sudden, we need to control EVERY electron, not just 'most' electrons. When we get too small, crazy quantum shit starts to happen.
Our electron is still in this well. However, instead of being 'trapped' by the walls, they have a certain probability of just going straight through! This is insane; it's like me saying that I'll throw a pingpong ball at a wall, and it passes straight through. Or, rather, I try to bounce the ping pong ball OVER the wall, but it's maximum height is half of the height of the wall, and it STILL goes through!!
Now, this messes with our transistor. All of a sudden, we have electrons flowing where we don't want them to flow. This is less of an 'unanswered question' as it is an 'unsolved problem'; we know what's happening, we just don't know the best way to stop it! Dammit, Quantum Physics!
oft. just oft. that is very very interesting, so the walls becoming to small increases the probabilty of electrons escaping and tunneling through - basically due to having a given energy that, when the walls are too small, they can move through? or just pre much teleport hahahaha? and this is a massive issue because the electrons won't do their jobs and can't be directed and controlled efficiently - but not all the electrons are tunnelling through right?
a very, probably silly question, but when we say energy of the wall, is that thickness or is there a required property? like the electrons are escaping because the walls get too thin? and you guys (gods) are basically working a way to decrease the probability of the electrons moving through the walls, but how do u (theoretically) decrease the probablity? basically what im tryna ask is what is the issue with the walls that allows the electrons to pass through, and how can it be fixed without widening the walls (theoretically)? because we say energy, so how do we increase the energy of the wall (sorry if ive missunderstood)
Post by: jakesilove on May 29, 2017, 01:01:11 pm
oft. just oft. that is very very interesting, so the walls becoming to small increases the probabilty of electrons escaping and tunneling through - basically due to having a given energy that, when the walls are too small, they can move through? or just pre much teleport hahahaha? and this is a massive issue because the electrons won't do their jobs and can't be directed and controlled efficiently - like not all the electrons are leaving the well right?
a very, probably silly question, but when we say energy of the wall, is that thickness or is there a required property? like the electrons are escaping because the walls get too thin? and you guys (gods) are basically working a way to decrease the probability of the electrons moving through the walls, but how do u decrease the probablity?
Yeaaaah so look you're pretty much right, obviously there's heaps of detail and maths that I'm not going to get into. Effectively, though, you're correct. When it comes to the walls themselves, there are two factors; the height of the wall (the higher the wall, the less likely Electrons are to jump through) and the thickness of the wall (the thicker the wall, the less likely Electrons are to jump through). However, smaller devices will have 'thinner' walls (although not necessarily 'less high' walls). So, tunnelling may become more likely! Which is bad, for all the reasons you described.
Yep, they pretty much teleport; they have to 'exist' inside the wall, somehow, but they're not 'physically' present there, they just sort of jump through.
Still, given how hard this is, you've got a pretty solid grasp aha
Post by: beau77bro on May 29, 2017, 01:05:36 pm
Yeaaaah so look you're pretty much right, obviously there's heaps of detail and maths that I'm not going to get into. Effectively, though, you're correct. When it comes to the walls themselves, there are two factors; the height of the wall (the higher the wall, the less likely Electrons are to jump through) and the thickness of the wall (the thicker the wall, the less likely Electrons are to jump through). However, smaller devices will have 'thinner' walls (although not necessarily 'less high' walls). So, tunnelling may become more likely! Which is bad, for all the reasons you described.
Yep, they pretty much teleport; they have to 'exist' inside the wall, somehow, but they're not 'physically' present there, they just sort of jump through.
Still, given how hard this is, you've got a pretty solid grasp aha
hahah omg that is reassuring. thankyou soo much, i can't imagine how complex this must get and i appreciate the analogies and nice ways in which you explained it, still confused (ofc) but i think i get it. so thanks very much - p.s. i added a mini weird question at the end of my last post.
Post by: jakesilove on May 29, 2017, 01:09:46 pm
hahah omg that is reassuring. thankyou soo much, i can't imagine how complex this must get and i appreciate the analogies and nice ways in which you explained it, still confused (ofc) but i think i get it. so thanks very much - p.s. i added a mini weird question at the end of my last post.
Hmm, good question. We do it in a number of ways; we can just increase the voltage (energy) between two regions, we can increase the physical distance between two regions, we can put substances/metals that are 'hard' to get through in between the regions... beyond that, it's sort of just a fundamental limit of nature! I haven't done extensive research on the 'application' of quantum theory; that's for Engineers to figure out
Post by: beau77bro on May 29, 2017, 01:17:41 pm
Post by: jakesilove on May 29, 2017, 01:33:21 pm
ohhhhh ok i see i see. i just watched this video, https://www.youtube.com/watch?v=MD_xiokVXwI and is this relevant or just like too far. would i explain it by saying, the issue becomes controlling electron spins, directing them so that they function accurately as a single integrated circuit? or is this just too far, and too different to what we just spoke about? im sorry to keep badgering you with these question, i am extremely interested, you are very patient ahhahah.
Probably a bit far :)
Post by: Bubbly_bluey on May 29, 2017, 07:32:45 pm
Hey! Essentially, it comes down to the fact that energy can be emitted/absorbed as discrete packets, rather than as waves. Planck solved the ultraviolet catastrophe by proposing that light (and thus energy) could come in phonons. This completely revolutionised our understanding of electromagnetic radiation, and a few years later Einstein would come and formalise the theory (think all-or-nothing principle, etc.). Additionally, Planck introduced the formula E=hf, from which we could accurately determine the energy/frequency of electromagnetic radiation. However, his calculation of the constant, h, was pretty far off (this would be refined later).Thank you! But I'm still not sure how Planck had revolutionised our understanding of electromagnetic radiation. What was the scientist's understanding of it before Planck's idea came along? Was it just how they thought that energy emitted from the black body would be a continuous wave by classical physics?
This is a broad overview of the topic area; basically, look at how Planck solved the Black Body problem. If you have any more specific questions, or want any clarification, let me know!
Post by: jamonwindeyer on May 29, 2017, 08:38:16 pm
Thank you! But I'm still not sure how Planck had revolutionised our understanding of electromagnetic radiation. What was the scientist's understanding of it before Planck's idea came along? Was it just how they thought that energy emitted from the black body would be a continuous wave by classical physics?
The Classical Law that dictated the shape of a black body radiation curve is called Rayleigh Jean's Law, but you don't need to know that. The specifics of the classical theory aren't assessable ;D
Post by: katnisschung on May 29, 2017, 08:55:13 pm
but i had like 1 point on it on my notes to explain the phenomenon for the magnet hover above
the superconductor like they actually had a 4 marker "explain the meissner effect" RIP PHYSICS MARK
RIP HSC RIP ATAR :'( :'( :'( :'( :'( :'( :'(
Post by: bsdfjnlkasn on May 29, 2017, 09:03:34 pm
I'm just a bit confused about p-type conductors. Does any conduction occur in the conduction band within p-type conductors as all i'm gathering is that conduction only occurs in the dopant/acceptor level. This acceptor level accepts electrons from the valence band (located just below it) - so what's the relevance of the conduction band? How does this new acceptor band just form? Does electron-hole conduction occur here? Is this where electron-hole conduction occurs in intrinsic conduction? Because I thought that intrinsic conduction occurred between the conduction and valence bands. So why does a whole new band form in this case of extrinsic/doped semiconductors? Furthermore, I'm just confused about how we can consider there being a valence band in semiconductors when we always discuss things in terms of the lattice and how electrons are shared intermolecularly. How can these interactions form a broader band? I'm not too sure about this whole band and atomic level business works as I don't see how they're related and how the two explanations are equally valid as one another - if in fact not related at all. In terms of considering the valence band for semiconductors, how is the band formed and is there an analogy or something that can help me picture it? In exams, should we discuss band theory or the intermolecular electron exchanges (particularly for positive holes in p-type conductors).
I know that's a heaps and heaps of questions, but I hope they're easy to address/understand. I'm still getting my head around this whole semiconductor business (we're being assessed on it as a part of a research task, so haven't actually been taught it). As always, any help would be super super appreciated.
Thank you!!
Post by: jamonwindeyer on May 29, 2017, 09:06:47 pm
cries are they allowed to explicitly ask u what the meissner effect like its not in the syllabus
but i had like 1 point on it on my notes to explain the phenomenon for the magnet hover above
the superconductor like they actually had a 4 marker "explain the meissner effect" RIP PHYSICS MARK
RIP HSC RIP ATAR :'( :'( :'( :'( :'( :'( :'(
Ouch! Really all they can ask you to explain is that it is the exclusion of magnetic flux from a superconductor below critical temperature - The exact workings of it for 4 marks certainly seems a bit much!! Don't worry though, it's not going to impact you in the long run! ;D
Post by: bsdfjnlkasn on May 30, 2017, 06:25:15 pm
Sorry if this is a really stupid question but what's the difference between a diode, resistor and transistor? All i'm seeing is these terms thrown around but not many definitions - it will really help me piece together the content haha
Thank you!!
Post by: jakesilove on May 30, 2017, 07:16:08 pm
Hey I'm back again,
Sorry if this is a really stupid question but what's the difference between a diode, resistor and transistor? All i'm seeing is these terms thrown around but not many definitions - it will really help me piece together the content haha
Thank you!!
Hey! This is definitely not necessary to understand in the course, but it's still easy to give a quick overview.
A diode is a semiconductor device that essentially 'rectifies' the current. In english, it ensures that current can only travel in one direction, not two.
Transistors are semiconductor devices that are used as 'switches' or 'amplifiers' of current. That's exactly what it sounds like.
A resistor is just anything with a high enough resistance to ensure the system doesn't short circuit. Nothing semiconductor related here.
All of these devices have different physical make-ups in terms of Semiconductors, but actually the course doesn't require you to have any sort of understanding of the physical build of transistors! Only society and the environment, and the fact that semiconductors led to transistors :)
Post by: bsdfjnlkasn on May 30, 2017, 07:30:21 pm
Hey! This is definitely not necessary to understand in the course, but it's still easy to give a quick overview.
A diode is a semiconductor device that essentially 'rectifies' the current. In english, it ensures that current can only travel in one direction, not two.
Transistors are semiconductor devices that are used as 'switches' or 'amplifiers' of current. That's exactly what it sounds like.
A resistor is just anything with a high enough resistance to ensure the system doesn't short circuit. Nothing semiconductor related here.
All of these devices have different physical make-ups in terms of Semiconductors, but actually the course doesn't require you to have any sort of understanding of the physical build of transistors! Only society and the environment, and the fact that semiconductors led to transistors :)
Hey thanks so much Jake! Do you think you could possibly help me out with the questions I posted last night? There's a lot so don't worry if you don't have time :)
EDIT: I'm just a bit confused here about the specifics of the depletion zone when a p-n junction is established. Is it the same as the potential barrier? If not, when does the depletion zone form? Or is it like the potential barrier prevents charges from moving across the junction which then results in a depletion zone?
I also read in my textbook that after the potential difference is established across a p-n junction (after charges drift across junction to opposite ends) that the semiconductor becomes electrically charged. Is this right? Because for a substance to become charged, shouldn't charges be removed/added, not moved? Further, could I get an explanation of the equilibrium business that occurs just after the p-n junction is formed as i've read two different things. I've read that some electrons move back to the n-type and the holes to the p-type and this will continue to happen until the depletion region no longer has any charges passing through. I've also read that charges will diffuse across the p-n junction until the number of electrons (which are now on the p-type’s boundary) have accumulated a large enough electrical charge to repel/prevent any more charge carriers from cross over the junction. Which is correct and why?
Finally is the following note correct? If a voltage is applied to the p-n junction, it will act as a diode, allowing current to flow from p --> n
Are we talking about an external voltage and when there is forward bias?
Thanks again :D :D
Post by: jamonwindeyer on May 31, 2017, 09:31:34 pm
I'm just a bit confused here about the specifics of the depletion zone when a p-n junction is established. Is it the same as the potential barrier? If not, when does the depletion zone form? Or is it like the potential barrier prevents charges from moving across the junction which then results in a depletion zone?
The depletion zone is the area within which charge carriers have all recombined with atoms (charge carriers have been depleted) - This indeed sets up a potential difference which eventually stops more charge carriers from moving and depleting! When this happens, we call the balance equilibrium ;D so I suppose it is the depletion region that sets up the potential barrier :)
Quote
I also read in my textbook that after the potential difference is established across a p-n junction (after charges drift across junction to opposite ends) that the semiconductor becomes electrically charged. Is this right? Because for a substance to become charged, shouldn't charges be removed/added, not moved?
Before the N and P type semiconductors are brought together, they aren't charged. P and N type semiconductors might have additional electrons/holes, but the charge is still neutral (the protons in the atoms balance everything out). Now, when we bring the two together, we've got negative and positive charges moving around. They were neutral to begin - So once everything has moved, their must now be regions of charge. Specifically, the electrons that drift to the P-type semiconductor make it negatively charged. The holes that drift to the N-type semiconductor make it positively charged.
Quote
Further, could I get an explanation of the equilibrium business that occurs just after the p-n junction is formed as i've read two different things. I've read that some electrons move back to the n-type and the holes to the p-type and this will continue to happen until the depletion region no longer has any charges passing through. I've also read that charges will diffuse across the p-n junction until the number of electrons (which are now on the p-type’s boundary) have accumulated a large enough electrical charge to repel/prevent any more charge carriers from cross over the junction. Which is correct and why?
I'd say the second is correct more than the first, because electrons move to the P-type, not the N-type. Does a better job explaining it. But it sounds like both are trying to say the same thing ;D
Quote
Finally is the following note correct? If a voltage is applied to the p-n junction, it will act as a diode, allowing current to flow from p --> n
Are we talking about an external voltage and when there is forward bias?
Yep - The junction itself is a diode, and yes, applying an external voltage to forward bias it will allow current to flow! ;D
Post by: katnisschung on June 01, 2017, 06:48:52 pm
Post by: jamonwindeyer on June 01, 2017, 07:58:46 pm
why does a charge that enters a magnetic field at a right angle start to move in a circular path?
Hey! Because by the right hand grip rule, the force on the charged particle is perpendicular to its motion. A force perpendicular to motion is what we get for uniform circular motion - Compare it to the direction of the force of gravity in an orbit ;D
Post by: kiwiberry on June 03, 2017, 11:25:41 pm
Post by: jamonwindeyer on June 04, 2017, 11:19:09 am
Hey guys, I'm kinda confused about how solar cells work. I get that at the p-n junction, electrons flow from n to p-type, setting up a potential difference which eventually prevents further flow of charges, forming the depletion zone. Does light then strike the electrons on the surface of the p-type or those in the n-type? I'm getting conflicting explanations from different sources and I'm really confused. Also, is only the n-type exposed to sunlight? If someone could give me a brief run down on how solar cells work that would be great 😅
Hey! So the electric field/potential difference set up by the depletion zone is directed from the N type to the P type, due to the excess positive charge in the N-type and the excess negative charge in the P-type. This sets up an electric field which pushes positive charges/holes towards the P-type, and negative charges/electrons towards the N-type. It doesn't really matter where it happens (just imagine it happening somewhere in the depletion zone), but when a photon of appropriate frequency strikes the diode, it frees an electron (thus also forming a hole). This is called an electron-hole pair. The electric field pushes electrons into the N-type, and holes towards the P-type - This constitutes the flow of current.
This video does a great job of explaining it ;D
This is a highly simplified version of what actually happens, mind you. Solar cells aren't actually just sticking a P and an N type semiconductor together and shining a light on it. But this is a perfectly acceptable description for HSC Physics ;D
Post by: kiwiberry on June 04, 2017, 12:05:32 pm
Hey! So the electric field/potential difference set up by the depletion zone is directed from the N type to the P type, due to the excess positive charge in the N-type and the excess negative charge in the P-type. This sets up an electric field which pushes positive charges/holes towards the P-type, and negative charges/electrons towards the N-type. It doesn't really matter where it happens (just imagine it happening somewhere in the depletion zone), but when a photon of appropriate frequency strikes the diode, it frees an electron (thus also forming a hole). This is called an electron-hole pair. The electric field pushes electrons into the N-type, and holes towards the P-type - This constitutes the flow of current.
This video does a great job of explaining it ;D
This is a highly simplified version of what actually happens, mind you. Solar cells aren't actually just sticking a P and an N type semiconductor together and shining a light on it. But this is a perfectly acceptable description for HSC Physics ;D
Ah thanks so much Jamon :D
Post by: seventeenboi on June 04, 2017, 05:43:36 pm
how do you explain the output (emf vs time) graph of a dc or ac motor in an explicit/detailed manner???
This is what I have in my notes so far:
AC:
Constant connection between rotating coil and external circuit due to split rings
As polarity of induced emf changes with every half-turn of the coil
Voltage varies like a sine wave
Current changes direction
DC:
Every half turn, a reversal of connection between coil and external circuit occurs due to split rings
As polarity of induced emf changes with every half turn
Voltage in external circuit fluctuates between 0 and maximum allowing current to flow in a constant direction
ALSO, for the bolded/ underlined parts "As polarity of induced emf changes with every half turn" is this applicable to both AC and DC generators? I'm not sure if i typed up my notes incorrectly :/
thanks!!
Post by: pikachu975 on June 04, 2017, 07:34:41 pm
Hi :)
how do you explain the output (emf vs time) graph of a dc or ac motor in an explicit/detailed manner???
This is what I have in my notes so far:
AC:
Constant connection between rotating coil and external circuit due to split rings
As polarity of induced emf changes with every half-turn of the coil
Voltage varies like a sine wave
Current changes direction
DC:
Every half turn, a reversal of connection between coil and external circuit occurs due to split rings
As polarity of induced emf changes with every half turn
Voltage in external circuit fluctuates between 0 and maximum allowing current to flow in a constant direction
ALSO, for the bolded/ underlined parts "As polarity of induced emf changes with every half turn" is this applicable to both AC and DC generators? I'm not sure if i typed up my notes incorrectly :/
thanks!!
For DC it uses split ring commutator not split rings. Also the polarities don't change that's just AC.
Firstly a motor doesn't have a purpose of inducing EMF, motors use supplied EMF to generate mechanical energy.
For a motor the current supplied just changes direction every half turn due to the split ring commutator
Post by: seventeenboi on June 05, 2017, 07:24:21 am
For DC it uses split ring commutator not split rings. Also the polarities don't change that's just AC.
Firstly a motor doesn't have a purpose of inducing EMF, motors use supplied EMF to generate mechanical energy.
For a motor the current supplied just changes direction every half turn due to the split ring commutator
oops sorry! yeup change my 'motor' to a 'generator' HAHAHHA sorry ;-;;
Post by: jamonwindeyer on June 05, 2017, 09:53:28 am
oops sorry! yeup change my 'motor' to a 'generator' HAHAHHA sorry ;-;;
Woop! Aha pretty much everything pikachu stays relevant, make sure you know:
- DC generators use split ring commutators, AC generators use slip rings (this results in the differences in output current patterns you describe)
- By default, the polarity of induced emf does swap every half turn. However, in DC, the split ring commutator maintains constant current direction
;D
Post by: Jyrgal on June 05, 2017, 10:36:02 am
Post by: jamonwindeyer on June 05, 2017, 10:58:46 am
hello, im pretty confused on the concept of voltage between electrical plates. Lets say one plate has 0V and the other has 100V. In my way of thinking, this means that when I place 1 coloumb of negative electric charge near the 100V plate, the charge will possess 100J of potential energy since 100V=100J/1c. This means the charge should flow from the 100V to 0V plates, meaning the 0V should be the positive terminal and 100V should be the negative. Is this correct?
Hey! You are close, except if it is a negative charge, it would move towards the 100V plate (the plate with higher potential). If you swapped to a positive charge, then you'd be 100% on the money ;D
Post by: Jyrgal on June 05, 2017, 12:08:20 pm
Hey! You are close, except if it is a negative charge, it would move towards the 100V plate (the plate with higher potential). If you swapped to a positive charge, then you'd be 100% on the money ;D
Ohhhh so electric potential for voltage relates with positive charge not negative charge. Thanks for clearing that up! hehe
Post by: jamonwindeyer on June 05, 2017, 12:26:09 pm
Ohhhh so electric potential for voltage relates with positive charge not negative charge. Thanks for clearing that up! hehe
Yeah exactly! Basically, wherever the higher voltage is, that is the place where you can think of a lot of positive charge being. So, positive charges will want to get away from it, negative charges will want to head towards it ;D
Glad I could help :)
Post by: Bubbly_bluey on June 06, 2017, 07:36:35 pm
Thanks :)
Post by: jamonwindeyer on June 06, 2017, 07:38:41 pm
Hi! If a question says to talk about application of photoelectric effect in solar cells,is that asking how the solar cell actually works(p-n junctions and electrons moving around to make electricity) or it is talking about applications as in solar panels and solar powered devices?
Thanks :)
The former - You don't need to know much at all about how solar cells are used!
Post by: Bubbly_bluey on June 06, 2017, 07:59:36 pm
The former - You don't need to know much at all about how solar cells are used!Thanks! Also, when a photon enters the solar cell, does it strike the n-type to release the excess electron to flow back to the hole in p-type?
Post by: jamonwindeyer on June 06, 2017, 08:04:33 pm
Post by: Bubbly_bluey on June 07, 2017, 06:35:31 pm
Thanks ;D
Post by: jakesilove on June 07, 2017, 06:37:25 pm
hi again! So I'm starting to get confuse: is a Solid State Device a fancy name for a transistor? (are they the same thing?) I have to talk about the invention of the transistor
Thanks ;D
Yep! Well, technically, a solid state device describes a broad range of electronic items, including transistors, but for the sake of the HSC they are one and the same :)
Post by: jamonwindeyer on June 07, 2017, 07:00:13 pm
hi again! So I'm starting to get confuse: is a Solid State Device a fancy name for a transistor? (are they the same thing?) I have to talk about the invention of the transistor
Thanks ;D
And PN diodes are ssd's as well!
Post by: Yagami Light on June 08, 2017, 11:12:37 pm
Post by: Bubbly_bluey on June 09, 2017, 10:20:01 pm
Thank you :)
Post by: pikachu975 on June 10, 2017, 12:46:58 am
Hi! What does it mean to discuss the possible future applications of superconductivity of development of the maglev train, supercomputers and electricity transmission? I just thought that they were already the applications of the superconductor, if you know what I mean.
Thank you :)
Maglev train isn't used everywhere in the world with superconductors. I can't really remember but either Japan or Germany use superconductors for their maglev trains and it's really hard because you have to cool the whole track. There's still lots of improvements to be made.
Supercomputers I haven't researched.
Electricity transmission doesn't use superconductors yet. Superconductors have zero electrical resistance when cooled below critical temperature so in the future they can be used to eliminate power loss altogether as Ploss = I^2 R and R = 0. Hasn't been used yet because I think it's be too costly to cool all transmission lines for now.
Post by: Bubbly_bluey on June 10, 2017, 09:00:58 am
Maglev train isn't used everywhere in the world with superconductors. I can't really remember but either Japan or Germany use superconductors for their maglev trains and it's really hard because you have to cool the whole track. There's still lots of improvements to be made.Oh thanks so much! So by "future applications" would I have to mention the limitations and suggest improvements for each application?
Supercomputers I haven't researched.
Electricity transmission doesn't use superconductors yet. Superconductors have zero electrical resistance when cooled below critical temperature so in the future they can be used to eliminate power loss altogether as Ploss = I^2 R and R = 0. Hasn't been used yet because I think it's be too costly to cool all transmission lines for now.
Post by: jakesilove on June 10, 2017, 10:00:24 am
Oh thanks so much! So by "future applications" would I have to mention the limitations and suggest improvements for each application?
Yeah pretty much! I would subheading like 'Supercomputers', 'Motors and generators', 'Power distribution' etc. then talk about how superconductors would benefit that technology, as well as potential limitations (eg Superconductors can only transmit DC, requires huge amounts of energy to keep below critical temperature etc.)
Post by: beau77bro on June 12, 2017, 12:02:51 am
we need to - say who developed the transistor,
- what they are
- how they work
- provide a brief timeline
- impacts of the development on society (+ve and -ve and how society has changed)
- how germanium was used due to a lack of availability
- describe difference in thermionic and solid state
- say why solid state replaced thermionic
- current research - and further questions to be answered
in 7 minutes - as well as providing an intro and conclusion - so somehow i need to get my 1150 words to like 900-1000. COULD YOU GUYS PLEASE READ THIS AND HIGHLIGHT ANY INFO YOU THINK ISNT NECESSARY - i need to cut down a lot
- cut doe coz i already got rid of the cool facts and extra info that made it interesting.
btw - green is like should i cut or nah?
i was also gonna add quantum computing:
Spoiler
Now we say the electron spin is pointing, up or down or anywhere in between. This is important because in classical computers we have up or down, on or off, a switch which determines 1s or 0s. But here we have electrons that can be in two possible states at the same time, meaning it can solve things exponentially faster, because as you add more quantum bits the processing power doubles. THAT IS RIDICULOUS. In your phone you have over a billion transistors, but take out your calculator and type in 2 to the 100, not billion, a hundred and it has well past the capability of your phone.
Basically the outer most electron of a phosphorus atom in a silicon chip can have information placed on it by rotating it’s spin with microwaves, as well as the spin of it’s nucleus. It’s really complicated and awesome, if you wanna know more about it there are great videos by veritasium, and there’s a whole list on the handout.
Basically the outer most electron of a phosphorus atom in a silicon chip can have information placed on it by rotating it’s spin with microwaves, as well as the spin of it’s nucleus. It’s really complicated and awesome, if you wanna know more about it there are great videos by veritasium, and there’s a whole list on the handout.
but i cant coz no space
sorry for long post - THANKYOU
Mod Edit: Added spoiler :)
Post by: jamonwindeyer on June 12, 2017, 10:58:36 am
we have an assessment tuesday, its a speech on transistors:
...
Hey! I had a peek and highlighted some stuff in yellow I'd remove (all the stuff in green, ditch it imo).
If you trim what I trimmed, then cut back the explanation of the Quantum Physics stuff at the end (see the comment inside, you need to consider the audience you are speaking to!), you should have room to put a brief explanation of Quantum Computing, and still be under 1000 words ;D
Post by: bluecookie on June 12, 2017, 06:59:57 pm
http://www.a-levelphysicstutor.com/images/fields/E-fields-unif.jpg
Because in yr 11 we were taught that a circuit must be connected to allow current to travel through it, but in this one there's no connection between the two plates? So how does electricity flow through them?
Post by: jakesilove on June 12, 2017, 07:05:19 pm
How does this work?
http://www.a-levelphysicstutor.com/images/fields/E-fields-unif.jpg
Because in yr 11 we were taught that a circuit must be connected to allow current to travel through it, but in this one there's no connection between the two plates? So how does electricity flow through them?
You're totally right to question this; it isn't something that's properly looked at in the HSC! Basically, a device like this is called a capacitor (you don't need to know this). The plates get 'charged up', and eventually have so much energy in them that they 'discharge', completing the circuit. The electrons essentially 'jump' the gap between the two plates, because whilst they really don't like to, electrons are capable of travelling through air. So, you can imagine that the circuit is (weakly) complete by the air between the plates, which the electrons are sometimes able to flow through.
In the HSC, you don't need to know any of this; you're just supposed to assume that the situation is all g. Really good question though. I haven't given you a huge amount of depth, because there's no point over complicating your HSC Physics course, but let me know if you have any more questions or if I can clarify anything!
Post by: bluecookie on June 12, 2017, 07:13:38 pm
You're totally right to question this; it isn't something that's properly looked at in the HSC! Basically, a device like this is called a capacitor (you don't need to know this). The plates get 'charged up', and eventually have so much energy in them that they 'discharge', completing the circuit. The electrons essentially 'jump' the gap between the two plates, because whilst they really don't like to, electrons are capable of travelling through air. So, you can imagine that the circuit is (weakly) complete by the air between the plates, which the electrons are sometimes able to flow through.
In the HSC, you don't need to know any of this; you're just supposed to assume that the situation is all g. Really good question though. I haven't given you a huge amount of depth, because there's no point over complicating your HSC Physics course, but let me know if you have any more questions or if I can clarify anything!
Thanks for the fast response :) It helped clarify a lot!
Post by: jakesilove on June 12, 2017, 07:15:29 pm
Thanks for the fast response :) It helped clarify a lot!
Glad it helped :) Great job picking up on something that I don't think I've ever thought about in the HSC context, and I don't think has ever been asked on the forums!
Post by: Aaron12038488 on June 12, 2017, 08:49:47 pm
is it similar to Electricity in home?
Post by: jamonwindeyer on June 12, 2017, 09:00:19 pm
how hard is the motors and generators topic for hsc?
is it similar to Electricity in home?
I'd say it is a little harder, there are some trickier concepts to grasp in there to compared to the Year 11 Topic :)
There is some crossover, and some required Y11 knowledge to really get the Year 12 stuff, but the two are pretty much completely different ;D
Post by: bluecookie on June 12, 2017, 09:25:55 pm
(Just a diagram of a plate with the positive plate at the top, and a positive charge placed in the field at the top, and then moving towards the bottom. No numerical values given for d, or v; they're just labelled as the letters).
The question states: Compare the kinetic energy of the Coulomb of charge at the top plate to the kinetic energy of the charge right before it collides with the bottom plate.
Halllp. How do you find the kinetic energy in terms of electricity?
Post by: jakesilove on June 12, 2017, 10:10:48 pm
Consider the movement of 1C of positive charge in a uniform electric field as shown below.
(Just a diagram of a plate with the positive plate at the top, and a positive charge placed in the field at the top, and then moving towards the bottom. No numerical values given for d, or v; they're just labelled as the letters).
The question states: Compare the kinetic energy of the Coulomb of charge at the top plate to the kinetic energy of the charge right before it collides with the bottom plate.
Halllp. How do you find the kinetic energy in terms of electricity?
Hey! I don't think you'd need to quantify any relationship. You'd rather need to discuss, generally, what the hell is going on.
In an electric field, a charged particle experiences a force. That force is the same, no matter where in the electric field you are, assuming that the electric field stays constant (which it does here). So, what can we say about the force? There is a constant force pushing the charge downwards!
Does that sound familiar? Well, it should; that's how gravity works as well! If there is a constant force applied, in the same direction, then the particle will accelerate downwards. So, if you just wanted to qualitatively discuss the kinetic energy, you'd say that it starts off at zero (as it is at rest), and then increases and an increasing rate the further down it is 'pushed'.
However, you could get a bit more technical than that. You could find that constant accelerating force, and label it 'a'. Then, you could literally just apply the same formulas that you use for projectile motion! For instance, we know that
However, the initial velocity is zero (assuming the particle at rest). So, the velocity of the particle will always be the accelerating force 'a', multiplied by time. Thus, we can plug this into our kinetic energy formula
None of this last part is necessary, just interesting the think about :)
Edit: A bit more 'precise'
We actually know that the force on a charged particle is
Here, q is equal to 1 C, so
But, force also equals mass times acceleration, so
Putting this into our kinetic energy formula
If I'm honest, we could also figure out a better value than t, using the known distance between the plates. ie. We could find how long it takes the particle to cover a distance, d, using projectile motion formulas. All of this is probably beyond the question, so I'll stop here.
Post by: Jyrgal on June 12, 2017, 10:38:23 pm
Determine the force he has to apply on the object
Is it right to consider the acceleration the person has to exert on the object to be 10.3ms^-2 upwards? If not, whats the proper way to figure out this question? Thanks :D
Post by: bluecookie on June 13, 2017, 05:00:06 pm
Thanks!
Post by: bluecookie on June 13, 2017, 05:41:58 pm
http://images.tutorvista.com/content/feed/u496/JJ_Thomson_exp2.png
In the other experiments the anode was always on the opposite end of the cathode, so I assume it was that that was pulling the electron stream towards the anode to complete the circuit. But in this one the anode is placed at the front, so wouldn't the electrons short circuit and not travel the remainder of the tube at all? Why does it continue travelling?
Post by: jakesilove on June 13, 2017, 05:45:32 pm
Halp I don't get this diagram:
http://images.tutorvista.com/content/feed/u496/JJ_Thomson_exp2.png
In the other experiments the anode was always on the opposite end of the cathode, so I assume it was that that was pulling the electron stream towards the anode to complete the circuit. But in this one the anode is placed at the front, so wouldn't the electrons short circuit and not travel the remainder of the tube at all? Why does it continue travelling?
Hey! So in the JJ Thomson experiment, electrons still get fired out of the cathode (left most piece of metal). They are accelerated towards the anode, based on the voltage difference between them (we call this the 'accelerating voltage'. However, there is basically an actual physical hole in the anode. So, some of the electrons flying towards the right hand side go through the hole, and continue in a straight path towards the rest of the experiment!
Again, great questions from you :)
Post by: bluecookie on June 13, 2017, 05:50:56 pm
Hey! So in the JJ Thomson experiment, electrons still get fired out of the cathode (left most piece of metal). They are accelerated towards the anode, based on the voltage difference between them (we call this the 'accelerating voltage'. However, there is basically an actual physical hole in the anode. So, some of the electrons flying towards the right hand side go through the hole, and continue in a straight path towards the rest of the experiment!
Again, great questions from you :)
So the anode has no physical affect on the motion of the cathode rays? Or is it a little, like the cathode rays travel in a slight parabolic arc towards the anode, so maybe the cathode rays towards the outer edge of the hole get influenced by the anode and sway towards the anode, where they touch the metal and get absorbed? But the ones inside (where the attraction of the anode is too weak) don't get enough influence to connect to the anode and thus go through?
Sorry, just want to iron things out here!
Post by: S200 on June 13, 2017, 05:57:03 pm
Sorry, just want to iron things out here!Good luck trying to iron out a parabolic ray... ;D :D
Post by: bluecookie on June 13, 2017, 06:01:10 pm
Good luck trying to iron out a parabolic ray... ;D :DFuck, haha XD
I hate physics. Physics is the death of me. In fact, all subjects are the death of me. I wish I didn't even have to do school at all hahahahah XD
Post by: jakesilove on June 13, 2017, 06:02:46 pm
So the anode has no physical affect on the motion of the cathode rays? Or is it a little, like the cathode rays travel in a slight parabolic arc towards the anode, so maybe the cathode rays towards the outer edge of the hole get influenced by the anode and sway towards the anode, where they touch the metal and get absorbed? But the ones inside (where the attraction of the anode is too weak) don't get enough influence to connect to the anode and thus go through?
Sorry, just want to iron things out here!
Totally fair enough. The electrons are fired, and accelerated, towards the anode. By the time they 'reach' the anode, they are going super super super fast. Some of them go through the hole. Like you say, the ones in the hole WILL be attracted the to inner-surface of the hole. Those that touch the surface will be absorbed. However, as they are going so so fast, they barely have time to 'bend' in any meaningful way. So, they continue on their merry way! Again, you're asking questions beyond the curriculum, but it's still great to have an understanding.
Post by: bluecookie on June 13, 2017, 06:05:19 pm
http://www.rfcafe.com/references/electrical/NEETS-Modules/images/16616imgF.gif
Post by: S200 on June 13, 2017, 06:06:44 pm
I hate physics. Physics is the death of me. In fact, all subjects are the death of me. I wish I didn't even have to do school at all hahahahah XDPhysics is my favourite subject right now...
But I'm def. with you for the rest!!
Post by: bluecookie on June 13, 2017, 06:08:24 pm
Totally fair enough. The electrons are fired, and accelerated, towards the anode. By the time they 'reach' the anode, they are going super super super fast. Some of them go through the hole. Like you say, the ones in the hole WILL be attracted the to inner-surface of the hole. Those that touch the surface will be absorbed. However, as they are going so so fast, they barely have time to 'bend' in any meaningful way. So, they continue on their merry way! Again, you're asking questions beyond the curriculum, but it's still great to have an understanding.
Thanks!
Post by: bluecookie on June 13, 2017, 06:12:44 pm
Physics is my favourite subject right now...Whoa, its awesome that you can enjoy a subject that I will never understand XD Props to you! (Should post on my profile under user comments if you want to keep chatting, don't wanna spam this thread)
But I'm def. with you for the rest!!
Post by: jakesilove on June 13, 2017, 06:12:59 pm
What's an electrostatic lens?
http://www.rfcafe.com/references/electrical/NEETS-Modules/images/16616imgF.gif
Wooooow this is probably the point where I need to put the breaks on; this is way way way beyond anything you need to understand. Forget about it aha
Post by: jakesilove on June 13, 2017, 06:13:38 pm
Whoa, its awesome that you can enjoy a subject that I will never understand XD Props to you! (Should post on my profile under user comments if you want to keep chatting, don't wanna spam this thread)
Feel free to spam the thread! Or, create a new thread, asking people how they're finding Physics! Or, private message each other! Or, chat on each other's profiles! This is your forum as much as ours :)
Post by: S200 on June 13, 2017, 06:21:03 pm
Feel free to spam the thread! Or, create a new thread, asking people how they're finding Physics! Or, private message each other! Or, chat on each other's profiles! This is your forum as much as ours :)Thanks Jake....
I might do the second...
How do you create a new thread??
Post by: jamonwindeyer on June 13, 2017, 06:41:25 pm
A person throws an object of 10kg upwards at 0.5ms^-1 on Earth's surface
Determine the force he has to apply on the object
Is it right to consider the acceleration the person has to exert on the object to be 10.3ms^-2 upwards? If not, whats the proper way to figure out this question? Thanks :D
Hey! So this question feels a little off to me, because it's a little vague. Do you mean \(0.5\text{ms}^{-2}\), so as in an acceleration? Or is that number an initial velocity? Just feels a little off to me - Perhaps snap a picture of the question? :)
Post by: johnk21 on June 13, 2017, 07:17:55 pm
Thanks in advance.
Post by: jamonwindeyer on June 13, 2017, 07:31:05 pm
Can someone please help me with this medical physics question.
Thanks in advance.
Hey! So basically this question is in four parts:
- Why a scan is urgent: This should be clear, they are bleeding from the head! You could go into stuff about how doing it quickly means you are more likely to spot where it is coming from, but really, pretty much anything sensible works here. You'd get a mark for it.
- Drawbacks of CT scans - You'd focus on heavy exposure to ionising radiation, imo. There are others, but this is the biggest.
- Drawbacks of MRI - Although better at visualising soft tissues (and so better at detecting this bleed, most likely), and don't involve ionising radiation, MRI's are far more expensive and usually take longer to perform/more preparation.
- Make the choice - There's no correct answer here, both would work! Just make a choice and justify your answer :)
Does this help? :)
Post by: johnk21 on June 13, 2017, 07:51:40 pm
Hey! So basically this question is in four parts:
- Why a scan is urgent: This should be clear, they are bleeding from the head! You could go into stuff about how doing it quickly means you are more likely to spot where it is coming from, but really, pretty much anything sensible works here. You'd get a mark for it.
- Drawbacks of CT scans - You'd focus on heavy exposure to ionising radiation, imo. There are others, but this is the biggest.
- Drawbacks of MRI - Although better at visualising soft tissues (and so better at detecting this bleed, most likely), and don't involve ionising radiation, MRI's are far more expensive and usually take longer to perform/more preparation.
- Make the choice - There's no correct answer here, both would work! Just make a choice and justify your answer :)
Does this help? :)
yes thanks so much!
Post by: Kle123 on June 14, 2017, 03:36:18 pm
Post by: jakesilove on June 14, 2017, 03:41:09 pm
Could anyone help me understand why the theory that energy is quantised fixed the ultraviolet catastrophe? THANK YOU
Hey! Simple answer is that you don't need to understand this. The important thing is: Planck sees the Ultraviolet Catastrophe. Planck makes up some Maths to explain the Ultraviolet Catastrophe, within which energy is quantised. He thought this was just a mathematical trick, however later Einstein would show that this 'trick' is actually how real life works!
Not in any way a satisfying answer, but the correct one for the HSC
Post by: katnisschung on June 14, 2017, 05:24:23 pm
Post by: jamonwindeyer on June 14, 2017, 05:26:40 pm
why is it with increasing frequency there is increasing energy for an em wave?
This is the old 'classical' way of considering EM waves, based on Rayleigh Jeans Law. Once you consider waves in terms of quanta, that is no longer necessarily the case!! You can read this guide for a bit of an explanation why ;D
Post by: katnisschung on June 14, 2017, 06:30:04 pm
my teacher was talking about him being able to locate the antinodes?
this is the paragraph i can't understand from the notes is got
"he found the wavelength by reflecting the waves back towards the emitter by using a piece of metal.
this set up a standing wave pattern?? between the emitter and reflector. He used (as a probe)
another detecting loop, he was able to locate the antinode. Sparking ocurred best at these places. From the
spacing of the antinodes, he worked out the wavelength.
i only understand it rudimentary term... he superimposed two waves, one directly sent
from the source to detector and another sent from the source to be reflected by a metal plate
and then the two superimpose and he somehow studied the interference patterns and found the wavelength.
Post by: itssona on June 16, 2017, 04:55:59 am
If i have a parallel circuit how wpuld i find voltage across a particular resistor which lies on the series branch?
Post by: S200 on June 16, 2017, 07:52:57 am
I would presume you would still use the 1/Rt = 1/R1+1/R2 etc...
Post by: itssona on June 16, 2017, 09:16:54 am
Could you give us an image??Oih okay makes sense thank youuu
I would presume you would still use the 1/Rt = 1/R1+1/R2 etc...
If i have a current vs voltage graph then do i find R by finding the slope and then inverse?
Also if its not linear then do we use tangent?
Mod Edit: Post merge :)
Post by: jakesilove on June 16, 2017, 09:51:32 am
so i don't fully understand how hertz found the wavelength.
my teacher was talking about him being able to locate the antinodes?
this is the paragraph i can't understand from the notes is got
"he found the wavelength by reflecting the waves back towards the emitter by using a piece of metal.
this set up a standing wave pattern?? between the emitter and reflector. He used (as a probe)
another detecting loop, he was able to locate the antinode. Sparking ocurred best at these places. From the
spacing of the antinodes, he worked out the wavelength.
i only understand it rudimentary term... he superimposed two waves, one directly sent
from the source to detector and another sent from the source to be reflected by a metal plate
and then the two superimpose and he somehow studied the interference patterns and found the wavelength.
You DEFINITELY don't need to know this stuff. DEFINITELY. All you (may) need to know is that he WAS investigating wavelength. Basically, found found areas of constructive and destructive interference, and back calculated the necessary wavelengths to cause that. Seriously, don't bother learning this stuff; no need to overcomplicate an already difficult topic area
Post by: jamonwindeyer on June 16, 2017, 10:42:11 am
If i have a current vs voltage graph then do i find R by finding the slope and then inverse?
Yep! Gradient would be \(\frac{I}{V}\), and by Ohm's Law resistance is \(\frac{V}{I}\), so take the reciprocal ;D
Quote
Also if its not linear then do we use tangent?
If the IV characteristics of a resistance are non-linear that means it is non-ohmic, meaning it doesn't obey Ohm's Law in the simplest sense. You can do resistance as the tangent to the curve if you want to, but you'll never touch that in HSC Physics (because it isn't a Calculus course) ;D
Post by: itssona on June 16, 2017, 04:44:34 pm
Yep! Gradient would be \(\frac{I}{V}\), and by Ohm's Law resistance is \(\frac{V}{I}\), so take the reciprocal ;DAh makes sense, thank you Jamon :)
If the IV characteristics of a resistance are non-linear that means it is non-ohmic, meaning it doesn't obey Ohm's Law in the simplest sense. You can do resistance as the tangent to the curve if you want to, but you'll never touch that in HSC Physics (because it isn't a Calculus course) ;D
Post by: winstondarmawan on June 16, 2017, 10:56:55 pm
Post by: jamonwindeyer on June 16, 2017, 11:18:09 pm
Hello! Just wanted to ask if semiconductors are useful on their own, or are they only useful when used together (p-n junctions) in applications? TIA
Hey! Really only useful when doped to form p and n types - Intrinsic semiconductors aren't very useful by themselves
Post by: Jyrgal on June 16, 2017, 11:29:36 pm
Can someone explain how solar cells work?
thanks!
Post by: jamonwindeyer on June 16, 2017, 11:51:56 pm
Hello ;D ;D
Can someone explain how solar cells work?
thanks!
Hey! That's a big question, understanding how they work requires a solid understanding of semiconductors (if you need a refresher, try this guide) :)
Assuming you have that, this is basically how it works. We have a PN-junction with a depletion zone, and that sets up an electric field. Then, we have some light hitting the junction, and this cause the photoelectric emission of electrons. These electrons experience a force due to the electric field, and this pushes them through an external circuit!! ;D
(If you need a refresher on the photoelectric effect, check this guide!) ;D
Basically, understanding solar cells is a test of a heap of the knowledge you gain in Ideas to Implementation. Read the guides, make sure you have that background knowledge, and understanding solar cells will be a lot easier :)
Post by: Jyrgal on June 17, 2017, 12:41:26 am
Hey! That's a big question, understanding how they work requires a solid understanding of semiconductors (if you need a refresher, try this guide) :)
Assuming you have that, this is basically how it works. We have a PN-junction with a depletion zone, and that sets up an electric field. Then, we have some light hitting the junction, and this cause the photoelectric emission of electrons. These electrons experience a force due to the electric field, and this pushes them through an external circuit!! ;D
(If you need a refresher on the photoelectric effect, check this guide!) ;D
Basically, understanding solar cells is a test of a heap of the knowledge you gain in Ideas to Implementation. Read the guides, make sure you have that background knowledge, and understanding solar cells will be a lot easier :)
Hey, thanks for answering!
I was just wondering, don't solar cells use the photovoltaic effect rather than the photoelectric effect? (should i mention this if a question asked about solar cells)
Another question, where exactly do the electrons get released from? From my understanding, electrons from the n-type semiconductor gets excited and jumps into the conduction band, creating a positive hole in the valence band. This will attract electrons from the p-type semiconductor which with the energy provided by attraction force will jump across the depletion zone and fill the n-type holes. However, my teacher says that the light knocks off electrons from within the depletion zone itself, and since there is an electric field with the positive side on the n-type, the electron will from to the n-type. If this is the case, how does the light ever reach the depletion zone to knock off electrons?
Sorry if its a bit of a long read, I have a slight interest in these type of stuff so I wanna have a good understanding :P.
Thanks!
Post by: jamonwindeyer on June 17, 2017, 12:51:57 am
Hey, thanks for answering!
I was just wondering, don't solar cells use the photovoltaic effect rather than the photoelectric effect? (should i mention this if a question asked about solar cells)
You are totally right! It is just that at this level the two terms are interchangeable - The two effects are the same, except in the photovoltaic effect, the electron isn't ejected but instead still contained within the material. You are okay to use either term, most HSC resources/marking schemes would just use the photoelectric effect so as to link to other bits of the syllabus ;D
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Another question, where exactly do the electrons get released from? From my understanding, electrons from the n-type semiconductor gets excited and jumps into the conduction band, creating a positive hole in the valence band. This will attract electrons from the p-type semiconductor which with the energy provided by attraction force will jump across the depletion zone and fill the n-type holes. However, my teacher says that the light knocks off electrons from within the depletion zone itself, and since there is an electric field with the positive side on the n-type, the electron will from to the n-type. If this is the case, how does the light ever reach the depletion zone to knock off electrons?
Hmm, both sound like sensible explanations to me!! It just sounds like you are going to an explanation based on energy levels, and your teacher is just being a little more abstract. The question of whether light reaches the depletion zone, you can really sort of extend to the N-type semiconductor too - Neither would be directly exposed in a solar cell. In practice, anti-reflective coatings are used to ensure the light is 'captured' and allowed to reach the semiconductors. Really, it will work wherever the electron is released :)
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Sorry if its a bit of a long read, I have a slight interest in these type of stuff so I wanna have a good understanding :P.
Thanks!
Don't be sorry! I was the same in the HSC, it is why I'm studying Electrical Engineering now ;D
Post by: beau77bro on June 17, 2017, 08:10:31 pm
Post by: jakesilove on June 18, 2017, 12:02:39 am
Hey so I'm kind of stuck at an impass. My teacher said we could do what ever option we wanted, but my head teacher has said we can't. She has said we r only doing medical in the trial because my teacher is pregnant and she might leave and the head only knows medical - but that still doesn't make sense because she wouldn't be teaching us Watever we choose in the first place... Anyway. I had cancer as a small child and I'm not very comfortable looking into radiotherapy and all that kind of medical stuff - which luckily won't be in the trial as we only cover first two dot points. But I'm still stuck with the decision of do I keep it for the HSC or do I do quanta - which I'm so interested in but I know it's quite difficult. I will have to learn some of medical for the trial anyway so I was wondering if I should just grit my teeth and do it, or if u guys think I could pull off quanta for the HSC? I do think I could dedicate time to it - and that will ultimately get me the results I need (ESP if it's interesting) but I do have to improve my English as well as carrying the immense load that is 4u maths Ahahha. Any and all advice is appreciated for this lil pickle.
Hey man,
Really sorry to hear about your previous health issues. I'm especially sorry to hear that those issues impact your wellbeing now, I can only imagine how tough that must be.
Generally, I would always recommend just to grit your teeth and do what your school teaches you. You'll have more resources to rely on, you'll have the benefit of them marking you, and you'll have to put less work in.
That being said, I've never given that advice in a circumstance like this. If you think learning that content will affect your mental health, then you should absolutely be discussing with you head teacher the option of doing another topic. Surely they would understand. I think, basically, you need to balance up your various interests here, and decide what you'd rather put yourself through. Unfortunately, this isn't a decision we can really help you with. Like I said, I would generally recommend just doing Medical Physics, but there are clearly real and important factors that make this study difficult. So, I guess I've said as much as I can. Feel free to keep chatting about this here, though; happy to help out in any way this community allows.
Post by: johnk21 on June 19, 2017, 09:05:26 pm
Hey so I'm kind of stuck at an impass. My teacher said we could do what ever option we wanted, but my head teacher has said we can't. She has said we r only doing medical in the trial because my teacher is pregnant and she might leave and the head only knows medical - but that still doesn't make sense because she wouldn't be teaching us Watever we choose in the first place... Anyway. I had cancer as a small child and I'm not very comfortable looking into radiotherapy and all that kind of medical stuff - which luckily won't be in the trial as we only cover first two dot points. But I'm still stuck with the decision of do I keep it for the HSC or do I do quanta - which I'm so interested in but I know it's quite difficult. I will have to learn some of medical for the trial anyway so I was wondering if I should just grit my teeth and do it, or if u guys think I could pull off quanta for the HSC? I do think I could dedicate time to it - and that will ultimately get me the results I need (ESP if it's interesting) but I do have to improve my English as well as carrying the immense load that is 4u maths Ahahha. Any and all advice is appreciated for this lil pickle.Sorry to hear about what you have been through, it really sucks :(
I am doing medical physics atm and i see that it mostly if not ALL memorisation.
My opinion would be if only the first two points are in your trials, literally just look at hsc past papers and memorise the answer because in Medical Physics the questions they can ask are VERY limited. Then do whatever option you want to do for the hsc. Worst case scenario is you just memorise everything for medical physics for the hsc, which is easily completable within a week imo.
Post by: beau77bro on June 19, 2017, 10:28:45 pm
Sorry to hear about what you have been through, it really sucks :(
I am doing medical physics atm and i see that it mostly if not ALL memorisation.
My opinion would be if only the first two points are in your trials, literally just look at hsc past papers and memorise the answer because in Medical Physics the questions they can ask are VERY limited. Then do whatever option you want to do for the hsc. Worst case scenario is you just memorise everything for medical physics for the hsc, which is easily completable within a week imo.
Thanks johnk21, appreciate the advice and will do - worried because the option I wanna do is quanta though haha- but I appreciate the support and I'm very reassured by the simplicity of medical as you described - if it's half as easy as that it will make it a lot easier to study even when I don't wanna do it?
Post by: jamonwindeyer on June 19, 2017, 11:04:28 pm
Thanks johnk21, appreciate the advice and will do - worried because the option I wanna do is quanta though haha- but I appreciate the support and I'm very reassured by the simplicity of medical as you described - if it's half as easy as that it will make it a lot easier to study even when I don't wanna do it?
Medical is a very memorisation heavy option, a few tricky bits and pieces to really understand how the scanning methods all work - I personally found Medical a really interesting Option myself though :)
Personally, your personal health aside, I'd recommend to do Medical. Purely to avoid double dipping. That said, you can absolutely learn Quanta yourself if you need to. Worst case, find a tutor who can spend a couple of weekends teaching it to you properly after you've learned Medical for your Trial.
You absolutely need to do what is best for you. If it puts your mind at ease, the way Medical is taught is very much Physics focused, as you'd imagine. The only relation to the human body is what sorts of afflictions each technique is good at spotting - Besides that, it is all on the science. Definitely no specific focus on the treatment and diagnosis end, it is more on how they work and a little bit of why you'd use one over the other :)
Post by: Maraos on June 19, 2017, 11:34:23 pm
I've got a practical examination this Thursday (worth 20% :-\ :-\) and I'm not really sure how i should prepare for it.
The notification is practically useless since the only information it gives is that the work covered involves 'all skills work as outlined by the syllabus'. My teacher wouldn't even tell us what topic the exam will be on >:(
This exam is different to my last 'skills' exam, which did not consist of a practical side to it. This exam will involve 20+ students partaking in pracs followed by questions etc. So I guess we are limited to only some pracs in the sense that 20 students will all have to do it.
Any advice, or ideas on what it could be would be greatly appreciated :) :)
Thanks! ;D
Post by: jamonwindeyer on June 19, 2017, 11:35:20 pm
Hey guys :D
I've got a practical examination this Thursday (worth 20% :-\ :-\) and I'm not really sure how i should prepare for it.
Hey Maraos! Which topic is this for? Ideas to Implementation or? :)
Post by: Maraos on June 19, 2017, 11:41:33 pm
Hey Maraos! Which topic is this for? Ideas to Implementation or? :)My teacher didn't specify, he won't even give us any hints haha.
So far my class has only done Motors/Generators, Ideas and Quanta. However my teacher did say that the prac will be on something we haven't done yet
Considering the entire class will need to do it I'm guessing it will have to involve a prac where alot of people can take part in it. So circuits? Motion pracs involving weights and retort stands?
Post by: jamonwindeyer on June 19, 2017, 11:48:13 pm
My teacher didn't specify, he won't even give us any hints haha.
So far my class has only done Motors/Generators, Ideas and Quanta. However my teacher did say that the prac will be on something we haven't done yet
Considering the entire class will need to do it I'm guessing it will have to involve a prac where alot of people can take part in it. So circuits? Motion pracs involving weights and retort stands?
Rats! Very strange - It could even be a Space prac with formulas given! You might be getting the pendulum experiment :)
In any case, try and read up on the following:
- The theory, if you can, but for you this is difficult because no hints!
- Accuracy, validity, reliability
- Variables (independent, dependent, controlled)
- Graphing and lines of best fit
- Discussions (what to include, how it is structured, etc)
There's not much else you can do unfortunately, especially with so little to go on! It might be worth seeing if there are any prac exams on THSC that you can use as a guide ;D
Post by: Maraos on June 19, 2017, 11:54:27 pm
Rats! Very strange - It could even be a Space prac with formulas given! You might be getting the pendulum experiment :)Thanks for the advice and quick reply! ;D
In any case, try and read up on the following:
- The theory, if you can, but for you this is difficult because no hints!
- Accuracy, validity, reliability
- Variables (independent, dependent, controlled)
- Graphing and lines of best fit
- Discussions (what to include, how it is structured, etc)
There's not much else you can do unfortunately, especially with so little to go on! It might be worth seeing if there are any prac exams on THSC that you can use as a guide ;D
I'll check out that site, also I'll see if i can squeeze anything out of my teacher this week haha :D
Post by: johnk21 on June 21, 2017, 01:54:01 pm
Post by: winstondarmawan on June 21, 2017, 06:30:11 pm
1. Can someone please explain Pauli's exclusion principle and electron's quantum numbers clearly? We went over it but I don't really understand it properly.
2. What are the things to look out for and include in a Prac Exam discussion? I lost many marks there and I'd like to avoid it in the future.
TIA
Post by: jamonwindeyer on June 21, 2017, 08:16:09 pm
Can someone please explain to me what Planck and Einsteins contribution are to the black body radiation curve? Thanks in advance :)
Hey John! I've written a couple of guides that cover this general topic area, see them here and here ;D Hope they help!
Post by: Maraos on June 21, 2017, 08:37:14 pm
I've got a question regarding the Pendulum experiment (when you are altering the mass to determine the acceleration due to gravity).
How would you improve the validity of this experiment?
My solution would be to ensure that the controlled variables are kept constant (even more constant then what they currently are).
ie: using a digital measuring gauge to ensure that the angular displacement (theta) of the pendulum is kept constant (as opposed to using a protractor which would increase the chances of human reading error).
Would this answer be fine? Also what else what you include to improve the validity of this particular experiment? And is my understanding of validity even correct? ;D ;D
Any help would be greatly appreciated ;D
Thanks :)
Post by: beau77bro on June 21, 2017, 08:40:30 pm
Hello!
1. Can someone please explain Pauli's exclusion principle and electron's quantum numbers clearly? We went over it but I don't really understand it properly.
2. What are the things to look out for and include in a Prac Exam discussion? I lost many marks there and I'd like to avoid it in the future.
TIA
Is this in ideas for superconductors? If so this is very very far from what you need to know. From what I understand of it- two electrons can't spin in the same state inside a quantum system - cooper pairs don't follow this and spin in the same state. This along with the lattice structure vibrating so little at the critical temperatures means these pairs can form. - I think. Quantum numbers I do not know and I look forward to a better explanation from the moderators.
Hope this helps a lil
Post by: winstondarmawan on June 21, 2017, 08:55:11 pm
Is this in ideas for superconductors? If so this is very very far from what you need to know. From what I understand of it- two electrons can't spin in the same state inside a quantum system - cooper pairs don't follow this and spin in the same state. This along with the lattice structure vibrating so little at the critical temperatures means these pairs can form. - I think. Quantum numbers I do not know and I look forward to a better explanation from the moderators.
Hope this helps a lil
No, this is for Quanta to Quarks (option topic). However, the help is appreciated. :)
Post by: kiwiberry on June 21, 2017, 09:38:57 pm
Hello!
1. Can someone please explain Pauli's exclusion principle and electron's quantum numbers clearly? We went over it but I don't really understand it properly.
2. What are the things to look out for and include in a Prac Exam discussion? I lost many marks there and I'd like to avoid it in the future.
TIA
Not 100% confident with this, but I'll give it a shot
The four quantum numbers are:
- Principal quantum number (n) - describes the energy shells and the electron's distance from the nucleus
- Angular momentum quantum number (L) - describes the shape of orbit
- Magnetic quantum number (ml) - describes the orientation of the electrons
- 'Spin' quantum number (ms) - describes the spin axis of the electrons
Pauli's exclusion principle states that no two electrons can have the same 4 quantum numbers. This provided the reason for the maximum number of electrons in each shell, thus explaining the position of elements on the periodic table.
For prac discussions, the main things to talk about are reliability, accuracy, validity, and any improvements you could make. Hope that helped :)
Post by: johnk21 on June 22, 2017, 09:20:21 pm
Explain the difference in the relaxation times of hydrogen in water and the relaxation time of hydrogen in other molecules. (3 marks)
Post by: winstondarmawan on June 24, 2017, 06:57:29 pm
Question:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19433791_1250773725048148_1099258653_n.jpg?oh=d51a25252a0cfc50107a2c6784586cd9&oe=59508E20
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19441301_1250773908381463_404155012_n.jpg?oh=95f7190931fa49c04e37e18e47b22826&oe=5950836F
Answer:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19244049_1250774138381440_1719650737_n.jpg?oh=5abcee2c34e5f38fd9843fa7176ddb7b&oe=5950D4BD
Post by: Jyrgal on June 24, 2017, 08:35:06 pm
I just need to get my head around length contraction & time dilation haha, unfortunately alot of past papers/questions get this wrong and im extremely confused on whether my thinking is right
So let's say from a stationary frame of reference, someone views a planet of length L. When this someone climbs into a spaceship and flies at 0.8C, he sees the L get contracted, while time also decreases (not legit but an example: 1 second on spaceship=2seconds on the stationary reference). They should both decrease in proportion as C(speed of light)=Distance/Time is constant
Is this correct?
Thanks :):)
Post by: jakesilove on June 25, 2017, 09:12:04 am
Can someone please help me with this HSC question for med physics.
Explain the difference in the relaxation times of hydrogen in water and the relaxation time of hydrogen in other molecules. (3 marks)
Hey! Hydrogen ions in water have a very long relaxation time (T1 is about 4000 ms, T2 is about 2000 ms). However, Hydrogen ions in other substances have a much shorter relaxation time. This makes testing for water concentration very easy, as the 'signature' of water is very distinct!
Post by: jakesilove on June 25, 2017, 09:16:10 am
Hello! Can someone please help me understand this question, I have the answer but I do not understand it.
Question:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19433791_1250773725048148_1099258653_n.jpg?oh=d51a25252a0cfc50107a2c6784586cd9&oe=59508E20
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19441301_1250773908381463_404155012_n.jpg?oh=95f7190931fa49c04e37e18e47b22826&oe=5950836F
Answer:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19244049_1250774138381440_1719650737_n.jpg?oh=5abcee2c34e5f38fd9843fa7176ddb7b&oe=5950D4BD
Hey! I can't really read the question or the answer based on that link, but I THINK you're talking about how we tell how much to shift a graph by, when a stopping voltage has initially been applied, and then taken away?
So, the stopping voltage was originally 4.1V. When it is removed, we can calculate exactly how many 'extra' energy the electrons will have (remember, stopping voltage is energy used to 'restrain' electrons to the nucleus. Thus, if you remove the stopping voltage, the electrons will be able to fire off more easily!). Convert 4.1V to energy (I think the method is on the formula sheet). Then, set that equal to hf, and find the new 'threshold frequency'. Finally, make sure the graph has a gradient of 'h', and you're done!
Sorry that this isn't more comprehensive, I've got an exam tomorrow :)
Post by: jakesilove on June 25, 2017, 09:24:38 am
Hello!
I just need to get my head around length contraction & time dilation haha, unfortunately alot of past papers/questions get this wrong and im extremely confused on whether my thinking is right
So let's say from a stationary frame of reference, someone views a planet of length L. When this someone climbs into a spaceship and flies at 0.8C, he sees the L get contracted, while time also decreases (not legit but an example: 1 second on spaceship=2seconds on the stationary reference). They should both decrease in proportion as C(speed of light)=Distance/Time is constant
Is this correct?
Thanks :):)
Hey! Pretty much everything you've said above is correct! Sounds like you have a good understanding of the Space topic area :) However, I don't think that the ratio of length to time is always constant. The formulas just don't indicate as much. You've set the speed of light as velocity; in the frame of the light particle, then yes length over time is always constant. However, I'm not certain that this is the case for an observer travelling at less than the speed of light.
Basically, it doesn't matter, there's no point knowing/talking about this because it is outside the curriculum. Just focus on being able to use the formulas, and explain the principle.
Post by: bsdfjnlkasn on June 28, 2017, 10:23:50 am
Just reviewing my Ideas to Implementation summary and have a few questions :)
Why do we use a high voltage source in cathode ray tubes? And why can it only be sourced from an induction coil? Why can't a transformer be used?
Do we need to know why/how Hertz was inaccurate in his experiments which supposedly proved the wave properties of light?
If so, how does a cathode tube with some gas remaining (not completely evacuated) cause no observable deflection when an electric field is applied?
Considering that F = qvBsin(theta) and F = qE, does E = vbson(theta)?
Is theta always 90° when the magnetic field is going in or out of the page and the particle is moving left/right/up/down, or in any direction around the page?
Thank you so much!!
(Expect some edits with more questions :) )
Post by: jamonwindeyer on June 28, 2017, 02:18:34 pm
Hi there,
Just reviewing my Ideas to Implementation summary and have a few questions :)
Sure thing!
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Why do we use a high voltage source in cathode ray tubes? And why can it only be sourced from an induction coil? Why can't a transformer be used?
A high voltage is required for the air in the tube (even the reduced amount we get by using an evacuated tube) to break down and allow a flow of electrons through it. This is called breakdown. Essentially, it is the point where even the resistance of the air in the tube isn't enough to stop a current flow between the cathode and anode.
Pretty sure you can get this voltage from anywhere you want to, but it needs to be DC. You can't use a transformer directly because that would be AC, but you could rectify the voltage coming out of a transformer if you wanted to :)
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Do we need to know why/how Hertz was inaccurate in his experiments which supposedly proved the wave properties of light?
I wouldn't put too much effort into it. Just say lack of accurate measuring equipment, that's what I'd do!
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If so, how does a cathode tube with some gas remaining (not completely evacuated) cause no observable deflection when an electric field is applied?
Not quite sure what you are asking here sorry, but no cathode ray tube is completely evacuated. There's always some air left inside. The amount of air will change the voltage you need across the plate, as well as the striation patterns, but within reason small differences in the amount of air in a tube won't affect much else :)
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Considering that F = qvBsin(theta) and F = qE, does E = vbson(theta)?
Not quite, those are two different forces. \(F=Bqv\sin{\theta}\) is the force on a moving charge due to a magnetic field, while \(F=Eq\) is the force on a charge due to an electric field. Together, we call the sum of these forces the Lorentz Force, but they aren't the same thing :)
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Is theta always 90° when the magnetic field is going in or out of the page and the particle is moving left/right/up/down, or in any direction around the page?
Yep :)
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Thank you so much!!
(Expect some edits with more questions :) )
You're welcome! Maybe ask the rest in a new post so we don't miss it ;D
Post by: Charlie_Sparkes on June 28, 2017, 06:05:45 pm
Just having a bit of trouble with the following question, and I cant seem to find the right resource...
A 1W beam of light transfers 1J per second from one point to another. With reference to the particle model of light, contrast a 1W beam of red light and a 1W beam of blue light.
Thanks in advance for the help :)
Post by: katnisschung on June 28, 2017, 06:26:05 pm
Post by: kiwiberry on June 28, 2017, 06:27:19 pm
Hi All,
Just having a bit of trouble with the following question, and I cant seem to find the right resource...
A 1W beam of light transfers 1J per second from one point to another. With reference to the particle model of light, contrast a 1W beam of red light and a 1W beam of blue light.
Thanks in advance for the help :)
Hey, welcome to the forums!!
Blue light has a higher frequency than red light. Since E=hf, the energy carried by a photon of blue light will be much higher than a photon of red light. Therefore, less photons are needed to produce a 1J/s or 1W beam of blue light than red light. :)
Post by: bsdfjnlkasn on June 28, 2017, 08:54:36 pm
You're welcome! Maybe ask the rest in a new post so we don't miss it ;D
You're a legend Jamon! Super detailed and hopeful :D
Still have a few more questions, if it's ok :)
Does the air in the tubes then ionise and because of this, charges are able to pass through the cathode ray tube?
To what detail should we know the striation patterns for the different pressures in discharge tube experiment?
Thank you :D
P.S. I'll just keep adding questions until things start to get answered - that way questions won't go unanswered.
Post by: jamonwindeyer on June 28, 2017, 09:53:20 pm
Does the air in the tubes then ionise and because of this, charges are able to pass through the cathode ray tube?
Yep, that's correct - The large potential differences ionises the air and allows the electrons to flow. Note that this is a little different to the thermionic emission method of electron ejection from a cathode ;D
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To what detail should we know the striation patterns for the different pressures in discharge tube experiment?
Not in huge detail - Understand why the pattern changes in the first place (changing air pressure, different amounts of air particles to collide with), and have a rough idea of how it changes (lower air pressures, less pattern/pattern focused close to the anode). As far as the names of the spaces and such, I don't think that is super important knowledge :)
Post by: bsdfjnlkasn on June 30, 2017, 07:40:55 pm
Some more questions :)
I'm just reading up on Hertz's experiments and how he observed the photoelectric effect but 'failed to investigate' it and was wondering if this part of the syllabus outcome just needs you to detail the experiments and state that he didn't provide an explanation?
I'm a bit confused with the results he observed (and am not sure if we need to know why he noted what he did, if we don't an explanation would still be cool 8) ). So when he enclosed his received loop to allow better observations, he saw the spark length (and intensity?) decrease. What's the relationship between spark length and intensity in this case? I imagine the more UV light the receiver is exposed to, the greater the photocurrent and so the longer the possible spark length. But how does intensity relate here? Should we just take for granted that as more photoelectrons are released, the brighter the stream is?
Thanks again! :D
Post by: jamonwindeyer on June 30, 2017, 07:45:51 pm
I'm just reading up on Hertz's experiments and how he observed the photoelectric effect but 'failed to investigate' it and was wondering if this part of the syllabus outcome just needs you to detail the experiments and state that he didn't provide an explanation?
Yep, that's correct! ;D
Quote
I'm a bit confused with the results he observed (and am not sure if we need to know why he noted what he did, if we don't an explanation would still be cool 8) ). So when he enclosed his received loop to allow better observations, he saw the spark length (and intensity?) decrease. What's the relationship between spark length and intensity in this case? I imagine the more UV light the receiver is exposed to, the greater the photocurrent and so the longer the possible spark length. But how does intensity relate here? Should we just take for granted that as more photoelectrons are released, the brighter the stream is?
Thanks again! :D
I don't have an exact mathematical relationship, but quantitatively you are correct! Enclosing the coil gets rid of some of your photons which reduces the photocurrent. The intensity of a spark is based on the size of the current (I believe), so it makes sense that intensity and length would decrease ;D
(And this isn't crucial to remember for HSC Physics) :)
Post by: bsdfjnlkasn on June 30, 2017, 07:57:28 pm
Yep, that's correct! ;D
I don't have an exact mathematical relationship, but quantitatively you are correct! Enclosing the coil gets rid of some of your photons which reduces the photocurrent. The intensity of a spark is based on the size of the current (I believe), so it makes sense that intensity and length would decrease ;D
(And this isn't crucial to remember for HSC Physics) :)
Awesome!!
Now for some questions about black body radiation, I'm just not sure what the relationship between temperature, intensity and wavelength are. I can make sense of it by picturing the curve (is it called a Rayleigh Jean curve, or is this just the law that predicted it's shape?) but don't know if I have to explain why increasing the temperature changes the peak intensity. And on that, why does the wavelength at which the maximum occurs decrease when temperatures increases?
Where is it relevant to mention Maxwell (because he is not mentioned in the syllabus) and his work with EMR?
Thanks again :D
Post by: jamonwindeyer on June 30, 2017, 09:04:29 pm
Now for some questions about black body radiation, I'm just not sure what the relationship between temperature, intensity and wavelength are. I can make sense of it by picturing the curve (is it called a Rayleigh Jean curve, or is this just the law that predicted it's shape?) but don't know if I have to explain why increasing the temperature changes the peak intensity. And on that, why does the wavelength at which the maximum occurs decrease when temperatures increases?
So that curve is just called a black body radiation curve, and the theoretical shape of it (the one that sparked the ultraviolet catastrophe) was governed by Rayleigh Jean's Law.
Here's a guide I wrote on Quantum Physics! It has a little on the black body curves and such - Basically, a black body of a higher temperature will have a smaller characteristic wavelength because the hotter a black body, the more likely it will be to have a large change in the energy of one of its electrons (more vibrations). Since the energy of a photon is dependent on some energy change within the atomic structure of the black body, this also means that the energy of emitted photons will increase. Energy is dependent on frequency, so characteristic frequency goes up (and characteristic wavelength goes down :)
This is a tough thing to explain, does this make sense? Happy to clarify ;D
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Where is it relevant to mention Maxwell (because he is not mentioned in the syllabus) and his work with EMR?
Thanks again :D
Never - You don't need Maxwell in HSC Physics ;D
Post by: winstondarmawan on July 01, 2017, 02:33:24 pm
My teacher said that the photocell only works when light strikes the p-type and not the n-type, because if electrons travel from the n-type through the circuit and fill a hole in the p-type, then the depletion layer becomes larger and larger as the n and p-type layers become more positive and negative respectively, stopping any more electrons from moving anywhere. This makes sense. However, other sources say that the electrons are released from the n-type. Just want to clarify this. TIA
Post by: jamonwindeyer on July 01, 2017, 02:41:27 pm
Hello! Just a querie about photocells:
My teacher said that the photocell only works when light strikes the p-type and not the n-type, because if electrons travel from the n-type through the circuit and fill a hole in the p-type, then the depletion layer becomes larger and larger as the n and p-type layers become more positive and negative respectively, stopping any more electrons from moving anywhere. This makes sense. However, other sources say that the electrons are released from the n-type. Just want to clarify this. TIA
Hey Winston! It can happen anywhere where the electric field has a substantial effect (I think most HSC sources just say it has to be in the depletion zone? In truth it doesn't matter too much). P-type or N-type, any released electrons will be forced through the external circuit by the electric field in the depletion layer ;D
This said, it is the P-type which has more electrons in the depletion layer (the negatively charged silicon ions formed by recombination) - I'd wager that is why your teacher said P-type!
I've honestly never considered the idea that electrons flowing through the circuit would recombine with holes further back in the P-type. I don't think that quite works because then you'd end up with an asymmetric distribution of charge - Just doesn't make sense in my head...
All this said, you don't need to stress about these sort of little details in the HSC. Light strikes solar cell, released electrons flow through circuit due to electric field in the depletion layer. Doneskies ;D
Post by: limtou on July 03, 2017, 08:14:18 pm
1. What properties of cathode ray are determined, when it fluoresces?
2. Can you explain the changing acceleration of a rocket during launch in terms of the Law of Conservation of Momentum? (straight from the dot point haha) What are we expected to know for this?
Thanks :)
Post by: JeffChiang on July 03, 2017, 10:15:35 pm
Hello! Have a few questions
1. What properties of cathode ray are determined, when it fluoresces?
2. Can you explain the changing acceleration of a rocket during launch in terms of the Law of Conservation of Momentum? (straight from the dot point haha) What are we expected to know for this?
Thanks :)
Hello, limtou
1) Cathode ray fluorsence is detected using a fluorescent display screen. This demonstrates that it induces fluorescence on phosphors which is a property of electromagnetic waves.
2) You basically need to be able to describe the different stages of a rocket during launch in terms of force. I don't think you have to go into a lot of detail for it.
e.g. A rocket sits on the launchpad experiencing 1g.
The rocket then launches upwards ejecting large amounts of exhaust gas downwards which opposes the direction of the rocket's motion. Due to the conservation of momentum, the rocket will go upwards as there is an equal and opposite reaction from the gas which propels the rocket.
Post by: jamonwindeyer on July 03, 2017, 10:18:49 pm
2. Can you explain the changing acceleration of a rocket during launch in terms of the Law of Conservation of Momentum? (straight from the dot point haha) What are we expected to know for this?
Thanks :)
In addition to the above great answer, let me link you to this guide on rocket launches! It goes into a few bits of info that will help with that dot point ;D
Post by: limtou on July 04, 2017, 02:32:15 pm
In addition to the above great answer, let me link you to this guide on rocket launches! It goes into a few bits of info that will help with that dot point ;D
Thanks Jamon!
Just another query, I have attached below the sample answer for a past school question on rocket launch and the momentum involved... and it states that the rocket is continually gaining momentum? Is this correct? I thought the rocket has a constant momentum so that with decreasing mass, there is increasing velocity.
Please explain, thanks!
Post by: limtou on July 04, 2017, 02:38:06 pm
Could you explain how to draw the loops of flux for a)? I don't particularly understand the marking criteria.
And please explain the answer for q5 multiple choice
Thank you~
Post by: Aaron12038488 on July 04, 2017, 03:09:11 pm
What will the prac likely be on, ahha sorry if this post is queer. thx. :o
Post by: jakesilove on July 04, 2017, 04:11:41 pm
Thanks Jamon!
Just another query, I have attached below the sample answer for a past school question on rocket launch and the momentum involved... and it states that the rocket is continually gaining momentum? Is this correct? I thought the rocket has a constant momentum so that with decreasing mass, there is increasing velocity.
Please explain, thanks!
Hey! 'Gaining' momentum is definitely a silly way to put it; however, it does get the point across, and is correct. Basically, there are two separate things going on here.
Firstly, remember that momentum must always be conserved. The gas has mass, and is moving at some velocity downwards. Therefore, the fuel has a momentum in the downwards direction (let's call this negative). By conservation of momentum, there must also be some sort of momentum UPWARDS to cancel out this downwards push. Therefore, the rocket accelerates upwards, and 'gains' momentum. Net change in momentum is zero; gas goes down, rocket goes up.
The second principle is the one you've described. As the rocket goes upwards, it loses mass, and thus increases its velocity.
But yeah, Rocket definitely increases its total momentum as fuel gets pushed out the bottom :)
Post by: jamonwindeyer on July 04, 2017, 04:28:47 pm
Have a few more questions,
Could you explain how to draw the loops of flux for a)? I don't particularly understand the marking criteria.
And please explain the answer for q5 multiple choice
Thank you~
Hey! You need to draw two loops, both circles going through the core and the bar. They cannot cross (magnetic field lines produced by a field in this fashion never do). They should also stay roughly the same distance apart as they travel around the coil, through the bar, back into the coil and to the end of the loop :)
For the multiple choice - When we do work putting something into orbit, that work is converted to two types of energy (neglecting losses). Kinetic energy (getting the object moving), and potential energy (raising the object in the field). In other words:
Post by: jamonwindeyer on July 04, 2017, 04:30:00 pm
im so nervous that in 1 more term im in year 12!. My weakest subject by far is Physics. Looking at the schedule of assessments (2016) it says I have a practical which is worth 15%. My teacher said it was on MOtors and Generators. Its held on Term 1 week 4 and it states 5% is based on knowledge and understanding, and 10% on skills in planning and conducting investigations, however 0 in Skills in communicating information, scientific thinking, problem solving and working as an individual/teamwork.
What will the prac likely be on, ahha sorry if this post is queer. thx. :o
Chances are it will be based on electromagnetic induction - At least, that is what makes the most sense in my head :) don't stress about it yet though! You don't know the content yet, and you've got Prelim exams before that - You seriously shouldn't be preparing for any Year 12 assessments just yet :)
Post by: left right gn on July 04, 2017, 09:27:17 pm
Any suggestions on the model?
Post by: winstondarmawan on July 05, 2017, 10:44:28 am
I have an upcoming assignment requires me to perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor.
Any suggestions on the model?
Try students in a classroom. N-type you can have an extra student who does not require seats to move around, and p-type you can have an extra vacant seat.
Post by: left right gn on July 05, 2017, 04:30:43 pm
Try students in a classroom. N-type you can have an extra student who does not require seats to move around, and p-type you can have an extra vacant seat.Any other ideas that do not require students as a part of my model?
Post by: seventeenboi on July 05, 2017, 05:00:27 pm
dumbass me is still struggling to understand the inconsistencies of the classical theory's interpretation of blackbody radiation and the UV catastrophe :( so far i know that they thought that emission was dependent on frequency, and that energy is absorbed is continuous ( that is, can occur at any amount and increases as wavelength becomes shorter except this true, but only applies to longer wavelengths)
could someone directly pinpoint and outline specifically ALL the things that were wrong with it?
cheers :)
Post by: winstondarmawan on July 05, 2017, 05:46:07 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19668099_1262289940563193_1371043497_n.png?oh=465b02f227d5f6f6ed9bc5344f18bb25&oe=595E44A2
Post by: pikachu975 on July 05, 2017, 05:54:31 pm
Hello! Need help with the following Q, TIA.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19668099_1262289940563193_1371043497_n.png?oh=465b02f227d5f6f6ed9bc5344f18bb25&oe=595E44A2
The particle model of light states that light is carried in photons which are quantised packets of energy with E = hf. You'd contrast them by saying they have different frequencies which means different energies by E = hf. You have to use the stimulus given so probably reference that they would have the same brightness because brightness of a light can be measured by Watts. This is because they give that they have the same intensity so Intensity = Wm^-2 and m^-2 and Intensity would be equal, so Watts are equal. That should be enough for 2 marks!
Post by: JeffChiang on July 05, 2017, 06:40:55 pm
Still remain somewhat confused from this dot point.
From my understanding, magnetic flux= number of magnetic field lines
magnetic flux density= number of magnetic field lines per unit area
So, if you were given a question to compare two coils in a magnetic field, would you literally count the amount of lines that run through the coil?
Is magnetic flux density the same as the strength of a magnetic field?
Thanks
Post by: jamonwindeyer on July 05, 2017, 09:19:33 pm
Any other ideas that do not require students as a part of my model?
You could perhaps do the same thing with a drawn grid and counters? The basic principle could probably be modelled in many ways (students, cars and parking spaces, counters/markers, etc) ;D
Post by: jamonwindeyer on July 05, 2017, 09:24:23 pm
HIHI :)
dumbass me is still struggling to understand the inconsistencies of the classical theory's interpretation of blackbody radiation and the UV catastrophe :( so far i know that they thought that emission was dependent on frequency, and that energy is absorbed is continuous ( that is, can occur at any amount and increases as wavelength becomes shorter except this true, but only applies to longer wavelengths)
could someone directly pinpoint and outline specifically ALL the things that were wrong with it?
cheers :)
Hey! You've pretty much spotted the two big issues, you won't need much else in the HSC! They thought intensity was related to frequency through a law called Rayleigh Jeans Law, which as you say, worked for longer wavelengths but started breaking down for shorter ones. Note that this classical theory also violated the conservation of energy - You can't have intensity (and thus, energy) approaching infinity at short wavelengths. That makes no sense. The idea of quantised energy was then the eventual solution ;D
Btw, definitely not a dumbass, this is tough stuff! And it seems you've got a solid understanding - Is anything particularly confusing you? :)
Post by: jamonwindeyer on July 05, 2017, 09:32:29 pm
What exactly is magnetic flux and magnetic flux density?
Still remain somewhat confused from this dot point.
From my understanding, magnetic flux= number of magnetic field lines
magnetic flux density= number of magnetic field lines per unit area
So, if you were given a question to compare two coils in a magnetic field, would you literally count the amount of lines that run through the coil?
Is magnetic flux density the same as the strength of a magnetic field?
Thanks
You are correct on every single point! ;D note though that magnetic flux does have a more rigorous mathematical definition than "the number of field lines." The field lines are just an easy way of representing magnetic flux, and we usually associate one line to \(1\text{Wb}\) of flux when we use it ;D
And yep, flux density is just flux per unit area, and that is how we define the strength of a magnetic field. Example, the earth's magnetic field, a butt tonne of magnetic flux, but low density - Hence, weak field ;D
Post by: JeffChiang on July 05, 2017, 10:29:12 pm
You are correct on every single point! ;D note though that magnetic flux does have a more rigorous mathematical definition than "the number of field lines." The field lines are just an easy way of representing magnetic flux, and we usually associate one line to \(1\text{Wb}\) of flux when we use it ;D
And yep, flux density is just flux per unit area, and that is how we define the strength of a magnetic field. Example, the earth's magnetic field, a butt tonne of magnetic flux, but low density - Hence, weak field ;D
Ahh right. Thanks for the clarification Jamon. It just seemed really confusing to me how simplified and undetailed the concepts are but I get it now.
Post by: jamonwindeyer on July 05, 2017, 10:31:44 pm
Ahh right. Thanks for the clarification Jamon. It just seemed really confusing to me how simplified and undetailed the concepts are but I get it now.
Welcome to HSC Physics ;) nah I totally get it, it's almost deceptive in how simplified it is - Always happy to give the extra little bit of knowledge if it helps ;D
Post by: RuiAce on July 05, 2017, 10:34:44 pm
Welcome to HSC Bullshitland ;)I think that's better Jamon 8)
Post by: jamonwindeyer on July 05, 2017, 10:37:22 pm
I think that's better Jamon 8)
The resentment is stroooooong ;)
Post by: seventeenboi on July 06, 2017, 02:10:13 pm
Hey! You've pretty much spotted the two big issues, you won't need much else in the HSC! They thought intensity was related to frequency through a law called Rayleigh Jeans Law, which as you say, worked for longer wavelengths but started breaking down for shorter ones. Note that this classical theory also violated the conservation of energy - You can't have intensity (and thus, energy) approaching infinity at short wavelengths. That makes no sense. The idea of quantised energy was then the eventual solution ;D
Btw, definitely not a dumbass, this is tough stuff! And it seems you've got a solid understanding - Is anything particularly confusing you? :)
ahhh okok thank you!!! glad to see that i'm on the right track! it's just that the entire dotpoint 2 is a little convoluted at times, but i'll keep working on it yay
Post by: AJ123 on July 07, 2017, 07:59:10 pm
Post by: winstondarmawan on July 07, 2017, 11:15:50 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19619126_1816906788625778_904743421_o.jpg?oh=0192829702f1c9aed574feea2a20b347&oe=5961D4B2
Post by: jamonwindeyer on July 07, 2017, 11:26:30 pm
How to nail those 5-7 markers and score in the highest band for those qs?
Hey! Big question, but a few tips:
- You need to know your syllabus, meaning, when they give you that one sentence request for information, you know the syllabus well enough to know exactly the information they want.
- Use your jargon; the proper terminology lets you express yourself more succinctly and communicate your knowledge to the marker more obviously. Make revision of these key terms an integral part of your study routine!
- Lots of practice, obviously ;D
- Respond to the verb - When asked for a judgement, give one. When asked for an explanation, be sure to link the effect of one thing to the cause of the text. Be efficient!
There's only a limited number of 5-7 markers they can ask, aim to have practiced all of them before exam time! ;D
Post by: pikachu975 on July 07, 2017, 11:28:26 pm
Would appreciate help with Q1. I got 8ms^-1, but apparently the answer is A. TIA
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19619126_1816906788625778_904743421_o.jpg?oh=0192829702f1c9aed574feea2a20b347&oe=5961D4B2
Your answer looks right to me.
Post by: jamonwindeyer on July 08, 2017, 12:20:36 am
Would appreciate help with Q1. I got 8ms^-1, but apparently the answer is A. TIA
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19619126_1816906788625778_904743421_o.jpg?oh=0192829702f1c9aed574feea2a20b347&oe=5961D4B2
Your answer looks right to me.
Ditto here, perhaps the answers are off? ;D
Post by: winstondarmawan on July 08, 2017, 11:32:19 am
Ditto here, perhaps the answers are off? ;DThought so... just found it weird bc that was from James Ruse. Thanks guys!
Post by: winstondarmawan on July 08, 2017, 02:17:47 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19808040_1265016066957247_1528084618_o.png?oh=e944fbaf2f5d58e0322503ffcea055db&oe=59633739
Post by: jamonwindeyer on July 08, 2017, 03:35:58 pm
Hello! Need help with the following, TIA.
Hey! My bet here would be that the answer is C -> Mostly because it doesn't really make much sense. DC input proportional to voltage - What? The wording seems off, so that would be my pick. Note that the spring is vital to make sure the pointer doesn't just spin wildly, the plane of the coil is always parallel to the field which removes any variation as it spins, and the area needs to be constant for the same reason. So, pretty sure it is C ;D
Post by: winstondarmawan on July 08, 2017, 04:43:15 pm
1. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19883468_1265097333615787_2044210444_n.png?oh=bcc97a8f4bd080378231c4a41352ea1d&oe=596225E2
Not sure how the answer can be A when the voltage needs to be stepped up... or does the orientation of the transformer not matter in this case?
2. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19904508_1265097566949097_1191709984_n.png?oh=6dbde37a3f515bad436a2a68ea64e3ce&oe=59632643
3. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19883825_1265098016949052_2064944303_n.png?oh=524fa20cdadf94347bdae0d48b2a68f9&oe=59624F62
Thought the answer was C but apparently it is B. Even after that, why is the lightbulb off?
4. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19894213_1265098310282356_1398760865_n.png?oh=e8c42473f96f19d88a151d6af7a12903&oe=59632482
5. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19866262_1265098473615673_1864449019_n.png?oh=23fcb08250bd8a080bd1b9228ee750a4&oe=59623D24
Again, sorry for the amount of questions! I really struggled with this paper and the solutions are not adequate. TIA
Post by: JeffChiang on July 08, 2017, 09:54:37 pm
This question is question 15 from the 2013 HSC.
The answer is A and I really don't understand how.
Post by: limtou on July 08, 2017, 11:47:13 pm
Hi,
This question is question 15 from the 2013 HSC.
The answer is A and I really don't understand how.
When calculating torque, it is the angle between the magnetic field and the plane WXYZ, in this case the labelled 30. For the force on WX, it the angle between the direction of the wire and the magnetic field; in this case the wire WX is vertical while the magnetic field/flux line is horizontal, hence they are perpendicular (i.e. 90 degree).
Hope this helps :)
Post by: kiwiberry on July 09, 2017, 12:37:46 am
4. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19894213_1265098310282356_1398760865_n.png?oh=e8c42473f96f19d88a151d6af7a12903&oe=59632482For 4), the semiconductor sensor needs to be able to detect radiation with wavelength of about 15 \(\mu m\) since the majority of radiation from Klingon is of this wavelength (from the graph). The semiconductor's band gap needs to be equal to the energy of a photon of Klingon radiation so that when the photon strikes the sensor, it gives an electron enough energy to promote it to the conduction band and make the sensor work (like a solar cell). So, calculating the energy of a photon:
5. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19866262_1265098473615673_1864449019_n.png?oh=23fcb08250bd8a080bd1b9228ee750a4&oe=59623D24
For 5), \(u_x = V\cos 60 = \frac{V}{2}\) and \(u_y = V\sin 60 = \frac{\sqrt{3}V}{2}\). The time is takes for the projectile to reach \(\Delta x = 55\) will equal the time it takes for it to reach \(\Delta y = -34 \)
Post by: winstondarmawan on July 09, 2017, 09:23:58 am
For 5), \(u_x = V\cos 60 = \frac{V}{2}\) and \(u_y = V\sin 60 = \frac{\sqrt{3}V}{2}\). The time is takes for the projectile to reach \(\Delta x = 55\) will equal the time it takes for it to reach \(\Delta y = -34 \)Not sure about the first 3, but hopefully someone else will step in! Hope this helps :)
Thank you! For this question, would it be appropriate to use maths methods?
Post by: jamonwindeyer on July 09, 2017, 10:39:00 am
Thank you! For this question, would it be appropriate to use maths methods?
As in, from Extension 1 Math? :) nope, you'll want to stick with methods taught in the course so that you can get marks for working if you make any small errors ;D
Post by: jamonwindeyer on July 09, 2017, 10:49:46 am
Hello again! Sorry for having so many questions, this paper has no solutions and just a vague marking criteria.
Again, sorry for the amount of questions! I really struggled with this paper and the solutions are not adequate. TIA
No need to be sorry! Let me give you a rundown:
1) Orientation doesn't matter, you can plug the input into the right and get the output from the left, no issue there. So the voltage jump from input to output would be 2.5x, so 10V becomes 25V. Then at 90% efficient, that is 22.5V - That is the best option to make sure you get your 22V without going ridiculously far over ;D
2) Recall that induced emf is proportional to the rate of change of flux. What this means, since you do 2U and will understand the terminology, is that the induced emf is proportional to the derivative of flux. Flux is a sine curve, so we'd expect the induced emf to be a cosine curve. The answer is (B). We could do this without the knowledge of derivatives (namely that \(\frac{d}{dx}\sin{x}=\cos{x}\)), but this is way easier, since we have it in our toolbelt ;D
3) Brutal question this one, the issue is that the voltmeter is in series, where it is supposed to be connected in parallel. Voltmeters have a very large internal resistance (essentially an open circuit) to prevent impacting the circuit when connected in parallel. In series, pfft, you done f-ed up. The huge resistance stops any current from flowing, so while you'll register the 12V in your circuit, the circuit stops working - No current (0A on the ammeter) and no light :(
Hope this helps ;D
Post by: winstondarmawan on July 09, 2017, 11:15:02 am
No need to be sorry! Let me give you a rundown:
1) Orientation doesn't matter, you can plug the input into the right and get the output from the left, no issue there. So the voltage jump from input to output would be 2.5x, so 10V becomes 25V. Then at 90% efficient, that is 22.5V - That is the best option to make sure you get your 22V without going ridiculously far over ;D
2) Recall that induced emf is proportional to the rate of change of flux. What this means, since you do 2U and will understand the terminology, is that the induced emf is proportional to the derivative of flux. Flux is a sine curve, so we'd expect the induced emf to be a cosine curve. The answer is (B). We could do this without the knowledge of derivatives (namely that \(\frac{d}{dx}\sin{x}=\cos{x}\)), but this is way easier, since we have it in our toolbelt ;D
3) Brutal question this one, the issue is that the voltmeter is in series, where it is supposed to be connected in parallel. Voltmeters have a very large internal resistance (essentially an open circuit) to prevent impacting the circuit when connected in parallel. In series, pfft, you done f-ed up. The huge resistance stops any current from flowing, so while you'll register the 12V in your circuit, the circuit stops working - No current (0A on the ammeter) and no light :(
Hope this helps ;D
Thank you so much! For the 3rd question, are we required by syllabus to know that? Because I never remember learning that in my life.
Post by: jamonwindeyer on July 09, 2017, 11:21:07 am
Thank you so much! For the 3rd question, are we required by syllabus to know that? Because I never remember learning that in my life.
Yeah that's a bit of a stretch imo - You learn it a bit in Year 11, but I doubt the HSC would assess you on it ;D
Post by: JeffChiang on July 09, 2017, 04:53:06 pm
When calculating torque, it is the angle between the magnetic field and the plane WXYZ, in this case the labelled 30. For the force on WX, it the angle between the direction of the wire and the magnetic field; in this case the wire WX is vertical while the magnetic field/flux line is horizontal, hence they are perpendicular (i.e. 90 degree).
Hope this helps :)
Yes. Thanks limtou. But does this mean that for F=BILsin(w), w=90 regardless of the wire's rotated position? so the force would be constant. I never really understood this part of the syllabus.
Post by: jakesilove on July 09, 2017, 04:55:25 pm
Yes. Thanks limtou. But does this mean that for F=BILsin(w), w=90 regardless of the wire's rotated position? so the force would be constant. I never really understood this part of the syllabus.
Yep, that's totally right! Imagine the direction of the arrows indicating the magnetic field, and the direction of the current in the wire. As the wire moves, it stays perpendicular to the arrows of the field lines. So, the angle between the field lines and the current doesn't change, and the force stays the same! Does that sort of make sense?
Post by: JeffChiang on July 09, 2017, 05:30:56 pm
Yep, that's totally right! Imagine the direction of the arrows indicating the magnetic field, and the direction of the current in the wire. As the wire moves, it stays perpendicular to the arrows of the field lines. So, the angle between the field lines and the current doesn't change, and the force stays the same! Does that sort of make sense?
Yes, it makes sense now. Thanks Jake!
Post by: justwannawish on July 10, 2017, 03:51:46 pm
This is more of a general question. Hope that's okay. I think our school is a bit behind and we've just started the third module for prelim physics and our teacher has estimated that we would have to gloss over Cosmic Engine at this rate. How relevant are the concepts taught in the module for HSC physics (is space an extension of cosmic engine)?
Post by: jakesilove on July 10, 2017, 03:54:18 pm
Hey,
This is more of a general question. Hope that's okay. I think our school is a bit behind and we've just started the third module for prelim physics and our teacher has estimated that we would have to gloss over Cosmic Engine at this rate. How relevant are the concepts taught in the module for HSC physics (is space an extension of cosmic engine)?
Hey!
In reality, year 12 content doesn't really build on year 11 knowledge. So, when you reach year 12, you could effectively forget the content you learned in Year 11, and still be absolutely ok.
The more important thing that you should get out of year 11 is the skills, and your study techniques. Practicals, writing the answers to extended response questions, using physics terminology; those are the sorts of skills that are invaluable when you get to Year 12. Similarly, if you get into good study habits now, and learn the ways in which you learn the best, you'll get a massive jump on Year 12.
However, overall, I wouldn't worry too much if you speed through some Year 11 content. You won't at all be at a disadvantage :)
Post by: jennifer.le11 on July 10, 2017, 10:37:32 pm
I'm having trouble pinpointing exactly why the answer to this is Q, is it because of the right hand grip rule? Could you explain in detail how the force acts towards Q?
Thank you very much! :)
Post by: kiwiberry on July 11, 2017, 12:11:23 am
Hi!
I'm having trouble pinpointing exactly why the answer to this is Q, is it because of the right hand grip rule? Could you explain in detail how the force acts towards Q?
Thank you very much! :)
Hey! In the coil, current flows from top to bottom at the front. Using the right hand grip rule, this induces a magnetic field towards the right - so there's a north pole at the right end of the coil. The copper ring experiences an increase of flux towards the right, and by Lenz's Law, a current will be induced such that it gives rise to a magnetic field which opposes this change of flux. This means that the ring will create a magnetic field towards the left, and there will be another north pole left of the ring. The two north poles repel each other, so the ring will be pushed towards Q. Does that make sense? :)
Post by: justwannawish on July 11, 2017, 09:08:21 am
Hey!
In reality, year 12 content doesn't really build on year 11 knowledge. So, when you reach year 12, you could effectively forget the content you learned in Year 11, and still be absolutely ok.
The more important thing that you should get out of year 11 is the skills, and your study techniques. Practicals, writing the answers to extended response questions, using physics terminology; those are the sorts of skills that are invaluable when you get to Year 12. Similarly, if you get into good study habits now, and learn the ways in which you learn the best, you'll get a massive jump on Year 12.
However, overall, I wouldn't worry too much if you speed through some Year 11 content. You won't at all be at a disadvantage :)
Ah that's good to know and eased my worries. Thank you!
Post by: limtou on July 11, 2017, 04:18:57 pm
Answers: 3. A and 13. C
(for 3, I managed to get the answer with the use of Earth's radius 6380km but it wasn't given in this question? Is there perhaps a way to get the answer without using Earth's radius?)
Also apologies for the split q. 13
Post by: winstondarmawan on July 11, 2017, 09:59:46 pm
Can someone explain the flaw in my thought because I confused myself when I thought about this..
Post by: kiwiberry on July 11, 2017, 10:20:33 pm
Hey, just a question about power loss. So the formula is P(loss)=(I^2)R, and thus power loss is directly proportional to the square of the current, and directly proportional to resistance. Thus, it is preferred to have high voltages as when voltage is stepped up current is stepped down. However, what happens if we substitute V^2/R^2 into I^2? P(loss) becomes V^2/R, and now power loss is directly proportional to the square of the voltage and inversely proportional to the resistance. Then why is voltage stepped up>
Can someone explain the flaw in my thought because I confused myself when I thought about this..
Hey! This is a pretty confusing idea - let me link you to this great explanation Jamon wrote a while back:
Step into my office ;)
Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.
The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need :)
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference ;D
Post by: austv99 on July 11, 2017, 11:27:10 pm
I dont understand the difference between A and C.
Post by: Aaron12038488 on July 12, 2017, 01:24:43 pm
Its in a table format,
and the first column states Helio- or Geocentric model?
It says 6 models, I've got Aristotle's, Aristarchus, Copernicus, Brahe's, Kepler, Galleli, Newton. Thats 7, which i found on someone else's notes. But when i searched it up, i got different answers.
Post by: seventeenboi on July 12, 2017, 07:29:23 pm
i've attached both question and answer below :)))
thanks heaps
Post by: jamonwindeyer on July 12, 2017, 08:07:05 pm
So I received an assignment for physics for which I RECEIVED A WHOLE TOPIC which was 'The Cosmic Engine'. I need assurance for a question. Complete the table below by summarising the historical models of the Universe.
Its in a table format,
and the first column states Helio- or Geocentric model?
It says 6 models, I've got Aristotle's, Aristarchus, Copernicus, Brahe's, Kepler, Galleli, Newton. Thats 7, which i found on someone else's notes. But when i searched it up, i got different answers.
Hey! Sorry, just a little unsure exactly what your question is? Are you confused about whether each model is heliocentric or geocentric? :)
Post by: bsdfjnlkasn on July 12, 2017, 09:35:35 pm
I know these are easy but I'm not getting any of the answers (B, A, C respectively). Would love some help - thank you!
Post by: kiwiberry on July 12, 2017, 09:52:57 pm
Hey there,
I know these are easy but I'm not getting any of the answers (B, A, C respectively). Would love some help - thank you!
Hey!
1)
2)
Post by: bsdfjnlkasn on July 12, 2017, 09:56:41 pm
Hey!
1)Remember that M is the mass of the central body ie. the earth, not the telescope
2)3)
Put those into a calculator and it should work out :)
Awesome, thanks so much! I realised I was adding the Earth's radius to every distance - why is this wrong?
Any ideas on the following?
Post by: winstondarmawan on July 12, 2017, 10:09:18 pm
Awesome, thanks so much! I realised I was adding the Earth's radius to every distance - why is this wrong?
Any ideas on the following?
You do not add the radius of Earth because the question already gives you the ORBITAL RADIUS, and not the distance from the surface of the Earth.
Also for the question, I'm pretty sure its B. Remember that EMF is directly proportional to the rate of change of flux.
Think about the situation where the magnet is closest to the coil. If you do 3U, you can think about it in terms of Simple Harmonic Motion and that is the position closest to the magnet is the "extreme point". At the extreme point, there is no velocity, and in this case, no change in magnetic flux, and thus no EMF.
So, if it is closest to the coil, no EMF should be present. Someone please correct me if i am wrong
Post by: jamonwindeyer on July 12, 2017, 10:37:50 pm
Hey, can I get help with this multiple choice question?
I dont understand the difference between A and C.
Hey austv, welcome to the forums!! ;D
I agree with you tbh! Without knowing which way the battery was connected you have no way to distinguish between A and C! :)
Post by: jamonwindeyer on July 12, 2017, 10:43:25 pm
HIHI i need help with this question ... why does the force/time graph of dropping a metallic sheet through a constant magnetic field look like this ????????
i've attached both question and answer below :)))
thanks heaps
Hey! Cool question, so:
- The sheet is falling under gravity, so there is a constant gravitational force on the sheet
- As the sheet enters the field, eddy currents form which act to oppose the motion. So, an upwards force is experienced, which subtracts from the gravitational force (since it is acting in the opposite direction). So, the overall force is reduced!
This is what causes the graph you see - The straight line due to gravity and the dips due to the opposing force caused by induced eddy currents! ;D
Post by: bsdfjnlkasn on July 12, 2017, 10:45:43 pm
I'm not sure where G and T have gone in this calculation - could it be a mistake?
Post by: Shadowxo on July 12, 2017, 10:52:14 pm
Hey there,
I'm not sure where G and T have gone in this calculation - could it be a mistake?
I believe it is. They should be there but aren't, and if you also calculate M using their calculation (second last line) you end up with a different answer. It's likely that line doesn't show it due to an error, but it's likely the final line is correct :)
Post by: bsdfjnlkasn on July 12, 2017, 11:16:08 pm
Was just wondering which competing theories we should discuss for the outcome: Discuss the role of the Michelson-Morley experiments in making determinations about competing theories
Thank you :)
Post by: jamonwindeyer on July 12, 2017, 11:42:12 pm
Hey there,
Was just wondering which competing theories we should discuss for the outcome: Discuss the role of the Michelson-Morley experiments in making determinations about competing theories
Thank you :)
Hey! It is basically discussing the Aether vs No Aether argument - How did the Michelson/Morley experiment impact on the view of whether there was an Aether or not? How did the results (or lack thereof) of the experiment contribute to our views on that dilemma? ;D
Post by: limtou on July 12, 2017, 11:53:30 pm
Please explain the multiple choice below :')
Answers: 3. A and 13. C
(for 3, I managed to get the answer with the use of Earth's radius 6380km but it wasn't given in this question? Is there perhaps a way to get the answer without using Earth's radius?)
Also apologies for the split q. 13
I think my post got missed, could someone please take a look at it :)
Post by: seventeenboi on July 13, 2017, 02:32:25 am
Hey! Cool question, so:
- The sheet is falling under gravity, so there is a constant gravitational force on the sheet
- As the sheet enters the field, eddy currents form which act to oppose the motion. So, an upwards force is experienced, which subtracts from the gravitational force (since it is acting in the opposite direction). So, the overall force is reduced!
This is what causes the graph you see - The straight line due to gravity and the dips due to the opposing force caused by induced eddy currents! ;D
ohh ok thanks!
Post by: bsdfjnlkasn on July 13, 2017, 09:30:46 am
I was a bit confused by the following notes I had - could someone explain/clarify if we even need to know this? It's regarding the deflection plates in electron guns :)
• If an alternating voltage were supplied, then the deflection pattern (across one plane) would form a straight line
- Voltage values corresponds to frequency of deflection
Is a collimator the same thing as the focusing anode/accelerating anode - as in they serve the same function?
Thank you!!
Post by: seventeenboi on July 14, 2017, 01:21:18 pm
I think my post got missed, could someone please take a look at it :)
oops i'm not sure about question 13 ;-; however for question 3, i do think that you do use the radius of the earth! I also got my answer this way.. for now i don't see any other way to do it. considering that the altitudes are provided, there arent any formulas in space that only require the altitude - in most cases you will always have to add the radius of the earth to be able to calculate a correct value (because it's asking for radius of orbit which is measured from the centre of mass)
+ radius of Earth is provided in the data sheet, so it's technically given to you despite not being provided in the question :) there will be many other things such as mass of electron, charge of electron etc which won't be outlined in the question that you will need to either memorise or just refer to given data sheet to be able to calculate stuff - this is because they need to test whether you know when to apply syllabus formulas!
Post by: Bubbly_bluey on July 14, 2017, 09:19:29 pm
Also: if there was a AC generator that we made it turn clockwise (for example), when find the direction of the current do we take the direction directly using the right hand palm rule or do we use the palm rule and flip the direction of our thumb.
Thanks :D
Post by: jamonwindeyer on July 14, 2017, 11:25:46 pm
Please explain the multiple choice below :')
Answers: 3. A and 13. C
(for 3, I managed to get the answer with the use of Earth's radius 6380km but it wasn't given in this question? Is there perhaps a way to get the answer without using Earth's radius?)
Also apologies for the split q. 13
I can help with Q13! So, the magnetic field created by the primary coil are roughly proportional to the input voltage (the waveform with the solid line). Now the magnitude of an induced voltage/EMF due to a changing magnetic field is given by Faraday's Law:
Now roughly speaking, this means the induced voltage is equal to the negative of rate of change of the input voltage. This is where the answer comes from - I know you do 3U, you might be able to deduce why the curve looks the way it does by considering the gradient of the solid curve at any given point? :)
Post by: jamonwindeyer on July 14, 2017, 11:29:25 pm
Hey there!
I was a bit confused by the following notes I had - could someone explain/clarify if we even need to know this? It's regarding the deflection plates in electron guns :)
• If an alternating voltage were supplied, then the deflection pattern (across one plane) would form a straight line
- Voltage values corresponds to frequency of deflection
Is a collimator the same thing as the focusing anode/accelerating anode - as in they serve the same function?
Thank you!!
Hey! That first point is definitely not necessary - The second you need but not in great detail. The collimator ensures there is a focused electron beam, rather than a broad scattering of electrons. The anode and accelerating anode are more just about the acceleration of electrons - But really not a hugely important thing to distinguish ;D
Post by: kiwiberry on July 14, 2017, 11:41:01 pm
oops i'm not sure about question 13 ;-; however for question 3, i do think that you do use the radius of the earth! I also got my answer this way.. for now i don't see any other way to do it. considering that the altitudes are provided, there arent any formulas in space that only require the altitude - in most cases you will always have to add the radius of the earth to be able to calculate a correct value (because it's asking for radius of orbit which is measured from the centre of mass)
+ radius of Earth is provided in the data sheet, so it's technically given to you despite not being provided in the question :) there will be many other things such as mass of electron, charge of electron etc which won't be outlined in the question that you will need to either memorise or just refer to given data sheet to be able to calculate stuff - this is because they need to test whether you know when to apply syllabus formulas!
The radius of the earth actually isn't on the data sheet for some reason, but yeah I don't see any other way of doing the question without using it! Usually it's given when you're expected to use it
Post by: jamonwindeyer on July 14, 2017, 11:48:58 pm
This might be a weird question but what is the significance of Simultaneity (more specifically the train thought experiment) in Special Relativity (what does it actually prove?) I understand the concept I just don't know what means.
Also: if there was a AC generator that we made it turn clockwise (for example), when find the direction of the current do we take the direction directly using the right hand palm rule or do we use the palm rule and flip the direction of our thumb.
Thanks :D
Hey! Basically it is just the idea that you can't say two events are simultaneous in a universal sense - You can only say they are universal to you, as an observer. Another observer may see different. The consequences from that are left to your imagination, but like, there's no super grand revelation here. If you understand this idea, then you are sweet ;D
For current direction you always use the rule directly, for direction of electron flow you always flip! :)
Post by: winstondarmawan on July 15, 2017, 05:56:20 pm
1. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20107046_1271654479626739_1256619788_o.jpg?oh=1ecaaeb15b7e9a0d570930a5b4605aa9&oe=596B78F4
9. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20133318_1271654596293394_1416995248_o.jpg?oh=8cef05660211826b19ef3cd8ce12fc22&oe=596B563C
Post by: pikachu975 on July 15, 2017, 08:47:12 pm
Would appreciate help with the Qs below, TIA.1) I'd put C because every emitter/metal has a different work function and since work function = Planck's constant times threshold frequency, then the threshold frequencies would be different.
1. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20107046_1271654479626739_1256619788_o.jpg?oh=1ecaaeb15b7e9a0d570930a5b4605aa9&oe=596B78F4
9. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20133318_1271654596293394_1416995248_o.jpg?oh=8cef05660211826b19ef3cd8ce12fc22&oe=596B563C
9) From first read it's a semiconductor so I'd probably put B or C. Since it seems that there is more holes (Bus seats) than electrons (passengers), I'd say it's a p-type so B.
Post by: winstondarmawan on July 15, 2017, 09:07:14 pm
1) I'd put C because every emitter/metal has a different work function and since work function = Planck's constant times threshold frequency, then the threshold frequencies would be different.
9) From first read it's a semiconductor so I'd probably put B or C. Since it seems that there is more holes (Bus seats) than electrons (passengers), I'd say it's a p-type so B.
Thank you! The explanation makes sense, however can you explain why it isn't B (for Q1)?
Post by: pikachu975 on July 15, 2017, 11:00:43 pm
Thank you! The explanation makes sense, however can you explain why it isn't B (for Q1)?
The frequency only determines if the electron is emitted, which Einstein proved. Classical physics thought frequency was related to photocurrent which is B, but quantum physics proved that intensity is related to photocurrent instead.
Post by: mercurry on July 16, 2017, 01:28:06 pm
(https://cdn.discordapp.com/attachments/248794056510013440/335985290420158464/help.png)
Answer: Spoiler
Force per unit length on P = 1.73 x 10-5 Nm up the page.
Force per unit length on Q = 2.40 x 10-5 Nm down the page.
Force per unit length on R = 0.69 x 10-5 Nm up the page.
Force per unit length on Q = 2.40 x 10-5 Nm down the page.
Force per unit length on R = 0.69 x 10-5 Nm up the page.
How is this worked out?
Post by: kiwiberry on July 16, 2017, 02:24:03 pm
Question: "Three current-carrying conductors, P, Q and R are set up as shown. Calculate the force per unit length on each wire due to the currents in the other two wires."(https://cdn.discordapp.com/attachments/248794056510013440/335985290420158464/help.png)Answer:SpoilerForce per unit length on P = 1.73 x 10-5 Nm up the page.
Force per unit length on Q = 2.40 x 10-5 Nm down the page.
Force per unit length on R = 0.69 x 10-5 Nm up the page.
How is this worked out?
Hey, welcome to the forums!
Remember that parallel conductors exert an attractive force towards each other and anti-parallel conductors exert a repulsive force, magnitude given by \(\frac{F}{l} = k\frac{I_1 I_2}{d}\). Note that l=1 in this question as we are calculating force per unit length.
On P, both Q and R will exert a repulsive force (up the page), so the total force on P will be
Hope that helps :)
Post by: mercurry on July 16, 2017, 06:05:26 pm
The book has x10-5 for FR but I also got the same answer as you, so it must have been a typo.
Thanks for your reply, it helped a lot!
Post by: jaskirat on July 16, 2017, 07:39:32 pm
Help!!
Post by: johnk21 on July 16, 2017, 07:48:16 pm
Okay so apparently the answer is C but in my KISS notes it says "whatever energy the spacecraft gains, the planet loses. Energy is conserved. The planets spin will be slowed down slightly by the transfer of energy to the spacecraft"You basically have answered your own question. I think you are just misinterpreting the question. It is stating the planet and spacecraft is one WHOLE SYSTEM together. Therefore if the planet loses energy, the spacecraft will gain it. Therefore the planet and spacecraft system does not lose energy.
Help!!
Hope this helped
Post by: limtou on July 16, 2017, 07:49:05 pm
I can help with Q13! So, the magnetic field created by the primary coil are roughly proportional to the input voltage (the waveform with the solid line). Now the magnitude of an induced voltage/EMF due to a changing magnetic field is given by Faraday's Law:
Now roughly speaking, this means the induced voltage is equal to the negative of rate of change of the input voltage. This is where the answer comes from - I know you do 3U, you might be able to deduce why the curve looks the way it does by considering the gradient of the solid curve at any given point? :)
Ahh thank you Jamon! I understand now :)
Post by: seventeenboi on July 17, 2017, 09:34:26 am
I'd like help with these multiple choice questions pls :)
Post by: jamonwindeyer on July 17, 2017, 10:25:25 am
HI sorry for the influx of questions HAHA
I'd like help with these multiple choice questions pls :)
Sure thing!
- First one, we don't get any induced currents initially, because the magnetic field caused by the current is constant. It needs to be a changing magnetic field to induce a current. So, initial current is zero. When the current in the outer ring is reduced, the inner ring will have a current induced to oppose the change. The way to oppose a current removed is to bring it back, so a clockwise current is induced. Answer is B.
- The answer is C, the induced currents in the metal will act to oppose its motion, attracting it as it moves away and repelling it as it moves towards the centre.
- On a quick look, I'd be leaning towards D, just because only A and C are definitely wrong because kinetic energy curve would look like a parabola, and the length of an object moving at relativistic speed gets smaller, not bigger. I don't think the shape of B quite matches, so I'm guessing D - But happy to be corrected :) you might want to plot a few points to check!
Post by: left right gn on July 17, 2017, 05:29:45 pm
Post by: blasonduo on July 17, 2017, 05:57:39 pm
hello can i get some on help on this trial question, I'm confused on how to determine the direction of the eddy current
Hello!
As you should know by Lenz's law, That eddy currents need to oppose the change that created them and looking at the diagrams, the only change occurring is A and C.
In A, it is entering a magnetic field, and thus wants to oppose it, so it will go clockwise.
In C, there is a decrease in magnetic flux and thus wants to mitigate this by increasing it by going anti-clockwise,
Post by: left right gn on July 17, 2017, 06:01:45 pm
Hello!is it possible to explain using the right hand palm rule?
As you should know by Lenz's law, That eddy currents need to oppose the change that created them and looking at the diagrams, the only change occurring is A and C.
In A, it is entering a magnetic field, and thus wants to oppose it, so it will go clockwise.
In C, there is a decrease in magnetic flux and thus wants to mitigate this by increasing it by going anti-clockwise,
Post by: blasonduo on July 17, 2017, 06:09:19 pm
is it possible to explain using the right hand palm rule?
Yes! I'd use the right hand thumb rule, quick and easy!
Post by: kiwiberry on July 17, 2017, 06:15:56 pm
is it possible to explain using the right hand palm rule?
Just as a note, don't get confused between the right hand palm rule and the grip/thumb rule - the palm rule is used to determine the direction of the force on a current-carrying conductor in a magnetic field, whereas the grip rule is used for the direction of current induced in a magnetic field (or the other way round)
Post by: BKO99 on July 18, 2017, 07:22:16 pm
How would you solve it.
Post by: jakesilove on July 18, 2017, 07:27:21 pm
Question 11 2008 Physics HSC.
How would you solve it.
Welcome to the forums! In future, please attach the question as an image/type out the question.
Question 11 looks at an electron moving through a magnetic field, with a certain velocity, and a certain 'direction'. However, the first thing to note is that the direction doesn't matter! The electron is moving perpendicular to the magnetic field (ie. in a different plane to the field). Thus, the 40 degree value you were given is not necessary in our calculation at all! This is probably the toughest part of the question.
From there, we know that the electron will experience a force due to the magnetic field.
This will result in a centripetal force, the formula for which includes the radius of the path of the electron in the field
Equate these two equations, sub in the values given, and solve for r! The answer you get will be D
Post by: Bubbly_bluey on July 18, 2017, 11:12:49 pm
Thanks for the help :D
Post by: blasonduo on July 18, 2017, 11:40:07 pm
For question 11, Both coils 1 and 3 will NOT produce a current as there is no change in magnetic flux. Coil 2 WILL produce a current as there is a change in surface area, as there is now a DECREASE in flux. So, for the answer, I would say D, BUT I COULD BE WRONG (someone please correct me?)
Question 12 is referring to back emf in motors, as when motors begin to rotate, there will be a change in flux, but due to lenz's law, an emf will be produced to OPPOSE the direction of current. This is why when you apply 8 Volts to a motor, it will not spin at 8 volts to say. When the motor is slowed down by a load, this back emf can no longer be induced, and thus more current can now flow, this extra current will produce more heat, and thus the answer is A.
Goodluck with your studies.
Post by: Bubbly_bluey on July 18, 2017, 11:53:35 pm
Question 12: So back emf is basically the emf that is induced by the coil to oppose the motion of the changing flux. When the drill enters a load it causes the slower cutting of flux and so less emf?
::)
Post by: blasonduo on July 19, 2017, 12:04:25 am
For question 12, just remember when a motor spins, it still follows the law of inductance, and will induce a current in the coil, but due to the laws of conservation, must oppose the current. :)
Post by: Bubbly_bluey on July 19, 2017, 12:10:27 am
Basically, as I think it goes, When the coil spins, the overall area it takes up changes, it is taking up less SPACE in the magnetic field, and thus the magnetic flux density in that area changes. (basically once it turns 90 degrees, there are less magnetic field lines passing through the coil, and thus a change in magnetic flux.) As for the other 2, yes they are moving, but there is no change in magnetic flux when they do so. (Remember back to braking systems, eddy currents were only produced when it was exiting and entering the magnetic field, but not in the middle)I get it! Thanks so much!! :D
For question 12, just remember when a motor spins, it still follows the law of inductance, and will induce a current in the coil, but due to the laws of conservation, must oppose the current. :)
Post by: Bubbly_bluey on July 19, 2017, 11:34:15 pm
Thanks ::)
Post by: jamonwindeyer on July 19, 2017, 11:36:51 pm
Hi back with another one! I only need help with part c). I don't know what you do with the work function being in volts. How do I change it to joules so that I can find the threshold frequency.
Thanks ::)
Hey again!! Not too bad this one, the conversion is on your reference sheet:
Edit: Woops, just caught what you meant!! They definitely mean electron-volts ;D
Post by: Bubbly_bluey on July 19, 2017, 11:42:27 pm
Hey again!! Not too bad this one, the conversion is on your reference sheet:woah speedy reply! Basically what I did was:
work function= 6.35 V
so i changed it to ev =6.35x 1.602x10-19
and i got: 1.01727x10-18eV
which is kinda fishy to me..
Post by: jamonwindeyer on July 19, 2017, 11:44:45 pm
woah speedy reply! Basically what I did was:
work function= 6.35 V
so i changed it to ev =6.35x 1.602x10-19
and i got: 1.01727x10-18eV
which is kinda fishy to me..
Ahaha too quick, totally missed it was VOLTS over electron volts - They definitely have just made a typo, treat it as electron volts and use the conversion as you did in the previous part I reckon ;D
Post by: samuels1999 on July 20, 2017, 05:57:16 pm
I was about to buy you physics topic test book and then when I went to finalise my payments it said that you do not accept PO Boxes as addresses.
My mail box has quite a lot of problems at the moment (hard to explain) and so we have changed to using a PO box, is there any way I can purchase it without giving my address.
Thanks,
Samuel
PS I would give my address if it was a another courier to Australia post who doesn't use the mail box.
Post by: itssona on July 20, 2017, 10:38:26 pm
So theres a question in which it says:
The graohs show the positions of the same wave 0.2s apart. Calculate max period of the wave
And the graohs are disp v time graphs with no values shown
How do i find the max period thank you :)
Post by: blasonduo on July 20, 2017, 10:42:05 pm
HiiiHello!
So theres a question in which it says:
The graohs show the positions of the same wave 0.2s apart. Calculate max period of the wave
And the graohs are disp v time graphs with no values shown
How do i find the max period thank you :)
Do you have a picture of this question? It'll make it easier for me to see.
Thanks!
Post by: itssona on July 20, 2017, 10:59:20 pm
Hello!http://i613.photobucket.com/albums/tt216/sssona09/20170720_224258_zpswifvzjkl.jpg
Do you have a picture of this question? It'll make it easier for me to see.
Thanks!
Srry for the delay - dodgy internet xD
Thank you!!! :)
Would u mind checking both questions btw ^ :)
Mod edit: Merged posts :)
Post by: blasonduo on July 20, 2017, 11:19:07 pm
Thank you!!
Okay, From the graphs, we can see that the wave has either moved a quarter of its period, or 3 quarters of its period, and because we want the MAXIMUM period, we want the 3 quarters one.
So if its moved 3 quarters of a wave in 0.2 seconds, we can easily work out how long it will take for 1 full period:
Period = 0.2/0.75 = 0.27s
Therefore the maximum period is 0.27 seconds
If you have any more questions (either about this or something else) feel free to ask!
Post by: itssona on July 20, 2017, 11:22:42 pm
Ahh u explained it so well!! I finally get it - thank you sOO much!! :)
Thank you!!
Okay, From the graphs, we can see that the wave has either moved a quarter of its period, or 3 quarters of its period, and because we want the MAXIMUM period, we want the 3 quarters one.
So if its moved 3 quarters of a wave in 0.2 seconds, we can easily work out how long it will take for 1 full period:
Period = 0.2/0.75 = 0.27s
Therefore the maximum period is 0.27 seconds
If you have any more questions (either about this or something else) feel free to ask!
I'll most likely post a heap on here tmr xD
Thank you for helping me!!
Post by: yattmoani on July 21, 2017, 09:50:18 pm
An ammeter connected in series with an electric motor is rotating at its maximum rate reads 2A. If the motor is slowed down, however, by applying a force to the spinning shaft, the current increases above 2A. Explain this observation.
Thanks again!
Post by: f_tan on July 21, 2017, 10:07:12 pm
Hi! I have a quick question that is worth three marks. My main problem is not understanding what the question is about so any pointers would be extremely helpful.
An ammeter connected in series with an electric motor is rotating at its maximum rate reads 2A. If the motor is slowed down, however, by applying a force to the spinning shaft, the current increases above 2A. Explain this observation.
Thanks again!
I think the question relates to back EMF and supply EMF - think about what happens when the motor is slowed down (less rate of change of magnetic flux) and what this means for back EMF, and then think about how the net current would change if back EMF is decreased.
Hope it helps!
Post by: yuriques on July 21, 2017, 10:24:12 pm
For example assessing advantages and disadvantages on society and environment for the AC system would have 5 points on benefits to society and 1 on environment and I've already used up all the lines writing advantages, and the environment side is left blank (and I still needed lines to write disadvantages and an assessment)
Was the table the wrong structure in this case? Should I cut out some of the advantages or maybe try to find more disadvatanges to even it out? Or would I just leave the spaces blank to show how much more advantageous it was?
Thank you in advanced!
Post by: jakesilove on July 21, 2017, 10:54:33 pm
Hi there, this isn't really just physics specfic but I remember Jake in his lecture talking about using tables for extended responses for questions about society and environment or advantages and disadvantages. I tried it out but I have a problem of having too much advantages compared to disadvantages so when I rule out my table, I'll have half of the lines blank in one of the boxes but would have used all the lines the advantages side when I need more to write an assessment.
For example assessing advantages and disadvantages on society and environment for the AC system would have 5 points on benefits to society and 1 on environment and I've already used up all the lines writing advantages, and the environment side is left blank (and I still needed lines to write disadvantages and an assessment)
Was the table the wrong structure in this case? Should I cut out some of the advantages or maybe try to find more disadvatanges to even it out? Or would I just leave the spaces blank to show how much more advantageous it was?
Thank you in advanced!
Hey! If you have a 5 mark question asking you to assess XXXXXXXX on Society and the Environment, I would make sure to split up your discussion of the two (Society and the Environment) approximately equally. Perhaps you'd give three advantages/disadvantages for Society, and two for the Environment, but like I wouldn't go overboard with one and minimise the other.
I would also ALWAYS have as many advantages as disadvantages. That's a method I've personally stuck with; would be interested to hear what others do.
When it comes to your table specifically, I would be drawing two: one for Society (advantages in one column, disadvantages in another, taking up 4 lines in total. One is the title, three are advantages/disadvantages) and one for the Environment (advantages in one column, disadvantages in another, taking up 3 lines in total. One is the title, two are advantages/disadvantages).
Does that sort of make sense? If you do the above, and are still struggling, post up a picture of your answer for us to mark!
Jake
Post by: Bubbly_bluey on July 22, 2017, 12:25:54 am
answers:
Spoiler
11) A and 18) B
Post by: kiwiberry on July 22, 2017, 02:22:09 am
Hi! not sure how to do these two multiple choices
answers:Thanks ;DSpoiler11) A and 18) B
Hey! :) For 11, if X is the positive terminal current will flow from X to Y, so downwards on the straight section where the compass is. Using the right hand grip rule, this induces a magnetic field which goes from left to right over the compass (because it's under the wire). The compass will point north in the direction of the magnetic field lines, so north will move to point right, meaning that it deflects clockwise. Sorry, not sure about Q18
Post by: Bubbly_bluey on July 22, 2017, 09:04:28 am
Hey! :) For 11, if X is the positive terminal current will flow from X to Y, so downwards on the straight section where the compass is. Using the right hand grip rule, this induces a magnetic field which goes from left to right over the compass (because it's under the wire). The compass will point north in the direction of the magnetic field lines, so north will move to point right, meaning that it deflects clockwise. Sorry, not sure about Q18Hi! Im still slightly confused. :) So when you use the right hand grip rule, is the thumb the induced current? How did you know that the direction of the resultant force on the compass with the grip rule?
Post by: kiwiberry on July 22, 2017, 11:55:25 am
Hi! Im still slightly confused. :) So when you use the right hand grip rule, is the thumb the induced current? How did you know that the direction of the resultant force on the compass with the grip rule?
Yeah, the thumb will be facing the direction of current. Pointing your thumb down, you'll see that the induced magnetic field will be going into the page on the left, and out of the page on the right - so anticlockwise around the wire. Because the compass is under the wire, it will be in the part of the magnetic field that goes from left to right. Compasses point in the direction of the magnetic field they're in, so in this case it will be pointing right. Let me know if this still doesn't make sense :)
Post by: itssona on July 22, 2017, 12:27:35 pm
Explain how curcuit breakers work
Electricity magnetises the electromagnet (in electromagnets a magnetic field is produced by an electric current). Moving charge i.e. current in a magnetic field experiences EMF and thus whrn the current is too high, the electromagnet has a strong EMF to pull down a lever connected to the switch.
Post by: beau77bro on July 22, 2017, 12:45:40 pm
Post by: kiwiberry on July 22, 2017, 01:14:18 pm
ive completely forgotten how to physics could someone help me. explanation would be much appreciated.
Jamon has explained these questions over here (Q1 and 3) :)
Post by: beau77bro on July 22, 2017, 01:25:20 pm
Jamon has explained these questions over here (Q1 and 3) :)
sorry I'm so tired completely forgot
Post by: kiwiberry on July 22, 2017, 01:42:06 pm
sorry I'm so tired completely forgot
No that's ok!! Didn't expect you to have read every question on this thread :)
Post by: jakesilove on July 22, 2017, 01:44:21 pm
Heey would someone be able to see if my answer makes sense?? Im in prelim so i dont think i have to go in depth with the motor effevt thing but here it is :D
Explain how curcuit breakers work
Electricity magnetises the electromagnet (in electromagnets a magnetic field is produced by an electric current). Moving charge i.e. current in a magnetic field experiences EMF and thus whrn the current is too high, the electromagnet has a strong EMF to pull down a lever connected to the switch.
Hey! Your answer is good; I would go back through and read it out loud, just to clean up the phrases etc. However, since I don't know how many marks this question is, it's sort of hard to judge whether you've included enough information. However, you've certainly got the correct information! You might add a sentence on why they are important?
Post by: itssona on July 22, 2017, 02:22:22 pm
Hey! Your answer is good; I would go back through and read it out loud, just to clean up the phrases etc. However, since I don't know how many marks this question is, it's sort of hard to judge whether you've included enough information. However, you've certainly got the correct information! You might add a sentence on why they are important?Ahh okay im glad i understand it well - thank you Jake!!:)
Ill certainly add that too :D
Post by: beau77bro on July 22, 2017, 05:21:01 pm
Also is this seriously what happens, I thought the "overlap" was just conceptual for band gaps so that we could understand it. Is this just worded poorly?(http://uploads.tapatalk-cdn.com/20170722/51e6ed5238796f597d3c7c127936e7c2.jpg)
Post by: jakesilove on July 22, 2017, 05:50:04 pm
Thanks jamon (from ages ago)
Also is this seriously what happens, I thought the "overlap" was just conceptual for band gaps so that we could understand it. Is this just worded poorly?(http://uploads.tapatalk-cdn.com/20170722/51e6ed5238796f597d3c7c127936e7c2.jpg)
Nope, the actual valence bands in metals overlap, insofar as ALL valence bands are connected, allowing the electrons orbiting around the nucleus to move freely from one nucleus to another.
Post by: Bubbly_bluey on July 22, 2017, 06:26:11 pm
answer:
Spoiler
B
Thanks ;D
Post by: yuriques on July 22, 2017, 10:19:14 pm
Hey! If you have a 5 mark question asking you to assess XXXXXXXX on Society and the Environment, I would make sure to split up your discussion of the two (Society and the Environment) approximately equally. Perhaps you'd give three advantages/disadvantages for Society, and two for the Environment, but like I wouldn't go overboard with one and minimise the other.
I would also ALWAYS have as many advantages as disadvantages. That's a method I've personally stuck with; would be interested to hear what others do.
When it comes to your table specifically, I would be drawing two: one for Society (advantages in one column, disadvantages in another, taking up 4 lines in total. One is the title, three are advantages/disadvantages) and one for the Environment (advantages in one column, disadvantages in another, taking up 3 lines in total. One is the title, two are advantages/disadvantages).
Does that sort of make sense? If you do the above, and are still struggling, post up a picture of your answer for us to mark!
Jake
Ok then! I guess I'll go look for some more disadvantages!
Can I ask how detailed should the points be to fit up a single line each? Are they just key words like "long distance transmission possible"? I'm worried that it wasn't a detailed enough answer so I would wrote "AC could be stepped up or down for more efficient, long distance transmission. By stepping up voltage, the current could be reduced (P=IV) and subsequently power loss was reduced (Ploss=I^2R). Thus price of electricity is dropped" and that went over 8 lines for one point...
Other than that, I'll redo it and see how it goes :)
Post by: winstondarmawan on July 22, 2017, 11:35:45 pm
Hey guys! having trouble with this question. I was able to deduce possible answers to B or D.Lenz Law states that the EMF will be induced in a manner that opposes the initial change.
answer:SpoilerB
In a generator, the initial change is the rotation. So the current (due to EMF) will be induced so that it opposes the initial change.
Let's look at B. As the generator rotates clockwise, the current generated will be generated so that it moves anticlockwise.
Using the right-hand palm rule, we can see that the current moves up and ends at terminal X, making thus making it have a "positive" voltage (remember it is the potential difference). This matches with the graph.
(If I'm wrong someone please correct me)
Post by: jamonwindeyer on July 23, 2017, 01:23:53 am
Lenz Law states that the EMF will be induced in a manner that opposes the initial change.
In a generator, the initial change is the rotation. So the current (due to EMF) will be induced so that it opposes the initial change.
Let's look at B. As the generator rotates clockwise, the current generated will be generated so that it moves anticlockwise.
Using the right-hand palm rule, we can see that the current moves up and ends at terminal X, making thus making it have a "positive" voltage (remember it is the potential difference). This matches with the graph.
(If I'm wrong someone please correct me)
I'm intrigued by this question and answer, because I think the answer is D (maybe it is 1:21am and I'm missing something but I'll throw this out there nonetheless) ;)
Your working is totally correct right until the end Winston (note we'll want B or D because those are slip rings and we need them for AC) - But if we have current flowing from Y to X, that means the voltage at Y is higher. Current flows from points of a higher potential to points of a lower potential (remember back to basic circuit theory). For this reason, I'd go for D!
Would love for someone to weigh in - Especially if it means calling me out on a mistake (people never do this, and I want them to!) ;D
Post by: jamonwindeyer on July 23, 2017, 01:29:35 am
Also quick question. If an exam questions asks to draw the transmission lines tower what would be a sufficient way of drawing it with all the relevant parts? Also what is it meant by "arcing"?
Thanks ;D
Ok then! I guess I'll go look for some more disadvantages!
Can I ask how detailed should the points be to fit up a single line each? Are they just key words like "long distance transmission possible"? I'm worried that it wasn't a detailed enough answer so I would wrote "AC could be stepped up or down for more efficient, long distance transmission. By stepping up voltage, the current could be reduced (P=IV) and subsequently power loss was reduced (Ploss=I^2R). Thus price of electricity is dropped" and that went over 8 lines for one point...
Other than that, I'll redo it and see how it goes :)
I think the detail you are providing there is good detail - The use of formula particularly is nice. Perhaps just adjust your phrasing, so I'd say that like:
Ability to transform AC power allows power loss to be minimised by minimising current (\(P=I^2R\)), thus increasing efficiency and reducing the cost of electricity
Though in truth, if the question is asking for impact on society and environment, even the formula may not be worth your space. Just saying:
Ability to transform allows reduced power loss, maximising efficiency, reducing energy cost and enabling distribution of power to remote regions to improve quality of life
This is more society focused, which is more your aim :)
Post by: ekhan_01 on July 23, 2017, 12:56:39 pm
I'm confused as to finding the value of max torque. The answers say it is 1.03x10^-2 Nm but i keep getting 5.89x10^-3 Nm.
Perhaps I've done something wrong but a second answer on this would much be appreciated
Thank You
Post by: jakesilove on July 23, 2017, 01:17:29 pm
Hi Jake :)
I'm confused as to finding the value of max torque. The answers say it is 1.03x10^-2 Nm but i keep getting 5.89x10^-3 Nm.
Perhaps I've done something wrong but a second answer on this would much be appreciated
Thank You
Hey! The formula for torque is
Here, our area is
Note that I have converted from cm to m: this is absolutely crucial to getting the answer right, and COULD be where your error comes from. Subbing all of this into our equation, we get
Whiiiiich is what you got. Great job on that conversion. I suspect the answers are wrong!
Post by: ekhan_01 on July 23, 2017, 03:22:40 pm
Thx for the prev answer gr8 reassurance
I'm just wondering how do you do b? (answer is anticlockwise) & I have no clue going abt this question
Ty
Post by: kiwiberry on July 23, 2017, 03:47:47 pm
Hey again!
Thx for the prev answer gr8 reassurance
I'm just wondering how do you do b? (answer is anticlockwise) & I have no clue going abt this question
Ty
Hello! :)
Here we have to use the right hand palm rule to determine the force on either side of the coil. Fingers represent the direction of the magnetic field, thumb points in the direction of current and your palm will face the direction of the force. On the left, current is going into the page and magnetic field is going right, so point your thumb into the page with your fingers pointing right - you'll end up with your palm facing downwards. This means that the left side of coil will experience a force downwards. Similarly, on the right, current is going out and the field is in the same direction, so using the same rule again you'll find that the right side will experience a force upwards! Together, these forces will make the coil spin anticlockwise.
Post by: yattmoani on July 23, 2017, 05:11:52 pm
Correct answers are:
8 - B
9 - D
19 - A
Post by: kiwiberry on July 23, 2017, 05:55:51 pm
Hey guys! I was doing a past paper today and I just couldn't wrap my head around these 3 questions. Loving this thread and thanks for the help guys!
Correct answers are:
8 - B
9 - D
19 - A
Hey!
8- The force on a charged particle in a magnetic field is given by F=qvBsintheta, where theta is the angle between the direction of the field and the particle's motion. Looking at B, the particle's motion is parallel to the magnetic field, hence theta=0 and F=0. Therefore this should be the one which exerts the least force because there is none! As a note, the force on a charged particle due to an electric field is given by F=qE, so the velocity of the particle will not affect the force of the electric field exerts on it.
9- First we need to find the force that P and Q exert on R. P and R are parallel, meaning that P will exert an attractive force on R towards the left
Hope this helps! :)
Post by: yattmoani on July 23, 2017, 06:36:26 pm
Post by: bsdfjnlkasn on July 25, 2017, 09:02:45 am
Was hoping to get some help with the following question :)
EDIT: Could I also get a thorough answer to this following 4 marker? I'm just not sure how Lenz's law is relevant here and the order of all the changes is a bit confusing...
Here is what I understand:
Because current now flows through the solenoid, it will produce a north pole at the bottom (RHCurlR). Then the ring also experiences a changing magnetic flux because there has been a south pole produced at the top. So in order for the ring to oppose this change, it will need to produce a north pole acting down to weaken/cancel the new magnetic field (Lenz's law) around the ring. Then using the curl rule again, the induced current will be clockwise (thumb down, fingers around). BUT I'm not sure what happens next to give me an upwards force. The provided answer is super vague so would love some clarity.
2. How would you solve the projectile part of this question? I need to find the time it will take for the stone to hit the water.
3. I'm not sure why for part b) you can't use vy2 = uy2 + 2aydeltaY?
4. Another projectile :)
I'm quite confused about what working out is happening after the initial velocity is calculated, because how does finding the angle to be 24° verify the table which states 22°? I'm also not really sure why they've used v = u+at
5. Lol really not getting these... Not sure what's happened after they found V (and also why it's negative)?
Any help would be super appreciated :)
Post by: beau77bro on July 27, 2017, 12:22:41 pm
for Q5, does take off mean it hasn't left the ground yet? I assumed the definition of G-force was always just an expression of appraent weight as a multiple of true weight. but the answer is D???
for 7 it says B and i jsut don't get that, because the experiment couldn't read the effects of the aether - it didn't disprove but it couldn't get tangible results.
8. Is B, but i dont get why it's not D? is it just because it's cutting more? isn't it awlays to do with change in flux not amount of?
12. I just don;t get this. doesn't back emf get produced until it completely negates the input - thus preventing acceleration. Can someone explain how Back Emf relates to current and voltage in a motor when it's running i feel like my understanding is very holey around there. like why is there still current when no Emf?
15. if the force is increasing but at a lesser rate, doesn't that mean the charge must be increasing. The Answer is D like dahel.
Post by: beau77bro on July 27, 2017, 12:23:26 pm
Post by: jamonwindeyer on July 27, 2017, 12:34:56 pm
Hey there!
Was hoping to get some help with the following question :)
Hey! This requires the formula:
This formula links magnetic flux to magnetic field strength (B) and area (A). Given the initial values of B and A, the initial flux is:
If we halve the area, we halve the flux, so the change in flux is roughly equal to 0.125 Webers. The induced EMF is therefore given by Faraday's Law:
Quote
EDIT: Could I also get a thorough answer to this following 4 marker? I'm just not sure how Lenz's law is relevant here and the order of all the changes is a bit confusing...
Here is what I understand:
Because current now flows through the solenoid, it will produce a north pole at the bottom (RHCurlR). Then the ring also experiences a changing magnetic flux because there has been a south pole produced at the top. So in order for the ring to oppose this change, it will need to produce a north pole acting down to weaken/cancel the new magnetic field (Lenz's law) around the ring. Then using the curl rule again, the induced current will be clockwise (thumb down, fingers around). BUT I'm not sure what happens next to give me an upwards force. The provided answer is super vague so would love some clarity.
You've got almost all of it - The induced current in the ring produces another magnetic field as you say. This interacts with the field produced by the coil and the ring is repelled ;D
====
Sorry, I'm a little confused as to how the rest of your screenshots line up with your questions (and there might be some missing?), you might need to clarify for me :)
Post by: jakesilove on July 27, 2017, 12:36:24 pm
Hey there!
Was hoping to get some help with the following question :)
EDIT: Could I also get a thorough answer to this following 4 marker? I'm just not sure how Lenz's law is relevant here and the order of all the changes is a bit confusing...
Here is what I understand:
Because current now flows through the solenoid, it will produce a north pole at the bottom (RHCurlR). Then the ring also experiences a changing magnetic flux because there has been a south pole produced at the top. So in order for the ring to oppose this change, it will need to produce a north pole acting down to weaken/cancel the new magnetic field (Lenz's law) around the ring. Then using the curl rule again, the induced current will be clockwise (thumb down, fingers around). BUT I'm not sure what happens next to give me an upwards force. The provided answer is super vague so would love some clarity.
2. How would you solve the projectile part of this question? I need to find the time it will take for the stone to hit the water.
3. I'm not sure why for part b) you can't use vy2 = uy2 + 2aydeltaY?
4. Another projectile :)
I'm quite confused about what working out is happening after the initial velocity is calculated, because how does finding the angle to be 24° verify the table which states 22°? I'm also not really sure why they've used v = u+at
5. Lol really not getting these... Not sure what's happened after they found V (and also why it's negative)?
Any help would be super appreciated :)
Maybe I'm looking at it wrong, but your questions don't seem to correspond to attachments. Please make this clearer, so we can better get to your questions. This could just be me failing to understand your explanation, but yeah.
Post by: jamonwindeyer on July 27, 2017, 12:47:04 pm
help please. I don't get these questions - explanation much appreciated
Hey!
Question 5: This is really just about knowing the definition. C is correct - G-force is about expressing apparent weight force under a certain acceleration in terms of weight force on earth :)
Question 7: Again, really just about knowing this fact. The Michelson-Morley experiment produced a null result, meaning it couldn't detect what it was looking for (which was the Aether Wind) :)
Question 8: Recall that the induced emf is proportional to the rate of change of magnetic flux, that is:
Now magnetic flux can be related to magnetic field strength through \(\phi=BA\). This should immediately suggest that a bigger coil rotating through a stronger magnetic field will give a larger flux, meaning more induced emf. More turns in the coil will have a similar effect, and spinning it with a greater frequency will make the change faster (increasing the rate of change). So, we want EVERYTHING to get bigger - I'd go for B :)
Question 12: I think the answer to this question depends a bit on if we assume the motor is ideal (frictionless). But given the answers, I think there is friction. In which case, it is fairly simple (I think, almost too easy?) - We need an AC voltage to drive an AC motor. So, B!
and this as i couldnt post all three.
The law for force experienced by a charge in an electric field:
Notice the similarity to a regular straight line, where \(y=F\) and \(E=x\). However, the line on the graph isn't straight. So clearly, the force isn't proportional to electric field - We can see that the line starts to taper off for larger fields - The force isn't changing much even though the field is. This means that the particle must be losing charge - The answer is D :)
Post by: beau77bro on July 27, 2017, 12:58:22 pm
Hey!
Question 5: This is really just about knowing the definition. C is correct - G-force is about expressing apparent weight force under a certain acceleration in terms of weight force on earth :)
Question 7: Again, really just about knowing this fact. The Michelson-Morley experiment produced a null result, meaning it couldn't detect what it was looking for (which was the Aether Wind) :)
Question 8: Recall that the induced emf is proportional to the rate of change of magnetic flux, that is:
Now magnetic flux can be related to magnetic field strength through \(\phi=BA\). This should immediately suggest that a bigger coil rotating through a stronger magnetic field will give a larger flux, meaning more induced emf. More turns in the coil will have a similar effect, and spinning it with a greater frequency will make the change faster (increasing the rate of change). So, we want EVERYTHING to get bigger - I'd go for B :)
Question 12: I think the answer to this question depends a bit on if we assume the motor is ideal (frictionless). But given the answers, I think there is friction. In which case, it is fairly simple (I think, almost too easy?) - We need an AC voltage to drive an AC motor. So, B!
The law for force experienced by a charge in an electric field:
Notice the similarity to a regular straight line, where \(y=F\) and \(E=x\). However, the line on the graph isn't straight. So clearly, the force isn't proportional to electric field - We can see that the line starts to taper off for larger fields - The force isn't changing much even though the field is. This means that the particle must be losing charge - The answer is D :)
5 was D, 7 was B, 8 i get now, so it's not that it's changing at a greater rate - that change is bigger meaning more EMF.
12 i still don't get because what about back emf?
15. OMG I SEE - YOU TREAT IT LIKE A LINE. SO q much be decreasing - should've written it as the relationship. thankyou
But yea 5 and 7 have me stumped and i don't quite get 12.
also i modified the last one - but don't worry i ask the same questions here.
Thanks jamon!!!!
Post by: jamonwindeyer on July 27, 2017, 01:05:10 pm
5 was D, 7 was B
No way either of those are correct - In honesty Q5 I was a little unsure because it is very much a technicality, but it can't be D, and no way 7 is B === 7 is definitely D :)
Quote
12 i still don't get because what about back emf?
At maximum speed there is definitely back EMF that opposes the supply, but maximum implies the back emf (plus friction) has balanced with the supply. The waveform looks the same but it has a smaller magnitude - That's my interpretation at least. What would you think Back EMF would do?
Post by: beau77bro on July 27, 2017, 01:28:37 pm
No way either of those are correct - In honesty Q5 I was a little unsure because it is very much a technicality, but it can't be D, and no way 7 is B === 7 is definitely D :)
At maximum speed there is definitely back EMF that opposes the supply, but maximum implies the back emf (plus friction) has balanced with the supply. The waveform looks the same but it has a smaller magnitude - That's my interpretation at least. What would you think Back EMF would do?
OK THANKGOD I WAS QUESTIONING MY EXISTENCE.
anyway I always thought Back emf just negated input, making it 0 - i never thought or i dont remember about frictions part? i see now Friction slows the rotor - reducing the back EMF induced? and thus leaving the input voltage supplying a lessened voltage. woudl that be correct - i just never thought of friction playing a part without reference.
thanks again
Post by: jamonwindeyer on July 27, 2017, 01:29:52 pm
OK THANKGOD I WAS QUESTIONING MY EXISTENCE.
anyway I always thought Back emf just negated input, making it 0 - i never thought or i dont remember about frictions part? i see now Friction slows the rotor - reducing the back EMF induced? and thus leaving the input voltage supplying a lessened voltage. woudl that be correct - i just never thought of friction playing a part without reference.
thanks again
Yep, so without friction the back emf would cancel the input and give us zero at max speed. But there is no blank graph option, so we assume friction plays a role, which means we must continually supply at least a small AC voltage to overcome that friction (even after back emf) :)
Post by: beau77bro on July 28, 2017, 08:27:50 am
Post by: Mymy409 on July 28, 2017, 06:59:42 pm
Post by: beau77bro on July 29, 2017, 09:00:37 am
Do we have to know about tension and other prelim concepts in HSC Physics?
they sometimes pop up in trials, like extra content to check your knowledge - usually it's pretty minor understanding though, and I haven't much if any in hscs but Im sure Jake and Jamon have a better idea
Post by: beau77bro on July 29, 2017, 09:02:25 am
Post by: jakesilove on July 29, 2017, 10:06:53 am
more help needed. sorry for being a hassle, but the answers say the wavelength should half for this, even though the speed is kept constant. I disagree, but Im not sure- the amplitude should double but yea
Hey!
If you double the number of turns, then the FORCE doubles. So, the rotor WILL rotate faster, halving the wavelength. Make sense?
Post by: statues on July 29, 2017, 10:25:26 am
17.
A proton moving at 99.999999% the speed of light has approximately as much kinetic energy as:
a. a 20g snail moving at 7 mms-1
b 100 kg car moving at 25 ms-1
c. 30 kg child running at 3 ms-1
d. 150 g ball thrown at 14 ms-1
Answer was A.
I don't know how they got it - I plugged the velocity into Einsteins mass dialation equation and then plugged that into the kinetic energy formula but none of them seemed to work out.
Thanks in advance :)
Post by: beau77bro on July 29, 2017, 10:42:35 am
Hey!
If you double the number of turns, then the FORCE doubles. So, the rotor WILL rotate faster, halving the wavelength. Make sense?
but it says the speed is kept the same? which is my problem.
Post by: jakesilove on July 29, 2017, 01:02:16 pm
but it says the speed is kept the same? which is my problem.
Ah yep I see; in that case, wavelength would stay the same, as you've mentioned. I would chalk it up to a badly worded/designed question
Post by: jakesilove on July 29, 2017, 01:10:41 pm
From the 2016 exam choice trial:
17.
A proton moving at 99.999999% the speed of light has approximately as much kinetic energy as:
a. a 20g snail moving at 7 mms-1
b 100 kg car moving at 25 ms-1
c. 30 kg child running at 3 ms-1
d. 150 g ball thrown at 14 ms-1
Answer was A.
I don't know how they got it - I plugged the velocity into Einsteins mass dialation equation and then plugged that into the kinetic energy formula but none of them seemed to work out.
Thanks in advance :)
Hey! So the kinetic energy of the proton will be
Plugging in relevant values, including the mass of the proton, gets us
Then, assuming A) was in mm per second
Which is pretty bloody close to the Proton kinetic energy!
Post by: Mymy409 on July 29, 2017, 06:45:06 pm
they sometimes pop up in trials, like extra content to check your knowledge - usually it's pretty minor understanding though, and I haven't much if any in hscs but Im sure Jake and Jamon have a better idea
ohhh, ive forgotten a lot of prelim stuff. welp, lets hope it doesnt come up in trials :D
Post by: statues on July 29, 2017, 09:16:37 pm
Hey! So the kinetic energy of the proton will be
Plugging in relevant values, including the mass of the proton, gets us
Then, assuming A) was in mm per second
Which is pretty bloody close to the Proton kinetic energy!
Hey so I can't get the same value for proton energy. Is this right?
`Ek = 0.5 \times(\frac{1.673\times10^{-27}}{\sqrt{1-0.99999999}})\times(0.99999999\times3\times10^{8})^{2}`
Post by: kiwiberry on July 29, 2017, 09:20:52 pm
Hey so I can't get the same value for proton energy. Is this right?
`Ek = 0.5 \times(\frac{1.673\times10^{-27}}{\sqrt{1-0.99999999}})\times(0.99999999\times3\times10^{8})^{2}`
The 0.99999999 under the square root needs to be squared, other than that you're all good :)
Post by: statues on July 29, 2017, 09:27:46 pm
The 0.99999999 under the square root needs to be squared, other than that you're all good :)
aahhhhh tysm :)
Post by: beau77bro on July 30, 2017, 08:26:35 am
help pls. I dont understand
could i get some help on this? the question is very strange - I don't think it gives u enough values but Im not great at medical so more help needed. thanks again guys u r life savers
could i also get help with this Question: i think they used √GM/r instead √2GM/r
so does that mean its wrong or am I missing something? The answer is A
with the second one the Answer is D - i can't see how u would get there treating the speed of light as 3x10^8
another one - this i get why the answer is A - but shouldn't having the frequency mean you are decreasing the rate of change of magnetic flux? So it shouldn't just double EMF because the input is doubled - but stay the same since the rate was halved?
Post by: katnisschung on July 30, 2017, 11:53:38 am
thanks :) (background this is for quanta to quarks)
Post by: jakesilove on July 30, 2017, 01:50:10 pm
could i get some help on this? the question is very strange - I don't think it gives u enough values but Im not great at medical so more help needed. thanks again guys u r life savers
could i also get help with this Question: i think they used √GM/r instead √2GM/r
so does that mean its wrong or am I missing something? The answer is A
with the second one the Answer is D - i can't see how u would get there treating the speed of light as 3x10^8
another one - this i get why the answer is A - but shouldn't having the frequency mean you are decreasing the rate of change of magnetic flux? So it shouldn't just double EMF because the input is doubled - but stay the same since the rate was halved?
2 (Moon Question)
Escape velocity is calculated as
Subbing in the values given
So, the answer is B. If it says that the answer is A, it's just wrong.
17 (Photon)
For some stupidly annoying reason, we aren't given the wavelength. We need to actually count the wavelengths, and divide the given length by that. Literally the biggest waste of time.
By my count, there are 12 wavelengths in the diagram. So,
We can calculate frequency by
Cool, so the answer is either C or D, and the photon energy is
This is GREATER than the work function, so a photon is released. The answer is D.
9 (Cathode Ray)
Yes, the voltage should stay the same, and it does. The initial graph has a max at 200V. The final graph has a max at 200V. So, you're correct!
Post by: jakesilove on July 30, 2017, 01:52:37 pm
why/how do accelerating charges produce emr?
thanks :) (background this is for quanta to quarks)
I didn't do Quanta, but I SUSPECT this is something you should just take on face value. It comes down to maths, relativity, and forces that I don't think you learn about. I don't want to write a confusing answer if it isn't needed, so I think I'll just leave it at that, but if someone who has done Quanta wants to jump in please do!
Post by: beau77bro on July 30, 2017, 02:01:53 pm
2 (Moon Question)OK SWEET THANKYOU JAKE I APPRECIATE YOUR HELP SO MUCH also wth why is the second one so annoying? that's like just an unnessecary trick to see if ur paying attention - rude. thanks so much tho.
Escape velocity is calculated as
Subbing in the values given
So, the answer is B. If it says that the answer is A, it's just wrong.
17 (Photon)
For some stupidly annoying reason, we aren't given the wavelength. We need to actually count the wavelengths, and divide the given length by that. Literally the biggest waste of time.
By my count, there are 12 wavelengths in the diagram. So,
We can calculate frequency by
Cool, so the answer is either C or D, and the photon energy is
This is GREATER than the work function, so a photon is released. The answer is D.
9 (Cathode Ray)
Yes, the voltage should stay the same, and it does. The initial graph has a max at 200V. The final graph has a max at 200V. So, you're correct!
Post by: Aaron12038488 on July 30, 2017, 06:31:59 pm
Describe 4 pieces of evidence which supports big bang theory
thx.
Post by: ekhan_01 on July 30, 2017, 06:35:17 pm
I'm sorry I am just plain clueless and cannot answer this question.
Can you please give a brief explanation
:)
Post by: kiwiberry on July 30, 2017, 06:45:17 pm
Hi Jake!!
I'm sorry I am just plain clueless and cannot answer this question.
Can you please give a brief explanation
:)
Hello!
Remember that the force on a charged particle moving in a magnetic field is given by:
Post by: statues on July 30, 2017, 06:49:31 pm
Why isn't induced emf max when the flux is max - isn't the most emf induced when it's moving through the b field?
Post by: ekhan_01 on July 30, 2017, 06:52:15 pm
Hello!
Remember that the force on a charged particle moving in a magnetic field is given by:Here, q, v and B are all constant (same charge, speed and magnetic field strength). The only variable is \(\theta\) - the angle between the particle's direction of motion and the magnetic field. X makes a 90o angle with the field, whereas Y makes a 30o angle. So the ratio of the force on X to the force on Y is:
Thx heaps!!
Also while we're on the topic of this why is it at times in a mag field;
- the angle is subtracted from 90deg eg. 90-30 = 60deg
- is the angle eg. 30deg
- or the angle is disregarded and is 90deg
Post by: Shadowxo on July 30, 2017, 07:12:29 pm
Could someone please help explain this multiple choice from girraween's 2015 trial.emf is dependent on the rate of change of flux.
Why isn't induced emf max when the flux is max - isn't the most emf induced when it's moving through the b field?
So EMF is at a max/min when the gradient is steepest. eg for D, at t=0 the rate of change of flux is very negative (steep gradient downwards) so the emf will be large and positive.
Post by: kiwiberry on July 30, 2017, 08:22:03 pm
Thx heaps!!
Also while we're on the topic of this why is it at times in a mag field;
- the angle is subtracted from 90deg eg. 90-30 = 60deg
- is the angle eg. 30deg
- or the angle is disregarded and is 90deg
It all comes down to the definition of \(\theta\) in the formula - the angle between the direction of the particle's motion and the direction of the mag field.
- In your question above, the angle marked on the diagram was exactly that, so we can plug \(\theta=30\) into the formula
- However, sometimes questions will give you the angle between the particle's motion and a line perpendicular to the field - this is where you'll have to subtract the angle from 90 to get \(\theta\)
- When a particle is travelling in a mag field that is directed into or out of the page, it will always be at an angle of 90o to the direction of the field no matter which direction it is travels in, so \(\theta\) is always 90. For example, if the mag field in your question was going into the page instead of to the right, to find the force on Y you would disregard the 30o and use \(\theta=90\) instead.
Let me know if anything doesn't make sense :)
Post by: beau77bro on July 31, 2017, 02:41:50 pm
Post by: beau77bro on August 01, 2017, 08:04:00 am
help please i hate these graph questions- i understand torque and force ones but its hard with voltage and current so any tips massively appreciated. q b btw
hello pls my exam is in 1hr - these graphs always kill me. I get the increasing and decreasing for EMF but i don't get current -is it just the reverse of emf?
Post by: beau77bro on August 01, 2017, 08:24:57 am
Post by: jakesilove on August 01, 2017, 10:33:02 am
Post by: left right gn on August 01, 2017, 06:13:55 pm
Post by: seventeenboi on August 01, 2017, 06:17:10 pm
For this question: 'assess the use of thought experiments in the development of our current understanding of time' how would you approach such a question?? thanks
Post by: daniel.strozek on August 01, 2017, 09:24:20 pm
Post by: kdawgs on August 01, 2017, 09:55:00 pm
Post by: jakesilove on August 01, 2017, 10:42:24 pm
Can someone explain donor and acceptor levels??
You do not need to understand anything to do with this concept. Just be able to draw them, and state which level belongs to each type of doped semiconductor. Use the diagram below, and just memorise it!
(http://www.physics.udel.edu/~watson/scen103/semi2.gif)
Hi :)
For this question: 'assess the use of thought experiments in the development of our current understanding of time' how would you approach such a question?? thanks
This question is asking about special relativity. So, you should first discuss the notion of relativity, and how the M-M experiment threw standard physics intro disarray. Using thought experiments, Einstein revolutionised modern physics. So, I would discuss a thought experiment, and show that if the speed of light is a constant, time must be relative. Then, I would show the results of this relativity; the formulas on your formula sheet, time slowing down at high speeds, the impact on space travel etc. Have a crack at the question, and post up your response!
outline, using a diagram and/or a clear description, your investigation demonstrating the principle of an induction motor and explain in terms of the physics involved, how the motor rotation is produced
Each school does a different prac of this dot point, so there's not much I can help you with. Recall an experiment where you caused something to turn (a coil? A piece of foil?). Use diagrams, and a clear description, and you'll get the marks.
How do you explain the meissner effect without using electromagnetic induction? like what is the alternate explanation?
The explanation I would use comes down the Lenz's law. As a magnet falls towards a superconductor, the superconductor experiences a changing magnetic field (due to the accelerating magnet). Thus, by the motor effect, and current will be produced on the superconductor. This will be in the form of Eddy currents, on the surface of the superconductor. By Lenz's law, the direction of these Eddy currents will be in opposition to the falling magnet; thus, the magnet will slow down. However, since the metal is a superconductor, the Eddy currents never dissipate! So, there is a constant upwards force resisting the moving magnet. Eventually, this force will equal the gravitational force, and the magnet will hover.
Post by: blasonduo on August 02, 2017, 01:25:28 pm
This question is asking about special relativity. So, you should first discuss the notion of relativity, and how the M-M experiment threw standard physics intro disarray. Using thought experiments, Einstein revolutionised modern physics. So, I would discuss a thought experiment, and show that if the speed of light is a constant, time must be relative. Then, I would show the results of this relativity; the formulas on your formula sheet, time slowing down at high speeds, the impact on space travel etc. Have a crack at the question, and post up your response!
Just to add on Jake's amazing work,
make sure to discuss the USE of thought experiments and the pros and cons with using them ;)
Post by: statues on August 02, 2017, 04:38:54 pm
Post by: kiwiberry on August 02, 2017, 04:52:40 pm
plz help I thought it was d. its a
GPE is given by:
Post by: Aaron12038488 on August 02, 2017, 07:51:32 pm
thx.
Post by: daniel.strozek on August 02, 2017, 10:04:22 pm
Post by: jamonwindeyer on August 02, 2017, 10:34:42 pm
i need help finding a retrogade motion diagram - and explaining the diagram.
thx.
Hey! Is this for the models of the universe for Year 11? I personally never dealt with them - Hopefully someone else can help! :)
How do you approach a massive extended response? like the general scaffold or something. if there is a general method
This guide written by Jake might be helpful for you! :)
Post by: bsdfjnlkasn on August 03, 2017, 08:14:30 pm
I have the following in my notes:
Bohr provided a theoretical/physical explanation for Balmer’s formula, making his model quite successful overall.
Do we have to know how Bohr explained it or what this explanation was? As i'm trying to answer a 6 marker on how mathematical models validated by experiments have aided our understanding of the atom and think that this info will be helpful.
Thank you!
Post by: kiwiberry on August 04, 2017, 12:18:42 am
Hey there!
I have the following in my notes:
Bohr provided a theoretical/physical explanation for Balmer’s formula, making his model quite successful overall.
Do we have to know how Bohr explained it or what this explanation was? As i'm trying to answer a 6 marker on how mathematical models validated by experiments have aided our understanding of the atom and think that this info will be helpful.
Thank you!
Hey! You don't need to know the exact mathematical derivation, but it may be worth mentioning that he combined the classical formulas of GPE and KE with his quantum formulas (like angular momentum) and was able to derive Balmer's equation and Rydberg's constant. This lent huge support to his model of the atom because it was able to explain the discrete nature of the spectral lines as well as the exact wavelengths of the lines. :)
Post by: AlphaGeek on August 04, 2017, 01:33:13 am
Post by: johnk21 on August 04, 2017, 11:46:03 am
Hey there, Could anyone please help me with solving this question, cheers.We can rule out C and D, as acceleration due to gravity on earth is greater than mars, so it wouldnt make sense for the ball to fall faster on mars compared to earth. but between A and B, im not sure how to do it. I have tried manipulating acceleration = distance over time squared but i am getting no where :( .
Sorry i couldnt be of that much help
Post by: kiwiberry on August 04, 2017, 01:10:08 pm
Hey there, Could anyone please help me with solving this question, cheers.
Hey, welcome to the forums!!
Because the only force on the ball is gravity, we can use projectile motion equations to find the time it takes for the ball to reach the ground:
Post by: devol on August 04, 2017, 02:46:55 pm
A: - mv^2
B: - Gm/rv
C: - Gmv/r
D: - (1/2) mv^2
Thanks :)
Post by: blasonduo on August 04, 2017, 02:51:36 pm
A satellite of mass m in orbit around the Earth has a velocity of v. The gravitational potential energy of the satellite is:
A: - mv^2
B: - Gm/rv
C: - Gmv/r
D: - (1/2) mv^2
Thanks :)
Hello! This question was answered just a few posts up! Go check it out, Kiwiberry did a terrific job at explaining it. :)
Post by: jamonwindeyer on August 04, 2017, 03:19:49 pm
A satellite of mass m in orbit around the Earth has a velocity of v. The gravitational potential energy of the satellite is:
A: - mv^2
B: - Gm/rv
C: - Gmv/r
D: - (1/2) mv^2
Thanks :)
Hello! This question was answered just a few posts up! Go check it out, Kiwiberry did a terrific job at explaining it. :)
Here to be precise ;D
Post by: Mymy409 on August 05, 2017, 01:45:37 am
Post by: beau77bro on August 05, 2017, 09:28:32 am
Hi, I'm having trouble with this question. I don't quite understand how we're meant to figure this out.G-force is described as a reactionary force - it's perceived weight, and we know how much we way from how hard we press against the floor. so here we would feel the heaviest at Q in the same way we would feel weightless at S, because at Q you are experiencing an acceleration change where the floor begins to push and move upwards, whilst you still have the tendency to go down(inertia) as such you feel the heaviest at the bottom. think of it as opposites, your weightless at the top (for the reverse reasons), heaviest at the bottom.
hope this helps goodluck with trials
Post by: kdawgs on August 05, 2017, 10:12:39 am
"Different tissues will have different T1 relaxation profiles or times. Large molecules and bound water molecules such as in the fat, liver and spleen have a short T1, while free water has a long T1."
"There are also different T2 profiles. Large molecules found in tendons and muscles have a short T2 while free water has a long T2 "
I don't understand how or why there is a difference in the relaxation times for t1 and t2 depending on the molecule.
Post by: winstondarmawan on August 05, 2017, 11:34:40 am
http://www.globalspec.com/ImageRepository/LearnMore/20153/superconductorbfc636ad4cee4704a85293b4e06adccb.gif
OR
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20643641_1291020447690142_1176167503_n.png?oh=22447a29bb4a5af096c3c7575b4960d9&oe=598732E4
And also this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20624073_1290386117753575_703406917_n.png?oh=07844f7c72598c6beb20572cb88b4fa3&oe=598846B3
TIA!
Post by: 12carpim on August 05, 2017, 12:52:36 pm
Thank you!
Post by: jakesilove on August 05, 2017, 12:57:19 pm
MEDICAL PHYSICS Q could someone please explain t1 and t2 relaxation times? Specifically these statements from my textbook:
"Different tissues will have different T1 relaxation profiles or times. Large molecules and bound water molecules such as in the fat, liver and spleen have a short T1, while free water has a long T1."
"There are also different T2 profiles. Large molecules found in tendons and muscles have a short T2 while free water has a long T2 "
I don't understand how or why there is a difference in the relaxation times for t1 and t2 depending on the molecule.
Hey! I wouldn't try to understand this point to any sort of high degree. Basically, it comes down to different quantum properties. Whilst you're probably imagining relaxation time as just an electron point one way, and then flipping around in a different direction, there are in reality a number of different impacts that a magnetic field has on a precessing Hydrogen atom.
T1 is just the time taken for one component to flip back, and T2 is just the time taken for another component to flip back. I wouldn't try to understand anything else.
Post by: jakesilove on August 05, 2017, 01:02:01 pm
Hello! For the graph of a superconductor (Temp vs Resistance), which shape would be correct?
http://www.globalspec.com/ImageRepository/LearnMore/20153/superconductorbfc636ad4cee4704a85293b4e06adccb.gif
OR
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20643641_1291020447690142_1176167503_n.png?oh=22447a29bb4a5af096c3c7575b4960d9&oe=598732E4
And also this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20624073_1290386117753575_703406917_n.png?oh=07844f7c72598c6beb20572cb88b4fa3&oe=598846B3
TIA!
Definitely use that first diagram.
As for the train question, note that the train must be accelerating to the left. If it were not, the ball should hang straight down, as the train would be in an inertial frame of reference.
So, since the train is accelerating to the left, if the string snaps the train will accelerate underneath the ball. So, it will move in a parabolic path downwards! Just think about the physical system, wave your arms around to try and understand it, and draw some diagrams :)
Post by: jakesilove on August 05, 2017, 01:09:16 pm
Please help!! I cant seem to get the answer right mine was off by 10000???
Thank you!
The information we have is
First, we always just look at the formulas and see if there's anything we can do. We can find an equation for time in terms of V, using the x-displacement equation
Now, we can use the y-displacement equation.
Since we know theta, this is an equation with only V in it! Solve for V, and we have our answer :)
Post by: 12carpim on August 05, 2017, 01:18:46 pm
Post by: armtistic on August 05, 2017, 01:25:59 pm
Post by: winstondarmawan on August 05, 2017, 01:51:54 pm
In HSC physics, given the formula R^3/T^2 = GM/4n^2. What is the unit for the period of orbit (T)? Is it seconds or hours or?
Everything in physics is done in SI, so T would be seconds.
Post by: johnk21 on August 05, 2017, 03:15:19 pm
In HSC physics, given the formula R^3/T^2 = GM/4n^2. What is the unit for the period of orbit (T)? Is it seconds or hours or?Strictly speaking in the HSC you should be using SI units, which in this case is seconds for time.
However, as keplers third law is moreso a RATIO, you can technically use and unit you put in. BUT by doing this you will get out that same unit.
e.g. if you put in time as hours, you get time out as hours if you are using the ratio of r^3/T^2 to find out periods and times of orbits around the same central body.
Post by: winstondarmawan on August 05, 2017, 04:58:05 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20631453_1291219204336933_771762729_n.png?oh=3373f6650396286eadf8c642e1f77b6b&oe=598705C8 Is A correct?
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20641460_1291217984337055_1025674672_o.png?oh=ebb05b043ab41190e96945d1a4eab658&oe=59870D2C Is C correct?
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20668463_1291218114337042_1787689365_n.png?oh=ab15922147f31c0a11bddfa516dc91a9&oe=598846A6 Part (ii)
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20668800_1291218261003694_586126526_n.png?oh=73f0df31101a4473339671d484db98ee&oe=598718B7 This would decrease output right? But I'm not too sure.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20631433_1291218487670338_710447928_n.png?oh=0104b363734be13cf451f8a1c2e33299&oe=59885469 Part (ii)
Thanks in advance!
Post by: blasonduo on August 05, 2017, 06:15:03 pm
Yes, I believe you are correct for Question 1, the further object will have a higher gravitational potential energy. Even though P is half the altitude of Q, It is not half the potential energy. This is due to the ratio being measured from the centre of the Earth, and thus the ratio is not 1:2.
For 2, Yes! I believe this is correct, fewer vibrations, the less the Cooper pairs break up, and temperatures reduce this!
For 3, Since an increase of Voltage increases the electric field (E = V/d) and the force of the electron is dependant on the Electric field (F = Eq) There will be a greater force on this electron. So for the diagram, as the force is greater, the parabolic motion will be smaller.
For 4, my understanding was that radial magnets gave a smoother output signal, as torque is always at a maximum as the angle between the plane of the coils and the magnetic field is constant. This will INCREASE efficiency, not decrease it :)
For 5, Faraday's law is heavily reliant on a CHANGE in magnetic flux, the faster the change, the more current is induced. As they are moving it very slowly, there is a much lower change in flux, and thus the lightbulb does not light up.
Hope this helped :) Goodluck with your studies!
Post by: winstondarmawan on August 05, 2017, 07:02:50 pm
Hello!For 3, would the range be affected?
For 3, Since an increase of Voltage increases the electric field (E = V/d) and the force of the electron is dependant on the Electric field (F = Eq) There will be a greater force on this electron. So for the diagram, as the force is greater, the parabolic motion will be smaller.
For 4, my understanding was that radial magnets gave a smoother output signal, as torque is always at a maximum as the angle between the plane of the coils and the magnetic field is constant. This will INCREASE efficiency, not decrease it :)
For 5, Faraday's law is heavily reliant on a CHANGE in magnetic flux, the faster the change, the more current is induced. As they are moving it very slowly, there is a much lower change in flux, and thus the lightbulb does not light up.
Hope this helped :) Goodluck with your studies!
For 4, I was thinking that since it was a radial magnetic field their would be a less change in flux, and thus a smaller output. I know with a motor torque increases but this Q isn't looking at speed of rotation..
For 5, wouldn't it just be a smaller flash? I don't get why there is no flash at all
Anyways, thanks for your response! And sorry for being so inquisitive lol
Post by: blasonduo on August 05, 2017, 07:26:55 pm
For 3, would the range be affected?
For 4, I was thinking that since it was a radial magnetic field there would be a less change in flux, and thus a smaller output. I know with a motor torque increases but this Q isn't looking at speed of rotation.
For 5, wouldn't it just be a smaller flash? I don't get why there is no flash at all
Anyways, thanks for your response! And sorry for being so inquisitive lol
When you say range, do you mean the horizontal distance travelled by the electron? If so, YES! As force is greater, it will reach the plate sooner, and thus will not travel as far.
If torque increases, there will be a faster change in flux right? This faster change in flux will produce more current!
It COULD be, depending on how slow it is going, most light bulbs won't light up with only 0.005 Volts!
Hope that clears it up ;)
Post by: Dante1091 on August 05, 2017, 09:30:17 pm
How do I do these questions
(from 2011 CSSA)
Post by: bsdfjnlkasn on August 05, 2017, 10:28:33 pm
Hello guys,
How do I do these questions
(from 2011 CSSA)
Hey there!
For Q12: I believe doubling the speed halves the period so the answer is A (thanks to the correction below :) ) as the EMF increases with increasing change of flux - the consequence of increased speed of rotation
Q13: Kiwiberry did some awesome working for a similar question here: https://atarnotes.com/forum/index.php?topic=164552.msg965221#msg965221 Hopefully the explanation helps as the method is exactly the same. This is a fairly common MC so now you'll nail it in your trial! :)
Q19: D
For this we need to remember that since Copper is a metal, it exists as a lattice. When energy is supplied, the particles begin to vibrate vigorously which increases the number of intermolecular collisions. This inhibits conductivity as it makes it harder for a clear flow of charge to be established i.e. resistance increases. Silicon is different because, being a semi-conductor, it's valence band and conduction bands are separated. Therefore, energy must be supplied in order for the electrons in the valence band to bridge the gap and so, the silicon to conduct. Energy in this case is supplied by heating the element which in turn increases conductivity as more charges are able to access the conduction band and move freely. The following diagram might be familiar/helpful to distinguish between the specified conductor and semi-conductor:
(http://quarkology.com/12-physics/94-ideas-implementation/images/94C-pic-band-theory.png)
Q16:
So after equating FB and FE, we deduce:
r=mv\qB (lol someone please correct this formatting)
Looking at all the variables, we know in order to increase r we have to increase the variables in the numerator. The only option this leaves us with is A :)
Hopefully this helped!
Post by: blasonduo on August 05, 2017, 10:46:41 pm
Hey there!
For Q12: I believe doubling the speed halves the period so the answer is C
Hello! Thanks for helping!
Just remember though, if something increases in speed, there is a larger change in flux, and thus more current can flow, So this will be A ;)
Post by: kiwiberry on August 05, 2017, 10:49:55 pm
Hello guys,In case you're still confused about Q13 :)
How do I do these questions
(from 2011 CSSA)
The force that X exerts on Y is
The current running through Z will be 4I1, so the force Z exerts on Y is
Therefore the total force on Y will be F+2F=3F N left!
Post by: Dante1091 on August 05, 2017, 10:57:31 pm
Post by: austv99 on August 06, 2017, 12:39:17 am
Confused with whether the radial magnetic field will allow for increased change in flux or none at all? So whether the current output increases or not?
Thanks.
Post by: bsdfjnlkasn on August 06, 2017, 06:19:54 am
Hi,
Confused with whether the radial magnetic field will allow for increased change in flux or none at all? So whether the current output increases or not?
Thanks.
Hey there!
We use a radial magnetic field to maintain the maximum possible torque in a generator. It ensues that the armature/current carrying conductor is always parallel to the magnetic field. Considering torque is proportional to cos(theta) and that cos0=1, we know this arrangement of the magnetic field and coil will give a maximum torque.
*theta = 0 because the two are parallel, meaning the coil is not inclined to the field
That is the main reason we use a radial magnetic field.
I'm not so sure about the question you asked though. Despite the coil cutting flux lines when it is rotating, I don't know whether this will induce an EMF because the field strength of the radial magnetic field may be uniform..
Post by: winstondarmawan on August 06, 2017, 07:45:05 am
15. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20677063_1291831764275677_1507629571_o.png?oh=918648afa568be439af7221113370957&oe=5988D33D Is this B?
19, 20. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20668365_1291831997608987_996377233_n.png?oh=f72244959b80493fa41debb899a2185f&oe=59889344 Are these C and B?
25. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20632754_1291832540942266_456187579_n.png?oh=71fb21ccdf5c1e3795d03945f8bd525b&oe=5988BBEE Distance goes on the x-axis right? Also need help with part b.
TIA
Post by: blasonduo on August 06, 2017, 10:47:52 am
19) Yep! C is correct, A control is a very easy way to identify any uncontrolled variables.
20) This one is tricky! But in my opinion, I would go A. B doesn't really fit as placing no data because of a potential bias won't increase reliability. For an experiment, data is needed. Having peer-reviewed articles does increase reliability and does reduce this bias (and hence why I think A is better suited) as peer-reviewers are experts in the field, they know what they are talking about. Also, multiple peer-reviewers are present will increase this. Overall resulting in a more reliable report.
25) The general rule of thumb here is that the INDEPENDENT variable goes on x-axis, In this case, we aren't specifically told, but it can be safe to assume, that distance was the independent variable here :P For part b, I will quickly draw up the graph and do b for you shortly :)
Post by: jakesilove on August 06, 2017, 11:11:35 am
15) I do believe this is correct, this question is trying to trick you into thinking the magnets are super conductors! (If they were the magnetic field lines wouldn't pass through them!)
19) Yep! C is correct, A control is a very easy way to identify any uncontrolled variables.
20) This one is tricky! But in my opinion, I would go A. B doesn't really fit as placing no data because of a potential bias won't increase reliability. For an experiment, data is needed. Having peer-reviewed articles does increase reliability and does reduce this bias (and hence why I think A is better suited) as peer-reviewers are experts in the field, they know what they are talking about. Also, multiple peer-reviewers are present will increase this. Overall resulting in a more reliable report.
25) The general rule of thumb here is that the INDEPENDENT variable goes on x-axis, In this case, we aren't specifically told, but it can be safe to assume, that distance was the independent variable here :P For part b, I will quickly draw up the graph and do b for you shortly :)
Absolutely legendary answer, thanks for all the incredible contributions!
Post by: blasonduo on August 06, 2017, 11:41:46 am
Post by: austv99 on August 06, 2017, 12:06:57 pm
Could someone explain both parts?
Post by: kiwiberry on August 06, 2017, 12:41:14 pm
Hey,
Could someone explain both parts?
Bohr proposed that electrons existed in discrete stationary states/energy levels, and that they would only absorb EMR when transitioning to a higher energy state and emit EMR when transitioning to a lower energy state. This would mean that only certain wavelengths of EMR would be emitted from electron transitions. The Balmer series showed that the emission spectrum of hydrogen was not a continuous rainbow but instead discrete lines at certain wavelengths - this lent support to Bohr's model.
The Balmer series shows the spectral lines that result from electron transitions from n>2 to n=2. So the four lines in the diagram show the transitions from n=3-->2, 4-->2, 5-->2 and 6-->2. The next line will be the transition from n=7-->2. We can find the wavelength of this line using Rydberg's equation with nf=2 and ni=7:
Hope this helped :)
Post by: mercurry on August 06, 2017, 02:01:18 pm
a) The currents are in opposite directions, therefore the forces experienced repel. (?? what's the best answer for two marks?)
b) ?? Something to do with weight?
c) yeahhh nah
Source: HSC 2005, Q21
Post by: kiwiberry on August 06, 2017, 02:27:59 pm
Heyy, how do I answer these questions? I tried but I'm totally lost lol.
a) The currents are in opposite directions, therefore the forces experienced repel. (?? what's the best answer for two marks?)
b) ?? Something to do with weight?
c) yeahhh nah
Source: HSC 2005, Q21
a) Yep :) Because the tubes carry current in opposite directions, they will exert a repulsive force on each other. However, because the bottom one is supported by the rack it can't move down, so only the top tube jumps up.
b) Yeah, for the top tube to jump up, the force due to the currents must be greater than its weight force.
c) Because transmission lines are very long and carry high currents, they will need to be placed far apart to ensure the force between them is at a minimum.
Hope this helped!
Post by: winstondarmawan on August 06, 2017, 03:49:20 pm
Hi,
Confused with whether the radial magnetic field will allow for increased change in flux or none at all? So whether the current output increases or not?
Thanks.
Gonna bump this because I am interested in the answer too
Post by: jamonwindeyer on August 06, 2017, 04:09:17 pm
Gonna bump this because I am interested in the answer too
The radial field will probably not increase the maximum level of current you get from the generator, but it will remove the sinusoidal variation in that current. You'll have a (relatively) constant output because the rate of change of flux will be constant! It MAY be greater, but no way to know for sure and not what that question would be targeting
So great to see everyone being so helpful in this thread! Seriously incredible! I'll be around to help tonight if anything still has people stumped
🏼
Post by: Aaron12038488 on August 06, 2017, 05:17:04 pm
Post by: blasonduo on August 06, 2017, 05:23:49 pm
how likely will hsc assessments (excluding exams) like practicals and take home assessments, will be a repeat from each year?
Coming from my school, they are NORMALLY the same thing every year, I remember in year 7 around term 2 a bunch of home made generators would pop up on the back bench and here I am, sitting at my desk with the generator above me.
Older years have also told me the similarity in assignments each year, with maybe a question tweaked here and there.
This is my school, and I don't know how other schools do it, but it just seems easier for teachers to give out the same assignment every year :)
Post by: f_tan on August 06, 2017, 07:05:28 pm
Thanks!
Post by: kiwiberry on August 06, 2017, 07:15:18 pm
When doing calculations to find GPE, do you include the negative?
Thanks!
Yes always!! Unless they ask for the magnitude of energy :)
Post by: left right gn on August 06, 2017, 11:00:26 pm
Can we someone please explain the graphs in motors and generators. Such as EMF, flux, voltage output.
I'm confused on all the graphs of all these, please help!
Post by: yattmoani on August 06, 2017, 11:05:04 pm
Post by: jamonwindeyer on August 06, 2017, 11:07:23 pm
Hello,
Can we someone please explain the graphs in motors and generators. Such as EMF, flux, voltage output.
I'm confused on all the graphs of all these, please help!
Hey! So these graphs are representations of some pretty important concepts - Electromagnetic induction, back EMF, Lenz's Law, etc etc. To understand the graphs you need to understand those concepts, there's no set pattern unfortunately! I've written some nice succinct guides on Motors and Generators, they are linked in the middle of this resource list. Perhaps they would be helpful? ;D
If there are any specific examples of graph questions you are confused about, upload em! Happy to help :)
Post by: jamonwindeyer on August 06, 2017, 11:23:58 pm
Hello! I've been doing some questions of BOSTES and I came across these questions. Help would be greatly appreciated :)
Hey! For the first one, we know a current will flow, and since the relative motion of the disc and the magnetic field is always the same, it will be a direct current, so we can cut out two answers straight away! By the right hand slap rule, the current induced will move towards Brush Y (remember fingers in direction of field, thumb in direction of movement, slap in direction of current - So I'm fairly sure the answer C?
Second one, we know the induced emf will be a maximum when the coil is parallel to the field, so we need to start at a peak value. There is one full rotation, so one complete oscillation. This matches Graph B! :)
This is a really tough one to explain online, but it again relies on right hand grip rule. Remember first that the force on a charged particle due to movement in a magnetic field is perpendicular to both the field AND the movement. So if we want it to experience a force in the same direction as current, it can't be moving in that direction initially. So that rules out A and B.
So if current goes upwards (let's say), that means the field will be into or out of the page around it. To get a force in the same direction as the current, we need it to be moving towards it to get an upwards force. In honesty, I'm thinking of ways to explain this, but I really can't without being there next to you showing it to you in real time, aha. Try it - Try the right hand slap rule and see what happens when the electron is moving towards the wire - It should all line up to give you a force in the same direction as the current :)
Post by: yattmoani on August 06, 2017, 11:41:15 pm
Post by: 12carpim on August 07, 2017, 09:51:51 am
This is a multiple choice question from a SBHS trial and i need some help in identifing the answer to the question :)
Post by: pikachu975 on August 07, 2017, 10:13:19 am
Hey,
This is a multiple choice question from a SBHS trial and i need some help in identifing the answer to the question :)
I think D since generators don't have back emf only motors.
Post by: jamonwindeyer on August 07, 2017, 10:20:36 am
Hey,
This is a multiple choice question from a SBHS trial and i need some help in identifing the answer to the question :)
That's a really dodgy question. Back EMF in motors is just the name given to the induced emf due to the rotation of the coil in the field. In a generator, that induced emf is the output voltage we want in the first place. The use of the terminology isn't appropriate here in my opinion :P
Assuming it just means induced emf, the answer is B. Induced emf is proportional to rate of change of flux, and the second graph matches that - It is a maximum when flux is changing the most ;D
Post by: kiwiberry on August 07, 2017, 12:28:49 pm
Oh my bad I forgot to add the answers to each of the questions. The answer for the first question according to BOSTES is actually B?
The current generated with induce a magnetic field which opposes the original field, so using the right hand palm rule (fingers pointing left, palm towards direction of movement) current will flow downwards from Y. Therefore current flows from Y to X through the globe :)
Post by: winstondarmawan on August 07, 2017, 01:02:54 pm
That's a really dodgy question. Back EMF in motors is just the name given to the induced emf due to the rotation of the coil in the field. In a generator, that induced emf is the output voltage we want in the first place. The use of the terminology isn't appropriate here in my opinion :PShouldn't it be a negative cos graph because it is the negative of the derivative? Or am I missing something here...
Assuming it just means induced emf, the answer is B. Induced emf is proportional to rate of change of flux, and the second graph matches that - It is a maximum when flux is changing the most ;D
EDIT: Also this question from Quanta to Quarks:
Explain why the spectroscope was important in the development of the Bohr model of the atom. (4 marks)
Post by: Mymy409 on August 07, 2017, 01:23:53 pm
I don't understand how in AC induction motors, the rotor follows the stator.
Also, do AC motors have brushes and commutators or not, coz I'm seeing different things in different books/sites.
Post by: pikachu975 on August 07, 2017, 01:56:31 pm
Shouldn't it be a negative cos graph because it is the negative of the derivative? Or am I missing something here...
EDIT: Also this question from Quanta to Quarks:
Explain why the spectroscope was important in the development of the Bohr model of the atom. (4 marks)
A spectroscope is a device that allowed the observation of the hydrogen emission spectrum. When the hydrogen atoms were excited through heating or passing a current through low density gas, it released visible light of specific wavelengths that were unexplained by the Rutherford model.
This led Bohr to develop his postulates:
1) Electrons exist in allowable energy states which explained the spectral lines rather than the continuous spectra.
2) The transition of electrons is accompanied by the emission or absorption of EMR, which explained why there were visible light being emitted when the hydrogen gas was excited.
3) Angular momentum of electron orbits are quantised in multiples of h/2pi.
Therefore the spectroscope was quintessential in the development of the Bohr model as the model was formed based on the inability of the Rutherford model in explaining the hydrogen emission spectrum; which was observed by the spectroscope.
Post by: pikachu975 on August 07, 2017, 02:00:26 pm
Hey,
I don't understand how in AC induction motors, the rotor follows the stator.
Also, do AC motors have brushes and commutators or not, coz I'm seeing different things in different books/sites.
In AC induction motors, the rotor wants to balance the change in flux. The only way to do this is to try and catch up to it so they rotate at the same speed. This would mean they have no relative motion and hence no change in flux. This is because the stator causes the apparent rotating magnetic field so the change in flux must be minimised by the rotor rotating. However it cannot reach the exact speed so the difference in speed is the slip speed.
AC motors DO have brushes. They have SLIP ring commutators, 2 of them, which each connect to one terminal. The brushes maintain electrical contact between the external circuit and the coil. The slip ring commutators reverse the polarity of the output current every half turn.
Post by: kiwiberry on August 07, 2017, 02:18:11 pm
AC motors DO have brushes. They have SLIP ring commutators, 2 of them, which each connect to one terminal. The brushes maintain electrical contact between the external circuit and the coil. The slip ring commutators reverse the polarity of the output current every half turn.
The slip rings don't reverse the polarity of the input current because AC already constantly changes direction, however the split-ring commutator in DC motors reverses the direction of current every half-turn to ensure unidirectional torque
Post by: winstondarmawan on August 07, 2017, 02:20:29 pm
A spectroscope is a device that allowed the observation of the hydrogen emission spectrum. When the hydrogen atoms were excited through heating or passing a current through low density gas, it released visible light of specific wavelengths that were unexplained by the Rutherford model.
This led Bohr to develop his postulates:
1) Electrons exist in allowable energy states which explained the spectral lines rather than the continuous spectra.
2) The transition of electrons is accompanied by the emission or absorption of EMR, which explained why there were visible light being emitted when the hydrogen gas was excited.
3) Angular momentum of electron orbits are quantised in multiples of h/2pi.
Therefore the spectroscope was quintessential in the development of the Bohr model as the model was formed based on the inability of the Rutherford model in explaining the hydrogen emission spectrum; which was observed by the spectroscope.
Thank you! How about this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20668088_1293149414143912_1322689573_n.jpg?oh=b4f5e827f3f630af98148796735dad4a&oe=598AE96A Part b.
TIA
Post by: kiwiberry on August 07, 2017, 02:37:41 pm
Thank you! How about this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20668088_1293149414143912_1322689573_n.jpg?oh=b4f5e827f3f630af98148796735dad4a&oe=598AE96A Part b.
TIA
You would draw something like this:
(https://i.stack.imgur.com/6raba.jpg)
But replace the values on the left with the values given in the table and only draw up to n=4. The energy levels for n>4 will be above n=4 :)
Post by: mercurry on August 07, 2017, 02:48:40 pm
Could someone explain these to me?
7. C
8. B
9. D
Cheers
Post by: pikachu975 on August 07, 2017, 02:59:55 pm
The slip rings don't reverse the polarity of the input current because AC already constantly changes direction, however the split-ring commutator in DC motors reverses the direction of current every half-turn to ensure unidirectional torque
My bad I think I'm confused with AC generators. For AC generators the slip rings change the polarity of the output current or something?
Post by: jamonwindeyer on August 07, 2017, 03:07:10 pm
Hey,
Could someone explain these to me?
7. C
8. B
9. D
Cheers
Sure!
7C - We consider the conservation of energy. So, the electrical energy we put in must come out as either kinetic energy, or losses in the form of heat. So the sum of those things is a constant for that reason ;D the others are incorrect - A violates the conservation of energy, B is wrong because supply voltage is constant, and D is wrong because current will decrease, not increase :)
8B - A single change in magnetic flux will cause a single direction of current flow. We are introducing extra field into the page, we need field out of the page to counter it. The current needs to flow anti-clockwise for this by the right hand grip rule :)
9D - If you read the other three answers, they are advantages! So it has to be D by default ;D
Post by: jamonwindeyer on August 07, 2017, 03:08:05 pm
My bad I think I'm confused with AC generators. For AC generators the slip rings change the polarity of the output current or something?
Even simpler - Slip rings literally just maintain contact. They change absolutely nothing about the output ;D
Post by: johnk21 on August 07, 2017, 03:13:18 pm
Compare and contrast the methods of X-rays and endoscopy. (4 marks)
THanks in advance
Post by: winstondarmawan on August 07, 2017, 03:45:39 pm
Assess the effectiveness of the Bohr-Rutherford model of the atom in accounting for experimental observations. (6)
Rutherford's original model of the atom (Mainly empty space with a minute, dense, positive nucleus and small negatively charged electrons) was sufficient in accounting for the results of the gold foil experiment, as the minute nucleus was able to explain the complete deflection of 1/20000 alpha particles and the empty space was able to explain the uninterrupted path of the majority of alpha particles. However, it had several flaws which were not explainable, such as the fact that an accelerating charged particle would give off EM radiation and thus lose energy, spiralling into the nucleus, yet this was not the case. It also offered no explanation for the line spectrum of hydrogen. Thus, Bohr introduced his three postulates to supplement Rutherford's model and could explain one of the above, the line spectrum of hydrogen. This model was known as the Bohr-Rutherford model. However, the model still had some flaws:
Could not explain why electrons had a stable orbit
Could not explain various intensities and broadness of spectral lines
Could not explain hyperfine structures
These were all experimental observations that the Bohr-Rutherford model could not explain.
Judgement: As such it is effective in accounting for some experimental observations, such as gold foil experiments, however failed to account for some other experimental observations as stated above.
Post by: blasonduo on August 07, 2017, 03:45:50 pm
Can someone please help me with this medical physics question:
Compare and contrast the methods of X-rays and endoscopy. (4 marks)
THanks in advance
Hello! This is a pretty chill question!
Just find 2 contrasts and 2 comparisons.
For example:
1) x-rays are non-invasive, endoscopy is.
2) They both use EMR to find internal problems in the body.
3) Endoscopy requires the patient to "flush out" their bodily wastes while x-rays do not
4) Both a very cheap method.
There are heaps! but 4 should give you the marks :)
Post by: kiwiberry on August 07, 2017, 04:12:28 pm
Hey! Can someone please look over my response for this question, and make sure I have all the relevant stuff:
Assess the effectiveness of the Bohr-Rutherford model of the atom in accounting for experimental observations. (6)
Rutherford's original model of the atom (Mainly empty space with a minute, dense, positive nucleus and small negatively charged electrons) was sufficient in accounting for the results of the gold foil experiment, as the minute nucleus was able to explain the complete deflection of 1/20000 alpha particles and the empty space was able to explain the uninterrupted path of the majority of alpha particles. However, it had several flaws which were not explainable, such as the fact that an accelerating charged particle would give off EM radiation and thus lose energy, spiralling into the nucleus, yet this was not the case. It also offered no explanation for the line spectrum of hydrogen. Thus, Bohr introduced his three postulates to supplement Rutherford's model and could explain one of the above, the line spectrum of hydrogen. This model was known as the Bohr-Rutherford model. However, the model still had some flaws:
Could not explain why electrons had a stable orbit
Could not explain various intensities and broadness of spectral lines
Could not explain hyperfine structures
These were all experimental observations that the Bohr-Rutherford model could not explain.
Judgement: As such it is effective in accounting for some experimental observations, such as gold foil experiments, however failed to account for some other experimental observations as stated above.
I would also state Bohr's three postulates (electrons exist in discrete stationary states, electrons only absorb/emit EMR when transitioning between these states, angular momentum is quantised) and explicitly say that this meant that only certain wavelengths of EMR could be emitted, thus explaining the hydrogen spectrum. Otherwise looks good!! :)
Post by: austv99 on August 07, 2017, 04:19:05 pm
Could someone explain Planck's Contribution to Bohr's model and postulates?
Also clarify the significance of the hydrogen spectrum and how this significance differs from the significance of the Balmer Series.
Post by: Mymy409 on August 07, 2017, 04:59:31 pm
Post by: winstondarmawan on August 07, 2017, 05:06:42 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20706467_1293245284134325_1582034743_n.png?oh=a0683f2b4127bb6b5f1996fafcea0021&oe=598ACF30
TIA
Post by: Bubbly_bluey on August 07, 2017, 05:07:58 pm
Can someone help me with this? The answer is C. Thanks!Hi! I remember doing this question and initially choosing D. But i think its C because if you increase the reistance too high then that would generate greater power loss by P=I^2 R. Its just a thought :)
Post by: Bubbly_bluey on August 07, 2017, 05:11:20 pm
Would appreciate help with the following question, part b).Hi! if you recall the formula for orbital velocity you can that the "m" is actually the mass of the moon not the spacecraft. So although the spacecraft has changed its mass it does not affect the orbital velocity. :)
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20706467_1293245284134325_1582034743_n.png?oh=a0683f2b4127bb6b5f1996fafcea0021&oe=598ACF30
TIA
Post by: kiwiberry on August 07, 2017, 05:13:18 pm
Can someone help me with this? The answer is C. Thanks!
I think it's because current and resistance are inversely proportional, so if current is high then resistance has to be low. Since P=I2R, having a high current and low resistance will result in higher heat losses and therefore faster heating than having low current and high resistance, so the answer is C :)
Post by: jamonwindeyer on August 07, 2017, 05:36:36 pm
Can someone help me with this? The answer is C. Thanks!
I think it's because current and resistance are inversely proportional, so if current is high then resistance has to be low. Since P=I2R, having a high current and low resistance will result in higher heat losses and therefore faster heating than having low current and high resistance, so the answer is C :)
This question caused a lot of contention - But kiwiberry is on it. If you've got a fixed AC voltage source (say the 230V we are supplied from mains), then \(I=\frac{230}{R}\). We want \(R\) to be low so current can be high, because joule heating is mostly determined by how much current you've got flowing ;D
Post by: kiwiberry on August 07, 2017, 05:45:22 pm
Hi,
Could someone explain Planck's Contribution to Bohr's model and postulates?
Also clarify the significance of the hydrogen spectrum and how this significance differs from the significance of the Balmer Series.
Planck proposed that energy emitted or absorbed by a black body is quantised, given by E=hf - this was the beginning of the concept of quantised energy and quantum physics. Bohr used Planck's theory to develop his model of the atom, proposing that the electrons exist in quantised energy levels (his 1st postulate). Bohr's second postulate was also based on Planck's theory, stating that electron transitions between these energy levels would result in the absorption/emission of a quantised amount of energy in the form of EMR, given by E=hf. His third postulate states that angular momentum of electrons in these stationary states is quantised and given by L=nh/2pi, a result of quantum physics. Thus Planck's contribution to quantum physics was significant in the development of Bohr's model and postulates.
The Balmer series is the visible part of the hydrogen spectrum. I'm not sure about the difference in their significance, but the Balmer series was significant because it showed that the hydrogen spectrum was not a continuous rainbow but rather lines at specific wavelengths. This suggested that electrons only existed in certain energy levels and influenced Bohr's atomic model and postulates. :)
Post by: pikachu975 on August 07, 2017, 06:12:13 pm
Even simpler - Slip rings literally just maintain contact. They change absolutely nothing about the output ;D
I thought it changes polarity of the brushes every half turn for an AC generator
Post by: kiwiberry on August 07, 2017, 06:15:11 pm
I thought it changes polarity of the brushes every half turn for an AC generator
The output will already be AC so the polarity doesn't need to be changed. In DC generators the output has to be changed every half turn to get a DC output, that's what the split-ring commutator does :)
Post by: pikachu975 on August 07, 2017, 06:44:23 pm
The output will already be AC so the polarity doesn't need to be changed. In DC generators the output has to be changed every half turn to get a DC output, that's what the split-ring commutator does :)
But in an AC generator how is it already AC?
Post by: winstondarmawan on August 07, 2017, 07:02:25 pm
But in an AC generator how is it already AC?
Imagine one side of the coil is "positive" and the other is "negative". As flux lines cut through that side of the coil, an EMF is produced. However, when flux lines cut through the other side of the coil, which will happen as it rotates, then EMF will be produced in the opposite direction. Thus, a generator with slip rings produces AC. Split rings correct this.
Post by: jamonwindeyer on August 07, 2017, 07:53:58 pm
But in an AC generator how is it already AC?
If you think about it, everything about an AC and DC generator is identical EXCEPT the slip rings vs split ring. If the split ring reverses polarity, and that gives DC, then the slip rings wouldn't want to do the same thing right? You'd still be getting DC ;D
Post by: winstondarmawan on August 07, 2017, 09:06:30 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20684659_1293396164119237_362264648_o.png?oh=8c6f6d113a92a264238faf7b26106b94&oe=598A9646
TIA!
Post by: blasonduo on August 07, 2017, 09:20:00 pm
Would appreciate help with the following quesiton:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20684659_1293396164119237_362264648_o.png?oh=8c6f6d113a92a264238faf7b26106b94&oe=598A9646
TIA!
This is all about Faradays Law and Lenz's law. When there is a change in magnetic flux, a current can be induced, but to stay in the laws of conservation of energy, it will flow in the direction to oppose that change.
When the power is initially turned on, current will flow, and in the plate, will flow in the direction to oppose that change (so clockwise) but then will stop. (I'm Assuming its DC current here) As there is no more change in flux. As the resistance becomes decreased, more current will flow and thus the plate will induce a current in the same direction again, but continuously this time.
Hope this helps :)
Final stretch before the exam :)
Post by: winstondarmawan on August 08, 2017, 06:34:16 am
Environmental impacts of transistors. (5)
Impact of transistors on communications. (6)
Struggling to find enough content for 5 and 6 marks. TIA!
EDIT: Do the sample answers on the BOSTES solutions get full marks?
Post by: blasonduo on August 08, 2017, 07:25:39 am
How would you answer a question that asks for the:
Environmental impacts of transistors. (5)
Impact of transistors on communications. (6)
Struggling to find enough content for 5 and 6 marks. TIA!
EDIT: Do the sample answers on the BOSTES solutions get full marks?
These Assess questions for transistor are almost identical for assess questions for generators
1) Communication has been improved, (eg mobile phones)
2) Improved imaging techniques for medical
3) However, an over reliance occurs, which leads to diseases such as obesity
4) A loss of jobs due to automation, but also allowed new unique jobs as well
5) Silicon has to be extracted from earth's crust
6) transistors are in computers, which then can overall, monitor the environment and any changes that occur
The list is endless :)
Also. the sample answers CAN be, but most of the time, no. They are not band 6 answers :)
Post by: left right gn on August 08, 2017, 05:01:40 pm
Post by: kiwiberry on August 08, 2017, 05:10:53 pm
For projectile motion questions, when subbing in values, does 'a' always have to be -9.8
Yes! \(a_y \) will always be -9.8 because the only force acting on a particle in projectile motion is gravity :)
Post by: Shadowxo on August 08, 2017, 05:46:47 pm
For projectile motion questions, when subbing in values, does 'a' always have to be -9.8What kiwiberry said is right but it also depends what direction you're taking to be positive. The (vertical) acceleration will always be -9.8 if you're taking "up" as being positive, but you could also say "down" is positive in which case a would be +9.8. Normally you'd say acceleration is -9.8 and take up as positive, just thought I'd mention it as some solutions may use down as positive :)
Post by: johnk21 on August 08, 2017, 05:55:11 pm
For projectile motion questions, when subbing in values, does 'a' always have to be -9.8Wanna also add that SOME but not ALL projectile motions are not on earth. BUT if it is on another planet, they will provide the acceleration due to gravity for that planet, and so a will equal what they gave you. But always remember to put in a NEGATIVE sign, assuming that you are setting up as positive.
Post by: winstondarmawan on August 08, 2017, 07:08:13 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20706627_1294328377359349_1149916873_n.png?oh=f2b6d4822245ea51975179b4bbea719d&oe=598C13B5
TIA
Post by: austv99 on August 08, 2017, 07:46:26 pm
Could someone help with these questions? The formula is missing for part f but need help for part g and h as well
Post by: sidzeman on August 08, 2017, 08:50:56 pm
Post by: blasonduo on August 08, 2017, 09:35:07 pm
pls halp
Hello! The answer is D
Looking at A, if one had 3 times the amount of coils, one graph should be 3 times the amplitude of the other, which is not the case
For B, It is similar to A where the strength of the magnetic field of twice should be twice the size
For C, It makes no sense how a ferrite core would flip the graphs
So it's D, the rotation results in a faster frequency with a higher amplitude
Hope this helps :)
Post by: blasonduo on August 08, 2017, 09:46:30 pm
Hey, would appreciate help with part d). No idea how to do angular velocity
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20706627_1294328377359349_1149916873_n.png?oh=f2b6d4822245ea51975179b4bbea719d&oe=598C13B5
TIA
I can't help you as much as there are no numbers for me to calculate! HOWEVER, this is the first time I've seen this certain question in HSC physics and I doubt this will be assessed,
These are the formulas I found regarding angular velocity:
Post by: jamonwindeyer on August 08, 2017, 11:16:43 pm
Hey, would appreciate help with part d). No idea how to do angular velocity
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20706627_1294328377359349_1149916873_n.png?oh=f2b6d4822245ea51975179b4bbea719d&oe=598C13B5
TIA
I can't help you as much as there are no numbers for me to calculate! HOWEVER, this is the first time I've seen this certain question in HSC physics and I doubt this will be assessed,
These are the formulas I found regarding angular velocity:
Just confirming angular velocity is definitely not HSC assessable :)
Post by: jamonwindeyer on August 08, 2017, 11:20:27 pm
Hi!
Could someone help with these questions? The formula is missing for part f but need help for part g and h as well
Hey! My best guess for this would be that you'd take the relative velocity from Part F, and use this velocity as \(v\) in the regular formulas we use for these sorts of questions in the subsequent part! For Part (g), recalculate the relative velocity then use that in the length contraction formula, I believe! :) feel free to post a solution if you want us to check it out? Just can't do much without the formula is all :)
Post by: sidzeman on August 09, 2017, 03:13:32 pm
Post by: sidzeman on August 09, 2017, 03:16:06 pm
I'm not quite sure how to apply the RHP/RHG rule to this
Post by: jamonwindeyer on August 09, 2017, 03:20:47 pm
I tried subbing the values into .5mv^2 but nothing worked out
Hey! You'll need to calculate the relativistic mass of the proton using the mass dilation formula:
You know the mass of the proton and its speed, which gives you \(m_v\). This is the mass you use to calculate the kinetic energy, and then you compare that with the kinetic energies of the other objects. You could calculate their relativistic masses as well, but the effects of their velocity are negligible ;D
Completely lost as to this question - thought the answer was C but apparently it's D - why?
A photon doesn't have mass so that formula wouldn't work! D makes sense, we can find the energy of proton using its rest mass (which is a known quantity) ;D
Reminder: Try to use the 'Modify' button to add questions to your previous post rather than double posting twice in a row! :)
Post by: naezeroth on August 09, 2017, 03:52:21 pm
Photons are massless particles and thus the formula E = mc^2 does not really apply to it. Photon energy is described by Planck's relationship E = hf where h is Planck's constant. Thus by logical deduction as it is not A/B/C it must be D
Post by: sidzeman on August 09, 2017, 05:12:45 pm
Post by: winstondarmawan on August 10, 2017, 02:41:55 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20747559_1529460377113486_1121039749_o.png?oh=2c0146af143515c0630bd04c5205e36f&oe=598E016E
Post by: blasonduo on August 10, 2017, 03:22:38 pm
Hey! Would appreciate help with this question. The answer is A, apparently.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20747559_1529460377113486_1121039749_o.png?oh=2c0146af143515c0630bd04c5205e36f&oe=598E016E
Hello! This is testing your knowledge on both Lenz's law and electromagnets.
When a south pole is moved towards a solenoid, to obey Lenz's law and the law of conservation of energy, it must oppose this change, and thus a south pole will be created.
Now we need to know in which direction the current will have to flow to make the left side of the solenoid a South pole. If the current is going CLOCKWISE, then a south pole will be produced. So from this, follow where the current flows where it must be going CLOCKWISE and you'll see it'll go through X first, to Y. So it HAS to be A OR B.
For point P, we must know how magnetic field lines must flow in an electromagnet, which is from SOUTH to NORTH. With this the force is going left to right, and thus point P will experience a force to the right so it is A!
Hope this helps!
Post by: aaronster on August 12, 2017, 12:40:07 pm
Post by: Shadowxo on August 12, 2017, 12:47:36 pm
Could you please explain the concept of ratios in Space questions and the example below as well Pls
For that question:
You should remember
This means the gravitational force will be the same on both objects - Newton's third law also says "Each action has an equal and opposite reaction" which confirms this. The gravitational force on object X from Y = gravitational force on object Y from X. But, the acceleration on each will be different due to their different masses.
Note: This applies to us and the earth too. The gravitational force on us from the earth is the same as the gravitational force the earth experiences from us! But, the earth's mass is far far greater than ours so it hardly moves while we feel it quite a bit.
Ratios in space questions: could you elaborate please?
If something is in orbit, Then the gravitational force is the centripetal force so Fg = Fc
Post by: jamonwindeyer on August 12, 2017, 01:17:59 pm
Could you please explain the concept of ratios in Space questions and the example below as well Pls
Hey! So as said above, the example you've provided is a trick. The force on each object is the same, the ratio is 1:1! :)
When you are asked to find a ratio of Quantity 1 to Quantity 2, that means dividing them:
So if for example looking at forces, \(Q_1=F\) but \(Q_2=2F\), then the ratio \(Q_1:Q_2\) would be 2:1 ;D
Often the expressions are more complex, not just \(F\), but that's the idea!
Post by: aaronster on August 12, 2017, 03:51:23 pm
Post by: itssona on August 13, 2017, 12:49:44 pm
light froma torch 200km away is 20 units. How bright is it 120 km away?
Post by: AlphaGeek on August 13, 2017, 01:47:48 pm
Post by: blasonduo on August 13, 2017, 02:58:53 pm
heeey how do i do this using inverse square law (thank you :))
light froma torch 200km away is 20 units. How bright is it 120 km away?
Post by: jamonwindeyer on August 13, 2017, 04:08:25 pm
Could anyone please help me by explaining the differences in graphs of Torque, Emf, Current, Voltage for AC & DC Motors/Generators. I keep seeing so many variations online when I search for them. Thank you.
Hey! So these graphs are representations of some pretty important concepts - Electromagnetic induction, back EMF, Lenz's Law, etc etc. To understand the graphs you need to understand those concepts, there's no set pattern unfortunately! I've written some nice succinct guides on Motors and Generators, they are linked in the middle of this resource list. Perhaps they would be helpful? ;D
If there are any specific examples of graph questions you are confused about, upload em! Happy to help :)
Post by: Sukakadonkadonk on August 13, 2017, 05:35:30 pm
Thanks.
Post by: jamonwindeyer on August 13, 2017, 06:06:34 pm
Hi, how would this be done?
Thanks.
Hey! So we use the centripetal force formula:
We have mass and we have radius (half the diameter, 0.25 metres) - We just need velocity! We know it completes 10 revolutions in 65.2 seconds, so we use that to calculate velocity:
Once you've got that, sub into your original formula for the answer!! Doing this I get:
Hmm. That's odd - Anyone see what I've done wrong here? Maybe we need to compensate for gravity with an upwards force as well to keep the circle horizontal but that seems a bit odd for how they've worded it :P
Surely it's not A, technically it is the closest but come on...
Post by: Sukakadonkadonk on August 13, 2017, 06:19:53 pm
Hey! So we use the centripetal force formula:
We have mass and we have radius (half the diameter, 0.25 metres) - We just need velocity! We know it completes 10 revolutions in 65.2 seconds, so we use that to calculate velocity:
Once you've got that, sub into your original formula for the answer!! Doing this I get:
Hmm. That's odd - Anyone see what I've done wrong here? Maybe we need to compensate for gravity with an upwards force as well to keep the circle horizontal but that seems a bit odd for how they've worded it :P
Surely it's not A, technically it is the closest but come on...
Thanks for helping but the answer they got was B, 8.5 newtons and yeah... I still don't get it haha.
Post by: blasonduo on August 13, 2017, 06:28:52 pm
Hi, how would this be done?
Thanks.
I think I have it, but I am about to leave home right now, i'll be back in 2 hours, and edit this if no one else has it ;) Is iit B? because my working said B
Sorry for the delay,
Post by: jamonwindeyer on August 13, 2017, 06:39:31 pm
I think I have it, but I am about to leave home right now, i'll be back in 2 hours, and edit this if no one else has it ;) Is iit B? because my working said B
Sorry for the delay,
It is indeed B and you are my saviour and hero ;D
Post by: pikachu975 on August 13, 2017, 06:40:23 pm
Hey! So we use the centripetal force formula:
We have mass and we have radius (half the diameter, 0.25 metres) - We just need velocity! We know it completes 10 revolutions in 65.2 seconds, so we use that to calculate velocity:
Once you've got that, sub into your original formula for the answer!! Doing this I get:
Hmm. That's odd - Anyone see what I've done wrong here? Maybe we need to compensate for gravity with an upwards force as well to keep the circle horizontal but that seems a bit odd for how they've worded it :P
Surely it's not A, technically it is the closest but come on...
I tried as well and got your answer. I tried using the 4U method of w = 2pi/T and then F = mrw^2 and also using your method and I got the same as you so I'm not sure what's wrong
Post by: blasonduo on August 13, 2017, 10:36:10 pm
Hi, how would this be done?
Thanks.
So, My way gave me 86.14 N (This was applying absolutely every force, and I think even here I could've done something VERY wrong)
HOWEVER
From just fiddling with potential errors with the answers I found that equating speed as 6.52 (YES this is 100% wrong) did give you the answer.
This gives the EXACT answer for B. From the answer of D having the exact decimals, I assume the answers are incorrect.
Post by: Sukakadonkadonk on August 13, 2017, 10:49:57 pm
So, My way gave me 86.14 N (This was applying absolutely every force, and I think even here I could've done something VERY wrong)
HOWEVER
From just fiddling with potential errors with the answers I found that equating speed as 6.52 (YES this is 100% wrong) did give you the answer.
This gives the EXACT answer for B. From the answer of D having the exact decimals, I assume the answers are incorrect.
Ohh, thank you very much.
Do you think the question is wrong then? Or is Jamon's method viable in solving this with the given info.
Post by: blasonduo on August 13, 2017, 10:54:19 pm
Ohh, thank you very much.
Do you think the question is wrong then? Or is Jamon's method viable in solving this with the given info.
Yes, it must be wrong. Follow Jamon's work ;)
I will sleep on this question and will post if anything changes (doubt it will) :P
Post by: Charlie_Sparkes on August 15, 2017, 05:30:49 pm
Hoping I can just get an explanation of how to tackle projectile motion questions... I have Physics trials tomorrow and I'm having some difficulty with separating Horizontal and vertical components and which formulas to use etc.
Thanks again, Charlie.
Post by: Shadowxo on August 15, 2017, 06:50:19 pm
Hey EveryoneThe motion questions should be on a formula sheet (pretty standard but I didn't do HSC so can't be sure). Note this is from a VCE perspective
Hoping I can just get an explanation of how to tackle projectile motion questions... I have Physics trials tomorrow and I'm having some difficulty with separating Horizontal and vertical components and which formulas to use etc.
Thanks again, Charlie.
You usually have to tackle the horizontal and vertical components separately.
If you're given the velocity at an angle, you have to figure out the magnitude of the velocity in each direction. I remember, if it's the adjacent side it's hypotenuse * cosx if it's opposite then hypotenuse *sine(x) (SOH CAH TOA, and sinx=O/H so O=Hsinx). Basically just resolve into each direction whichever way you remember
Vertically, gravity is the force acting on the object, the acceleration of which is g (9.8 ). Remember to take into account directions, if you're taking up as positive gravity will be negative and vice versa. This is typically the only force acting on it so g is the only acceleration you have to worry about.
So vertically, you know a, and you should know 2 other variables (typically initial velocity and distance - remember to take into account directions). From this you can find the vertical component of its velocity as it hits the ground, or the time it's in the air (common), or whatever other variable they want you to calculate. The vertical component is what determines the time in the air, as the object will continue travelling horizontally until it hits the ground.
Horizontally, there's no force acting on it (usually). This means its speed won't change! So you can calculate the horizontal distance it travelled (s=vt), and the velocity will stay constant. This is usually the only horizontal thing you'll have to do.
If they want you to find the velocity it hits the ground at, you can use the horizontal velocity and vertical velocity and use pythag to find the overall velocity (V2=Vhorizontal2+Vvertical2)
Also, there's some symmetry in projectile motion. Keep in mind if you throw a ball up, it'll fall back in your hand with the negative of that velocity, so if an object is thrown at an angle at a certain velocity, it will land at the negative of that velocity (provided it doesn't travel any distance vertically)
Hope this helps and feel free to ask for anything more :)
Post by: jamonwindeyer on August 15, 2017, 09:04:22 pm
Hey Everyone
Hoping I can just get an explanation of how to tackle projectile motion questions... I have Physics trials tomorrow and I'm having some difficulty with separating Horizontal and vertical components and which formulas to use etc.
Thanks again, Charlie.
Awesome response from Shadow above, I also wrote this guide which includes a big section on projectile motion. It might be helpful! ;D
Good luck for tomorrow!! :)
Post by: Mymy409 on August 18, 2017, 10:52:05 am
What is currently used to define the standard metre?
(A) The speed of light
(B) The signals from GPS satellites
(C) The wavelength of light from a krypton lamp
(D) The distance between two lines on a platinum iridium bar
Post by: Shadowxo on August 18, 2017, 11:34:35 am
This question has totally stumped me.I'm not sure whether you're expected to know this (we didn't for VCE Physics) but if not you can just look it up. According to Wikipedia
What is currently used to define the standard metre?
(A) The speed of light
(B) The signals from GPS satellites
(C) The wavelength of light from a krypton lamp
(D) The distance between two lines on a platinum iridium bar
"The metre is defined as the length of the path travelled by light in a vacuum in
1/299 792 458 seconds."
So, a)
But using the process of elimination you should be able to see that b) and d) are unlikely, as they're not really well defined or constant. Wavelengths are often not precise (usually we get a range for different kinds of light. Even a laser has a narrow range for example) so c) is unlikely. a) we know that c is a constant and well defined and so that can be used to define a meter :)
Post by: jamonwindeyer on August 18, 2017, 02:23:34 pm
This question has totally stumped me.
What is currently used to define the standard metre?
(A) The speed of light
(B) The signals from GPS satellites
(C) The wavelength of light from a krypton lamp
(D) The distance between two lines on a platinum iridium bar
Ditto above, though you don't necessarily need to know the exact value - This is one of those 'rote learn' dotpoints ;D
Post by: Bubbly_bluey on August 19, 2017, 05:27:39 pm
Thanks ;D
Post by: kiwiberry on August 19, 2017, 05:56:09 pm
Hey guys! I got B using the right hand palm rule but the answer is A. Does anyone know why??
Thanks ;D
By Lenz's law, the current generated will induce a magnetic field which opposes the external field, so you have to point your fingers in the opposite direction of the external field when using the right hand palm rule. On the top of the coil, palm will face left, fingers pointing up and current will flow out of the page - so it will flow from X to Y in the conductor :)
Post by: Maraos on August 19, 2017, 09:53:08 pm
I was just wondering what would be the correct extended response answer to the following dot point in the option topic Quanta to Quarks?
Dot point 2.4: Explain the stability of the electron orbits in the Bohr atom using de Broglie's hypothesis.
What would be the main issues i need to raise?
Thanks!
Post by: kiwiberry on August 19, 2017, 10:57:00 pm
Hi! :)
I was just wondering what would be the correct extended response answer to the following dot point in the option topic Quanta to Quarks?
Dot point 2.4: Explain the stability of the electron orbits in the Bohr atom using de Broglie's hypothesis.
What would be the main issues i need to raise?
Thanks!
Hello! :))
So, de Broglie proposed that all moving particles have wave properties, with wavelength given by \(\lambda=\frac{h}{mv}\). He suggested that electrons must exist in orbits made of an integer number of wavelengths of the moving electrons (\(2\pi r = n\lambda \)) so that they would form standing waves - these don't propagate and are therefore stable and will not lose energy. If electron orbits contained a non-integer number of wavelengths, the electrons would destructively interfere with each other and lose energy. Hence de Broglie's hypothesis was able to provide a quantum explanation of the stability of electron orbits in Bohr's atomic model.
Hope this helped!
Post by: Maraos on August 19, 2017, 11:46:03 pm
Hello! :))Thanks so much for the info! :D
So, de Broglie proposed that all moving particles have wave properties, with wavelength given by \(\lambda=\frac{h}{mv}\). He suggested that electrons must exist in orbits made of an integer number of wavelengths of the moving electrons (\(2\pi r = n\lambda \)) so that they would form standing waves - these don't propagate and are therefore stable and will not lose energy. If electron orbits contained a non-integer number of wavelengths, the electrons would destructively interfere with each other and lose energy. Hence de Broglie's hypothesis was able to provide a quantum explanation of the stability of electron orbits in Bohr's atomic model.
Hope this helped!
Post by: Mymy409 on August 20, 2017, 10:25:46 am
Post by: Shadowxo on August 20, 2017, 10:38:26 am
In projectile motion, if we find, say, the initial horizontal velocity in the first part of the question and round it up, do we use that value to find other values? Or do we use the unrounded value?Use the unrounded value when you can
Post by: jamonwindeyer on August 20, 2017, 11:02:24 am
In projectile motion, if we find, say, the initial horizontal velocity in the first part of the question and round it up, do we use that value to find other values? Or do we use the unrounded value?
You'd be marked correctly for both (I'd wager), but if it is easy to use the unrounded value, use it. Otherwise, take a few extra decimal places. If they ask to find a rounded value in Part (a), you can use the rounded value in Parts (b), (c) etc (for example) ;D
Post by: Maraos on August 21, 2017, 10:57:14 pm
Apparently the answer is A but I don't see how.
Isn't A moving in a clockwise direction??, isn't the left side a south pole and the right side a north pole, according to the right hand rule?
Any help would be great!
Thanks :)
Post by: jamonwindeyer on August 21, 2017, 11:01:34 pm
This question is driving me insane. ;D
Apparently the answer is A but I don't see how.
Isn't A moving in a clockwise direction??, isn't the left side a south pole and the right side a north pole, according to the right hand rule?
Any help would be great!
Thanks :)
You are right about the polarity of the magnetic field, but watch the current, it goes in the opposite side - The right hand side of the coil has current going upwards (into the page?), use the right hand slap rule and you will get anti-clockwise! :)
Would you have picked another option? Will do my best to explain over text, aha ;D
Post by: Maraos on August 21, 2017, 11:27:39 pm
You are right about the polarity of the magnetic field, but watch the current, it goes in the opposite side - The right hand side of the coil has current going upwards (into the page?), use the right hand slap rule and you will get anti-clockwise! :)Ohhhhhh....
Would you have picked another option? Will do my best to explain over text, aha ;D
I was using my right hand palm rule wrong...... I didn't point my fingers in the direction of the field, I was pointing towards the North pole........
I probably shouldn't be studying for trials this late at night hahaha ;D
Thanks for the help!
Post by: arunasva on August 22, 2017, 02:41:01 pm
t = fd = mgl to get to my answer. Thanks :)
Post by: kiwiberry on August 22, 2017, 03:31:34 pm
Hey can someone please tell me how to find the length in part b of this question so that i can do
t = fd = mgl to get to my answer. Thanks :)
I think it's 0.1m (the radius of the pulley) because the force is applied at the circumference, not 100% sure though!
Post by: jamonwindeyer on August 22, 2017, 04:32:45 pm
I think it's 0.1m (the radius of the pulley) because the force is applied at the circumference, not 100% sure though!
Vouching ;D
Post by: arunasva on August 23, 2017, 12:43:41 am
I think it's 0.1m (the radius of the pulley) because the force is applied at the circumference, not 100% sure though!
you're a ledge(end) im hangin on to you
Post by: 12carpim on August 23, 2017, 06:31:29 pm
I'm stuck on question B from the physics textbook. I continue to try and find where I am wrong because my answer is never the one they have in the book, if someone can have a crack at it and validate my answer.
Thanks :) :) :)
Post by: jamonwindeyer on August 23, 2017, 10:04:24 pm
Hey,
I'm stuck on question B from the physics textbook. I continue to try and find where I am wrong because my answer is never the one they have in the book, if someone can have a crack at it and validate my answer.
Thanks :) :) :)
Does this match your answers? :)
Post by: 12carpim on August 24, 2017, 02:51:54 pm
Post by: jamonwindeyer on August 24, 2017, 08:24:53 pm
Post by: Adammurad on August 24, 2017, 10:40:27 pm
Post by: bsdfjnlkasn on August 27, 2017, 10:07:52 am
I was just wondering what the best way to prepare for the HSC Physics exam was? Which schools have the 'best' papers and until which year of the HSC should we go back to?
Should we be doing anything extra? Apart from learning and possibly rewriting summaries?
Any advice would be super appreciated!!
Post by: jamonwindeyer on August 27, 2017, 11:13:20 am
Hey there!
I was just wondering what the best way to prepare for the HSC Physics exam was? Which schools have the 'best' papers and until which year of the HSC should we go back to?
Should we be doing anything extra? Apart from learning and possibly rewriting summaries?
Any advice would be super appreciated!!
There isn't a 'perfect' method to preparing for any HSC Exam, you really just need to do what works best for you - Hopefully your Trials revealed a bit about what works well for you, and what doesn't work quite as well. Build on the strengths and get rid of the weaknesses because only you know how you best learn :)
That said, obviously practice makes perfect, and you've got about 7 Physics papers from the current syllabus now (2010 and beyond). Try and do as many of those as you can (and really, that might be all the practice exams you need, add a few from THSC if you want more). Supplement that with anything you like - Brainstorms around the big syllabus dot points, teaching your friends the content and having them teach you, anything you think will be helpful for you ;D
Post by: katnisschung on August 27, 2017, 02:57:16 pm
why did annealing of the nickel lead to diffraction? in other words
why didn't diffraction of the electrons occur in their results before the nickel annealed?
i assume it has something due to the nickel forming larger, single atoms??
thanks :)
Post by: bsdfjnlkasn on August 27, 2017, 04:58:44 pm
Could I please get some help with the following questions? Can't seem to find the answer online anywhere.. thank you in advance!
(http://uploads.tapatalk-cdn.com/20170827/c51a113594b8f919754bfbd714ede0b8.jpg)
Post by: A TART on August 27, 2017, 05:16:30 pm
Hey there!
Could I please get some help with the following questions? Can't seem to find the answer online anywhere.. thank you in advance!
(http://uploads.tapatalk-cdn.com/20170827/c51a113594b8f919754bfbd714ede0b8.jpg)
Types of transmuations:
Alpha Decay: 2 Neutrons and 2 Protons (Nucleus of an Helium atom, often denoted by He)
Beta Decay: 1 electron (or if Beta positve, a Positron)
Gama Deacay: A photon (light) is emitted from an unstable element (often denoted by *)
Neutron Capture: A neutron is added
(I'm not too sure about the HSC course, but I think you need to remember them)
So for the first one, a neutron is added to the reactant, hence it's neutron capture
For the second one, one reactant has resulted in 2 products, one of which is [He] (2 protons and 2 neutrons making a mass number of 4 and atomic number of 2). Hence it is alpha decay.
For the last one it's also neutron capture since a neutron has been added. It might be something else but I'm not sure. (Part of fission? Idk)
b) Since you don't find indivdual neutrons in nature it can be assumed that it's not the first or last one, hence, through elmination, it can be said the second one is a natural process. (which is true since Uranium is unstable and will decay)
c) I would say the last transmutation since it's a part of fission (when a heavier element is split into smaller ones and releases energy). This is how Nuclear power plants work.
Post by: A TART on August 27, 2017, 05:50:41 pm
for the davisson and germer experiment (quanta to quarks focus area 2)
why did annealing of the nickel lead to diffraction? in other words
why didn't diffraction of the electrons occur in their results before the nickel annealed?
i assume it has something due to the nickel forming larger, single atoms??
thanks :)
This is coming from a chemical POV. The size of an atom doesn't change when you treat it, however, the size of a crystal (a bunch of atoms) does. When you anneal a metal, the crystals become larger and often, more uniform. (due to the slow cooling process where atoms have the time to join to form larger crystals as opposed to quenching-where the metal is rapidly cooled and forms small crystals).
A more uniforum arrangement of atoms will leave less gaps for electrons to go straight through, hence, lead to more defraction.
The nickel prior to heat treatment might've been (polyatomic/amorphous, just messy in general). When it was treated, it resulted in a crystalline strucutre: (see pic)
https://upload.wikimedia.org/wikipedia/commons/thumb/2/2c/Crystalline_polycrystalline_amorphous.svg/150px-Crystalline_polycrystalline_amorphous.svg.png
I don't think you'll need the chem vocab since this is physics afterall (which is superior to chem).
Post by: jakesilove on August 28, 2017, 10:50:55 am
I don't think you'll need the chem vocab since this is physics afterall (which is superior to chem).
So true
Post by: 12carpim on August 29, 2017, 12:26:19 pm
I'm not sure if it is necessary to understand but why do some protons not align with a magnetic field in the MRI topic of medical physics.
Post by: austv99 on August 29, 2017, 02:58:36 pm
Could i get help this question? I dont understand why it's B.
Post by: pikachu975 on August 29, 2017, 05:07:42 pm
Hey,
Could i get help this question? I dont understand why it's B.
Emf is the negative of the rate of change of flux. This means if you imagine the flux graph, the stationary points on the flux graph will be the x-intercepts of the emf graph (basic 2 unit maths). The maximum flux passing through the coil is at Q and S so there's x intercepts here so it's B or C.
There is maximum emf when the torque is maximum, i.e. coil parallel to magnetic field (theta = 0 in T = nBILcostheta) which is at point P etc, so it must be B then since at P the emf is maximum.
Post by: jamonwindeyer on August 29, 2017, 05:51:07 pm
Hey,
I'm not sure if it is necessary to understand but why do some protons not align with a magnetic field in the MRI topic of medical physics.
Pretty sure that's beyond the syllabus - And I'd probably butcher the Physics if I tried to explain it ;) it's fairly complicated as far as I know! :P
Post by: itssona on August 29, 2017, 07:06:25 pm
a 60pkg drag car completes the 400m with a final speed of 200kmhr. whats the net force on the car assuming the acceleration of the car is constant
my teacher used the fact that v+u/2 =average velocity
Post by: Shadowxo on August 29, 2017, 07:25:04 pm
hiii pls help :/Hi :)
a 60pkg drag car completes the 400m with a final speed of 200kmhr. whats the net force on the car assuming the acceleration of the car is constant
my teacher used the fact that v+u/2 =average velocity
So, as the acceleration is constant, the average velocity would in fact be (u+v)/2, but I'd recommend using a different formula
You know u =0, v=200/3.6 m/s , s=400m and you want to know a.
Use v2=u2+2as to find a, then F=ma to find the force
Post by: itssona on August 29, 2017, 07:48:56 pm
Hi :)ohh!! thank you sooo much!! :D
So, as the acceleration is constant, the average velocity would in fact be (u+v)/2, but I'd recommend using a different formula
You know u =0, v=200/3.6 m/s , s=400m and you want to know a.
Use v2=u2+2as to find a, then F=ma to find the force
Post by: julies on August 30, 2017, 01:27:04 pm
Would someone be able to explain how forward biased p-n junctions work?
I don't understand how the depletion zone decreases when the positive terminal is connected to the p type and the negative terminal is connected to the n type...
Post by: austv99 on August 31, 2017, 08:59:10 am
"describe some medical and industrial applications of radio-isotopes
identify data sources, and gather, process, and analyse information to describe the use of: – a named isotope in medicine – a named isotope in agriculture – a named isotope in engineering"
Is Cobalt-60 for Agriculture or Engineering?
I'm confused on how Heisenberg and Pauli contributed to atomic theory
Can anyone clarify this?
Mod edit: Merged posts
Post by: kiwiberry on August 31, 2017, 09:38:36 am
What are the ideal radioisotopes for this dotpoint (Quanta to Quark)?
"describe some medical and industrial applications of radio-isotopes
identify data sources, and gather, process, and analyse information to describe the use of: – a named isotope in medicine – a named isotope in agriculture – a named isotope in engineering"
Is Cobalt-60 for Agriculture or Engineering?
I'm confused on how Heisenberg and Pauli contributed to atomic theory
Can anyone clarify this?
Mod edit: Merged posts
Hey! Co-60 is used in agriculture for food irradiation. For medicine, Tc-99m (used in diagnostic imaging) is probably the best one to use. For engineering, you could use Na-24 (used to detect leaks) or Ir-192 (used to detect structural faults).
Heisenberg developed matrix mechanics, which was a completely mathematical model of the atom - this was able to provide an explanation of the limitations of Bohr's model (different intensities of spectral lines, hyperfine lines, Zeeman effect, emission spectra of atoms larger than H) and more. He also proposed the uncertainty principle, which states that there is a fundamental limit to how accurately an electron's position and momentum can be determined:
Pauli proposed the exclusion principle, which states that no two electrons can have the same 4 quantum numbers (principle, orbital, magnetic, spin) - this explained the electron distribution in atoms and the position of elements in the periodic table. He also proposed the existence of the neutrino to explain the differing energies of beta decay.
Hope this helps! :)
Post by: jamonwindeyer on August 31, 2017, 11:19:50 am
hey there =)
Would someone be able to explain how forward biased p-n junctions work?
I don't understand how the depletion zone decreases when the positive terminal is connected to the p type and the negative terminal is connected to the n type...
Hey! This is right on the borderline of what you are expected to know, but roughly:
- When you connect a voltage source such that the positive terminal is connected to the P-type, and the negative terminal is connected to the N type, this pushes the positive holes in the P-type towards the junction, and the negative electrons in the N-type towards the junction. This is a forward bias connection. This reduces the size of the depletion region. Another way to think of it is that the applied voltage sort of 'cancels' the potential difference caused by the depletion region in the first place. This allows current to flow!
- If we connect it the other way around, the opposite occurs! We don't take away from the voltage, we make it worse, so to speak! This increases the size of the region and thus makes it very difficult for current to flow. Practically, eventually, you would get current - It happens when the diode goes "nope" and electrons start being ripped away from their atoms. But this is reverse bias, essentially a no current scenario!
Post by: blasonduo on August 31, 2017, 03:32:15 pm
An AC motor, when COMPARED to a DC motor:
a) will spin at a rate determined by the electricity supply frequency.
b) will spin at a rate determined by the electricity supply voltage.
c) will produce more torque than a similar DC motor.
d) will spin faster than DC motor.
thanks :)
Post by: winstondarmawan on September 01, 2017, 12:27:32 am
7. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21267582_1313522785439908_359337286_o.png?oh=e3810f91e5c593c5ebdcbd6cb50703e5&oe=59AA295D
15. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21245454_1313523275439859_1895111845_n.png?oh=cfd8d8d15289dc9308a4256bd8c2a820&oe=59AA21CC
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21268713_1313673448758175_1133813723_n.png?oh=0122454ffa0cd5e02612106c8e5c44f6&oe=59AB27C9
Thanks in advance!
Post by: jamonwindeyer on September 01, 2017, 12:48:42 am
Hello, this was the multiple choice question I got wrong in my trial paper, could you please help?
An AC motor, when COMPARED to a DC motor:
a) will spin at a rate determined by the electricity supply frequency.
b) will spin at a rate determined by the electricity supply voltage.
c) will produce more torque than a similar DC motor.
d) will spin faster than DC motor.
thanks :)
Hey! Before I explain it, I want to make sure I'm correct because I'm not 100% on how to interpret it - Is the answer A? ;D
Post by: jamonwindeyer on September 01, 2017, 12:52:07 am
Hello! Would appreciate help with the following:
7. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21267582_1313522785439908_359337286_o.png?oh=e3810f91e5c593c5ebdcbd6cb50703e5&oe=59AA295D
15. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21245454_1313523275439859_1895111845_n.png?oh=cfd8d8d15289dc9308a4256bd8c2a820&oe=59AA21CC
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21268713_1313673448758175_1133813723_n.png?oh=0122454ffa0cd5e02612106c8e5c44f6&oe=59AB27C9
Thanks in advance!
Hey! Q7, Weber is Magnetic Flux, related to flux density by the formula:
The units on the RHS there are \(T\) and \(m^2\), so the answer is C :)
For Q15, the change in voltage between the plates is linear. The proton is two thirds the distance from the bottom plate that the top plate is, so the answer is \(\frac{2}{3}\times600=400V\) :)
Hopefully someone who did Quanta can jump in for the last one :)
Post by: blasonduo on September 01, 2017, 08:08:09 am
Hey! Before I explain it, I want to make sure I'm correct because I'm not 100% on how to interpret it - Is the answer A? ;D
Yes sir :)
Post by: jamonwindeyer on September 01, 2017, 10:55:18 am
Yes sir :)
Cool - So we know that in an AC Motor, the changing direction of current is what maintains the constant direction of torque (the job that the split ring commutator does in a DC motor). What this means is that the AC frequency (how fast it switches) will determine how quickly the motor will spin. It will complete one half spin, then the current will switch direction, so it can complete another half spin, and so on. In a DC motor, it is purely the size of the current that determines how quickly it will spin - In an AC motor this is more or less irrelevant, because the frequency will apply the strictest conditions on its motion! :)
It's a little hard to explain admittedly, does that make sense though? :)
Post by: kiwiberry on September 01, 2017, 01:19:12 pm
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21268713_1313673448758175_1133813723_n.png?oh=0122454ffa0cd5e02612106c8e5c44f6&oe=59AB27C9
Thanks in advance!
Hey there! de Broglie hypothesised that all moving particles have a wave nature with wavelength given by \(\frac{h}{mv}\). I reckon they want you to talk about Davisson and Germer's experiment and how it provided support for de Broglie's proposal. So D&G fired electrons onto annealed nickel, and found that they produced a diffraction pattern - diffraction is a wave property, so they were able to show that electrons also have a wave nature. Using this pattern and Bragg's equation they were able to calculate the electron's wavelength, and this corresponded to that predicted by de Broglie's equation \(\lambda = \frac{h}{mv}\). Thus, D&G's experiment provided support for de Broglie's hypothesis and led to the new matter-wave theory. I definitely think this question should be worth more than 2 marks haha - hope this helps! :)
Post by: winstondarmawan on September 01, 2017, 06:14:18 pm
Hey there! de Broglie hypothesised that all moving particles have a wave nature with wavelength given by \(\frac{h}{mv}\). I reckon they want you to talk about Davisson and Germer's experiment and how it provided support for de Broglie's proposal. So D&G fired electrons onto annealed nickel, and found that they produced a diffraction pattern - diffraction is a wave property, so they were able to show that electrons also have a wave nature. Using this pattern and Bragg's equation they were able to calculate the electron's wavelength, and this corresponded to that predicted by de Broglie's equation \(\lambda = \frac{h}{mv}\). Thus, D&G's experiment provided support for de Broglie's hypothesis and led to the new matter-wave theory. I definitely think this question should be worth more than 2 marks haha - hope this helps! :)So would you talk about how it is an exception to the flowchart as de Broglie did not test the hypothesis himself?
And thank you! :)
Post by: kiwiberry on September 01, 2017, 06:57:03 pm
So would you talk about how it is an exception to the flowchart as de Broglie did not test the hypothesis himself?
And thank you! :)
Hm, I don't think it would be an exception, the flowchart was still followed even though his hypothesis was tested by different people :)
Post by: blasonduo on September 01, 2017, 07:44:03 pm
Cool - So we know that in an AC Motor, the changing direction of current is what maintains the constant direction of torque (the job that the split ring commutator does in a DC motor). What this means is that the AC frequency (how fast it switches) will determine how quickly the motor will spin. It will complete one half spin, then the current will switch direction, so it can complete another half spin, and so on. In a DC motor, it is purely the size of the current that determines how quickly it will spin - In an AC motor this is more or less irrelevant, because the frequency will apply the strictest conditions on its motion! :)
It's a little hard to explain admittedly, does that make sense though? :)
Cool, Thank you, but this does make me ask;
Let's just go to the extreme, and assume we supply a voltage of 1million hertz. The change of flow in the AC motor is WAYY too fast for it to spin correctly. (as once the forces due to the flow of current let's say make it spin clockwise, but before it can really move, the current swaps so it is now spinning anti-clockwise, so it can't spin unidirectional)
From this, how can I deduce that B is incorrect?
I appreciate your help!
Post by: julies on September 01, 2017, 09:34:07 pm
Hey! This is right on the borderline of what you are expected to know, but roughly:thanks jamon, this was super helpful :D
- When you connect a voltage source such that the positive terminal is connected to the P-type, and the negative terminal is connected to the N type, this pushes the positive holes in the P-type towards the junction, and the negative electrons in the N-type towards the junction. This is a forward bias connection. This reduces the size of the depletion region. Another way to think of it is that the applied voltage sort of 'cancels' the potential difference caused by the depletion region in the first place. This allows current to flow!
- If we connect it the other way around, the opposite occurs! We don't take away from the voltage, we make it worse, so to speak! This increases the size of the region and thus makes it very difficult for current to flow. Practically, eventually, you would get current - It happens when the diode goes "nope" and electrons start being ripped away from their atoms. But this is reverse bias, essentially a no current scenario!
Post by: jamonwindeyer on September 01, 2017, 09:41:30 pm
Cool, Thank you, but this does make me ask;
Let's just go to the extreme, and assume we supply a voltage of 1million hertz. The change of flow in the AC motor is WAYY too fast for it to spin correctly. (as once the forces due to the flow of current let's say make it spin clockwise, but before it can really move, the current swaps so it is now spinning anti-clockwise, so it can't spin unidirectional)
From this, how can I deduce that B is incorrect?
I appreciate your help!
Yeah cool, so let's go with your scenario. The motor won't spin, it will basically vibrate at 1MHz (I think you'd basically turn your motor into a radio antenna at that frequency lol), but you are right, too quick. My response is that, even if you turned up the voltage, it still wouldn't spin. So what that tells us is that whether the motor spins or not is dependent on frequency in that scenario, not voltage! So, that leads you back to A. You'd conclude B is incorrect from this because voltage has no effect on whether the motor spins or not at that frequency :)
Post by: blasonduo on September 01, 2017, 10:46:00 pm
Yeah cool, so let's go with your scenario. The motor won't spin, it will basically vibrate at 1MHz (I think you'd basically turn your motor into a radio antenna at that frequency lol), but you are right, too quick. My response is that, even if you turned up the voltage, it still wouldn't spin. So what that tells us is that whether the motor spins or not is dependent on frequency in that scenario, not voltage! So, that leads you back to A. You'd conclude B is incorrect from this because voltage has no effect on whether the motor spins or not at that frequency :)
Hmm, There's still something I don't quite get, sorry for being a bother. What is wrong with my logic?
So, when I have a set AC frequency which causes the motor to rotate, If I change the voltage, it WILL spin slower/faster, but it will cause the motor to "be out of whack" and not spin, So a new frequency will need to be set to fix the motor, setting this new frequency will always make the motor spin again.
HOWEVER, If I have a differing frequency which causes the motor to not rotate, fixing the voltage WILL fix the motor spin, as Increasing the voltage does increase the speed of a motor, (i.e its RPM)
So, whats wrong?
Thanks :)
Post by: jamonwindeyer on September 01, 2017, 11:12:29 pm
Hmm, There's still something I don't quite get, sorry for being a bother. What is wrong with my logic?
No bother at all! This is a little tough - First, remember that voltage (more appropriately, current) isn't directly related to speed. It is related to torque, \(\tau=BAIn\). While torque can translate into greater speed, it does depend a little what we've got the thing attached to. You can actually analyse what's called the torque/speed characteristics of a motor, the torque it can provide at a given speed. This is beyond syllabus but the idea is that we can't assume a direct relationship between voltage and speed, especially for an AC motor.
The logic you've used is definitely correct in principle. I think the simplest way to think about how to apply it to this question is this. Consider the simple AC motor that we like to draw (which we never use in the real world, mind you, which is why this all seems a bit wishy washy):
- If the AC frequency is 50Hz, we literally can't make the motor spin at more than 50 revolutions per second. We can up the voltage all we like, but 50 revs per second is the limit.
- Now what if we dropped to 45Hz without changing the voltage. Would we still spin at all? Maybe, maybe not - But what's the new limit? 45 revolutions per second. If we do spin, that's what we spin at.
I suppose the point I'm driving is, you can't go faster than your AC frequency allows. It sets the speed limit. You need to be in tune with it to get your spins. Varying voltage means nothing if you aren't working with your AC frequency.
I liken it to running on an electronic treadmill (where the belt runs at a speed you set on a control panel). Once you set that speed, that determines how quickly you are running. Sure, you can run 'harder,' pump harder with your legs and do more work - That's kind of like increasing your voltage. But you aren't going to move any faster - If you try to use that energy to move faster, you crash into the front of the treadmill and you stop. So it doesn't work!
Again, I know this is wish-washy. You are critiquing a model of an AC motor, and you are right to do it, because the model you learn isn't practical and doesn't actually make sense. Don't worry, the AC motors that actually move stuff around in our world are far more sophisticated than a coil in between a couple of bar magnets ;) it might not rest quite right, and that's okay. Just try and remember that frequency is king/queen for an AC motor :)
Post by: blasonduo on September 01, 2017, 11:30:18 pm
No bother at all! This is a little tough - First, remember that voltage (more appropriately, current) isn't directly related to speed. It is related to torque, \(\tau=BAIn\). While torque can translate into greater speed, it does depend a little what we've got the thing attached to. You can actually analyse what's called the torque/speed characteristics of a motor, the torque it can provide at a given speed. This is beyond syllabus but the idea is that we can't assume a direct relationship between voltage and speed, especially for an AC motor.
The logic you've used is definitely correct in principle. I think the simplest way to think about how to apply it to this question is this. Consider the simple AC motor that we like to draw (which we never use in the real world, mind you, which is why this all seems a bit wishy washy):
- If the AC frequency is 50Hz, we literally can't make the motor spin at more than 50 revolutions per second. We can up the voltage all we like, but 50 revs per second is the limit.
- Now what if we dropped to 45Hz without changing the voltage. Would we still spin at all? Maybe, maybe not - But what's the new limit? 45 revolutions per second. If we do spin, that's what we spin at.
I suppose the point I'm driving is, you can't go faster than your AC frequency allows. It sets the speed limit. You need to be in tune with it to get your spins. Varying voltage means nothing if you aren't working with your AC frequency.
I liken it to running on an electronic treadmill (where the belt runs at a speed you set on a control panel). Once you set that speed, that determines how quickly you are running. Sure, you can run 'harder,' pump harder with your legs and do more work - That's kind of like increasing your voltage. But you aren't going to move any faster - If you try to use that energy to move faster, you crash into the front of the treadmill and you stop. So it doesn't work!
Again, I know this is wish-washy. You are critiquing a model of an AC motor, and you are right to do it, because the model you learn isn't practical and doesn't actually make sense. Don't worry, the AC motors that actually move stuff around in our world are far more sophisticated than a coil in between a couple of bar magnets ;) it might not rest quite right, and that's okay. Just try and remember that frequency is king/queen for an AC motor :)
<3 It has finally clicked! Thank you so much!
Post by: jamonwindeyer on September 02, 2017, 12:05:26 am
<3 It has finally clicked! Thank you so much!
Sahweeet - No worries at all!! ;D
Post by: Dragomistress on September 02, 2017, 04:52:43 pm
A heater uses 2.2x10^6 J of energy in three hours and 20 minutes. How much energy, in kilowatt-hours, is used by this heater?
Post by: Shadowxo on September 02, 2017, 05:10:01 pm
May someone help me with this question?First you have to figure out how many joules a kilowatt hour is.
A heater uses 2.2x10^6 J of energy in three hours and 20 minutes. How much energy, in kilowatt-hours, is used by this heater?
It's basically if you ran a 1000 Watt (1 kilowatt) device for an hour. Remember a Watt is 1 Joule per Second
So 1 kWh = 1000 * 60 * 60 (as an hour is 60*60 seconds) = 3.6*106J
So, it uses
Post by: austv99 on September 03, 2017, 05:09:42 pm
Can I get help on answering part ii? (Quanta to Quark)
Post by: Iminschool on September 06, 2017, 09:27:04 am
Post by: jamonwindeyer on September 06, 2017, 10:32:41 am
I said D, the answers say B. How'd they get B?
Hey! So without the coil and magnet, we'd expect the speed to be:
Now obviously it ends up moving more slowly than that - Some of the kinetic energy has been taken and converted into electrical energy. How much? Well it is just the kinetic energy corresponding to the difference in speed:
Now since the resistance is small, the light will convert all of that to other forms of energy - Issue of course being this doesn't match the answers :( hmm, can't spot what I did wrong right this second (multitasking, lol) - Could you upload your working to get D? Could anyone else chime in? :)
Post by: kiwiberry on September 06, 2017, 04:24:47 pm
I said D, the answers say B. How'd they get B?
Hey! So without the coil and magnet, we'd expect the speed to be:
Now obviously it ends up moving more slowly than that - Some of the kinetic energy has been taken and converted into electrical energy. How much? Well it is just the kinetic energy corresponding to the difference in speed:
Now since the resistance is small, the light will convert all of that to other forms of energy - Issue of course being this doesn't match the answers :( hmm, can't spot what I did wrong right this second (multitasking, lol) - Could you upload your working to get D? Could anyone else chime in? :)
You forgot the square the difference in velocities in the KE formula! But even then, you get 0.0198 J :-\ I used GPE=mgh to find the difference in GPE after the 1m drop and then subtracted the final KE of the magnet from that, and I got D - I'm stumped too haha
Post by: jamonwindeyer on September 06, 2017, 04:34:02 pm
You forgot the square the difference in velocities in the KE formula! But even then, you get 0.0198 J :-\ I used GPE=mgh to find the difference in GPE after the 1m drop and then subtracted the final KE of the magnet from that, and I got D - I'm stumped too haha
Ahhhh thank you so much. So it should have been:
Yep, I'm feeling like the answer is D! :)
Post by: kiwiberry on September 06, 2017, 04:53:38 pm
Hi,
Can I get help on answering part ii? (Quanta to Quark)
Hi there, sorry for the late reply, I must have missed this!
So the binding energy is the amount of energy required to separate a nucleus into its individual nucleons, and is equal to the mass defect (mass of individual nucleons - mass of nucleus) using E=mc2. Hence, atoms with a higher binding energy per nucleon will have a larger mass defect, and each nucleon in the nucleus will have less mass. When two elements below iron undergo nuclear fusion, they produce an element with a higher binding energy per nucleon as shown by the graph - each nucleon in this larger nucleus will have less mass than ones in the reactants, so the extra mass is converted into energy and released. Similarly, when an element above iron undergoes nuclear fission, it produces two elements with higher binding energies per nucleon, producing energy. Let me know if that doesn't make sense - hope this helps! :)
Post by: 12carpim on September 06, 2017, 08:04:34 pm
Thanks :) :)
Post by: austv99 on September 06, 2017, 08:26:34 pm
I thought it was D since electrical resistance is required to dissipate heat?
Post by: kiwiberry on September 06, 2017, 08:47:08 pm
Why is the answer C?
I thought it was D since electrical resistance is required to dissipate heat?
Assuming that voltage is constant, current and resistance are inversely proportional by V=IR - so if current is low, resistance must be high and vice versa. Since Ploss=I2R, the amount of heat is determined by both current and resistance, but mostly current. So, we want resistance to be low so current can be high, in order to maximise power losses and therefore have faster heating :)
Post by: winstondarmawan on September 09, 2017, 07:00:09 pm
5. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21557295_1321052091353644_1671431094_o.jpg?oh=0055bf0754276894538f81a3e740e1da&oe=59B53747
21. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21557082_1321052058020314_955487187_o.jpg?oh=14d3385cf665d072aba1485eac630b1e&oe=59B61DD6
TIA
Post by: Bri MT on September 09, 2017, 08:39:43 pm
Would appreciate help with the following:F=ma Therefore, acceleration will be greatest at maximum force (NOT maximum change in force)
5. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21557295_1321052091353644_1671431094_o.jpg?oh=0055bf0754276894538f81a3e740e1da&oe=59B53747
Post by: austv99 on September 09, 2017, 08:50:32 pm
Post by: austv99 on September 09, 2017, 09:37:55 pm
TIA
Post by: jamonwindeyer on September 09, 2017, 10:57:19 pm
Is b referring to the fact that without a split ring, the coil alternates? Since ac induction motors dont use the motor effect?
Hey! So this is for the second one right? Yeah you've got it, so this is for AC motors that use the slip rings, the ones very similar to DC motors - Not induction motors like the first part of the question. Weird, I know!
So you'd identify what the motor effect actually is for a mark, describe the effect is has on the coil in such a motor for a mark, then explain how the alternating current maintains a constant direction of torque for a mark ;D
Can someone help with this question?
TIA
The two cannonballs, despite being launched at different angles, reach the same maximum height in the same amount of time. That is, the horizontal motion of Cannonball Q hasn't affected its vertical motion - The horizontal and vertical components are independent. This is what supports Galileo's analysis! :)
Post by: beau77bro on September 10, 2017, 02:03:12 pm
Post by: pikachu975 on September 10, 2017, 02:08:19 pm
how does magnetic levitation work??
Are you referring to the prac or maglev trains?
Post by: beau77bro on September 10, 2017, 02:54:34 pm
Are you referring to the prac or maglev trains?"Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting" a secondary dot point in the syllabus - it was not covered well in class so i don't really get it.
Post by: pikachu975 on September 10, 2017, 03:32:31 pm
"Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting" a secondary dot point in the syllabus - it was not covered well in class so i don't really get it.
Surface currents (cooper pairs) prevent any magnetic flux from penetrating the superconductor known as the Meissner effect. This means an equal and opposite magnetic field is produced to oppose the magnet's magnetic field and to oppose it's weight force downwards, causing it to levitate.
Post by: beau77bro on September 10, 2017, 03:55:28 pm
Surface currents (cooper pairs) prevent any magnetic flux from penetrating the superconductor known as the Meissner effect. This means an equal and opposite magnetic field is produced to oppose the magnet's magnetic field and to oppose it's weight force downwards, causing it to levitate.
elaborate please? how do cooper pairs prevent the magnetic field? how is the opposing magnetic field produced - i was told eddy currents aren't responsible. also i just had a trial question based upon it rising initially and what causes it - what happens there?
Post by: pikachu975 on September 10, 2017, 04:13:13 pm
elaborate please? how do cooper pairs prevent the magnetic field? how is the opposing magnetic field produced - i was told eddy currents aren't responsible. also i just had a trial question based upon it rising initially and what causes it - what happens there?
It's not eddy currents, it's surface currents i.e. cooper pairs. Also we don't need to know how meissner effect works I think that's beyond HSC it probably involves some quantum theory. What causes it to rise is basically what I said - the cooper pairs (surface currents) prevent flux entering the superconductor known as meissner effect, then they produce a magnetic field equal in magnitude and opposite in direction to balance the magnet's weight force (make sure to include it balances the weight force or opposes it).
Post by: beau77bro on September 10, 2017, 06:02:09 pm
It's not eddy currents, it's surface currents i.e. cooper pairs. Also we don't need to know how meissner effect works I think that's beyond HSC it probably involves some quantum theory. What causes it to rise is basically what I said - the cooper pairs (surface currents) prevent flux entering the superconductor known as meissner effect, then they produce a magnetic field equal in magnitude and opposite in direction to balance the magnet's weight force (make sure to include it balances the weight force or opposes it).
Is that all we have to know? thanks pikchu. still interested in how it works beyond the course tho, maybe a certain quantum physicist could send some info our way?
Post by: jamonwindeyer on September 10, 2017, 06:26:48 pm
Is that all we have to know? thanks pikchu. still interested in how it works beyond the course tho, maybe a certain quantum physicist could send some info our way?
Yep, so provided you (roughly) understand the idea of how Cooper Pairs form and such, you don't need to go into the Meisner Effect a whole lot. Important to know that superconductors also have the property of exactly cancelling out any changing magnetic fields with eddy currents - But as above, this is separate to the Meisner Effect.
The Meisner Effect is a very complex quantum mechanical effect. I've sort of had it explained to me I think (I could never possibly do it justice lol), but I'm pretty sure it is beyond what can be handled without properly studying quantum mechanics. Indeed, it's also something we are still investigating - This isn't like semiconductors, that we understand quite well. Superconductors are a huge topic of research right now - We're always learning more :)
Post by: winstondarmawan on September 10, 2017, 10:54:18 pm
Would appreciate help with the following:
21. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21557082_1321052058020314_955487187_o.jpg?oh=14d3385cf665d072aba1485eac630b1e&oe=59B61DD6
TIA
Believe my post got missed. :(
What is the best approach for the above question?
Also for this question:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21534597_1846449502338173_1092053360_o.jpg?oh=36b00078af6bf1bf34c14bbef3561e49&oe=59B790FD
How is the answer B? An explanation would be greatly appreciated. :)
Thanks in advance.
Post by: jamonwindeyer on September 10, 2017, 11:26:31 pm
Believe my post got missed. :(
What is the best approach for the above question?
Also for this question:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21534597_1846449502338173_1092053360_o.jpg?oh=36b00078af6bf1bf34c14bbef3561e49&oe=59B790FD
How is the answer B? An explanation would be greatly appreciated. :)
Thanks in advance.
Hey! For the earlier post, you'd be focusing on Edison/Westinghouse and the role financial motivations played in the competition for which distribution system to use. In honesty, I'd probably be picking space exploration there - A lot of the space scientists you cover in that topic developed their technology due to funding from the World Wars. That's an easier link in my opinion :)
For the latter, the time we are given is \(t_v\) to use in the formula. We know this because time passes by more slowly for a body in motion - That 10 year timeframe is the value that has been 'stretched' relative to earth. So to find the original:
Post by: kiwiberry on September 10, 2017, 11:27:25 pm
21. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/21557082_1321052058020314_955487187_o.jpg?oh=14d3385cf665d072aba1485eac630b1e&oe=59B61DD6
TIA
Sorry for missing this!! For the electricity distribution systems, I would agree and talk about the rivalry between Westinghouse (AC) and Edison (DC). At first, Edison's DC electricity was favoured because technology using it was well established. However, because DC could only be distributed at the voltage it was to be used, there were large, expensive power losses and it could only be distributed over short distances. AC electricity, introduced by Westinghouse, was able to overcome these shortcomings through the use of transformers. Recognising the threat this posed to his business, Edison tried to ban the use of AC and prove that it was dangerous by electrocuting animals and using it in the first electric chair, despite knowing the benefits of AC to society.
Post by: winstondarmawan on September 10, 2017, 11:36:52 pm
We know this because time passes by more slowly for a body in motion - That 10 year timeframe is the value that has been 'stretched' relative to earth. So to find the original:
If time passes more slowly in a body of motion, wouldn't the time elapsed on Earth be a greater value?
Post by: jamonwindeyer on September 11, 2017, 09:26:23 am
If time passes more slowly in a body of motion, wouldn't the time elapsed on Earth be a greater value?
Time passing more slowly results in a smaller value - That is, if you watch a clock on a body moving very quickly for 1 year, less than 1 year will pass on the clock. A smaller value.
Hmm, on Googling this question apparently it's a bit controversial, I understand why.
The issue is that we are taking both measurements, the 10 years and the one we calculate, from the relativistic frame of reference. I see where your answer comes from, but consider it this way. The way the question is worded, the spaceship is the observer. The earth is the thing moving away at 0.8c (remember, relativity, both are valid). So, the time on earth should be slower compared to the spaceship, purely because we've chosen the spaceship as our frame of reference.
Now we are measuring a time of 10 years as the earth moves away - That is \(t_v\), because we are the observer watching the earth move away at speed, our value is affected by special relativity. \(t_0\) is the unknown, the value unaffected by special relativity :)
I wouldn't stress about this question - It's poorly framed at best :P
Post by: winstondarmawan on September 11, 2017, 05:05:52 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21584700_1322520041206849_361794579_n.png?oh=2863b97b71c254113417eadfe759b600&oe=59B8BA81
Here are the values if it is not readable from the above pic:
Mass of block: 30g
Hangs 4.25cm from the axle.
Dimensions of armature: 4cm x 8.5cm
Current: 2.47x10^-3 A
Magnetic Field Strength: 0.12T
Thanks in advance!
Post by: kiwiberry on September 11, 2017, 08:28:34 pm
Would appreciate help for the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21584700_1322520041206849_361794579_n.png?oh=2863b97b71c254113417eadfe759b600&oe=59B8BA81
Here are the values if it is not readable from the above pic:
Mass of block: 30g
Hangs 4.25cm from the axle.
Dimensions of armature: 4cm x 8.5cm
Current: 2.47x10^-3 A
Magnetic Field Strength: 0.12T
Thanks in advance!
Hey!
Because the motor does not rotate, the torque due to the weight force of the block must equal the torque due to the motor effect:
Post by: gilliesb18 on September 13, 2017, 04:46:23 pm
I have an exam tomorrow for physics and I just have a few questions from a past paper that I would appreciate some help on....
1. Explain how Einstein's famous equation E=mc^2 relates to the Big Bang theory.
2. If crumple zones help reduce injuries to people during collisions, why then is the cabin of the vehicle made so rigid??
Thanks!!!!
Post by: blasonduo on September 13, 2017, 09:38:21 pm
Hello :)
I have an exam tomorrow for physics and I just have a few questions from a past paper that I would appreciate some help on....
1. Explain how Einstein's famous equation E=mc^2 relates to the Big Bang theory.
2. If crumple zones help reduce injuries to people during collisions, why then is the cabin of the vehicle made so rigid??
Thanks!!!!
Hello! I am a bit rusty on these questions, (especially 1) so apologies if you wanted more
1) The big bang theory suggested that energy expanded at an extremely fast rate before slowing down to a slower pace while producing matter. E=mc^2 is a way to vaguely explain this. From this, we see that the big bang produced matter instead of producing energy.
2) We need crumple zones to significantly decrease the energy input into the car. The more it "crumples", the more energy dissipates before entering the human. HOWEVER, if this cabin was not sturdy, it would also crumple, but into the passengers! This crumpling could be more dangerous than the initial impact as the crumpled zone is now being "intruded" into you!
I know these explanations aren't the best, but I hope they help ;)
Post by: winstondarmawan on September 14, 2017, 05:57:53 pm
What is the weak nuclear force and it's role in holding nucleons together?
TIA.
Post by: austv99 on September 14, 2017, 06:34:44 pm
All from 2005 HSC
Can someone clarify what Enrico Fermi did for beta decay? (Quanta to Quark)
Mod Edit: Post merge :)
Post by: jakesilove on September 14, 2017, 08:16:52 pm
Quanta to Quarks:
What is the weak nuclear force and it's role in holding nucleons together?
TIA.
Hey! The weak nuclear force is one of four fundamental forces in our universe. Relative to the strong nuclear force, it has a strength of 10^(-6). The weak nuclear force operates across distances 0.1% of the diameter of the Proton, and so operates at REALLY, REALLY small scales, such as the interaction of nucleons! They are also important in the interaction of quarks and leptons. Not too sure if you would need to know more detail than this; I definitely don't!
Post by: kiwiberry on September 14, 2017, 11:26:41 pm
Quanta to Quarks:
What is the weak nuclear force and it's role in holding nucleons together?
TIA.
Another thing to remember is the fact that the weak nuclear force is carried by the intermediate vector bosons W/Z. But yeah, I'm pretty sure we don't need to know this in much detail, what Jake said is probably more than enough! :)
Would appreciate help in understanding these three questions
All from 2005 HSC
Can someone clarify what Enrico Fermi did for beta decay? (Quanta to Quark)
Mod Edit: Post merge :)
Hey!
6) The induced current is proportional to the rate of change of flux. The speed that the wire rotates will definitely change this, so C is wrong. When the wire is north-south, it will be parallel to the Earth's magnetic field, and it will cut much less flux than if it was east-west, so D is also wrong. I'm not sure why B is the answer though - can someone else lend a hand? :)
15) First of all, the electrons and holes won't speed up or slow down, because the force due to the mag field will be at right angles to the mag field and the particle's velocity, so we can eliminate A and B. Due to the current, the electrons will move to the closer end and the holes to the further end. Using the right hand grip rule, both electrons and holes will move to he bottom of the rod (remember to flip the direction of the force for negative particles, so the answer is C
5) Maximum range will be when the angle of elevation is 45o, so since 40o and 50o are both 5o out from this, both will have the same range. When the angle is 50o, the vertical component of the particle's velocity will be larger, so it will have a longer time of flight, hence C
Fermi was the one who formally proposed the theory that a neutron is transformed into an electron, proton and antineutrino during beta decay.
Hope this helps :)
Post by: austv99 on September 15, 2017, 02:16:46 am
15) First of all, the electrons and holes won't speed up or slow down, because the force due to the mag field will be at right angles to the mag field and the particle's velocity, so we can eliminate A and B. Due to the current, the electrons will move to the closer end and the holes to the further end. Using the right hand grip rule, both electrons and holes will move to he bottom of the rod (remember to flip the direction of the force for negative particles, so the answer is CJust to clarify, for 15, the holes move in same way as electrons because they are relatively positive but not considered positively charged? And they wont speed up or slow down since to do so, the force must be in the same direction of motion? In this case it's down?
Fermi was the one who formally proposed the theory that a neutron is transformed into an electron, proton and antineutrino during beta decay.
If a question was about how fermi contrubuted to understanding of beta decay, would pauli be explained first?
Post by: jamonwindeyer on September 15, 2017, 09:24:13 am
Just to clarify, for 15, the holes move in same way as electrons because they are relatively positive but not considered positively charged? And they wont speed up or slow down since to do so, the force must be in the same direction of motion? In this case it's down?
Close - They are relatively positive so we do consider them as positive charges. Normally, the force on an electron and hole would be in opposite directions - If they were stationary. Here, they are moving in opposite directions, so the fact they are moving in opposite directions and opposite in charge cancels out, and they move in the same direction. You can deduce all this with the right hand slap rule, but it is faster to just use intuition ;D and yes, to speed them up or slow them down you would need forces parallel to motion - It doesn't really make sense as an answer in this context ;D
Post by: katnisschung on September 16, 2017, 11:31:25 am
question for quanta to quarks regarding nuclear decay/radiation
how do i determine what type of radiation is emitted by the element given its altered atomic weight.
i.e. potassium-40
thorium-232
radium-226
iodine-131
so in my notes i have written the rules to determine what type of radiation emitted to be the following:
alpha decay--> occurs for elements too big, atomic number greater than 83
beta minus decay--> occurs for elements where there are too many neutrons compared to protons, in larger
atoms where neutron:proton ratio is 1.5:1
so i did the calculations and the rules all work as i checked my answers but radium-40 undergoes beta minus decay according to the answers but the ratio is no where near 1.5:1 its more 1.1....... could someone clarify thanks :)
Post by: kiwiberry on September 16, 2017, 11:51:34 am
hi :)
question for quanta to quarks regarding nuclear decay/radiation
how do i determine what type of radiation is emitted by the element given its altered atomic weight.
i.e. potassium-40
thorium-232
radium-226
iodine-131
so in my notes i have written the rules to determine what type of radiation emitted to be the following:
alpha decay--> occurs for elements too big, atomic number greater than 83
beta minus decay--> occurs for elements where there are too many neutrons compared to protons, in larger
atoms where neutron:proton ratio is 1.5:1
so i did the calculations and the rules all work as i checked my answers but radium-40 undergoes beta minus decay according to the answers but the ratio is no where near 1.5:1 its more 1.1....... could someone clarify thanks :)
Hiya! :)
You're right about alpha decay. Beta minus decay does occur when the n:p ratio is too high, however there is a different limit to how high this can get depending on the size of the atom
(https://katherinetchem2010.wikispaces.com/file/view/Nuclear_chemistry_2.png/194838730/Nuclear_chemistry_2.png)
From this graph, an atom will be unstable if its n:p ratio is above about:
- 1:1 for atomic number<20
- 1.3:1 for atomic number\(\approx \)50
- 1.5:1 for atomic number\(\approx \)80
Did you mean potassium-40? Assuming you did, potassium has an atomic number of 19, so it will undergo beta decay with an n:p ratio of above 1:1 :)
Hope this helps!
Post by: austv99 on September 16, 2017, 07:27:44 pm
Post by: jamonwindeyer on September 16, 2017, 08:21:18 pm
What's the best appraoch for this question? And in general? Excluding the example.
You could discuss pretty much any of the experiments/scientists you've studied that lead to changes in our understanding of how things work. Particularly:
- Einstein's Special Theory of Relativity (particularly this one)
- Michelson and Morley
- Thompson's Experiment with Cathode Rays
You could discuss those, explaining what the model was, how the experiment was tested and how it validated/altered the model :)
Post by: pikachu975 on September 16, 2017, 08:30:14 pm
What's the best appraoch for this question? And in general? Excluding the example.
This is how they validate models:
Observation -> Problem Raised -> New hypothesis -> Experiment -> Collection of data -> Analysis of results -> Results support hypothesis (if not then back to new hypothesis) -> New theory or law
from 2007 HSC
Post by: austv99 on September 16, 2017, 09:47:41 pm
With 27a, would i talk about the uv catastrophe and explain the existence of the peak and the graph going towards zero?
Post by: kiwiberry on September 16, 2017, 10:35:24 pm
Would appreciate help with 26c
With 27a, would i talk about the uv catastrophe and explain the existence of the peak and the graph going towards zero?
26) The power loss will be the difference between the input power and the output power:
27) Yep! Make sure to mention Planck's hypothesis in there somewhere :)
Post by: pikachu975 on September 17, 2017, 10:29:50 am
Would appreciate help with 26c
With 27a, would i talk about the uv catastrophe and explain the existence of the peak and the graph going towards zero?
Yep for 27 first speak about what classical thought, basically the UV catastrophe how at shorter wavelengths there would be faster oscillations hence infinite energy which wasn't possible according to conservation of energy. Then introduce Planck's quantisation of energy emitted by a black body cavity as E = hf and explain how experiments showed definite peaks, which was due to some transitions being more probable than others. Speak about how he explained the UV catastrophe because there is a maximum amount of shells an atom can jump (quantised energy levels).
Post by: beau77bro on September 17, 2017, 11:29:50 am
Post by: jamonwindeyer on September 17, 2017, 11:44:54 am
hey so in medical, it talks about x, y, z axes. the z axis has a gradient magnetic field which alters larmor frequency, but the x-axis changes the precession frequency. what is the difference between the larmor frequency and precession frequency?
Hey! So I wouldn't worry about this in toooo much detail, it's a complicated bit of Physics we're dealing with here. Basically, the magnetic fields in the MRI make it so that every single position in the body is subjected to a magnetic field that is slightly different in terms of either strength or direction - This will slightly change the frequency and phase of radio waves emitted by the resonating hydrogen nuclei. This is what they mean by frequency (I believe, my understanding here isn't much greater than yours!). Incidentally, the y-axis is used to control phase :)
Probably not the best answer, and someone else might have done wider reading than I and be able to help more!! But understanding broadly was always enough for me and will absolutely be enough to get the marks in an MRI question ;D
Post by: katnisschung on September 19, 2017, 05:22:40 pm
I forgot the rule for how many significant figures u need to take in your answer??
was it sth like u take the most in your answer as the most in the question?
Post by: Natasha.97 on September 19, 2017, 05:32:24 pm
hi :)
I forgot the rule for how many significant figures u need to take in your answer??
was it sth like u take the most in your answer as the most in the question?
Yep! You take the least accurate value given in the question (e.g. if given 0.3836 and 0.40, do 2 s.f.) :)
Post by: katnisschung on September 19, 2017, 05:41:46 pm
I got this rather simple question wrong when revising content (probably becos i don't know quanta to quarks well enough)
calculate the frequency of the matter wave when a neutron is made to move at 53.6x10^3 m/s
so why is it that they take 'c' as the speed. do all matter waves travel at c
Post by: austv99 on September 20, 2017, 05:09:40 pm
2013 HSC 27b
Post by: JuliaPascale123 on September 20, 2017, 07:09:47 pm
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.
(http://i67.tinypic.com/jkbyn6.png)
Thanks <3,
Julia
Post by: gilliesb18 on September 20, 2017, 08:04:52 pm
Just wondering how much harder physics gets in yr 12? Cause i'm considering dropping it now that some of my marks for this year haven't been so great...
Can someone give me their thoughts on this?
Thanks heaps :)
Post by: jamonwindeyer on September 20, 2017, 10:13:31 pm
Would appreciate help with this question
2013 HSC 27b
Hey! So the things to discuss:
- The electric vehicle will propel itself using the motor effect - Explain this effect for a mark.
- The electric vehicle will brake itself using the principle of electromagnetic induction, as the movement of the wheels is used to induce a current in coil windings. This will brake the vehicle, because the induced currents create a magnetic field to interact with some other permanent magnetic field, to oppose the motion. Essentially, just typical stuff for a generator. Explain this for a mark.
- Analyse how these two processes are related in terms of energy, that is, electrical energy being converted into kinetic energy by the motor effect, and kinetic energy being converted to electrical energy by induction, for a mark.
- Last mark for identifying that these aren't perfect energy transformations, and so you'll always end up with less energy than you started with at the start of the process ;D
(You might get marks for slightly different things, but this is an indicator) ;D
Post by: jamonwindeyer on September 20, 2017, 10:14:42 pm
Hi Guys,
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.
(http://i67.tinypic.com/jkbyn6.png)
Thanks <3,
Julia
Hi Julia! This doesn't look like a HSC question to me - Where did you find it? :)
Post by: jamonwindeyer on September 20, 2017, 10:16:25 pm
Hello :)
Just wondering how much harder physics gets in yr 12? Cause i'm considering dropping it now that some of my marks for this year haven't been so great...
Can someone give me their thoughts on this?
Thanks heaps :)
Hey! In my opinion, Physics does get a little harder in Year 12. The concepts get a lot 'stranger,' and you do need to invest more time to wrap your head around them.
You can always improve!! But you'll need to want to put that extra effort in - Is Physics something you enjoy? ;D
Post by: JuliaPascale123 on September 20, 2017, 10:24:46 pm
Hi Julia! This doesn't look like a HSC question to me - Where did you find it? :)
Hi Jamon,
I think I accidently submitted it into the maths questions aswell though its suppose to be for physics only. So our teacher gave us this and said if we can solve it we should be fine with any other question they throw at us regarding p =f/a
Do you have any idea how to solve it?
Thanks,
Julez
Post by: jamonwindeyer on September 20, 2017, 10:49:26 pm
Hi Jamon,
I think I accidently submitted it into the maths questions aswell though its suppose to be for physics only. So our teacher gave us this and said if we can solve it we should be fine with any other question they throw at us regarding p =f/a
Do you have any idea how to solve it?
Thanks,
Julez
\(P=\frac{F}{A}\) isn't a HSC assessable formula!! Perhaps your teacher is extending you? I've done a little bit on this at uni but I probably wouldn't be comfortable trying to answer it properly, sorry! :)
Post by: gilliesb18 on September 21, 2017, 10:23:30 am
Hey! In my opinion, Physics does get a little harder in Year 12. The concepts get a lot 'stranger,' and you do need to invest more time to wrap your head around them.Ok thanks for that... I dont know if I do entirely enjoy it actually!!! But all the same its a pretty cool subject....
You can always improve!! But you'll need to want to put that extra effort in - Is Physics something you enjoy? ;D
Anymore ideas??
Post by: jamonwindeyer on September 21, 2017, 11:33:26 am
Ok thanks for that... I dont know if I do entirely enjoy it actually!!! But all the same its a pretty cool subject....
Anymore ideas??
If you really do enjoy it then I'd keep it, if you are less sure than I'd consider whether you want less units or not. This guide and this one might be useful for you ;D
Post by: gilliesb18 on September 21, 2017, 04:03:46 pm
But just thinking about it now, if I did drop Physics, and my other subjects are: Business Services, Mathematics(2U), Std English, Geography and German, would I still be able to get an atar? I can't seem to find much on it online for some reason...
Post by: jamonwindeyer on September 21, 2017, 07:22:43 pm
Ok thanks thats super helpful:):
But just thinking about it now, if I did drop Physics, and my other subjects are: Business Services, Mathematics(2U), Std English, Geography and German, would I still be able to get an atar? I can't seem to find much on it online for some reason...
You definitely can, Business Services is a Category B course, you are allowed 2 units of Category B (along with 8 units of Category A), so you qualify ;D
Here is a category list if you're interested :)
Post by: gilliesb18 on September 21, 2017, 07:29:20 pm
Post by: Mymy409 on September 22, 2017, 01:22:33 pm
Perform a first-hand investigation,
gather information and analyse data to
calculate initial and final velocity,
maximum height reached, range and
time of flight of a projectile for a
range of situations by using
simulations, data loggers and
computer analysis
Post by: arunasva on September 22, 2017, 11:03:51 pm
Does anyone have any good notes under this dotpoint?
Perform a first-hand investigation,
gather information and analyse data to
calculate initial and final velocity,
maximum height reached, range and
time of flight of a projectile for a
range of situations by using
simulations, data loggers and
computer analysis
in the excell book yeah. Plus they never ask you anything too hard on that. There was a question in accuracy in that experiment and presentation of graphical data in the CSSA trial. Other an' that ive seen one question with a football, but that was mostly math. As long as you remember what you did you should be fine.
Post by: Mymy409 on September 23, 2017, 07:29:43 pm
in the excell book yeah. Plus they never ask you anything too hard on that. There was a question in accuracy in that experiment and presentation of graphical data in the CSSA trial. Other an' that ive seen one question with a football, but that was mostly math. As long as you remember what you did you should be fine.
Thanks. :)
I've got another question: In the topic of escape velocity, why does the object have 0 velocity and 0 energy at infinity?
Post by: arunasva on September 23, 2017, 11:07:19 pm
Thanks. :)
I've got another question: In the topic of escape velocity, why does the object have 0 velocity and 0 energy at infinity?
With Gravitational Potential energy, the value of the energy itself at a point is not relevant as it does not give you any important information. But... the difference in energy after something went from one point to another is. So if you take the potential energy as 0 on the surface of the earth (mgh) or take 0 a 10^5km from earth it does not matter. You can define the 0 point yourself. Since gravitational fields only exert 0 force at infinity GPE is considered 0 at infinity. As for velocity even Idk :'(
Post by: Shadowxo on September 23, 2017, 11:32:39 pm
Thanks. :)Just to elaborate on arunasva's post,
I've got another question: In the topic of escape velocity, why does the object have 0 velocity and 0 energy at infinity?
The escape velocity is the minimum velocity required to escape the planet's gravitational field. So, at infinity, all the kinetic energy has been used up to get away from the gravitational field / has been converted into grav potential energy, so velocity and k.e. are 0. And at infinity, gravitational potential energy is zero (as at an infinite distance, you don't feel the effects of it). Note that all the kinetic energy was converted into grav potential energy, so the grav potential energy increased up to 0, as gpe is negative.
Hence, at infinity, both velocity and energy are zero.
If the velocity were greater than the escape velocity however, this would not be the case as at infinity both velocity and k.e. would be positive.
Post by: arunasva on September 24, 2017, 02:02:44 pm
Just to elaborate on arunasva's post,
The escape velocity is the minimum velocity required to escape the planet's gravitational field. So, at infinity, all the kinetic energy has been used up to get away from the gravitational field / has been converted into grav potential energy, so velocity and k.e. are 0. And at infinity, gravitational potential energy is zero (as at an infinite distance, you don't feel the effects of it). Note that all the kinetic energy was converted into grav potential energy, so the grav potential energy increased up to 0, as gpe is negative.
Hence, at infinity, both velocity and energy are zero.
If the velocity were greater than the escape velocity however, this would not be the case as at infinity both velocity and k.e. would be positive.
ohhh thanks :)
Post by: austv99 on September 24, 2017, 05:37:35 pm
TIA
Post by: katnisschung on September 24, 2017, 05:39:28 pm
what is the speed for matter waves?
I got this rather simple question wrong when revising content (probably becos i don't know quanta to quarks well enough)
calculate the frequency of the matter wave when a neutron is made to move at 53.6x10^3 m/s
so why is it that they take 'c' as the speed. do all matter waves travel at c
Post by: arunasva on September 25, 2017, 01:48:33 am
Might be a rookie question but some clarification for this question is appreciated -part ii
TIA
From my chemistry knowledge, I'm guessing cos the first reaction releases a neutron, so that neutron can bombard atoms to bring about the reaction. In the subsequent fission more neutrons are produced these have the capacity of bombarding even more atoms to bring about fusion and it goes on and on the next fission will produce more neutrons than the previous one which will keep bombarding atoms to release energy. That's why power plants have an absorber thingy to absorb neutrons and control the rate
Post by: austv99 on September 25, 2017, 12:58:40 pm
All from 2015 HSC
Post by: blasonduo on September 25, 2017, 04:01:47 pm
Help is appreciated for 14, 18, 19.
All from 2015 HSC
Hello! I hope I can explain this well :)
14) For these questions especially, I'd like to picture myself in these situations and go from there. Imagine you are driving a car, speeding it up or slowing it down will NOT make you experience a force to the left or right (there is no circular motion!), so this eliminates both B and D. Now referring back to the car analogy, when you turn the car, you'd always experience a force opposite to the way you are turning. The car is beginning its turning motion while you continue in a straight line path. This is because of Newton's first law.
So back to the question, the ball began to roll down the page, and from what was said before, it must mean the train was turning up the page, or to the RIGHT, so it is C.
18) For this, a bit of calculation is best for the description.
Let's assume a couple things for the equation
19) This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why, i'll be happy to!)
I hope this helps! :)
Post by: bsdfjnlkasn on September 25, 2017, 07:39:07 pm
I just have a few questions from the Quanta to Quarks option module and was hoping to get some help :)
1. What was it about Thomson’s model that led Rutherford to believe no large angle deflections would occur?
2. Where did the nitrogen atoms come from in Chadwick’s experiment? The following dot point was in a summary I found online and it's just really confusing to me. It's for the dot point which asks how Chadwick used conservation laws to discover the neutron. I get nuclei of hydrogen atoms because that's simply referring to the protons that were dislodged from the paraffin wax as a result of the neutron's colliding with them. But Nitrogen? It's a mystery to me.
• Chadwick measured the recoil of the nuclei of hydrogen and nitrogen atoms after interacting with the natural radiation.
3. Do we need to know the details of the specific violations which led Pauli to propose the existence of neutrino? There are two conservation violations that i've read up on, but I don't know how relevant the actual details are. Here is the paragraph that specifies these:
• During beta decay, initially scientists thought only beta particles were emitted. When they evaluated the energies involved, they came up with a figure for the maximum kinetic energy that a beta particle should have. All beta particles should have been emitted with this velocity, but this wasn’t the case. Instead, almost none were emitted with the full amount of kinetic energy, and most of them were emitted with significantly less. This meant that the slow beta particles were missing kinetic energy, leading to a violation of conservation of energy. Also, the sum of the momentums before and after beta decay was not equal - assuming the nucleus starts off stationary, the sum of momentums should be zero. However, when the momentums of the beta particle and the remainder of the nucleus were added, it was not zero, so conservation of momentum was being violated.
Thank you so much!! :D :D
Post by: Sukakadonkadonk on September 25, 2017, 07:58:36 pm
I have a quick question, could you say that the magnitude of torque increases as motor speed increases (of course, regarding a motor)?
Thanks.
Post by: winstondarmawan on September 25, 2017, 08:50:40 pm
18) For this, a bit of calculation is best for the description.
Let's assume a couple things for the equation
I hope this helps! :)
Don't galvanometers measure current?
Post by: bsdfjnlkasn on September 25, 2017, 08:55:47 pm
Hi,
I have a quick question, could you say that the magnitude of torque increases as motor speed increases (of course, regarding a motor)?
Thanks.
Hey! If we look at the formula for torque: t = nBIA cos(a) we can deduce that torque is only affected (considering they are proportional) by:
1. The number of turns on the coil
2. Strength of the magnetic field
3. The size of the current travelling through the coil
4. Area of the solenoid (multiply the side lengths of the coil)
5. Angle between the plane of the coil and the magnetic field
Because of this, the speed won't affect the torque acting on the coil as a result of the magnetic field in which it's in :)
Post by: bsdfjnlkasn on September 25, 2017, 08:59:24 pm
Might be a rookie question but some clarification for this question is appreciated -part ii
TIA
Hey, i'll give this a shot :)
Energy is released as a part of radioactive decay as it assists in stabilising the radioactive nucleus. This is also explained by the fact that new elements are being formed (i.e. nuclear transmutation).
Post by: Sukakadonkadonk on September 25, 2017, 09:26:00 pm
Hey! If we look at the formula for torque: t = nBIA cos(a) we can deduce that torque is only affected (considering they are proportional) by:Hey! Thanks for response.
1. The number of turns on the coil
2. Strength of the magnetic field
3. The size of the current travelling through the coil
4. Area of the solenoid (multiply the side lengths of the coil)
5. Angle between the plane of the coil and the magnetic field
Because of this, the speed won't affect the torque acting on the coil as a result of the magnetic field in which it's in :)
I don't know, I just find it confusing because if I picture myself manually rotating a motor faster, it just seems like the torque would increase.
How do I wrap my head around this? Just remember the torque equation?
Post by: bsdfjnlkasn on September 25, 2017, 09:36:07 pm
Hey! Thanks for response.
I don't know, I just find it confusing because if I picture myself manually rotating a motor faster, it just seems like the torque would increase.
How do I wrap my head around this? Just remember the torque equation?
No worries :)
I think keeping in mind the formula is a good thing to do. Perhaps you're trying to link increasing the rate of rotation with the concept of magnetic flux. Because indeed, when you start to turn the coil faster, the rate of change in magnetic flux increases, and the more it increases, the greater the current that is induced. But other than that, torque is only dependent on the variables I listed above because torque is the turning effect of a force acting on an object. The speed will not change how the coil will turn in terms of the forces experienced :)
Let me know if you have anymore questions :)
Post by: jamonwindeyer on September 25, 2017, 09:56:32 pm
Hey! Thanks for response.
I don't know, I just find it confusing because if I picture myself manually rotating a motor faster, it just seems like the torque would increase.
How do I wrap my head around this? Just remember the torque equation?
Picture how fast a car is moving versus the 'revs' of the car. You can have low revs but high speed - Or, high revs but low speed. It shows that torque and speed aren't necessarily going to go up and down together - It depends on the load attached to the motor :) you don't need to worry about this much at HSC level! :)
Post by: julies on September 25, 2017, 10:16:28 pm
Hey there!
I just have a few questions from the Quanta to Quarks option module and was hoping to get some help :)
1. What was it about Thomson’s model that led Rutherford to believe no large angle deflections would occur?
Hey, so Thomson's plum pudding model consisted of negatively charged particles dispersed in a diffuse positively charged "dough". This dough is uniformly charged, and large in size, meaning that the charge at any particular point of the "dough" is quite weak. This is in opposition to Rutherford's model, where the small size of the nucleus essentially lends itself to carrying a very concentrated positive charge. If positive alpha particles are fired at a weakly positive region, then the deflection is expected to be small.
Hope this helped haha
Post by: Sukakadonkadonk on September 25, 2017, 10:36:59 pm
No worries :)
I think keeping in mind the formula is a good thing to do. Perhaps you're trying to link increasing the rate of rotation with the concept of magnetic flux. Because indeed, when you start to turn the coil faster, the rate of change in magnetic flux increases, and the more it increases, the greater the current that is induced. But other than that, torque is only dependent on the variables I listed above because torque is the turning effect of a force acting on an object. The speed will not change how the coil will turn in terms of the forces experienced :)
Let me know if you have anymore questions :)
Alright, I think I get it. Just gotta remember formula.
Cheers!
Post by: katnisschung on September 26, 2017, 11:19:52 am
I talked quite vaguely in my answer....ability to make variations in electric currents visible and measure, analyse...
any specific examples?
Post by: clarence.harre on September 26, 2017, 03:10:04 pm
The marking guidelines don't really give an understanding of the depth I needed to go to, so I'm hoping one of you could please check my answer.
***
Question:
'Important fundamental discoveries in physics often lead to applications which
have a significant effect on society.’
Evaluate this statement, with reference to the contributions of Rutherford,
Einstein and Fermi to the development of the atomic bomb.
***
Response:
Rutherford's analysis of 1909 Geiger-Marsden gold foil experiment catalysed the development of the Rutherford-Bohr model of the atom. It depicted the atom as consisting mainly of empty space, which separates orbiting electrons from a very small, dense, positively charged central mass (nucleus). His work pioneered the exploration of nuclear physics, and was thus significant in developing the bomb.
Einstein's 1905 'Theory of Special Relativity' proposed that a large amount of energy could be released from a small quantity of matter, commonly expressed in the equation E = mc2
where the energy of matter in its rest state is equal to the its mass multiplied by the speed of light squared. While this work was integral to the bomb's operations, his greatest contribution to its development was signing a letter in 1939 to President Roosevelt warning of German's advancements towards their own nuclear bomb.
Fermi's 1942 Chicago experiment utilised the work of Rutherford and Einstein to create the first controlled nuclear chain reaction. His nuclear pile consisted of a graphite moderator, cadmium control rods and fissile Uranium-235 pellets. After firing a neutron into the reaction chamber, he used the control rods to absorb neutrons, and thus controlled the reaction. This demonstration was integral in developing the uncontrolled chain reaction used in the bomb.
n+ 235U >> 236U >> 139Ba + 94Kr + 3n
The bomb negatively impacted society, as it was used to systematically wipe out entire cities (Hiroshima and Nagasaki). These areas are still radioactive, and the descendants of the survivors suffer from radiation poisoning. A positive impact of the bomb was a peace based on Mutually Assured Destruction.
The statement is valid, as the exploration of nuclear physics authorised important fundamental discoveries which resulted in the development of the atomic bomb: an application of nuclear physics that had a major impact on society.
Post by: yattmoani on September 26, 2017, 05:19:31 pm
Post by: blasonduo on September 26, 2017, 07:24:47 pm
Hi, guys, just got a few quick questions here. Would be great to have some help :)
Hello! Let's see what I can do!
1) For this one, we can use the simple motor effect, palm downwards, fingers into the page, leaving current going towards the LEFT. HOWEVER!!!!! This current is being induced! So it must resist the change in flux (due to the law of conservation of energy) So, current travels to the RIGHT (to R). Now, electrons will always flow in the OPPOSITE direction to the flow of current, and therefore electrons will flow to P, and thus there will be a higher concentration at P. So the answer is A.
2) The question states that the INDEPENDENT (the one we change) variable is on the x-axis. In this experiment, they changed the HEIGHT, so this eliminates both C and D instantly. The next step is to identify which graph has correctly used a "line of best fit". The two graph's points are clearly beginning to plateau, but graph B is clearly ignoring this, so it must be A.
Also, we know that doubling the height of an object dropped does not double the time falling, so it can't be linear ;)
Post by: yattmoani on September 26, 2017, 07:28:21 pm
EDIT: Just saw blasonduo's reply. Thanks for the help! Really cleared things up for me
Post by: bsdfjnlkasn on September 26, 2017, 07:28:50 pm
Hello! Let's see what I can do!
1) For this one, we can use the simple motor effect, palm downwards, fingers into the page, leaving current going towards the LEFT. HOWEVER!!!!! This current is being induced! So it must resist the change in flux (due to the law of conservation of energy) So, current travels to the RIGHT (to R). Now, electrons will always flow in the OPPOSITE direction to the flow of current, and therefore electrons will flow to P, and thus there will be a higher concentration at P. So the answer is A.
2) The question states that the INDEPENDENT (the one we change) variable is on the x-axis. In this experiment, they changed the HEIGHT, so this eliminates both C and D instantly. The next step is to identify which graph has correctly used a "line of best fit". The two graph's points are clearly beginning to plateau, but graph B is clearly ignoring this, so it must be A.
Also, we know that doubling the height of an object dropped does not double the time falling, so it can't be linear ;)
Hey :)
With your answer to the first question, why are we considering an induced current? Isn't the constant velocity going to ensure a constant rate of change of magnetic flux?
Post by: blasonduo on September 26, 2017, 08:01:46 pm
Hey :)
With your answer to the first question, why are we considering an induced current? Isn't the constant velocity going to ensure a constant rate of change of magnetic flux?
Hey! This is a REALLY great question!
This is now linking to ideas to implementation, as for this question you need to understand how electrons experience forces in a magnetic field.
The formula for the FORCE on a charged particle is:
F = qvB.
Even if the velocity and magnetic field are constant, it just means that the force the electron experiences is also constant!
So, the electrons do move!
I hope this helps! :) If there are any more questions about this, (I know! This part is very tricky!) I'll be happy to help :)
Post by: jamonwindeyer on September 26, 2017, 11:53:37 pm
Hey :)
With your answer to the first question, why are we considering an induced current? Isn't the constant velocity going to ensure a constant rate of change of magnetic flux?
And just to respond further, if induced EMF is proportional to the rate of change of magnetic flux, then a "constant rate of change of magnetic flux" means a constant induced EMF, which is exactly what we get here ;D
So like, you and blasonduo are definitely saying the same thing. Your statement is totally correct - You might have just misinterpreted in your head :)
Post by: justwannawish on September 27, 2017, 07:08:41 am
1. Does centrifugal motion cause work to be done?
2. Does the atmosphere have mass? I know it's not exactly physics but just a curiosity
Post by: jamonwindeyer on September 27, 2017, 07:57:20 am
Hey I just have a few questions
1. Does centrifugal motion cause work to be done?
2. Does the atmosphere have mass? I know it's not exactly physics but just a curiosity
1. Centripetal (circular) motion requires acceleration, so yes, work is done!
2. Yep, ozone particles have mass! ;D
Post by: justwannawish on September 27, 2017, 12:05:54 pm
Post by: jamonwindeyer on September 27, 2017, 12:12:41 pm
So when we calculate the mass of the earth to be 6.0 x 10^24 does it include the mass of the atmosphere? Or is it too small to be included?
Seems the mass of the atmosphere is about \(5\times10^{18}\text{kg}\), give or take, so that would be inconsequential compared to the mass of the earth - Kind of like taking your earphones out before getting on the scales maybe ;)
Post by: justwannawish on September 27, 2017, 12:51:39 pm
Seems the mass of the atmosphere is about \(5\times10^{18}\text{kg}\), give or take, so that would be inconsequential compared to the mass of the earth - Kind of like taking your earphones out before getting on the scales maybe ;)
Yep, that makes sense. another question, sorry, I'm just trying to learn space before school does so it makes sense to me :)
Is the acceleration at the top of a point (maximum height) 0 or -9.8? Because what I'm currently understanding is that the vertical velocity be 0 as well, so would the acceleration would be 0 as well??
Btw, bought the atarnotes practice papers book and it's looking great :D
Post by: jamonwindeyer on September 27, 2017, 12:57:58 pm
Yep, that makes sense. another question, sorry, I'm just trying to learn space before school does so it makes sense to me :)
Is the acceleration at the top of a point (maximum height) 0 or -9.8? Because what I'm currently understanding is that the vertical velocity be 0 as well, so would the acceleration would be 0 as well??
Btw, bought the atarnotes practice papers book and it's looking great :D
All good! Good on you for getting ahead :)
The vertical acceleration is constant, \(-9.8\text{m/s}^2\) - Doesn't matter what point in the motion you are at, that is always the value :)
So glad you are liking the topic tests ;D
Post by: sidzeman on September 27, 2017, 12:58:32 pm
Yep, that makes sense. another question, sorry, I'm just trying to learn space before school does so it makes sense to me :)
Is the acceleration at the top of a point (maximum height) 0 or -9.8? Because what I'm currently understanding is that the vertical velocity be 0 as well, so would the acceleration would be 0 as well??
Btw, bought the atarnotes practice papers book and it's looking great :D
No, acceleration towards the centre of the Earth remains constant at 9.8 (or -9.8 ) during the entirety of the projectiles motion.
Post by: sidzeman on September 27, 2017, 12:59:30 pm
Post by: clarence.harre on September 27, 2017, 01:01:48 pm
Q11: What is the wavelength, in metres, of a photon with an energy of 3.5eV?
Working:
3.5 * 1.602*10^-19 = 5.607*10^-19 Joules
Since E = hc/λ,
then λ = hc/E
Therefore, λ = hc/(5.607*10^-19)
= 3.54 * 10^-7
This corresponds with option B, but the answer is C, i.e. λ = 1.18 * 10^-15
Post by: justwannawish on September 27, 2017, 01:10:28 pm
All good! Good on you for getting ahead :)
The vertical acceleration is constant, \(-9.8\text{m/s}^2\) - Doesn't matter what point in the motion you are at, that is always the value :)
So glad you are liking the topic tests ;D
No, acceleration towards the centre of the Earth remains constant at 9.8 (or -9.8 ) during the entirety of the projectiles motion.
Okay, thank you guys. Will probably be back with many more questions soon! But a huge thank you to the community here for answering them
Post by: itssona on September 27, 2017, 01:29:41 pm
the gravitational field vector g has an average value on the surface of earth, of 9.8Nkg^-1 or ms^-2. Show that the two alternative units quoted are equivalent
Post by: Shadowxo on September 27, 2017, 01:38:35 pm
hiii lil question from space module,, thank you :DSo 1N = 1kg.ms-2 (F=ma)
the gravitational field vector g has an average value on the surface of earth, of 9.8Nkg^-1 or ms^-2. Show that the two alternative units quoted are equivalent
Hence Nkg-1=kgms-2kg-1=ms-2
Post by: itssona on September 27, 2017, 01:45:13 pm
So 1N = 1kg.ms-2 (F=ma)thank you Shadow :)
Hence Nkg-1=kgms-2kg-1=ms-2
Post by: itssona on September 27, 2017, 02:01:48 pm
my textbook talked about how earth would actually needto rotate every 20 seconds in order for objects to fling off the surface.. but i dont get where theyre going at?
also is thrust of a rocket basically its initial force, so Net Force PLUS weight since it has to overcome gravity?
thank you so muchhh
Post by: pikachu975 on September 27, 2017, 02:07:56 pm
sorry for coming again buuut, having trouble understanding how earth's rotation affects g?
my textbook talked about how earth would actually needto rotate every 20 seconds in order for objects to fling off the surface.. but i dont get where theyre going at?
also is thrust of a rocket basically its initial force, so Net Force PLUS weight since it has to overcome gravity?
thank you so muchhh
Earth's rotation creates a centrifugal force that goes outwards, which reduces the EFFECTIVE VALUE of g.
Thrust = mv I think (can someone pls confirm) but using the formula, am = T - mg, T = F + mg so yes you're correct
Post by: itssona on September 27, 2017, 02:11:00 pm
Earth's rotation creates a centrifugal force that goes outwards, which reduces the EFFECTIVE VALUE of g.ohh makes sense now, thank youuu :D
Thrust = mv I think (can someone pls confirm) but using the formula, am = T - mg, T = F + mg so yes you're correct
Post by: jamonwindeyer on September 27, 2017, 02:58:09 pm
Question - do geostationary satellites not experience orbital decay?
Nope - Well, a negligible amount at least :)
Please help with this Multiple Choice question from the 2016 paper.
Q11: What is the wavelength, in metres, of a photon with an energy of 3.5eV?
Working:
3.5 * 1.602*10^-19 = 5.607*10^-19 Joules
Since E = hc/λ,
then λ = hc/E
Therefore, λ = hc/(5.607*10^-19)
= 3.54 * 10^-7
This corresponds with option B, but the answer is C, i.e. λ = 1.18 * 10^-15
I also get B when I calculate! Perhaps it's an error in the supplied solutions? :) (happy for someone to correct me!)
Post by: pikachu975 on September 27, 2017, 05:50:35 pm
Please help with this Multiple Choice question from the 2016 paper.
Q11: What is the wavelength, in metres, of a photon with an energy of 3.5eV?
Working:
3.5 * 1.602*10^-19 = 5.607*10^-19 Joules
Since E = hc/λ,
then λ = hc/E
Therefore, λ = hc/(5.607*10^-19)
= 3.54 * 10^-7
This corresponds with option B, but the answer is C, i.e. λ = 1.18 * 10^-15
HSC answers have B, maybe you looked at the wrong one!
Post by: itssona on September 27, 2017, 06:20:09 pm
Post by: austv99 on September 27, 2017, 11:42:50 pm
Thanks.
Post by: justwannawish on September 28, 2017, 07:12:37 pm
Also do astronauts in space actually experience gravity? My textbook says they are in freecall and thus experience acceleration due to gravity, but I don't get the correlation... sorry!
Post by: bsdfjnlkasn on September 28, 2017, 08:10:51 pm
I'm VERY confused about this question, what knowledge should I be applying here? ???
I know it's space, but that's about it!
Post by: blasonduo on September 28, 2017, 08:27:55 pm
Hey there!
I'm VERY confused about this question, what knowledge should I be applying here? ???
I know it's space, but that's about it!
Just for clarification, is it B?
If so, I'll be able to help! (if not oh no!)
Post by: mary123987 on September 28, 2017, 08:39:05 pm
Hey guys, what exactly is g-force and when do you start experiencing it?Hey in simple terms g- forces refers to a term used to express a person's apparent weight as a multiple of their true weight
Also do astronauts in space actually experience gravity? My textbook says they are in freecall and thus experience acceleration due to gravity, but I don't get the correlation... sorry!
that is g forces = apparent weight /normal true weight
now apparent weight = mg +ma
and normal true weight = 9.8m (as w=mg where g = 9.8 )
now when subbing that in the m cancels and you are left with g + a /9.8
to answer your question g forces are first experienced during lift off or more appropriately it is apparent at this time why ? well when your on the earth accroding to the formula you should still experience a g-force of 1 (9.8+a/9.8 =1 as you are not accelerating whilst on the earth) why you dont experience though is because at this point F= W ( as f=mg and w=mg). so appropriately when f>w or f<w it is apparent .
whilst in space astronauts do experience gravity to a very small extent depending on the distance you are away from the radius of the earth the further away the smaller the value of g experienced .
Now an important thing to note is when in free fall a sense of weightlessness is experienced this is because the rocket is orbiting aound the earth horizontally and thus gforces = T/9.8m(now considering it is in free fall T =0 as no fuel is being ejected and thus thrust =0) . If the rocket is falling its gravitational poential energy is inceasing and thus yes it is subject to downward acceleration due to gravity.
Hope this helps and makes sense if ur confused let me know
Post by: justwannawish on September 28, 2017, 08:49:50 pm
Hey in simple terms g- forces refers to a term used to express a person's apparent weight as a multiple of their true weight
that is g forces = apparent weight /normal true weight
now apparent weight = mg +ma
and normal true weight = 9.8m (as w=mg where g = 9.8 )
now when subbing that in the m cancels and you are left with g + a /9.8
to answer your question g forces are first experienced during lift off or more appropriately it is apparent at this time why ? well when your on the earth accroding to the formula you should still experience a g-force of 1 (9.8+a/9.8 =1 as you are not accelerating whilst on the earth) why you dont experience though is because at this point F= W ( as f=mg and w=mg). so appropriately when f>w or f<w it is apparent .
whilst in space astronauts do experience gravity to a very small extent depending on the distance you are away from the radius of the earth the further away the smaller the value of g experienced .
Now an important thing to note is when in free fall a sense of weightlessness is experienced this is because the rocket is orbiting aound the earth horizontally and thus gforces = T/9.8m(now considering it is in free fall T =0 as no fuel is being ejected and thus thrust =0) . If the rocket is falling its gravitational poential energy is inceasing and thus yes it is subject to downward acceleration due to gravity.
Hope this helps and makes sense if ur confused let me know
Yeah, it really did help :)
Just a follow up question, why does space not have gravity? Is it because it's a vacuum? Or is there gravity but it's very minute (if so, what type of gravity would it be)?
Post by: mary123987 on September 28, 2017, 09:05:58 pm
Yeah, it really did help :)That is ight there is most definitely gravity in space however it is so minute why?
Just a follow up question, why does space not have gravity? Is it because it's a vacuum? Or is there gravity but it's very minute (if so, what type of gravity would it be)?
well simply look at the law of universal gravitation where : F = Gm1m2/d
from this formula two things are very important
1) this gravity or attraction depends on distance
2) secind to that it depends on mass
for example consider the sun it is so huge (great mass) but consider how far it is , it is so far that this attreaction is almost 0 (you clearly dont see humans floating casually to the sun)
My point is gravity exists however it is incredibly minute , right now humans are attracted to anything : humans , particles anything but when compared to mass and distance attraction to the earth is greater .
hope that helps :)
Post by: justwannawish on September 28, 2017, 09:27:38 pm
That is ight there is most definitely gravity in space however it is so minute why?
well simply look at the law of universal gravitation where : F = Gm1m2/d
from this formula two things are very important
1) this gravity or attraction depends on distance
2) secind to that it depends on mass
for example consider the sun it is so huge (great mass) but consider how far it is , it is so far that this attreaction is almost 0 (you clearly dont see humans floating casually to the sun)
My point is gravity exists however it is incredibly minute , right now humans are attracted to anything : humans , particles anything but when compared to mass and distance attraction to the earth is greater .
hope that helps :)
Yep, that really did help :) thank you so much!!! It really helped me out
Post by: mary123987 on September 28, 2017, 09:44:40 pm
Yep, that really did help :) thank you so much!!! It really helped me outall good !!
Post by: jamonwindeyer on September 28, 2017, 10:36:00 pm
anyone know how galileo's analysis of projectile motion showed evidence to support his heliocentric model?
I don't think you'd be expected to answer this in a HSC exam, if it helps - And I've honestly got no clue how you'd link them! :)
Hey there!
I'm VERY confused about this question, what knowledge should I be applying here? ???
I know it's space, but that's about it!
So I'm pretty sure the logic is this. \(1.49\times10^{11}\) metres is the current distance and that gives a period of 1 year (we orbit the sun once a year, ish) and that gives a winter of 0.25 years. We need a winter of 10 years, meaning a 40x increase in the period, to 40 years.
By Kepler's Law of Periods, therefore:
Solving that should give B - So yep, you were on it blasonduo (increase the confidence my friend) ;D
Post by: raymatar on September 29, 2017, 10:19:45 am
Thanks
Post by: winstondarmawan on September 29, 2017, 10:31:08 am
Would appreciate help with this question, even after reading the solutions I am confused:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22070347_1853706338279156_1387821905_o.jpg?oh=0c3940feff466f0fe0c62a9317c8a023&oe=59CFE3C5
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22095316_1853706898279100_1567201710_o.jpg?oh=55cabfc2a79f0599cc9cc6b5aa09370d&oe=59CFF432
Post by: blasonduo on September 29, 2017, 10:51:59 am
Can someone explain this because I dont understand the solutions
Thanks
Hello! The question is asking how it is possible for a telescope to have the same orbital period than the Earth's.
From Kepler's laws of periods, it gives us a ratio as shown:
Given that this is a RATIO, when one variable changes, the other variable HAS to change to keep the equation true, which is shown in the question. We know that the telescope has a larger radius than Earth's so theoretically, it should have a LARGER period.
However, the question states that they both have the same orbital period, which contradicts Kepler's law, and this is because Kepler's law only applies to TWO objects, and doesn't take into account other forces.
For Earth, It has all of its force coming from the sun, HOWEVER, the telescope has forces from BOTH the sun AND the earth, meaning the telescope is experiencing more FORCE than Earth's. As force is equal to the centripetal force, it has a HIGHER orbital velocity.
Hope this helps ;)
Post by: itssona on September 29, 2017, 03:33:54 pm
<br>
so basically, thrust force remains constant in the rocket, and because mass is decreasing due to fuels burning, acceleration increases. (due to f=ma). And due to increasinh acceleration, astronauts experience an increasing force (why??). And this increasing force is a g force.<br>
<br>
this means we need to decrease thrust (according to textbooks and ppl) ... but i dont get how thrust even relates to force on the ASTRONAUT. Plus, they say that this is why we should decrease thrust but first they say that thrust is constant.<br>
<br>
so lost, <br>
sorry and thank you all so much for the continuous help :)
Post by: Zainbow on September 29, 2017, 06:05:30 pm
heey kinda confused about rockets :/ someone please check this?<br>
<br>
so basically, thrust force remains constant in the rocket, and because mass is decreasing due to fuels burning, acceleration increases. (due to f=ma). And due to increasinh acceleration, astronauts experience an increasing force (why??). And this increasing force is a g force.<br>
<br>
this means we need to decrease thrust (according to textbooks and ppl) ... but i dont get how thrust even relates to force on the ASTRONAUT. Plus, they say that this is why we should decrease thrust but first they say that thrust is constant.<br>
<br>
so lost, <br>
sorry and thank you all so much for the continuous help :)
Hey! I'll try my best with explaining this one
Ok so, yes, the thrust force remains constant (in the first stage of rocket launch anyway) and due to the law of conservation of momentum the acceleration increases as mass decreases, which is what you've said. Due to 'a' increasing however, it is G-forces that increase and not the thrust force. So for this :
Quote
astronauts are experiencing an increasing force (why??), they're not really. Keep in mind that G's are not determined by force but by acceleration (hence the formula G = 1 + a/9.8 ).
I'm not really sure what you mean by decreasing thrust (hopefully someone else can explain this better), but basically, when the rocket nears the end of its first stage of launch and the first rockets run out of fuel, the thrust force pretty much disappears as the rockets stop firing and detach themselves. Here, because there is a momentary absence of thrust force, the acceleration drops dramatically and astronauts thus feel a momentary lack of G-forces, or weightlessness. This is until the second set of rockets fire up.
Post by: austv99 on September 29, 2017, 06:10:31 pm
Assess the contributions made by Heisenberg and Pauli to the development of atomic theory.
Heisenberg and Pauli both made significant contributions to the development of atomic theory.
Heisenberg’s introduction of the ‘Uncertainty Principle’, where the position and momentum of a particle cannot be simultaneously measure to exact precision contributed to addressing a fundamental property of the quantum theory. This, therefore, led to Heisenberg applying this to atomic theory where he explained that the wave nature of electrons in stable orbits proposed by de Broglie’s standing wave represented the probabilities of an electron’s position. From this, he devised matrix mechanics to explain the quantum probabilities of the electron’s position (electron clouds), leading to an atomic theory entirely based on quantum physics rather than Bohr’s combination of classical and quantum physics.
Pauli’s introduction of the ‘Exclusion Principle’ where no two electrons can exist in the same quantum state developed the understanding of energy states that electrons can take within the atomic theory. He also introduced the 4th quantum number (magnetic spin) which was useful in addressing the existence of hyperfine spectral lines in the hydrogen spectrum which proved as a limitation to Bohr’s model of the atom. Further, Pauli also suggested the existence of the neutrino to account for the distribution of energies observed in beta decay. This was later experimentally confirmed. Thus, he contributed to our understanding by suggesting the existence of another subatomic particle in atomic theory.
Judgement: Heisenberg made significant contributions as his ‘Uncertainty Principle led to him developing de Broglie’s proposal of electron waves through the introduction of quantum probabilities (electron clouds) of its position, leading to an atomic theory entirely based upon quantum physics.
Pauli made significant contributions as his ‘Exclusion Principle’ explained the quantum states of electrons. He also resolved a limitation of the Bohr model and suggested the existence of the neutrino.
Post by: itssona on September 29, 2017, 06:28:14 pm
Hey! I'll try my best with explaining this oneohh i finally get it!! oml thank you a million asdfhjkll
Ok so, yes, the thrust force remains constant (in the first stage of rocket launch anyway) and due to the law of conservation of momentum the acceleration increases as mass decreases, which is what you've said. Due to 'a' increasing however, it is G-forces that increase and not the thrust force. So for this : , they're not really. Keep in mind that G's are not determined by force but by acceleration (hence the formula G = 1 + a/9.8 ).
I'm not really sure what you mean by decreasing thrust (hopefully someone else can explain this better), but basically, when the rocket nears the end of its first stage of launch and the first rockets run out of fuel, the thrust force pretty much disappears as the rockets stop firing and detach themselves. Here, because there is a momentary absence of thrust force, the acceleration drops dramatically and astronauts thus feel a momentary lack of G-forces, or weightlessness. This is until the second set of rockets fire up.
Post by: pikachu975 on September 29, 2017, 08:42:03 pm
Would appreciate if someone could check my response to this question. Just to make sure if what im saying is accurate and/or if im missing on some bits of info. TIA
Assess the contributions made by Heisenberg and Pauli to the development of atomic theory.
Heisenberg and Pauli both made significant contributions to the development of atomic theory.
Heisenberg’s introduction of the ‘Uncertainty Principle’, where the position and momentum of a particle cannot be simultaneously measure to exact precision contributed to addressing a fundamental property of the quantum theory. This, therefore, led to Heisenberg applying this to atomic theory where he explained that the wave nature of electrons in stable orbits proposed by de Broglie’s standing wave represented the probabilities of an electron’s position. From this, he devised matrix mechanics to explain the quantum probabilities of the electron’s position (electron clouds), leading to an atomic theory entirely based on quantum physics rather than Bohr’s combination of classical and quantum physics.
Pauli’s introduction of the ‘Exclusion Principle’ where no two electrons can exist in the same quantum state developed the understanding of energy states that electrons can take within the atomic theory. He also introduced the 4th quantum number (magnetic spin) which was useful in addressing the existence of hyperfine spectral lines in the hydrogen spectrum which proved as a limitation to Bohr’s model of the atom. Further, Pauli also suggested the existence of the neutrino to account for the distribution of energies observed in beta decay. This was later experimentally confirmed. Thus, he contributed to our understanding by suggesting the existence of another subatomic particle in atomic theory.
Judgement: Heisenberg made significant contributions as his ‘Uncertainty Principle led to him developing de Broglie’s proposal of electron waves through the introduction of quantum probabilities (electron clouds) of its position, leading to an atomic theory entirely based upon quantum physics.
Pauli made significant contributions as his ‘Exclusion Principle’ explained the quantum states of electrons. He also resolved a limitation of the Bohr model and suggested the existence of the neutrino.
It's pretty good but in the judgement I'd recommend making it more explicit that BOTH Heisenberg and Pauli's work led to the develop of the CURRENTLY USED electron cloud model, hence revealing the significant impact of their work as it still resonates today.
Post by: winstondarmawan on September 29, 2017, 08:54:55 pm
Quanta to Quarks:Bump
Would appreciate help with this question, even after reading the solutions I am confused:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22070347_1853706338279156_1387821905_o.jpg?oh=0c3940feff466f0fe0c62a9317c8a023&oe=59CFE3C5
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22095316_1853706898279100_1567201710_o.jpg?oh=55cabfc2a79f0599cc9cc6b5aa09370d&oe=59CFF432
Post by: austv99 on September 29, 2017, 10:41:37 pm
Would appreciate help with part ii for both pictures- the natural radioactivity one and the nuclear equation onebump :)
Thanks.
Post by: pikachu975 on September 29, 2017, 11:08:14 pm
bump :)
Natural Radioactivity one - Transmutations that involve natural radioactivity include alpha, beta minus, and beta plus decay. These all involve the transmutation of one element into another. This change in binding energy due to a different element with differing amount of nucleons means that the extra energy is converted to mass via the energy mass equivalence. For example in beta minus decay, a neutron changes to a proton and an electron and anti-neutrino are released.
Nuclear Equation one - Energy is released as there is a difference in the binding energies between the left and right hand side of the equation. This difference in binding energy means the extra energy is released in the nuclear reactor, e.g. as heat to be used by the coolant to do work.
Post by: itssona on September 30, 2017, 10:38:00 am
Post by: blasonduo on September 30, 2017, 11:26:10 am
heyy do we need to know advantage/disadvantage of solid fuel/liquid fuel enginses in rockets
Are you referring the Goddard and the named scientist dot point? If so, Identify why solid fuel was insuffiecient for space exploration, what Goddard was able to do with oxygen and hydrogen, and how this improved space exploration and why it was significant. (able to explore further)
Post by: itssona on September 30, 2017, 11:34:08 am
Are you referring the Goddard and the named scientist dot point? If so, Identify why solid fuel was insuffiecient for space exploration, what Goddard was able to do with oxygen and hydrogen, and how this improved space exploration and why it was significant. (able to explore further)yes thank you so much!!
Post by: sidzeman on September 30, 2017, 12:58:35 pm
I know that the application of a strong magnetic field causes protons to align with the direction of the mag field (parallel or anti parallel), and then later a radio pulse wave knocks them out of alignment.
However, where does the protons precessing in phase come in? Is that due to the strong magnetic field as well, or as a result of the radio pulse?
Also, how do the gradient coils ensure every voxel is unique?
Post by: bsdfjnlkasn on September 30, 2017, 02:52:47 pm
Could someone please explain to me why the maximum wavelength occurs when the energy of a photon = minimum energy required to eject an electron from a metal surface?
Post by: sidzeman on September 30, 2017, 04:03:21 pm
Hey there!Energy of a photon is inversely proportional to its wavelength (E = hc/lamda) - meaning as wavelenght goes up total energy goes down. Therefore it makes sense that max wavelength of a photon that can still cause emission is given by the min energy to eject the electron (work function) - with wavelength any higher, the energy of the photon would not be enough to eject the electron.
Could someone please explain to me why the maximum wavelength occurs when the energy of a photon = minimum energy required to eject an electron from a metal surface?
Post by: jamonwindeyer on October 01, 2017, 05:17:27 pm
Hey regarding MRI in Med Phys,
I know that the application of a strong magnetic field causes protons to align with the direction of the mag field (parallel or anti parallel), and then later a radio pulse wave knocks them out of alignment.
However, where does the protons precessing in phase come in? Is that due to the strong magnetic field as well, or as a result of the radio pulse?
Also, how do the gradient coils ensure every voxel is unique?
Hey! Basically, precession is just something nuclei do in a magnetic field. You don't really need to explain why. There is actually a formula for the precession frequency from the properties of the particle, and the applied magnetic field called the Larmor relationship:
But this isn't assessable - You just need to know what precession is, really ;D
As for unique voxels, picture every slice in the Z-direction assigned a number 1-255: We do this by varying the magnetic field strength at each point along the axis. Then, do the same for the X-Y directions. So then, each voxel is unique because there is a varying field strength in that direction. It becomes like a coordinate system, so (22,43,176) or (54,197,105) become specific points due to the differing fields in each direction ;D
Post by: Aaron12038488 on October 02, 2017, 01:53:36 pm
Thx :o ;D
Post by: winstondarmawan on October 02, 2017, 02:30:06 pm
11. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192872_1340184369440416_505939024_n.png?oh=951d4909cc99aaa481a3dab26b06e4c1&oe=59D3E216
Answer is B, but not sure how this makes sense. Since wavelength is inversely proportional to energy, don't all low wavelength photons have high energy?
12. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207574_1340181356107384_294692406_n.png?oh=c96794f9307d05aa00a60c4426ac77dc&oe=59D4C8C4
Answer is B, but I got A.
My method was:
F=qvBsin(theta)
F=ma
a=(qvBsin(theta))/m
Because from proton to alpha particle mass (electron mass is negligible) is doubling and so is charge, there would be no effect on the acceleration of the alpha particle. Can someone clarify this and explain why the answer is B?
21. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207213_1340181849440668_499896216_n.png?oh=b29ca888cccec63415bb04481f6a0762&oe=59D393E1
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22184809_1340181916107328_1024926263_n.png?oh=6b585ab564eb36b31536643abce2f1d0&oe=59D39E15
I don't remember learning the third point: Superconductors only transmit DC. Can someone please explain this?
27. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156703_1340182206107299_1451835380_n.png?oh=5f8f623e6e65c0da3a2a5b214d3b13d1&oe=59D3AD90
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192973_1340182382773948_1424324833_n.png?oh=fe5852336b67e0da077f9b2559c0efa5&oe=59D3DD8F
Is it okay if from 10-40 minutes my gradient was a little sloped down (due to orbital decay)?
Can someone also give me a mark indication for the following question:
Explain how the adoption of AC as the dominant electricity supply benefits society in terms of the advantages of AC over DC. (6)
In the 18th century, Westinghouse triumphed over Edison for electrical distribution in Washington D.C, campaigning for the benefits over AC electricity over DC electricity. This is indicative of AC electricity's dominance over DC, greatly benefiting society as we know it today.
- AC electricity allowed for long distance transmission, through the use of transformers. Due to the large power losses associated with DC, power stations were created closer to the city, resulting in urban clutter. Transformers operate via Faraday's Law, which states that a conductor experiencing a change in magnetic flux will have an induced EMF. This is only achievable through AC, where oscillating charges produce a continuous change in flux. Transformers allow for the stepping-up of voltage, in turn reducing current due to the Law of Conservation of Energy (V1I1=V2I2), thus minimising power loss (Ploss=I^2R) and benefiting society through more efficient energy. Furthermore, this allowed power stations to be built further apart from major cities, reducing urban clutter and exposure of society to pollution.
- The use of AC and transformers had allowed for the use of a wide array of electronic devices, whilst minimising excessive cabling. If DC was used, a different supply cable would be required for each output. However through AC, simple transformers and rectifiers can be used to allow access to electrical devices such as TVs, phone chargers - all of which increase quality of life and benefit society immensely.
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156994_1340194479439405_1611052904_n.png?oh=a50fb621978bb8f8b9290520cbb64c8f&oe=59D3CDE7
Part (ii).
Thanks in advance!
Post by: mary123987 on October 02, 2017, 05:56:11 pm
Would appreciate clarification with the following:Hey I am answering the first question so far :
11. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192872_1340184369440416_505939024_n.png?oh=951d4909cc99aaa481a3dab26b06e4c1&oe=59D3E216
Answer is B, but not sure how this makes sense. Since wavelength is inversely proportional to energy, don't all low wavelength photons have high energy?
12. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207574_1340181356107384_294692406_n.png?oh=c96794f9307d05aa00a60c4426ac77dc&oe=59D4C8C4
Answer is B, but I got A.
My method was:
F=qvBsin(theta)
F=ma
a=(qvBsin(theta))/m
Because from proton to alpha particle mass (electron mass is negligible) is doubling and so is charge, there would be no effect on the acceleration of the alpha particle. Can someone clarify this and explain why the answer is B?
21. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207213_1340181849440668_499896216_n.png?oh=b29ca888cccec63415bb04481f6a0762&oe=59D393E1
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22184809_1340181916107328_1024926263_n.png?oh=6b585ab564eb36b31536643abce2f1d0&oe=59D39E15
I don't remember learning the third point: Superconductors only transmit DC. Can someone please explain this?
27. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156703_1340182206107299_1451835380_n.png?oh=5f8f623e6e65c0da3a2a5b214d3b13d1&oe=59D3AD90
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192973_1340182382773948_1424324833_n.png?oh=fe5852336b67e0da077f9b2559c0efa5&oe=59D3DD8F
Is it okay if from 10-40 minutes my gradient was a little sloped down (due to orbital decay)?
Can someone also give me a mark indication for the following question:
Explain how the adoption of AC as the dominant electricity supply benefits society in terms of the advantages of AC over DC. (6)
In the 18th century, Westinghouse triumphed over Edison for electrical distribution in Washington D.C, campaigning for the benefits over AC electricity over DC electricity. This is indicative of AC electricity's dominance over DC, greatly benefiting society as we know it today.
- AC electricity allowed for long distance transmission, through the use of transformers. Due to the large power losses associated with DC, power stations were created closer to the city, resulting in urban clutter. Transformers operate via Faraday's Law, which states that a conductor experiencing a change in magnetic flux will have an induced EMF. This is only achievable through AC, where oscillating charges produce a continuous change in flux. Transformers allow for the stepping-up of voltage, in turn reducing current due to the Law of Conservation of Energy (V1I1=V2I2), thus minimising power loss (Ploss=I^2R) and benefiting society through more efficient energy. Furthermore, this allowed power stations to be built further apart from major cities, reducing urban clutter and exposure of society to pollution.
- The use of AC and transformers had allowed for the use of a wide array of electronic devices, whilst minimising excessive cabling. If DC was used, a different supply cable would be required for each output. However through AC, simple transformers and rectifiers can be used to allow access to electrical devices such as TVs, phone chargers - all of which increase quality of life and benefit society immensely.
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156994_1340194479439405_1611052904_n.png?oh=a50fb621978bb8f8b9290520cbb64c8f&oe=59D3CDE7
Part (ii).
Thanks in advance!
Well to answer your question intensity determines the number of photons I.e high intensity corresponds with high levels of photons and vice versa)
the question tells you it has a lower intensity and hence a lower number of photons from looking at this it is b but to further justify this :
E=hf
and c=fλ
and f=c/λ
by equating these two formulas we get E =h(c/λ)
if it has a very short wavelength as menitioned in the question you must consider the equation E =h(c/λ)
so think about h is a constant and c is a constant so the only thing determining energy is λ so the wavelength is small sub in a value say 10 you get :
1.9878 x 10^-26
ok now sub in a high value say 10000 you get :
1,9878 x 10^-29
when you compare these the relationship is evident E is highe as λ decreases thus the answer must be B hope this makes sense let me know if you are lost
Post by: justwannawish on October 02, 2017, 06:00:50 pm
Post by: winstondarmawan on October 02, 2017, 06:16:37 pm
Hey I am answering the first question so far :
Well to answer your question intensity determines the number of photons I.e high intensity corresponds with high levels of photons and vice versa)
the question tells you it has a lower intensity and hence a lower number of photons from looking at this it is b but to further justify this :
E=hf
and c=fλ
and f=c/λ
by equating these two formulas we get E =h(c/λ)
if it has a very short wavelength as menitioned in the question you must consider the equation E =h(c/λ)
so think about h is a constant and c is a constant so the only thing determining energy is λ so the wavelength is small sub in a value say 10 you get :
1.9878 x 10^-26
ok now sub in a high value say 10000 you get :
1,9878 x 10^-29
when you compare these the relationship is evident E is highe as λ decreases thus the answer must be B hope this makes sense let me know if you are lost
Ohhh okay thank you I get it. I misinterpreted the answer and thought it meant that photons can have different energies given the same wavelength. Thanks :)
Post by: Shadowxo on October 02, 2017, 06:49:13 pm
12. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207574_1340181356107384_294692406_n.png?oh=c96794f9307d05aa00a60c4426ac77dc&oe=59D4C8C4I'll just answer a couple of your questions :)
Answer is B, but I got A.
My method was:
F=qvBsin(theta)
F=ma
a=(qvBsin(theta))/m
Because from proton to alpha particle mass (electron mass is negligible) is doubling and so is charge, there would be no effect on the acceleration of the alpha particle. Can someone clarify this and explain why the answer is B?
27. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156703_1340182206107299_1451835380_n.png?oh=5f8f623e6e65c0da3a2a5b214d3b13d1&oe=59D3AD90
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192973_1340182382773948_1424324833_n.png?oh=fe5852336b67e0da077f9b2559c0efa5&oe=59D3DD8F
Is it okay if from 10-40 minutes my gradient was a little sloped down (due to orbital decay)?
12. (Q17) The thing you missed is that the alpha particle has 4x the mass of the proton (it has 2 protons and 2 neutrons of similar mass, but the proton just has 1 proton)
27. I believe orbital decay would be negligible - it's only in the air for 40m (orbiting for 30m) so your graph should be horizontal.
Post by: jamonwindeyer on October 02, 2017, 07:47:51 pm
my school will likely select the pendulum experiment as our practical exam. I was wondering if someone could elaborate on their practical exam and how they went about it. E.g. Was a method provided and which you carried out? Or did u have to design and carry out your own.
Thx :o ;D
At my school we had equipment, some background info and an aim provided and had to devise the method ourselves :)
Could I please have a quick introduction to what to expect for medical physics?
You'll learn, for various medical imaging techniques (ultrasound, endoscopy, X-rays, MRI, etc):
- The principles behind its operation
- How these are applied to obtain an image of the body
- The strengths and limitations of these images
- What sorts of ailments each technique is best at detecting and for what reason
:)
Post by: jamonwindeyer on October 02, 2017, 08:00:51 pm
21. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207213_1340181849440668_499896216_n.png?oh=b29ca888cccec63415bb04481f6a0762&oe=59D393E1
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22184809_1340181916107328_1024926263_n.png?oh=6b585ab564eb36b31536643abce2f1d0&oe=59D39E15
I don't remember learning the third point: Superconductors only transmit DC. Can someone please explain this?
Yep, so superconductors can't really transmit AC effectively (this is massively simplifying and is perhaps a tad incorrect, but it's what we say at the HSC level). It's because cooper pairs can only form and travel without resistance if they are moving in a single direction. It doesn't work if they are zipping back and forth :)
Quote
Can someone also give me a mark indication for the following question:
I don't think you are quite at full marks yet, you spend a lot of time explaining the principles behind transformers, which isn't necessary here. Use that time to cover other advantages of AC, extrapolate your ideas further (allowing long distance transmission reduces urban pollution, allowing a larger range of electronic devices to be used has allowed increased technological innovation, increased efficiency makes energy more accessible, etc) :)
Post by: Sukakadonkadonk on October 02, 2017, 09:08:47 pm
Why would the line of best fit look like that? ie, what are the steps in determining the line of best fit for a graph again? I'm always getting conflicting answers.
Thanks.
Post by: justwannawish on October 02, 2017, 11:38:52 pm
Hi, could someone help me with part a) ?
Why would the line of best fit look like that? ie, what are the steps in determining the line of best fit for a graph again? I'm always getting conflicting answers.
Thanks.
The tips are:
-The line doesn't have to go through any particular data points or the origin
- should have equal number of data points on both sides except for obvious outliers, which I think are meant to be removed
Not sure if there is anything else besides always draw with a ruler
Post by: Aaron12038488 on October 03, 2017, 11:34:11 am
for 1.6.2 why is the answer a +4.61x10^9J instead of a -4.61x10^9J.
Also for simple questions such as F=mg and W=mg do i always give a direction?
Post by: winstondarmawan on October 03, 2017, 02:42:38 pm
Can someone please check my answer for this question and give a mark indication. :)
A photon is incident on a hydrogen atom in the ground state.
Explain, using de Broglie's hypotehsis, why the photon is not absorbed by the hydrogen atom.
Answer: de Broglie's hypothesis stated that anything with momentum had a wavelength, also known as the wave-particle duality (i.e. all waves exhibited particle behaviour and vice versa). Bohr's second postulate stated that electrons must absorb/emit a quantised amount of of energy when performing quantum jumps up/down respectively in electron shells. Considering this in light of the de Broglie hypothesis, it is evident that the photon of energy was not an integral multiple of the ground state electron wavelength, and as such, was not absorbed.
Thanks in advance!
Post by: Sukakadonkadonk on October 03, 2017, 05:04:02 pm
The tips are:
-The line doesn't have to go through any particular data points or the origin
- should have equal number of data points on both sides except for obvious outliers, which I think are meant to be removed
Not sure if there is anything else besides always draw with a ruler
Hey, thanks for the answer. But I saw this as the suggested solution. Would it be accurate? It doesn't really have equal points on both sides.
*Edit: Sorry, needs to be rotated right for image to be right side up. :)
Post by: hinakamishiro on October 03, 2017, 05:16:26 pm
Post by: sidzeman on October 03, 2017, 06:12:51 pm
Hey guys could I have some help with these questions? I just don't understand the mathematical working out. Thanks! :)Not 100% sure if correct but,
For the motorcycle question, there are 2 forces you need to consider. For moving in a circular path, the first force you should immediately think about is centripetal - mv^2 / r. The other force is Weight (mg)
Now just equate the 2 forces (mv^2 / r = mg), and solve for v. You should get an answer that rounds to 5.9
2nd Q:
Normally when you see 2 orbits and orbital periods you think of equating keplars law - however that is only when they are equating the same object.
For this, we have to stick with the usual speed = distance/time. Rearrange to get T = D/S, then sub in the relevant equations. Distance is the circumference of the planets which is given by 2piR. Velocity is simply the orbital velocity. Rearrange that into a nicer form, and that will give you the orbital period for the Earth orbiting object. Now just sub in whats different about the object orbiting Xerus - you'll see its only the mass, and you'll get your answer of half the period.
Post by: blasonduo on October 03, 2017, 06:25:25 pm
Hey guys could I have some help with these questions? I just don't understand the mathematical working out. Thanks! :)
Hey!
For the first one, sidzeman is 100% correct (if you need any clarification, I'll be happy to help)
2) For this question, we know the radius is the same while the planets mass changes, and we need to know how fast it is going.
Orbital velocity is the formula which pops to mind here.
As we know Xerus is 4 times the weight of earth, we will assume earth's mass = 1, and Xerus's mass = 4
For Earth:
As the other terms are constant for both equations, we can just ignore them :)
For Xerus;
From this, we can tell that the satellite around Xerus is going TWICE as fast than the satellite around Earth.
and as we know, If something is going twice as fast, it will take HALF the time. (IE its period is half)
So from this, it is B :)
EDIT: when posting, I was not alerted of sidzeman's edit :P Well done to you sir!
Post by: justwannawish on October 03, 2017, 07:30:42 pm
At my school we had equipment, some background info and an aim provided and had to devise the method ourselves :)
You'll learn, for various medical imaging techniques (ultrasound, endoscopy, X-rays, MRI, etc):
- The principles behind its operation
- How these are applied to obtain an image of the body
- The strengths and limitations of these images
- What sorts of ailments each technique is best at detecting and for what reason
:)
Ooh 😀 I've heard that it was an unpopular option and was wondering why but it sounds good to me
Also could someone explain how the slingshot effect works? How do the satellites/rockets use the planet to gain velocity? I don't get it!!
Post by: arunasva on October 03, 2017, 08:00:10 pm
Ooh 😀 I've heard that it was an unpopular option and was wondering why but it sounds good to me
Also could someone explain how the slingshot effect works? How do the satellites/rockets use the planet to gain velocity? I don't get it!!
Imagine you are flying close to a tree towards well Dwayne Johnson who is running around the tree, he holds your hand, spins you around and tosses you away. You're gonna fly faster relative to the tree cos you have his speed around the tree added to your speed. Similarly a spacecraft (you) comes and latches onto the gravitational field of a planet(Dwayne) by making an elastic collision. The planet takes the spaceship around and its orbital velocity relative to the sun (the tree) is gained by the spacecraft and the planet slows down a lil'. However the speed relative to the planet is still the same, as you gained the planet's velocity relative to the sun and your velocity relative to the planet didn't slow down.
Post by: jamonwindeyer on October 03, 2017, 08:01:30 pm
Also could someone explain how the slingshot effect works? How do the satellites/rockets use the planet to gain velocity? I don't get it!!
^^ Love the explanation above!! I think of it like riding a bike and grabbing onto a moving car - The car drags you along and increases your speed without their own speed being affected in any significant way. Except for the slingshot effect, it is a planet pulling a probe with a gravitational field - The principle is the same, but of course there is the orbital aspect which is explained better above ;D
I wrote some nifty summaries of all the course content if it helps as you are working through, they are about halfway down in this list :)
Post by: justwannawish on October 03, 2017, 10:44:41 pm
I'm going to remember that analogy now haha! Thank you so much.
Imagine you are flying close to a tree towards well Dwayne Johnson who is running around the tree, he holds your hand, spins you around and tosses you away. You're gonna fly faster relative to the tree cos you have his speed around the tree added to your speed. Similarly a spacecraft (you) comes and latches onto the gravitational field of a planet(Dwayne) by making an elastic collision. The planet takes the spaceship around and its orbital velocity relative to the sun (the tree) is gained by the spacecraft and the planet slows down a lil'. However the speed relative to the planet is still the same, as you gained the planet's velocity relative to the sun and your velocity relative to the planet didn't slow down.
^^ Love the explanation above!! I think of it like riding a bike and grabbing onto a moving car - The car drags you along and increases your speed without their own speed being affected in any significant way. Except for the slingshot effect, it is a planet pulling a probe with a gravitational field - The principle is the same, but of course there is the orbital aspect which is explained better above ;D
I wrote some nifty summaries of all the course content if it helps as you are working through, they are about halfway down in this list :)
Just checked it out (the maths post was so amazing :)) and rockets are finally starting to make more sense now :D
Post by: Mymy409 on October 05, 2017, 08:24:28 am
Post by: mary123987 on October 05, 2017, 09:38:45 am
When an electron is passing through a magnetic field, a centripetal force acts on it, right? What other forces? What needs to be equated?hey to explain this simply If A partlicle with charge q is moving with velocity v perpendicularly to a magnetic field of strength B, the particle will experience a ,magnetic force given by F=qvBsinθ .Now from my understanding you are not required to explain where this formula is from or even how to derive it(its currently beyond our understanding) . To answer your question with respect to the force acting on it , it really depends on the circumstance as it will not always be experiencing centripetal force however if it is orbiting something like a nucleus (which to be honest i dont think ive ever seen a past hsc paper ask) then yeah you can say the force keeping it in orbit is centripetal force .However in general terms the force experienced is from the interaction with the magnetic field . Also centripetal force generally applies in the circumstance above and if this force is perpendicular to the direction of motion (I.e. it is a centripetal force) and will cause the particle to undergo uniform circular motion (if no other forces are in play).
if not perpendicular and the particle enters at an oblique angle, it will follow a helical path. hope that makes sense let me know if your'e confused
Post by: Dragomistress on October 05, 2017, 01:47:59 pm
Post by: hinakamishiro on October 05, 2017, 05:57:57 pm
Post by: sidzeman on October 05, 2017, 06:12:15 pm
Hey guys! Can I please have some please with these two questions? Any explained working out would be really helpful. Thanks! :)For question 19 - we know that length contracts for objects travelling at near light speeds. The key thing to recognise here is "when observed from the trains frame of reference" - therefore we know that the length of the train will be contracting. If it was from the trains frame of reference, the platform would appear to be contracting. Therefore we know lo (200) and lv (160 which it contracts to be the same length as the platform), and the rest is just simple subbing into the formula.
Post by: blasonduo on October 05, 2017, 06:23:47 pm
Hey guys! Can I please have some please with these two questions? Any explained working out would be really helpful. Thanks! :)
Sidzeman is right! a simple substitution! If you need further help, I'll be happy to!
20) This will be a bit tricky to explain, but it uses triangles!!
It is launched horizontally, so the initial VERTICAL acceleration is 0. We also know the Horizontal velocity is constant.
Knowing that after 2 seconds, the angle is 45 degrees, we know that this triangle in ISOSCELES, which means that the horizontal AND vertical components have to be the same!
As we know gravitational acceleration is a constant 9.8 PER SECOND, after TWO seconds, the vertical velocity of the ball is:
9.8 X 2 = 19.6
And as we know the horizontal and vertical velocities HAVE to be equal, it means the horizontal velocity is ALSO 19.6.
The find the resultant velocity, we must find the hypotenuse so
(19.6)^2 + (19.6)^2 = v^2
v = 27.72
Which is D!!
Post by: Shadowxo on October 05, 2017, 07:15:07 pm
I would like to ask, why is the change in potential energy 52J since the work done on the object is translated to the potential energy gained during this time. So if I was to move it parallel to the ground, it will also gain potential energy?It's because it's on a slope, and the surface is frictionless. The work done on it results in it gaining g.p.e. and k.e. with the k.e. being converted into g.p.e. as it continues travelling up the slope. So the final g.p.e. = work done. If it was travelling parallel to the ground, the g.p.e. would never change and all the work done would be converted into k.e. instead.
Post by: justwannawish on October 05, 2017, 10:41:07 pm
The rest length of a train is 200 m and the rest length of a railway platform is 160 m. The train rushes past the platform so fast that, when observed in the platform’s frame of reference, the train and the platform are the same length.
Post by: pikachu975 on October 06, 2017, 12:18:32 am
How do I tackle question 19 from the 2014 paper?
The rest length of a train is 200 m and the rest length of a railway platform is 160 m. The train rushes past the platform so fast that, when observed in the platform’s frame of reference, the train and the platform are the same length.
so Lv = 160m as the person on the platform sees the train's length contract from 200 to 160. Hence sub in Lv = 160 and Lo = 200
Post by: Bubbly_bluey on October 06, 2017, 10:57:57 am
Just a quick question about "other consequences of Special Relativity": Does this mean time,length and mass dilations? What does it mean by some other consequences?
Thank you!:D
Post by: felix12345 on October 06, 2017, 06:07:25 pm
Thanks
Post by: pikachu975 on October 06, 2017, 08:22:37 pm
Hey, I was wondering if anyone could provide an explanation to this question - answer is b.
Thanks
p = mv so since velocity is constantly slowed down by friction F, then since p is proportional to m then p will decrease linearly therefore B
Post by: jamonwindeyer on October 06, 2017, 09:05:26 pm
Hey guys!
Just a quick question about "other consequences of Special Relativity": Does this mean time,length and mass dilations? What does it mean by some other consequences?
Thank you!:D
Definitely just means time dilation, length contraction and mass dilation - As well as the relativity of simultaneity! ;D
Post by: f_tan on October 06, 2017, 10:50:18 pm
(https://i.imgur.com/5YfEWjl.png)
Thank you!
Post by: sidzeman on October 07, 2017, 04:11:36 pm
For the transformer question - the MG calculate power loss with the equation of P = I^2 R - I attempted it by calculating the power of the primary coil (VpxIP) and the power of the secondary coil (VsIs), and then subtracting the 2. Why doesn't this approach work as well?
Thanks in advance!
Post by: winstondarmawan on October 07, 2017, 05:30:16 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22290522_1344113975714122_1473090637_n.png?oh=5b467e24114901787583256d3160a4fa&oe=59DB1CEA
As the motor becomes faster, the back EMF reduces the current being fed into the motor and as torque is directly proportional to current, the answer is D.
However, I'm just wondering what happened to the traditional sin/cos curve that is used for torque.
Thanks in advance.
Post by: Iminschool on October 07, 2017, 05:36:11 pm
Question 12 - Shouldn't the force on the parallel field be a curved graph, F = qvbsintheta - and theta is slowly rising to 90?The C.C.C is perpendicular to the magnetic field lines at all times and is thus at a maximum at all points in the journey (reverses every half cycle). Thus the answer should be C.
For the transformer question - the MG calculate power loss with the equation of P = I^2 R - I attempted it by calculating the power of the primary coil (VpxIP) and the power of the secondary coil (VsIs), and then subtracting the 2. Why doesn't this approach work as well?
Thanks in advance!
Post by: Iminschool on October 07, 2017, 05:44:48 pm
Hi, can anyone help me with this question? I'm not sure what else to write, other than to check that information is consistent across a range of sources, but the question is a 3 marker:I'd probably add that MRIs produce functional images where certain abnormalities may be detected ehich woukd otherwise not be viewable through structural images (such as those produced in X-rays)
(https://i.imgur.com/5YfEWjl.png)
Thank you!
Post by: armtistic on October 07, 2017, 06:06:48 pm
Just want to check if my understanding of this question is correct:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22290522_1344113975714122_1473090637_n.png?oh=5b467e24114901787583256d3160a4fa&oe=59DB1CEA
As the motor becomes faster, the back EMF reduces the current being fed into the motor and as torque is directly proportional to current, the answer is D.
However, I'm just wondering what happened to the traditional sin/cos curve that is used for torque.
Thanks in advance.
nothing's happened to the sine curve normally used, because that curve represents "Torque vs Time", in this case we have "torque vs motor speed" thats all
Post by: Maraos on October 07, 2017, 07:03:19 pm
I would just like to make sure that my understanding of the wilson cloud chamber track paths is correct. My teacher didn't actually teach us this in class so I've had to research it.
Alpha particles: Leave a thick straight track
Beta Particles: Leave a thin line that curves at the end
Gamma rays: Leave short swiggly lines
I used this image as a reference: http://1.bp.blogspot.com/_FAgA6-CHYr8/SoggUi2sWYI/AAAAAAAAAKA/316jS1V70s0/s400/track+patten.bmp
Is this correct?
Thanks! :)
** forgot to add this is for quanta to quarks**
Post by: blasonduo on October 07, 2017, 07:24:48 pm
Hi, can anyone help me with this question? I'm not sure what else to write, other than to check that information is consistent across a range of sources, but the question is a 3 marker:
(https://i.imgur.com/5YfEWjl.png)
Thank you!
Your point is 100% correct, I would add checking the quality of the information, If it is from reputable sources such a .gov sites or peer-reviewed scientific journals, you'd know the information is accurate and reliable!
Also, a point would be to know when the article was published (ie sources from 1999 might not be as reliable as sources from 2017)
Hope this helps :)
Post by: sidzeman on October 07, 2017, 09:30:02 pm
Thanks!
Post by: blasonduo on October 07, 2017, 10:01:29 pm
Question 12 - Shouldn't the force on the parallel field be a curved graph, F = qvbsintheta - and theta is slowly rising to 90?
For the transformer question - the MG calculate power loss with the equation of P = I^2 R - I attempted it by calculating the power of the primary coil (VpxIP) and the power of the secondary coil (VsIs), and then subtracting the 2. Why doesn't this approach work as well?
Thanks in advance!
Hello!
For the transformer, Its because you aren't calculating power, you are showing a RATIO.
Which is odd, because, when you minus one from the other you should get 0. (when I calculated this, I got that)
Just remember that you are showing that one formula equals another;
VpIp = VsIS
If I have not made this clear, just remember equals sign means exactly the same! So it is the equivalent of
24 = 24
and if you minus one from the other, you will always end with 0 ;)
Post by: bsdfjnlkasn on October 07, 2017, 10:22:14 pm
Hey can someone explain stopping voltage to me and its relevant formula (max KE = qv) or something
Thanks!
Hey I just learnt this :D
Stopping voltage (Vs) - voltage required to stop current.
Remember when you were investigating the photoelectric effect at school (or if not, you can still read the following explanation and hopefully it'll make sense) you had a set up like this?
(http://physicsopenlab.org/wp-content/uploads/2015/12/setupFotoelettrico.png)
So the cathode here is the stopping electrode. We are able to make the cathode more and more negative, by increasing the voltage supplied by the external supply. This then begins to repel the photoelectrons (because like charges repel) and eventually, the flow of charge will be stopped. Hence the name stopping electrode. When it is high enough, the voltage will be such that electrons will be able to just reach the cathode but not flow into the external circuit. When this happens, the voltage being applied is known as the stopping voltage i.e. V = Vs.
Now onto making sense of: Ek(max) = qVs
Recall that
Ek(max) refers to the kinetic energy of the fastest moving photoelectron. The electrons which are furthest away from the nucleus are held the weakest and so require the least amount of energy to be released when a photon strikes the metal surface. When an incoming photon of light (carrying E = hf) has sufficiently high frequency, and so energy, these outermost electrons will be emitted with the greatest kinetic energy.
Spoiler
For further clarity and so, to continue: This is because the photon's energy is larger than the energy required to overcome the electrostatic attraction acting between the electron and nucleus. Because not all of the photon's energy is absorbed in the process of liberating the electron (it only needs enough to overcome the attractive force) it still has some energy left over which is then converted into kinetic energy. The outermost electrons will have the greatest amount of energy left after interacting with the incoming photon of light because the attractive force is smallest and so the energy required to overcome this force is the smallest.
At the stopping voltage, Vs, only the fastest moving electrons are reaching the (stopping) anode (note, they're just reaching, not flowing into external circuit so current reading is 0), because all the slower moving ones don't have enough Ek to overcome the large potential difference (i.e. they are repelled). These emitted electrons (photoelectrons) have the highest kinetic energy as:
Really hoped that helped convey a more conceptual explanation - it really puts the formulas into context and saves you the time of memorising the formula (plus it'll have more meaning to you too)
Let me know if you have any more questions :D
If you need me to, i'm happy to also explain the formula:
Post by: bsdfjnlkasn on October 08, 2017, 10:42:53 am
I'm still a bit confused about how Bohr's model of the atom helped explained the hydrogen emission spectra so would like some clarification with close reference to his 3 postulates.
I'm struggling with piecing the 3rd postulate into it all (without de Broglie's wavelength equation)
Post by: hobocop on October 08, 2017, 10:45:15 am
Post by: pikachu975 on October 08, 2017, 11:01:04 am
Hi, I am having trouble understanding this question. Could I get some help?
Using right hand grip rule the magnetic field is out of the page in the loop. Since the current is decreasing constantly then the magnetic field strength is decreasing, hence a magnetic field will be induced out of the page due to the loop to balance it. Hence the current will be anticlockwise using right hand grip rule.
Hey there,
I'm still a bit confused about how Bohr's model of the atom helped explained the hydrogen emission spectra so would like some clarification with close reference to his 3 postulates.
I'm struggling with piecing the 3rd postulate into it all (without de Broglie's wavelength equation)
1st postulate was that electrons existed in stationary states and possessed an unexplained stability. This means that the specific orbits of electrons meant that there is a line spectra rather than a continuous spectra which would occur if the electrons could orbit at any radii.
2nd postulate is that the transition between stationary states is accompanied by the emission or absorption of EMR. This basically explains how the spectral lines are produced as hydrogen is excited by heat/current causing emission of EMR which shows in the spectrum.
3rd postulate is the quantisation of angular momentum in multiples of h/2pi. This also shows how orbits are at fixed radii, explaining the line spectra rather than a continuous one.
Post by: winstondarmawan on October 08, 2017, 12:32:42 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22368857_1345010415624478_564241496_o.png?oh=30c594a949bd76361f09b328f04f86f6&oe=59DBD1CE
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22345179_1345046995620820_683957314_o.png?oh=3813959ef0fe3f95977c336eef6c3d7d&oe=59DBC189
Quanta:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22330778_1345047115620808_1229680097_n.png?oh=9812596fc2e4c9286ffd492c6a0eec6b&oe=59DB8039
This is from the sample answers, not sure how Bohr used LCM and LCE for his postulates. Can someone please explain?
Post by: blasonduo on October 08, 2017, 02:19:21 pm
Would appreciate help with the following, TIA.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22368857_1345010415624478_564241496_o.png?oh=30c594a949bd76361f09b328f04f86f6&oe=59DBD1CE
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22345179_1345046995620820_683957314_o.png?oh=3813959ef0fe3f95977c336eef6c3d7d&oe=59DBC189
Quanta:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22330778_1345047115620808_1229680097_n.png?oh=9812596fc2e4c9286ffd492c6a0eec6b&oe=59DB8039
This is from the sample answers, not sure how Bohr used LCM and LCE for his postulates. Can someone please explain?
Hello! I can help you with the first one, but I don't do quanta to quarks :(
For the first one, the huge hint here is that they are both in geostationary orbits. As tower B is with along the equator, it follows the motion of earth (which geostationary satellites do!) So due to this orbital velocity, when the mass is dropped at tower B, It will not fall but however orbit around Earth.
Tower A is slightly different, It is in a geostationary height, but it is currently experiencing no orbital velocity, so when It drops, It will fall. As we have to explain the motion, we need to include that its velocity will continuously fall faster, However, its gravitational acceleration will not be 9.8 when dropped but as it gets closer to the earth's surface, will approach an acceleration of 9.8.
This is all I can think of, and I'm pretty sure this will give full marks :)
Post by: bsdfjnlkasn on October 08, 2017, 04:23:18 pm
Would appreciate help with the following, TIA.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22368857_1345010415624478_564241496_o.png?oh=30c594a949bd76361f09b328f04f86f6&oe=59DBD1CE
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22345179_1345046995620820_683957314_o.png?oh=3813959ef0fe3f95977c336eef6c3d7d&oe=59DBC189
Quanta:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22330778_1345047115620808_1229680097_n.png?oh=9812596fc2e4c9286ffd492c6a0eec6b&oe=59DB8039
This is from the sample answers, not sure how Bohr used LCM and LCE for his postulates. Can someone please explain?
Hey there,
I do Q2Q so hopefully can help you out :)
So Bohr's second postulate is that the energy between the electron's quantised orbits is equal to E = hf. As the energy of the orbits are exact (quantised) they require exact amounts of energy to move from one shell to another. This amount of energy is quantified by E = hf. Bohr was able to account for the hydrogen emission spectra because specific frequencies of light are emitted when electrons move from one shell to another. And all the energy is conserved when it is absorbed from an incoming source and converted into kinetic energy (so velocity and light) hence, LCE.
LCM also comes into it when we consider how incoming energy is completely absorbed by the electron if it is the amount it requires to move between electron shells (all or nothing principle). This interaction is considered an elastic collision in which the energy's (e.g. a photon) momentum is transferred without any losses. This energy then allows it to move to a higher electron shell, but because the mass and velocity were conserved, we satisfy the LCM.
Hopefully that helped :)
Post by: bsdfjnlkasn on October 08, 2017, 05:45:03 pm
Although the rest were incorrect, shouldn't D (which is the answer) say that all of the external magnetic field is excluded? Instead of some?
Thank you! It could be that my understanding still needs to be refined...
Post by: blasonduo on October 08, 2017, 06:04:49 pm
Hey guys!
Although the rest were incorrect, shouldn't D (which is the answer) say that all of the external magnetic field is excluded? Instead of some?
Thank you! It could be that my understanding still needs to be refined...
Hey! It's a good thing the rest were clearly wrong, as I probably would've gotten this wrong! :P
I'm pretty sure it says some, as not all magnetic field lines could be influenced (see picture)
So assuming the magnet makes a large enough area, some of the further out magnetic field lines might not be influenced?
As I said before, If this question had an answer that said "all" then the question becomes silly. I'm pretty sure they were just testing for the "exclusion" part :)
Post by: austv99 on October 09, 2017, 10:18:26 am
TIA
Post by: blasonduo on October 09, 2017, 10:33:41 am
Hey, would appreciate help with this question
TIA
Hey, I LOVED this question!
Now to start with this question, look where the wires are going, they act as electromagnets as well as being the actual coil for the motor!
From the diagram alone, we can work out the poles of the electromagnets by following the direction of the current. With the one on the right, we work out using North (anti-clockwise) or South (clockwise) and we can clearly see that the closer side is a SOUTH pole.
Using the same method, we can figure out the pole on the left magnet, and it will show that the closer side is NORTH. So the magnetic field is going left to right. Using the right-hand palm rule, we can figure out that the motor will move clockwise So A and C are wrong.
Here comes the fun part! :D
Now this is AC, so the current will swap direction, so take the diagram and swap the + with a - and vice versa, and DO THE EXACT SAME THING. You will notice that the poles have swapped! Where the magnetic field is now going RIGHT to LEFT. Follow the current and use the right-hand palm rule, and you'll see that it is........... still going clockwise!
So the answer will be B.
This is really cool concept and I love the set-up :D
Post by: imda.beast on October 09, 2017, 11:11:31 am
Post by: blasonduo on October 09, 2017, 11:16:04 am
hi, is physics getting hsc revision videos. i hope it can be uploaded soon because im struggling with some of the ideas to implementation concepts
Yes it is! It might still be a while, but if there is something you don't quite understand, just post it here!
EDIT: Check here for details :) https://atarnotes.com/forum/index.php?topic=173138.0
EDIT2: THEY ARE OUT NOW! CHECK EM OUT
Post by: seventeenboi on October 09, 2017, 11:26:26 am
"In your study of Q2Q you have performed an investigation to observe radiation emitted from a nucleus using a Wilson Cloud Chamber or a similar detection device. Describe how you carried out your investigation."
What is the most preferable and common answer to this question ????????
thanks :^)
Post by: mary123987 on October 09, 2017, 10:35:02 pm
thanks a lot
Mary :)
Post by: bsdfjnlkasn on October 09, 2017, 10:53:37 pm
HEY guys just had a quick question ; do you need to know the pauli exclusion principle for ideas to implementation ? or is it just for quanta to quarks ( and if you are wondering my option is medical physics so no i dont need to know it for my otpion)
thanks a lot
Mary :)
Hey there!
The first exposure I had to Pauli's exclusion principle was in Quanta to Quarks so I think it is safe to say, that no you don't need to know it :D
Post by: bsdfjnlkasn on October 10, 2017, 08:14:48 am
The answer listed for Q7 is B and I was 99% sure that the answer was D when I was answering..
Also for Q16, I just wanted some clarification. When is the angle that they've given you a distractor? Is it only when the field is into/out of the page. And when, in this situation, do they give you the wrong angle? What does that look like?
Would love some help on these, thank you :D
Post by: sidzeman on October 10, 2017, 12:13:10 pm
Could someone also explain the other questions to me please
Post by: Savas_P on October 10, 2017, 01:00:49 pm
For this question, does GPE being halved mean the satellite is brought closer to Earth or further away?. I thought it meant closer, since GPE rises as you get further away, but apparently this is not the case. Is it do the with the fact that GPE gets closer to 0 when moved away? I always get confused when they say GPE has increased as I don't know if it means its gotten closer to 0 and further away
Could someone also explain the other questions to me please
1. The GPE is halved use this equation E=-G(m1*m2)/R to see that the radius is doubled, then plug into the force equation F=Gm1m2/r^2 to see that the force will become a quarter of the initial force.
2. Use the right hand grip rule with your thumb pointing down to Y to see the direction of the force.
3. The force is the calculated from the angle between the wire and the magnetic field, you can see that WX is perpendicular to the poles of magnets. The torque is read off the information because it is the angle of the plane and the magnetic field.
Post by: sidzeman on October 10, 2017, 01:25:00 pm
1. The GPE is halved use this equation E=-G(m1*m2)/R to see that the radius is doubled, then plug into the force equation F=Gm1m2/r^2 to see that the force will become a quarter of the initial force.So for 2 the rotation is clockwise right?
2. Use the right hand grip rule with your thumb pointing down to Y to see the direction of the force.
3. The force is the calculated from the angle between the wire and the magnetic field, you can see that WX is perpendicular to the poles of magnets. The torque is read off the information because it is the angle of the plane and the magnetic field.
I understand the angle between the wire and the mag field for force - but what do you mean by "the plane" for torque?
Post by: Savas_P on October 10, 2017, 01:29:46 pm
Hey guys!
The answer listed for Q7 is B and I was 99% sure that the answer was D when I was answering..
Also for Q16, I just wanted some clarification. When is the angle that they've given you a distractor? Is it only when the field is into/out of the page. And when, in this situation, do they give you the wrong angle? What does that look like?
Would love some help on these, thank you :D
1. The experiment had a null result, so the answer should definitely be D lol.
2. Think you need to look at whether the electron is perpendicular to the field, in your example the electron is entering at 30° which is what you put in F=qvBsinθ and in the equation where you put in 90° is when the magnetic field lines are going into or out of the page.
Post by: Savas_P on October 10, 2017, 01:42:04 pm
So for 2 the rotation is clockwise right?
I understand the angle between the wire and the mag field for force - but what do you mean by "the plane" for torque?
yep clockwise, you put your thumb down and then your fingers curl clockwise.
sorry, I meant the torque depends on the angle of plane WXYZ and the magnetic field. It will change as the coil rotates but in the picture they have marked it at 30°. for the force imagine XY as a conductor and it will always be 90° because it will always be perpendicular.
Post by: sidzeman on October 10, 2017, 03:08:19 pm
yep clockwise, you put your thumb down and then your fingers curl clockwise.Hmmmm sorry this is the one part of physics I always struggle with. So the force individual sections of the coil will always be a max because it will always be perpendicular really?
sorry, I meant the torque depends on the angle of plane WXYZ and the magnetic field. It will change as the coil rotates but in the picture they have marked it at 30°. for the force imagine XY as a conductor and it will always be 90° because it will always be perpendicular.
Post by: Savas_P on October 10, 2017, 03:43:36 pm
Hmmmm sorry this is the one part of physics I always struggle with. So the force individual sections of the coil will always be a max because it will always be perpendicular really?
hahaha I am confused as well, hopefully someone can confirm. would have thought that is right, then the force of the rotation produces the toque. the torque considers the area of the plane WX*XY so how much flux lines the shape cuts would change the torque as it rotates.
Post by: SesquipedalianIndividual on October 10, 2017, 08:23:02 pm
I have a lot of trouble with understanding torque and back emf and i dont want it to hold me back
can i get another attempt at an explanation please
Post by: Savas_P on October 10, 2017, 09:41:55 pm
first time posting so im not entirely sure how its supposed to go but anyways
I have a lot of trouble with understanding torque and back emf and i dont want it to hold me back
can i get another attempt at an explanation please
I am so glad I am not the only person that struggled with this, I also have no idea how to graph the emf, sidzeman I only just realized that the force is a flat line because it is always perpendicular. I still don't understand how to tell when is the emf/force on the graph positive or negative if someone could explain this please.
Post by: bsdfjnlkasn on October 10, 2017, 10:35:40 pm
Was hoping to get some explanation on these two multiple choice questions :)
Thank you!!
Post by: bsdfjnlkasn on October 10, 2017, 10:49:43 pm
first time posting so im not entirely sure how its supposed to go but anyways
I have a lot of trouble with understanding torque and back emf and i dont want it to hold me back
can i get another attempt at an explanation please
Hey there!
Welcome to the forums :D What in particular do you not understand about torque and back EMF?
I can give you some quick definitions, and if there are some conceptual things you want me to expand on, let me know.
Torque: Turning force which acts a certain distance from the axis of rotation. Has two formulas (both of which are listed on the data sheet)
Back EMF: Induced EMF which seeks to oppose the change that caused it as a consequence of the law of conservation of energy (otherwise known as Lenz's Law)
Post by: bsdfjnlkasn on October 10, 2017, 10:56:27 pm
I am so glad I am not the only person that struggled with this, I also have no idea how to graph the emf, sidzeman I only just realized that the force is a flat line because it is always perpendicular. I still don't understand how to tell when is the emf/force on the graph positive or negative if someone could explain this please.
Hey there :)
In order to graph EMF, you will need to graph the change in magnetic flux over time. In exam questions, this is usually given to you. A quick trick (if you do 2 unit maths or higher) would be to just sketch the derivative because the gradient of the flux/time graph is our definition for induced EMF!
Force is defined as positive when it is acting upwards, and yes the lines are always straight because the force acts in a constant direction. This is why we need a split ring commutator to ensure that the current's direction changes allowing a constant force to act down, hence allowing the coil to rotate in a consistent direction.
Post by: Savas_P on October 11, 2017, 11:11:33 am
Hey there :)
In order to graph EMF, you will need to graph the change in magnetic flux over time. In exam questions, this is usually given to you. A quick trick (if you do 2 unit maths or higher) would be to just sketch the derivative because the gradient of the flux/time graph is our definition for induced EMF!
Force is defined as positive when it is acting upwards, and yes the lines are always straight because the force acts in a constant direction. This is why we need a split ring commutator to ensure that the current's direction changes allowing a constant force to act down, hence allowing the coil to rotate in a consistent direction.
wow thanks!! you are legend. that's a good trick, now i just need to practice my derivatives.
Post by: winstondarmawan on October 11, 2017, 12:33:22 pm
13. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22414729_1347448845380635_323509929_n.jpg?oh=134c4ab9f911a5324356ef6198cfc21c&oe=59DFE736
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22386637_1347448898713963_942589714_n.jpg?oh=2ea0b0e92ae1cdc8e1af5165ca2a0c3c&oe=59DFB08E
14. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22447103_1347448998713953_512491506_n.jpg?oh=356849cd6400b6bee39669ea6fc0fb11&oe=59E0A675
And also, can someone please tell me that this question is impossible (if not how do I do it):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22471704_1347448518714001_1038035986_n.jpg?oh=18c9c49a3b603443015d3a57eddae686&oe=59DFA088
Thanks in advance!
Post by: jamonwindeyer on October 11, 2017, 06:44:34 pm
Hey there,
Was hoping to get some explanation on these two multiple choice questions :)
Thank you!!
For the first one, the disc spinning in the magnetic field will induce a current in the disc. The discs movement is in the one direction, so it is a direct current - So the question is, what direction does it flow? If you consider the right hand slap rule, the disc is moving into the page, and the field is directed to the right - This results in a current flow directed downwards, through Y and to X! So I'm pretty sure the answer is B ;D
For your second, it is definitely the wire that will move - The solution moving makes no sense and the magnet definitely isn't going anywhere either. Use the right hand slap rule for the force on a current carrying conductor - Field is sort of directed perpendicular to the wire, to the upper left corner. Current goes down and to the left. Force will be into the page, which would cause the wire to rotate ;D
Hard to stuff to explain in text, but I hope this helps - You might need to just try the rules a bit and work through it a few times ;D
Post by: 3unitgang on October 12, 2017, 03:06:31 am
Post by: jamonwindeyer on October 12, 2017, 10:54:47 am
Hi, I have just recently bought the physics atar notes textbook and it had the question, "A projectile has a range of 1240 metres and a flight time of 5 seconds. At what velocity is it projected?" but when finding the initial velocity in terms of y, is used the formula r=u+1/2at^2 isn't it r=ut+1/2at^2?
Hey! Great catch, that looks like an error! That working should be:
So your final answer should in fact be the answer given, divided by 5 -> 24.5 metres per second :) sorry for the error! First one I know of in those tests, I'll make sure it is fixed in the future ;D
Post by: Zainbow on October 12, 2017, 02:54:17 pm
Post by: Savas_P on October 12, 2017, 04:56:23 pm
Hey! Great catch, that looks like an error! That working should be:
So your final answer should in fact be the answer given, divided by 5 -> 24.5 metres per second :) sorry for the error! First one I know of in those tests, I'll make sure it is fixed in the future ;D
as pointed out on another forum, i am not sure if this is the final answer
delta x = Ux * t
1240 = Ux *(5)
Ux = 248 m/s
delta y = Uy*t - 1/2 *gt^2
0 = 5Uy - 1/2 * 9.8*25
Uy=24.5 m/s as jamon has said
but U = sqrt (Uy^2 + Ux^2)
which means the FINAL answer is U = 249.2072.. m/s @ 5°37'
hope this is right
Post by: johnk21 on October 12, 2017, 05:48:13 pm
For extended responses, if you opt to answer in dot points how exactly would you do that? How can I make sure it's clear enough and that there is flow to my answer? I've always answered in full sentences, and although I still prefer to do that in a few cases, there are some questions where dot points would be easier and favourable but I don't know the best method to approach that kind of response.To be honest, its completely fine to always write in dotpoint form; however its NOT okay to do what they call an 'info dump'.
Here is an example how dotpoints could be used in say the BCS theory:
Conditions for BCS Theory to work:
- Must be a Type I superconductor
- Temperature must be below critical temperature where lattice vibrations are at a minimum
Once these set conditions are achieved, the BCS Theory is accountable for superconductivity.
- Firstly an electron causes the positive lattice to be distorted -> creates a net positive region
- Another electron is attracted to this region
- A phonon is released due to the lattice distortion, and is absorbed by the electron to overcome electrostatic force of other electron
- Both electrons form a cooper pair, allowing for zero resistance as it is unimpeded
Note: Not saying dotpoints are desirable in this specific question (i personally wouldnt), but dotpoints are good for even 6 and 7 markers such as quantum theory ones were you want to briefly talk about planck and einstein, and show their relation as well as providing a judgement.
Hope i helped :)
Post by: winstondarmawan on October 12, 2017, 06:02:13 pm
Would appreciate help with the following MCs:Bump
13. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22414729_1347448845380635_323509929_n.jpg?oh=134c4ab9f911a5324356ef6198cfc21c&oe=59DFE736
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22386637_1347448898713963_942589714_n.jpg?oh=2ea0b0e92ae1cdc8e1af5165ca2a0c3c&oe=59DFB08E
14. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22447103_1347448998713953_512491506_n.jpg?oh=356849cd6400b6bee39669ea6fc0fb11&oe=59E0A675
And also, can someone please tell me that this question is impossible (if not how do I do it):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22471704_1347448518714001_1038035986_n.jpg?oh=18c9c49a3b603443015d3a57eddae686&oe=59DFA088
Thanks in advance!
Post by: jamonwindeyer on October 12, 2017, 07:20:35 pm
as pointed out on another forum, i am not sure if this is the final answer
delta x = Ux * t
1240 = Ux *(5)
Ux = 248 m/s
delta y = Uy*t - 1/2 *gt^2
0 = 5Uy - 1/2 * 9.8*25
Uy=24.5 m/s as jamon has said
but U = sqrt (Uy^2 + Ux^2)
which means the FINAL answer is U = 249.2072.. m/s @ 5°37'
hope this is right
Yeah you've got it - The working he spotted the error in was an extract from the full solution in the Notes he referenced, but thanks for popping the full thing up!
Post by: Iminschool on October 12, 2017, 09:47:40 pm
Post by: jamonwindeyer on October 12, 2017, 10:02:16 pm
Bump
Q14: Consider the direction of holes first! They are moving into the page, the magnetic field is to the right, the resultant force is downwards.
Next, electrons. They are moving out of the page (opposite direction to current flow), magnetic field is to the right, the resultant force (negative charged particle) is also downwards by the right hand slap rule. So, they move in the same direction!
You might need to fiddle with the right hand slap rule (or similar) to make sense of this - Let me know if I can clarify!
For that other question, 33 right? Definitely possible, it looks like the method shown in your photo is correct! Energy of emitted photoelectron will be the energy per photon minus the work function, in other words:
Is this the question you meant? Any part of the solution in the image particularly giving you trouble? :)
Post by: winstondarmawan on October 12, 2017, 10:14:07 pm
Q13 Winston
Q14: Consider the direction of holes first! They are moving into the page, the magnetic field is to the right, the resultant force is downwards.
Next, electrons. They are moving out of the page (opposite direction to current flow), magnetic field is to the right, the resultant force (negative charged particle) is also downwards by the right hand slap rule. So, they move in the same direction!
You might need to fiddle with the right hand slap rule (or similar) to make sense of this - Let me know if I can clarify!
For that other question, 33 right? Definitely possible, it looks like the method shown in your photo is correct! Energy of emitted photoelectron will be the energy per photon minus the work function, in other words:
Is this the question you meant? Any part of the solution in the image particularly giving you trouble? :)
Thank you guys!
For 33, the resulting KE is negative, so I thought it was an error. Is it possible for KE to be negative?
Post by: pikachu975 on October 12, 2017, 10:16:03 pm
Thank you guys!
For 33, the resulting KE is negative, so I thought it was an error. Is it possible for KE to be negative?
No since KE = 1/2 mv^2 and m is positive and v^2 is positive
Post by: jamonwindeyer on October 12, 2017, 10:33:21 pm
Thank you guys!
For 33, the resulting KE is negative, so I thought it was an error. Is it possible for KE to be negative?
No since KE = 1/2 mv^2 and m is positive and v^2 is positive
Ahhh didn't catch that, well if the answer ends up negative what that means is that you in fact don't have enough energy to free any electrons (indeed, can't have negative energy, it just means that nothing happens). So, the maximum kinetic energy is 0J - Trick question! :)
Post by: bsdfjnlkasn on October 12, 2017, 11:54:10 pm
Q14: Consider the direction of holes first! They are moving into the page, the magnetic field is to the right, the resultant force is downwards.
Next, electrons. They are moving out of the page (opposite direction to current flow), magnetic field is to the right, the resultant force (negative charged particle) is also downwards by the right hand slap rule. So, they move in the same direction!
You might need to fiddle with the right hand slap rule (or similar) to make sense of this - Let me know if I can clarify!
For that other question, 33 right? Definitely possible, it looks like the method shown in your photo is correct! Energy of emitted photoelectron will be the energy per photon minus the work function, in other words:
Is this the question you meant? Any part of the solution in the image particularly giving you trouble? :)
Hey there!
I was just wondering if for that 2 marker, you could find the energy of the photoelectrons emitted because you're given the wavelength? Then you can just plug it into E = hf? Then by comparing what the value of the emitted electrons are and what the minimum required is (Ek min), then you can deduce that because the calculated value is less than the minimum, that no photoelectrons will be emitted?
Post by: jamonwindeyer on October 13, 2017, 10:31:49 am
Hey there!
I was just wondering if for that 2 marker, you could find the energy of the photoelectrons emitted because you're given the wavelength? Then you can just plug it into E = hf? Then by comparing what the value of the emitted electrons are and what the minimum required is (Ek min), then you can deduce that because the calculated value is less than the minimum, that no photoelectrons will be emitted?
Yeah absolutely, that would be the approach! I suppose it is just that a HSC exam would never throw you a trick question like that, but still very easy to deduce that you won't get any photoelectrons ;D
Post by: itssona on October 15, 2017, 10:36:11 pm
in the conclusion of the prac report, should we include how we can minimise errors and improve our method?
also im confused,
we always learnt that reliability is associated with repetition, so how come all answers to questions about reliability, associate with accuracy? why do they all overlap :/
thank you :)
Post by: arunasva on October 16, 2017, 06:01:46 pm
hiii question about the pendulum prac (and all prac tho);
in the conclusion of the prac report, should we include how we can minimise errors and improve our method?
also im confused,
we always learnt that reliability is associated with repetition, so how come all answers to questions about reliability, associate with accuracy? why do they all overlap :/
thank you :)
Yes, you should mention how you avoided error which includes taking precise mearuerments and from the centre each time (accuracy). Ensure only the variable you are testing gets changed WHICH IS LENGTH OF THE STRING and angle of displacement on each side is constant. (validity) and doing more repetitions to see if the result is reproducible (reliability). Reliability and accuracy both are two separate things that need to be maintained in a dataset hence they are asked together but separately. Its like find the area of a triangle and find the length of all its sides. two separate things. :)
Post by: itssona on October 16, 2017, 07:03:33 pm
Omg thank you!!!
Yes, you should mention how you avoided error which includes taking precise mearuerments and from the centre each time (accuracy). Ensure only the variable you are testing gets changed WHICH IS LENGTH OF THE STRING and angle of displacement on each side is constant. (validity) and doing more repetitions to see if the result is reproducible (reliability). Reliability and accuracy both are two separate things that need to be maintained in a dataset hence they are asked together but separately. Its like find the area of a triangle and find the length of all its sides. two separate things. :)
Post by: bsdfjnlkasn on October 17, 2017, 06:03:58 pm
I got this question wrong and I really don't know how to reason the answer - let me know if you can help :)
Post by: jamonwindeyer on October 17, 2017, 06:10:33 pm
Hey there,
I got this question wrong and I really don't know how to reason the answer - let me know if you can help :)
Hey! Go to Kepler's Law of Periods:
By moving to the moon instead of the earth, \(M\) decreases \(r\) stays the same, so therefore, the satellite must have a longer orbital period to maintain the equality.
So compared to the satellite orbiting earth, the satellite orbiting the moon has a longer orbital period. Answer should be D ;D
Post by: shaner on October 17, 2017, 07:04:34 pm
1. A space shuttle of mass 7.5x10^4 kg is in orbit around the Moon at a height of 100 km. Calculate the gravitational potential energy of the space shuttle.
I've used the mgh formula however it's given me a pretty large value and I'm not sure whether it's right.
2. At the end of its life, our Sun will lose half its mass and shrink to the size of the Earth. Calculate the gravitational potential energy possessed by a body of mass 1.0kg on its surface.
How would I tackle a question like this?
Post by: pikachu975 on October 17, 2017, 10:05:35 pm
hiii question about the pendulum prac (and all prac tho);
in the conclusion of the prac report, should we include how we can minimise errors and improve our method?
also im confused,
we always learnt that reliability is associated with repetition, so how come all answers to questions about reliability, associate with accuracy? why do they all overlap :/
thank you :)
Nah conclusion ONLY answers the aim. The reducing error thing is in discussion
Post by: jamonwindeyer on October 17, 2017, 10:34:52 pm
I've got some Gravitational Potential Energy questions.
1. A space shuttle of mass 7.5x10^4 kg is in orbit around the Moon at a height of 100 km. Calculate the gravitational potential energy of the space shuttle.
I've used the mgh formula however it's given me a pretty large value and I'm not sure whether it's right.
You'd maybe want to use the other formula:
Or, if you are using \(E=mgh\), then make sure you use the value of \(g\) for the moon, either by calculation or provision. The two values would be different but both probably quite large ;D
Quote
2. At the end of its life, our Sun will lose half its mass and shrink to the size of the Earth. Calculate the gravitational potential energy possessed by a body of mass 1.0kg on its surface.
How would I tackle a question like this?
Again, use the longer formula for gravitational potential energy:
We will take the suns mass and halve it to get \(M\), \(m=1kg\), and \(d\) would be the radius of the earth! You can calculate from there ;D
Post by: julia9102 on October 18, 2017, 09:01:56 am
How did Rutherford deduce that the electrons orbited the nucleus of an atom?
Post by: bsdfjnlkasn on October 18, 2017, 09:21:02 am
Hello :) I just have a question on quanta,
How did Rutherford deduce that the electrons orbited the nucleus of an atom?
Hey there!
He didn't! That's an assumption he made which then resulted in a lot of confusion, he literally just assumed that they took on a planetary-like orbit. And as we know, there were serious complications because of this assumption (i.e. leading the nucleus to collapse if you were to follow his reasoning). Happy to explain that last bit if you want me to? :)
Post by: uusunny on October 18, 2017, 10:10:07 am
But the graph of momentum of the rocket vs time shows that momentum of the rocket increases? I am not really sure how these two ideas align. :/
Post by: julia9102 on October 18, 2017, 11:22:49 am
Hey there!
He didn't! That's an assumption he made which then resulted in a lot of confusion, he literally just assumed that they took on a planetary-like orbit. And as we know, there were serious complications because of this assumption (i.e. leading the nucleus to collapse if you were to follow his reasoning). Happy to explain that last bit if you want me to? :)
Yes please! Thank you :)
Post by: bsdfjnlkasn on October 18, 2017, 12:44:52 pm
Yes please! Thank you :)
Sure thing :)
So we know from Maxwell's theory of electromagnetic radiation, that an accelerating charge will emit EMR.
In Rutherford's model, the charges are orbiting around the nucleus and as the direction of their motion is constantly changing, they are considered accelerating (remembering that velocity is a vector quantity which accounts for speed AND direction of motion).
Since the electrons (charges) are accelerating, they would be emitting EMR. As a result, they would be losing energy (because it's being radiated out). This energy loss will slow their motion (remembering the equation for kinetic energy which is proportional to the velocity). This means they will slowly spiral into the nucleus, causing the atom to collapse which we know doesn't happen (just look at the world around us :P)
In summary :) :
Electrons orbit nucleus --> electrons change direction --> electrons said to be accelerating --> electrons emit EMR --> electrons lose energy --> electrons move at a slower speed --> altitude drops even more --> electron slowly spirals into nucleus --> atom collapses --> but wait, we're still here...
Let me know if you want anything clarified!
Post by: bsdfjnlkasn on October 18, 2017, 12:52:28 pm
I have some questions :)
I'm confused with the first one, because I thought apparent weight = mg + ma. Would this be a correct explanation?
If we're moving in the same direction as the gravitational field, there is no net force acting on the body since it's not doing work against the field to move upwards?
It seems that i've just tried to join 3 different concepts in that explanation and I don't really understand..
And for Q19, I thought it was d?
Post by: bsdfjnlkasn on October 18, 2017, 12:54:45 pm
I thought the answer for Q16 was B? Especially because of the way the coils are orientated?
Post by: jamonwindeyer on October 18, 2017, 01:34:13 pm
I agree with the explanation that the total momentum of the gas and rocket is conserved. Since the momentum of the gas is constant, momentum of the rocket at any time should be constant.
But the graph of momentum of the rocket vs time shows that momentum of the rocket increases? I am not really sure how these two ideas align. :/
Hey! I'm guessing that graph doesn't assume that the thrust is constant, which would mean the momentum of the rocket does change. It's important to understand that the way we look at it in HSC Physics is a huge simplification.
So I suppose the answer is, the graph and the explanation don't match. The explanation is an idealisation, whereas the graph probably isn't, if that makes sense? :)
Post by: jamonwindeyer on October 18, 2017, 01:47:15 pm
Hey there!
I have some questions :)
I'm confused with the first one, because I thought apparent weight = mg + ma. Would this be a correct explanation?
If we're moving in the same direction as the gravitational field, there is no net force acting on the body since it's not doing work against the field to move upwards?
It seems that i've just tried to join 3 different concepts in that explanation and I don't really understand..
And for Q19, I thought it was d?
So for the first one, go back to intuition. I'm assuming the answer is B? If that's right, it is because when you move the bag downwards, in that motion you could consider it sort of 'weightless,' like in free-fall. Moving it upwards quickly, you'd feel the bag pull back down on you - The apparent weight has increased. Moving it downwards, the apparent weight has decreased... I think?
For Q19, you can't explain the levitation with eddy currents because the magnet was resting on the superconductor to begin with. Therefore, no change in flux, therefore, no induced emf, therefore, no eddy currents. So every explanation with eddy currents is immediately incorrect (and a good time to remind, the Meisner Effect cannot be fully explained using eddy currents.
That leaves C - You are looking for the words 'exclusion' or 'repulsion' of the magnetic field, because this is what characterises the Meisner Effect (and the word diamagnetic is another one) ;D
I wasn't able to attach any more photos so sorry for spamming!
I thought the answer for Q16 was B? Especially because of the way the coils are orientated?
The current carrying coils act as electromagnet that induce a field going up and down the page (consider each end of the coil as a north/south pole). Use your right hand grip rule and you'll see the induced force on the moving electron due to the magnetic field is actually horizontal ;D
Post by: uusunny on October 18, 2017, 01:57:18 pm
Hey! I'm guessing that graph doesn't assume that the thrust is constant, which would mean the momentum of the rocket does change. It's important to understand that the way we look at it in HSC Physics is a huge simplification.
So I suppose the answer is, the graph and the explanation don't match. The explanation is an idealisation, whereas the graph probably isn't, if that makes sense? :)
Thank you for your reply! So in realistic situations, momentum isn't entirely conserved?
How would you answer this multiple choice??
The rocket travelling out of earth's atmosphere turns its engines off over a period of 5 seconds, and then immediately ejects the first stage fuel tanks. How does the momentum of the rocket change during this procedure?
a) the momentum of the rocket remains constant
b) the momentum of the rocket remains constant then drops
c) the momentum of the rocket decreases
d) the momentum of the rocket increases and drops
Post by: jamonwindeyer on October 18, 2017, 02:02:11 pm
Thank you for your reply! So in realistic situations, momentum isn't entirely conserved?
Exactly, I mean it is conserved but it won't appear to be, you'll lose to rotational motion and other stuff ;D
Quote
How would you answer this multiple choice??
The rocket travelling out of earth's atmosphere turns its engines off over a period of 5 seconds, and then immediately ejects the first stage fuel tanks. How does the momentum of the rocket change during this procedure?
a) the momentum of the rocket remains constant
b) the momentum of the rocket remains constant then drops
c) the momentum of the rocket decreases
d) the momentum of the rocket increases and drops
I believe the answer would probably be B - If we ignore friction, the momentum of the rocket is constant while the engines are off. But when it releases the first stage tanks, it has lost mass, so the momentum should drop! :)
Post by: austv99 on October 18, 2017, 02:26:43 pm
2002 HSC
23bi
21a
10
Post by: owidjaja on October 18, 2017, 04:34:39 pm
So today my teacher was trying to explain gravitational potential energy moving away from Earth and why it reaches zero (or something like that) but I still don't understand it. Can someone please explain it to me?
Thanks :)
Post by: KiNSKi01 on October 18, 2017, 04:44:19 pm
Hey guys,
So today my teacher was trying to explain gravitational potential energy moving away from Earth and why it reaches zero
Thanks :)
Lol another question to add to this. If gravitational field strength follows an inverse square law,how could GPE technically reach zero?
Post by: winstondarmawan on October 19, 2017, 01:19:14 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22664053_1354425641349622_941418257_o.png?oh=e98f61955b94845d8bd99f4edd3b454a&oe=59EA1936
Also, how did Chadwick know that the neutron was neutral simply by LCE and LCM?
TIA.
Post by: fireives1967 on October 19, 2017, 07:40:49 pm
My physics teacher made an excuse to avoid solving it ;D
Post by: jamonwindeyer on October 19, 2017, 07:48:47 pm
Would appreciate help with these questions, thanks.
2002 HSC
23bi
21a
10
For 23, End X will be negative. Use the right hand slap rule - With your fingers pointed to the right and your thumb upwards to match the info given, you are slapping into the page. So, positive charges move into the page and negative charges move out of the page - X is negative ;D
For 21, we need acceleration such that the capsule goes from rest to the given speed, \(v\), over the given distance, \(r\). Substitute into the formula from your reference sheet:
For 10, the induced emf will peak when the rate of change of magnetic flux is largest. This happens when the coil is aligned horizontally, parallel to the field. So we need to start at a peak - It must be B or D. We complete one revolution so we should get one induced AC waveform: Answer is B ;D
Post by: dannimoussa on October 19, 2017, 07:51:38 pm
How would you solve this?
My physics teacher made an excuse to avoid solving it ;D
I think I got it but do you answers so I can confirm
Post by: jamonwindeyer on October 19, 2017, 07:53:34 pm
Hey guys,
So today my teacher was trying to explain gravitational potential energy moving away from Earth and why it reaches zero (or something like that) but I still don't understand it. Can someone please explain it to me?
Thanks :)
Hey! This guide could be worth a read, but it is basically a matter of convention.
Think of it this way - We know that an object in a gravitational field has GPE. So, the only way to have ZERO GPE, would be to NOT BE in a gravitational field. However, gravitational fields are infinite in size, so you can never actually leave one. So we use a trick - If you go infinitely far away, that will get you out of the field and thus set your GPE to zero.
We want GPE to increase as we move further away from the centre of a field, so that necessitates GPE taking a negative value, so it can increase to zero as defined ;D
Lol another question to add to this. If gravitational field strength follows an inverse square law,how could GPE technically reach zero?
GPE and the force experienced due to a gravitational field (the strength of the field) are two different, but related, quantities. However, neither can actually be zero. Again, it's a mathematical trick to set GPE=0 at an infinite distance (and this would imply \(F_g=0\) as well) - Just something we do ;D
Post by: bsdfjnlkasn on October 19, 2017, 08:16:41 pm
How would you solve this?
My physics teacher made an excuse to avoid solving it ;D
Hey! Hope this is clear :)
(http://uploads.tapatalk-cdn.com/20171019/eac15f3382c7510ed0319fa721197dea.jpg)
Post by: bsdfjnlkasn on October 19, 2017, 08:35:36 pm
Would appreciate help with the following (Q2Q HSC 2009):
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22664053_1354425641349622_941418257_o.png?oh=e98f61955b94845d8bd99f4edd3b454a&oe=59EA1936
Also, how did Chadwick know that the neutron was neutral simply by LCE and LCM?
TIA.
Hey there!
I'll help with your second question :) Chadwick was able to determine the mass of the neutron to be similar to a proton by using conservation laws. He already knew that it was neutral because the highly penetrating 'radiation' observed by Joliot and Curie remained undeflected in both electric and magnetic fields - this was the property that made them so difficult to be detected in the first place! We don't need to know how the conservation laws were applied but, basically when they were firing the alpha particles at the beryllium, the 'radiation' initially thought to be gamma rays was directed at a sheet of paraffin wax. Because it is proton-rich, when the radiation (neutrons) from the beryllium reached the wax, protons were being ejected. The velocity at which this occurred would have been equal to the incoming velocity (as on such a small scale, the collisions would have been elastic :) ). From there, the mass could be determined (pf=pi=mv) and indeed it confirmed the neutral particle Rutherford proposed when he discovered the nucleus in 1919 :D
If you could please send through a plan of your 6 mark response that would be good because then I can give some more proactive feedback in terms of structure and content (it wouldn't be useful for either of us if I were to straight up give you an answer ;) )
Post by: austv99 on October 19, 2017, 09:39:10 pm
For 23, End X will be negative. Use the right hand slap rule - With your fingers pointed to the right and your thumb upwards to match the info given, you are slapping into the page. So, positive charges move into the page and negative charges move out of the page - X is negative ;DIm confused about 23. Since positive charges go into the page, isnt Y negative? Since conventional current runs from positive to negative?
For 21, we need acceleration such that the capsule goes from rest to the given speed, \(v\), over the given distance, \(r\). Substitute into the formula from your reference sheet:
For 10, the induced emf will peak when the rate of change of magnetic flux is largest. This happens when the coil is aligned horizontally, parallel to the field. So we need to start at a peak - It must be B or D. We complete one revolution so we should get one induced AC waveform: Answer is B ;D
Post by: jamonwindeyer on October 19, 2017, 09:43:01 pm
Im confused about 23. Since positive charges go into the page, isnt Y negative? Since conventional current runs from positive to negative?
So positive charges are going into the page, towards Y, which would imply that Y is positively charged. So X must be negative. This isn't really a situation to analyse in terms of current, because there isn't a complete path to flow through. Instead, try to think of where the positive charges and negative charges go ;D
Post by: winstondarmawan on October 19, 2017, 10:31:13 pm
Hey there!
I'll help with your second question :) Chadwick was able to determine the mass of the neutron to be similar to a proton by using conservation laws. He already knew that it was neutral because the highly penetrating 'radiation' observed by Joliot and Curie remained undeflected in both electric and magnetic fields - this was the property that made them so difficult to be detected in the first place! We don't need to know how the conservation laws were applied but, basically when they were firing the alpha particles at the beryllium, the 'radiation' initially thought to be gamma rays was directed at a sheet of paraffin wax. Because it is proton-rich, when the radiation (neutrons) from the beryllium reached the wax, protons were being ejected. The velocity at which this occurred would have been equal to the incoming velocity (as on such a small scale, the collisions would have been elastic :) ). From there, the mass could be determined (pf=pi=mv) and indeed it confirmed the neutral particle Rutherford proposed when he discovered the nucleus in 1919 :D
If you could please send through a plan of your 6 mark response that would be good because then I can give some more proactive feedback in terms of structure and content (it wouldn't be useful for either of us if I were to straight up give you an answer ;) )
Thank you heaps!
My response would include how the standard model was created in an attempt to explain the nature and composition of matter and it's interaction at it's most fundamental level. This included an explanation of the behaviour of the four fundamental forces which were: x,y,z,w and was explained through the exchange of bosons, a fundamental particle which was postulated to carry force between matter, for example strong nuclear was carried by gluons and gravitational attraction by gravitons. Standard model also deconstructed electrons, protons and neutrons into even smaller fundamental particles that were quarks and leptons. There were 6 types of each (list types) and protons comprised of up,up,down quarks and neutrons consisted of down,down,up quarks which explained their elementary charge. (Here, would also talk about the +2/3 and -1/3 charge of up/down quarks). Electrons were a special case of leptons, particles with negligible rest mass that interacted through weak nuclear and EM attraction.
I'd make a judgement about the impact of this on understanding, then say 'despite this there are flaws and questions that arise':
* Could not explain why there were 6 types of quarks and leptons, as well as their charge.
* Gravitons were only predicted, and not yet discovered. This was similar case with the Higgs Boson, which was only predicted by the standard model and found using extensive research and experimentation.
Please tell me what I am missing and how to improve this response. :)
Also, I would appreciate a rundown on all the scientists which contributed to the Manhattan Project and their explicit contributions.
Thanks again!
Post by: bsdfjnlkasn on October 19, 2017, 11:16:28 pm
Thank you heaps!
My response would include how the standard model was created in an attempt to explain the nature and composition of matter and it's interaction at it's most fundamental level. This included an explanation of the behaviour of the four fundamental forces which were: x,y,z,w and was explained through the exchange of bosons, a fundamental particle which was postulated to carry force between matter, for example strong nuclear was carried by gluons and gravitational attraction by gravitons. Standard model also deconstructed electrons, protons and neutrons into even smaller fundamental particles that were quarks and leptons. There were 6 types of each (list types) and protons comprised of up,up,down quarks and neutrons consisted of down,down,up quarks which explained their elementary charge. (Here, would also talk about the +2/3 and -1/3 charge of up/down quarks). Electrons were a special case of leptons, particles with negligible rest mass that interacted through weak nuclear and EM attraction.
I'd make a judgement about the impact of this on understanding, then say 'despite this there are flaws and questions that arise':
* Could not explain why there were 6 types of quarks and leptons, as well as their charge.
* Gravitons were only predicted, and not yet discovered. This was similar case with the Higgs Boson, which was only predicted by the standard model and found using extensive research and experimentation.
Please tell me what I am missing and how to improve this response. :)
Also, I would appreciate a rundown on all the scientists which contributed to the Manhattan Project and their explicit contributions.
Thanks again!
Hey there!
Awesome stuff :D There are just a few extra things that I would like to mention, hopefully they make sense!
First things first, we need to figure out what the question's asking: it's asking us to link theories (i.e. predictions) with experiments (i.e. results from particle accelerators) as they relate to the SM of M.
So you're right by starting with a definition of the standard model and you have all the right stuff for it:
a) Explained the composition of matter on a fundamental level
b) Accounted for the interactions between particles
Then I'd go into how it organised matter into (would help cut your response down a bit. After you quickly run past the structure of the SM of M, then we go into the next part of the question which wants us to talk about experiments)
1. Fermions
- Leptons: Don't experience strong nuclear force
(can list if you want: electron, muon, tau and their neutrino pair)
- Quarks: Experience all four fundamental forces (i.e. strong/weak nuclear force, electromagnetic and gravity)
(Hadrons --> Baryons and Mesons)
2. 'Force Carrier' Bosons (particles through which forces are mediated)
- Zo, X+, X- (I personally haven't come across W or Y.. what are they? :) )
3. And all their anti-particles
Some annotations on your response:
This included an explanation of the behaviour of the four fundamental forces which were: x,y,z,w and was explained through the exchange of bosons, a fundamental particle which was postulated to carry force between matter, for example strong nuclear was carried by gluons and gravitational attraction by gravitons. Standard model also deconstructed electrons, protons and neutrons into even smaller fundamental particles that were quarks and leptons Be careful here. Standard model didn't 'deconstruct' electrons, they're a fundamental particle. What the model did do, was classify the electron as leptons. Also, I would rephrase and say "nucleons were also predicted to be comprised of smaller fundamental particles called quarks". There were 6 types of each (list types) and protons comprised of up,up,down quarks and neutrons consisted of down,down,up quarks which explained their elementary charge. (Here, would also talk about the +2/3 and -1/3 charge of up/down quarks) Although this detail really shows you know your stuff, you need to be including more of the experimental aspect of the question to really be addressing it. So we could do a bit more in terms of ensuring that all the info is relevant. Electrons were a special case of leptons, particles with negligible rest mass that interacted through weak nuclear and EM attraction Good, but unfortunately irrelevant.
Here we need to mention the use of particle accelerators :)
The standard model is the product of predictions and experimental results (i.e. research). Linear accelerators and cyclotrons were initially used to 'probe' the structure of matter as new particles were created from these high speed collisions. These technologies were very effective, but now we have even better synchrotrons and particle accelerators (LHC) which are used more widely.
I'd make a judgement about the impact of this on understanding, then say 'despite this there are flaws and questions that arise': This is perfect! Super important to mention
* Could not explain why there were 6 types of quarks and leptons, as well as their charge.
* Gravitons were only predicted, and not yet discovered. This was similar case with the Higgs Boson, which was only predicted by the standard model and found using extensive research and experimentation.
Important to also note that the SM of M is a quantum-mechanical model which makes it incompatible with Einstein's general theory relativity. This prevents gravity from unifying the forces
Post by: winstondarmawan on October 20, 2017, 12:08:57 am
Hey there!
Awesome stuff :D There are just a few extra things that I would like to mention, hopefully they make sense!
First things first, we need to figure out what the question's asking: it's asking us to link theories (i.e. predictions) with experiments (i.e. results from particle accelerators) as they relate to the SM of M.
So you're right by starting with a definition of the standard model and you have all the right stuff for it:
a) Explained the composition of matter on a fundamental level
b) Accounted for the interactions between particles
Then I'd go into how it organised matter into (would help cut your response down a bit. After you quickly run past the structure of the SM of M, then we go into the next part of the question which wants us to talk about experiments)
1. Fermions
- Leptons: Don't experience strong nuclear force
(can list if you want: electron, muon, tau and their neutrino pair)
- Quarks: Experience all four fundamental forces (i.e. strong/weak nuclear force, electromagnetic and gravity)
(Hadrons --> Baryons and Mesons)
2. 'Force Carrier' Bosons (particles through which forces are mediated)
- Zo, X+, X- (I personally haven't come across W or Y.. what are they? :) )
3. And all their anti-particles
Some annotations on your response:
This included an explanation of the behaviour of the four fundamental forces which were: x,y,z,w and was explained through the exchange of bosons, a fundamental particle which was postulated to carry force between matter, for example strong nuclear was carried by gluons and gravitational attraction by gravitons. Standard model also deconstructed electrons, protons and neutrons into even smaller fundamental particles that were quarks and leptons Be careful here. Standard model didn't 'deconstruct' electrons, they're a fundamental particle. What the model did do, was classify the electron as leptons. Also, I would rephrase and say "nucleons were also predicted to be comprised of smaller fundamental particles called quarks". There were 6 types of each (list types) and protons comprised of up,up,down quarks and neutrons consisted of down,down,up quarks which explained their elementary charge. (Here, would also talk about the +2/3 and -1/3 charge of up/down quarks) Although this detail really shows you know your stuff, you need to be including more of the experimental aspect of the question to really be addressing it. So we could do a bit more in terms of ensuring that all the info is relevant. Electrons were a special case of leptons, particles with negligible rest mass that interacted through weak nuclear and EM attraction Good, but unfortunately irrelevant.
Here we need to mention the use of particle accelerators :)
The standard model is the product of predictions and experimental results (i.e. research). Linear accelerators and cyclotrons were initially used to 'probe' the structure of matter as new particles were created from these high speed collisions. These technologies were very effective, but now we have even better synchrotrons and particle accelerators (LHC) which are used more widely.
I'd make a judgement about the impact of this on understanding, then say 'despite this there are flaws and questions that arise': This is perfect! Super important to mention
* Could not explain why there were 6 types of quarks and leptons, as well as their charge.
* Gravitons were only predicted, and not yet discovered. This was similar case with the Higgs Boson, which was only predicted by the standard model and found using extensive research and experimentation.
Important to also note that the SM of M is a quantum-mechanical model which makes it incompatible with Einstein's general theory relativity. This prevents gravity from unifying the forces
Thank you so much for your response!
Don't worry about the W and Y stuff, I just wrote those letters as placeholders for the four fundamental forces. They don't mean anything. :)
Also, I'm a little bit confused on what is meant by 'This prevents gravity from unifying the forces'.
Post by: bsdfjnlkasn on October 20, 2017, 08:48:28 am
Thank you so much for your response!
Don't worry about the W and Y stuff, I just wrote those letters as placeholders for the four fundamental forces. They don't mean anything. :)
Also, I'm a little bit confused on what is meant by 'This prevents gravity from unifying the forces'.
Hey there!
Just wanted to make a quick correction, I meant to write W+ and W- minus instead of X because those force-carriers are the ones responsible for changing the charge on a quark/lepton. I think the main issue with the standard model is that it only currently accounts for 3 of the fundamental forces (electromagnetic, strong and weak nuclear force). But we also need a particle which explains the existence of gravity which we have already predicted, the graviton. But since there hasn't been any experimental evidence for it, we can't unify the forces under the standard model. Does that make a bit more sense? It's because the force of gravity is preventing all the forces from being united under one explanation/model.
Post by: julies on October 20, 2017, 12:40:47 pm
2014 HSC answers say that superconductors only transmit DC but I always thought it was compatible with AC as well....
:/
Post by: jamonwindeyer on October 20, 2017, 02:14:40 pm
hey guys, are superconductors able to conduct AC electricity?
2014 HSC answers say that superconductors only transmit DC but I always thought it was compatible with AC as well....
:/
Superconductors can transmit AC but nowhere near as effectively as they can transmit AC (as is my understanding)!
At HSC level, alternating current doesn't allow the formation of cooper pairs to travel through the lattice. Hence, no superconductivity. I don't think this is 100% correct but I think it is what we roll with in this course ;D
Post by: JuliaPascale123 on October 21, 2017, 01:41:51 am
https://imgur.com/a/Z1oBO (Graph)
Could I please have some assistance with calculating the pressure, volume and temp of each of the four points. (question 2)
I already did Question 1 so the efficency of the diesel motor is:
https://imgur.com/a/uesqN
Notice: This is physics im just not sure how to find the values from what they have supplied
Post by: jamonwindeyer on October 21, 2017, 02:34:58 pm
https://imgur.com/a/8rUw1 (question)
https://imgur.com/a/Z1oBO (Graph)
Could I please have some assistance with calculating the pressure, volume and temp of each of the four points. (question 2)
I already did Question 1 so the efficency of the diesel motor is:
https://imgur.com/a/uesqN
Notice: This is physics im just not sure how to find the values from what they have supplied
Hey! This looks like tertiary physics or perhaps another high school curriculum, I don't think you'll find much help here in the HSC section, sorry Julia! :)
It could be worth asking here :)
Post by: Iminschool on October 21, 2017, 08:51:19 pm
Post by: itssona on October 21, 2017, 11:15:55 pm
so if we have an apple and earth, according to law of gravitation, we can find the force exerted on them. And it's the same for both, so this means both the apple and earth exert an equal force on eachother (newton's third law). Also we can say that since there's a force, there must be acceleration (so thats why they accelerate towards each other????). We can use f=ma to see that the earth's acceleration is negligible.
thank you
Post by: blasonduo on October 21, 2017, 11:24:31 pm
Guys can somebody explain Q.20 of the 2016 HSC exam pls?
Hey, I LOVED this question!
Now to start with this question, look where the wires are going, they act as electromagnets as well as being the actual coil for the motor!
From the diagram alone, we can work out the poles of the electromagnets by following the direction of the current. With the one on the right, we work out using North (anti-clockwise) or South (clockwise) and we can clearly see that the closer side is a SOUTH pole.
Using the same method, we can figure out the pole on the left magnet, and it will show that the closer side is NORTH. So the magnetic field is going left to right. Using the right-hand palm rule, we can figure out that the motor will move clockwise So A and C are wrong.
Here comes the fun part! :D
Now this is AC, so the current will swap direction, so take the diagram and swap the + with a - and vice versa, and DO THE EXACT SAME THING. You will notice that the poles have swapped! Where the magnetic field is now going RIGHT to LEFT. Follow the current and use the right-hand palm rule, and you'll see that it is........... still going clockwise!
So the answer will be B.
This is really cool concept and I love the set-up :D
I hope this makes sense :)
Post by: jamonwindeyer on October 22, 2017, 02:16:57 pm
Hey, I LOVED this question!
Same here, and your explanation is awesome. Love your work ;D
just wanna check my understanding :)
so if we have an apple and earth, according to law of gravitation, we can find the force exerted on them. And it's the same for both, so this means both the apple and earth exert an equal force on eachother (newton's third law). Also we can say that since there's a force, there must be acceleration (so thats why they accelerate towards each other????). We can use f=ma to see that the earth's acceleration is negligible.
thank you
Yep, that's correct ;D
Post by: bundahboy on October 22, 2017, 05:26:19 pm
HSC Q12 2016: The correct answer is supposed to be B, but my interpretation of the right-hand palm rule leads me to A). Can anyone explain what's happening here?
Post by: jamonwindeyer on October 22, 2017, 05:31:51 pm
Hey guys,
HSC Q12 2016: The correct answer is supposed to be B, but my interpretation of the right-hand palm rule leads me to A). Can anyone explain what's happening here?
Cathode rays are negatively charged particles, not positive! So you need to swap your final direction - If you remember that I reckon you'll get B! ;D
Post by: skullcandy on October 23, 2017, 10:17:10 am
Post by: jamonwindeyer on October 23, 2017, 10:30:26 am
(http://i66.tinypic.com/e12dd3.png)
Hey! This doesn't quite look like a HSC Physics question, where did you find it? :)
Post by: edvinat on October 23, 2017, 01:34:09 pm
Can someone please give me a quick summary of how back emf works? No matter how many times I go through it I just cant seem to get it
Thank you ! :)
Post by: bsdfjnlkasn on October 23, 2017, 10:35:21 pm
Hiii,
Can someone please give me a quick summary of how back emf works? No matter how many times I go through it I just cant seem to get it
Thank you ! :)
Hey there, back-EMF isn't all that tricky :)
Hopefully this explanation helps:
I just thought to go through the basic to make sure you know why back EMF is induced :)
When a current carrying conductor experiences a change in magnetic flux, an EMF is induced causing a current to flow (if a closed circuit is formed). The interaction between the magnetic fields from the moving charges and external field interact cause the ends of the coil to experience a force. This force (Torque) causes the magnetic flux to continue to change around the coil.
We need a Back EMF to be induced otherwise the coil will continue to turn faster and faster. Think of it this way:
1. The coil experiences a change in flux
2. Current induced in coil
3. Torque experienced
4. Rotation causes increases in magnetic flux
5. Current in coil increases
-------> Cycle continues, with the coil getting faster and faster due to the rate of change in flux continuously increasing. This implies that without adding energy, the coil is gaining speed and so energy. This contravenes the LCE!! :O
This leads us to Lenz's law: According to Lenz's law, which is really just an extension of the LCE, a back EMF will be induced in the coil to oppose the change in flux and ensure that the coil isn't gaining energy without it being supplied. This induced current is otherwise known as induced emf.
For example, if there is an increase in magnetic flux due to the introduction of an external magnetic field, then the back EMF in the coil will be such that it produces a magnetic field in the opposite direction to the incoming field lines.
Or, if there is a decrease in the number of magnetic field lines (flux), then the induced EMF (current, because there's a closed circuit) will be such that a magnetic field will be directed in the way of the decreased magnetic field (so that the magnetic flux can be restored to it's amount before the change).
The idea is that back EMF will always be induced to counteract the change in flux which caused it to be 'created' in the first place. It wants to restore the magnetic flux to the state it was before the change occurred.
If you have any examples that you'd want us to look at, that might be a more proactive way for us to really help refine your understanding.
P.S. Not sure if you do chem, but it's really similar to Le Chatelier's principle :)
Post by: bsdfjnlkasn on October 24, 2017, 07:46:37 am
Does anyone know how I might answer this? I thought of using that change in gravitational potential = -Ek + heat, friction, etc... But still didn't get any of the answers
Post by: edvinat on October 24, 2017, 11:52:31 am
Hey there, back-EMF isn't all that tricky :)
Hopefully this explanation helps:
I just thought to go through the basic to make sure you know why back EMF is induced :)
When a current carrying conductor experiences a change in magnetic flux, an EMF is induced causing a current to flow (if a closed circuit is formed). The interaction between the magnetic fields from the moving charges and external field interact cause the ends of the coil to experience a force. This force (Torque) causes the magnetic flux to continue to change around the coil.
We need a Back EMF to be induced otherwise the coil will continue to turn faster and faster. Think of it this way:
1. The coil experiences a change in flux
2. Current induced in coil
3. Torque experienced
4. Rotation causes increases in magnetic flux
5. Current in coil increases
-------> Cycle continues, with the coil getting faster and faster due to the rate of change in flux continuously increasing. This implies that without adding energy, the coil is gaining speed and so energy. This contravenes the LCE!! :O
This leads us to Lenz's law: According to Lenz's law, which is really just an extension of the LCE, a back EMF will be induced in the coil to oppose the change in flux and ensure that the coil isn't gaining energy without it being supplied. This induced current is otherwise known as induced emf.
For example, if there is an increase in magnetic flux due to the introduction of an external magnetic field, then the back EMF in the coil will be such that it produces a magnetic field in the opposite direction to the incoming field lines.
Or, if there is a decrease in the number of magnetic field lines (flux), then the induced EMF (current, because there's a closed circuit) will be such that a magnetic field will be directed in the way of the decreased magnetic field (so that the magnetic flux can be restored to it's amount before the change).
The idea is that back EMF will always be induced to counteract the change in flux which caused it to be 'created' in the first place. It wants to restore the magnetic flux to the state it was before the change occurred.
If you have any examples that you'd want us to look at, that might be a more proactive way for us to really help refine your understanding.
P.S. Not sure if you do chem, but it's really similar to Le Chatelier's principle :)
Hey! Thank you so so much! I think i've just never understood it before because no ones ever properly explained it to me, but it makes a lot more sense now. Thanks again, you're a legend :)
Post by: jamonwindeyer on October 24, 2017, 03:11:33 pm
Hey there!
Does anyone know how I might answer this? I thought of using that change in gravitational potential = -Ek + heat, friction, etc... But still didn't get any of the answers
Hey! So at this height, probably okay to use simplified GPE formula to calculate the energy lost:
Kinetic energy gained is:
Difference between these is about \(5.34\text{MJ}\), so the answer would be C. Slightly off, again, because of using the simplified version of GPE (we can see it didn't really make a huge difference) ;D
Post by: Baylsskool on October 24, 2017, 05:27:00 pm
Post by: Baylsskool on October 24, 2017, 07:01:11 pm
Post by: samorchard on October 25, 2017, 10:04:48 am
A 0.05 kg mass is lifted at a constant speed by a DC motor. The motor has a coil of 100 turns in a 0.1 T magnetic field. The area of the coil is 0.0012 m2. The motor shaft has a radius of 0.004 m
a) F= 0.49N
b) Calculate the minimum current required in the coil to lift the mass.
t=Fd
Post by: samorchard on October 25, 2017, 10:13:44 am
Are control rods a type of moderator?? Cause like deuterium is heavy water, a moderator, but absorb neutrons to slow them down?? Is that not what control rods do?
Moderators slow down neutrons so that they can interact with the fissile material and cause fission. They don't abosrb, only slow down the neutron through inelastic collisions. Control rods absorb the neutrons and hence can control the reaction, slowing down or stopping the chain reaction whereas the moderators can't do this.
Hope that helps
Post by: Baylsskool on October 25, 2017, 10:39:46 am
Moderators slow down neutrons so that they can interact with the fissile material and cause fission. They don't abosrb, only slow down the neutron through inelastic collisions. Control rods absorb the neutrons and hence can control the reaction, slowing down or stopping the chain reaction whereas the moderators can't do this.Yeah that helps, so is it fair to say that through inelastic collisions, deuterium momentarily absorbs a neutron becoming tritium but the collision is too great so it has to let go of the neutron but does so at a slower speed?
Hope that helps
Post by: ajd938 on October 25, 2017, 11:30:31 am
Post by: Baylsskool on October 25, 2017, 11:38:00 am
Aim/purpose of the michelson morley experiment??? Was it to prove the existence of the aether or measure the relative velocity of earth through the aether?? All websites I have checked give a different answer. pls helpShould've been just to prove the existence of the aether wind by doing the light and mirror experiment where the light came back to the eye in equal increments after being separated, because they didn't come back at different times, there was the conclusion of no aether, since the conclusion answers the aim, the conclusion was there was no aether and therefore the aim was proving the existence of the aether.
Post by: jamonwindeyer on October 25, 2017, 11:56:55 am
Hi, I can't figure out why radius equals the distance in t = fd for this question (29b 2013)
A 0.05 kg mass is lifted at a constant speed by a DC motor. The motor has a coil of 100 turns in a 0.1 T magnetic field. The area of the coil is 0.0012 m2. The motor shaft has a radius of 0.004 m
a) F= 0.49N
b) Calculate the minimum current required in the coil to lift the mass.
t=Fd
Hey! The force due to the mass pulling down on the pulley is what generates the torque - \(F\) generating \(\tau\). The formula linking those is the distance from where the force is applied, and the centre of the rotation the torque is causing. That, in this case, is the radius! ;D
Post by: samorchard on October 25, 2017, 01:34:34 pm
Yeah that helps, so is it fair to say that through inelastic collisions, deuterium momentarily absorbs a neutron becoming tritium but the collision is too great so it has to let go of the neutron but does so at a slower speed?
I'm not too sure about that reaction, i think it would be safer to say that the neutron transfers some of its kinetic energy to the deuteriumin the collision.
The deuterium shouldn't absorb the neutron at all, as moderators are made from materials with light nuclei. If it did absorb the neutron then the most likely reaction would be a fission reaction and that is not the case.
Post by: samorchard on October 25, 2017, 01:56:03 pm
Hey! The force due to the mass pulling down on the pulley is what generates the torque - \(F\) generating \(\tau\). The formula linking those is the distance from where the force is applied, and the centre of the rotation the torque is causing. That, in this case, is the radius! ;D
MAkes sense, thankyou!
Post by: Dante1091 on October 25, 2017, 03:32:16 pm
From the dotpoint: Identify data sources, gather, process, analyse information and use available evidence to assess the impact
of the invention of transistors on society with particular reference to their use in microchips and
microprocessors
How much of the uses in microchips and microprocessors would need to be known, since I only know of the impacts of transistors onto society
Post by: Baylsskool on October 25, 2017, 03:48:46 pm
I'm not too sure about that reaction, i think it would be safer to say that the neutron transfers some of its kinetic energy to the deuteriumin the collision.Makes sense , thanks sam
The deuterium shouldn't absorb the neutron at all, as moderators are made from materials with light nuclei. If it did absorb the neutron then the most likely reaction would be a fission reaction and that is not the case.
Post by: blasonduo on October 25, 2017, 05:43:34 pm
Hello guys,
From the dotpoint: Identify data sources, gather, process, analyse information and use available evidence to assess the impact
of the invention of transistors on society with particular reference to their use in microchips and
microprocessors
How much of the uses in microchips and microprocessors would need to be known, since I only know of the impacts of transistors onto society
Hello! Describing what transistor are and how they benefit/disadvantage society IS most of it, as when you think about it, the impact as transistors on society was mainly due to the creations of microprossessors and microchips! To be sure, just name a technology that uses a microchip and a technology that uses a microprossessor (and what they do!!)
From that, you will be 100% sweet!!
Goodluck! You'll smash that exam!
Post by: Mymy409 on October 25, 2017, 06:33:29 pm
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Thanks in advance.
Post by: CyberScopes on October 25, 2017, 08:21:13 pm
Can somebody help me with Q9 of the 2015 paper?
http://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf
Thanks in advance.
P and Q are fixed in place. Since P and R are in the same direction, the force between them pulls R to the left. But since Q and R are in opposite directions, the force between them pushed R to the right. So pretty much find both forces and add them:
Force : F ∝ I1I2L / d
Since I and L are constant; F = 1 / d
Now you're given the force between Q and R, thats F, so:
F = 1 / 10 to the right between Q and R
d is 20mm between P and R so force between P and R = 1 / 20
Since F = 1/10 from before and youre tryna get an answer in terms of F, you can sub F as 1/10:
F between P and R = 1/10(2) = F/2
So FQR to the right (positive) plus FPR to the left (negative)
= F + - F/2 = F/2
Positive means to the right, so its B
(There is probably an easier way for this but this is what makes most sense to me :D)
Post by: Baylsskool on October 25, 2017, 09:22:52 pm
I mean with the ultraviolet catastrophe,why can't there be an infinite source of energy for an infinitly small wavelength, cause it checks out mathematically, we just havnt discovered any frequencies higher than gamma, of course it can't be infinite but surely there's more to discover so how can the peak wavelength be so?? Shouldn't it technically, if we did find the smallest wavelength possible at least hit the wall of the graph and decrease from there???
Mod Edit: Post merge :)
Post by: Jayden Nguyen on October 25, 2017, 09:33:07 pm
Could someone please help me with this question: account for the properties of diffraction and interference by electrons.
ps: this site looks so cool!
Post by: winstondarmawan on October 25, 2017, 11:22:33 pm
Would appreciate help with the question below:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22854697_1360395137419339_998506963_n.png?oh=0a008b316557edb40a47baa2393cc263&oe=59F2AF12
TIA!
Post by: statues on October 26, 2017, 11:38:51 am
Is it important to memorise the six types of leptons?
I've got the types of quarks down, and the fundamental forces I'm just not sure if it's important to memorise also the leptons - there's a lot to remember and I don't want to have information I won't need.
Many THanks
Post by: sidzeman on October 26, 2017, 12:18:44 pm
Post by: beau77bro on October 26, 2017, 02:42:36 pm
Post by: CyberScopes on October 26, 2017, 03:18:10 pm
For Med Phys, B scans produce a 2D cross sectional image by moving the transducer around to view the body from different angles correct? But aren't sector scans a series of B scans that are taken at different angles? I'm not exactly sure what the difference is now could someone clarify please
As far as I know, B scans themselves do not actually produce a 2D image, they only produce the set of "brightness dots" which shows the distance between organs. When there are multiple transducers producing B scans in a convex shape (as a sector scan), or if the transducer is moved like you said, then the 2D image is produced. B scans refers to more of how the different points of the organs are found (as compared to A scans), while sector scans is a way in which an image can be formed from these B scans. So you aren't wrong, but B scans and sector scans shouldnt be comparable cause they refer to different things (the type of scan verses how the scan is used to produce an image, if that makes sense).
Post by: bsdfjnlkasn on October 26, 2017, 04:28:10 pm
In the Quanta to Quarks Option:
Is it important to memorise the six types of leptons?
I've got the types of quarks down, and the fundamental forces I'm just not sure if it's important to memorise also the leptons - there's a lot to remember and I don't want to have information I won't need.
Many THanks
Hey there!
There's actually only 3 lepton names that you have to remember:
Electron, muon and tau
The other 3 are there antineutrinos so not too much to remember!
You should learn some extra information in case they ask you to compare leptons with quarks. I would say that leptons experience all four fundamental forces except the SNF (which quarks do experience) and that they are fundamental particles (aren’t formed of anything smaller than leptons).
Let me know if you need more help :)
Post by: bsdfjnlkasn on October 26, 2017, 04:41:42 pm
Hi there
Could someone please help me with this question: account for the properties of diffraction and interference by electrons.
ps: this site looks so cool!
Hey there, what topic is this for? :)
Post by: bsdfjnlkasn on October 26, 2017, 04:44:08 pm
Quanta to Quarks:
Would appreciate help with the question below:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22854697_1360395137419339_998506963_n.png?oh=0a008b316557edb40a47baa2393cc263&oe=59F2AF12
TIA!
Hey there,
They would experience an attractive force - not sure what the second mark is for :/
Post by: gawdn on October 26, 2017, 04:52:48 pm
This is a 2013 HSC multiple choice question. You see I recognise it as one of faraday's first motors but I don't understand how to determine the direction of rotation. Would appreciate some help.
(https://preview.ibb.co/hYkKvm/faradaysmotor.png)
Post by: blasonduo on October 26, 2017, 05:07:25 pm
Hey,
This is a 2013 HSC multiple choice question. You see I recognise it as one of faraday's first motors but I don't understand how to determine the direction of rotation. Would appreciate some help.
(https://preview.ibb.co/hYkKvm/faradaysmotor.png)
Hello! This is more applying your knowledge of the Motor Effect.
For this, we have to use our good ol' right hand!
We know the current is going down the page (thumb)
We know its a north pole and the magnetic field lines are going away from the magnet (fingers)
We then can see where our palm is facing, (which is into the page!) but as we move backwards our direction of magnetic field changes VERY slightly! (as we see in the diagram) It is this slight chance which causes the wire to rotate CLOCKWISE around the magnet ;)
Post by: gawdn on October 26, 2017, 05:15:06 pm
Hello! This is more applying your knowledge of the Motor Effect.
For this, we have to use our good ol' right hand!
We know the current is going down the page (thumb)
We know its a north pole and the magnetic field lines are going away from the magnet (fingers)
We then can see where our palm is facing, (which is into the page!) but as we move backwards our direction of magnetic field changes VERY slightly! (as we see in the diagram) It is this slight chance which causes the wire to rotate CLOCKWISE around the magnet ;)
Ah! I’m a bit daft turns out I had the right hand rule and everything but I got the direction a clock goes wrong. Whoops :( Thank you - seems I always figure out these things after I post them for the world to see.
Post by: jamonwindeyer on October 26, 2017, 07:33:51 pm
help please... like i know the information but i dont know how to direct it to answering the question properly, could someone give me an example and explanation on how to approach this?
I think you might have forgotten the attachment? :)
Post by: julies on October 26, 2017, 09:48:56 pm
I don't understand how to distinguish the theory of general relativity from special relativity... :/
thanks in advance :D
Post by: jamonwindeyer on October 26, 2017, 11:20:54 pm
Hey!
I don't understand how to distinguish the theory of general relativity from special relativity... :/
thanks in advance :D
- Relativity, the idea in general, is the idea that we have to make measurements with respect to a frame of reference. That is, we can only measure velocity of one thing with respect to another
- Special relativity was Einstein's theory on how this worked for objects moving at relativistic speeds (time dilation, length contraction, etc)
- There is also Einstein's Theory of General Relativity, but that is beyond the syllabus ;D
Post by: winstondarmawan on October 26, 2017, 11:24:21 pm
Hey there,
They would experience an attractive force - not sure what the second mark is for :/
Yes, my question is more when will the SNF balance with the electromagnetic? I think that's where the second mark is allocated and it sucks that this paper doesn't have solutions.
Post by: ProfLayton2000 on October 27, 2017, 11:23:01 am
What other things do I have to mention other than:
Definition of semiconductor
Doping to get n/p type
N has electron as main charge carrier
P has hole (area of positive charge) as main charge carrier
The movement of these carriers allows for improved conductivity in the semiconductor
BOSTES marking criteria was reallllllllllll helpful.
Post by: blasonduo on October 27, 2017, 01:03:47 pm
HSC 2001 q26: "In the context of semiconductors, explain the concept of electrons and holes" , 8 marks
What other things do I have to mention other than:
Definition of semiconductor
Doping to get n/p type
N has electron as main charge carrier
P has hole (area of positive charge) as main charge carrier
The movement of these carriers allows for improved conductivity in the semiconductor
BOSTES marking criteria was reallllllllllll helpful.
These are some great points!!
Just to add to it, I would add diagrams of the doped semiconductors and how the energy band gap decreases due to this (see diagram attached ;))
and to add a VERY brief explanation how the knowledge of semi-conductors allowed technologies to benefit society such as solar cells. (but this probably isn't even needed :)))
Goodluck for your exam!
Post by: Anfar3 on October 27, 2017, 03:10:35 pm
Can I get a response to the operation of solar cells and how they work for example if it was a 4 marker. i always seem to lose marks with solar cells as I do not really understand it/ cant describe it.
Thanks
Post by: blasonduo on October 27, 2017, 03:48:03 pm
Hey guys,
Can I get a response to the operation of solar cells and how they work for example if it was a 4 marker. i always seem to lose marks with solar cells as I do not really understand it/ cant describe it.
Thanks
Hey! Let's see what we can do! :))
Firstly, we have 2 semiconductors, a N-type and a P-type and what we do is we "stick" them together, and this does something very peculiar.
As an N-type semiconductor has extra free electrons, these electrons will want to fall into the holes into the p-type semiconductor. This will cause something we call a Depletion zone
This depletion zone will act as an electric field, This is because electrons have a negative charge and as the electrons moved towards the p-type semiconductor, the P-type side of the depletion zone is now slightly negative, and the N-type side of the depletion zone is slightly positive. Electrons will continue to jump the gap from the N-type to the P-type, making the electric field stronger and stronger until the electrons can no longer jump the gap (As the electric field applies a force to the electrons!)
Now, we attach a circuit to this combined semiconductor and add the power of the sun!! The sun here is just an application of the photoelectric effect, where all it does is pass energy to the electrons, making the electrons able to move but due to the depletion zone, they cannot pass into the P-type, meaning the only way they can move is through the external circuit (which it does!!) Due to this flow of electrons through the circuit, a current is produced!
annnnnnnd voilà! A solar cell!
Just a HUGE reminder! Don't get this mixed up with the photocell! :P
I hope this helps! If you have any questions about this! Feel free to ask! :))
Post by: Anfar3 on October 27, 2017, 04:42:39 pm
9.4- describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices
Do we need to know abt the operation of npn junctions and forward/reversed bias for this dotpoint?
Thanks
Post by: blasonduo on October 27, 2017, 05:03:07 pm
Legend!
9.4- describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices
Do we need to know abt the operation of npn junctions and forward/reversed bias for this dotpoint?
Thanks
Not really! That is past the course!
All you really need to do is identify the function of solid state devices, (not really the physics behind it) and why they were superior to thermionic devices :P
Post by: blasonduo on October 27, 2017, 05:07:35 pm
Next exam, physics! :D Study hard and study well, we will get through this together ;)
Post by: sidzeman on October 27, 2017, 05:09:53 pm
Post by: blasonduo on October 27, 2017, 05:15:08 pm
For the BCS theory and superconductors - do superconductors posses 0 resistance AT their critical temperature or BELOW it. Also, the BCS theory only accounts for DC current correct?
Yes! AT that temperature!! Just remember the graph, the critical temperature is when it hits 0 resistance and falls on the x-axis! :)
Yep! DC only! (at our level anyway :P)
You are spot on :)
Post by: winstondarmawan on October 27, 2017, 06:23:33 pm
Can someone help explain the 1st and 3rd point in the answers? I'm not used to seeing DC motors like this so I am very confused.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22851172_1362180510574135_1756100432_o.jpg?oh=c99e6aa9f3e2042b617df10595f4c77b&oe=59F488F0
A: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22906523_1362180517240801_1608008686_o.jpg?oh=63611363f57c0174f90fb5a1afcdc067&oe=59F56D36
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22851579_1362180543907465_554963234_o.jpg?oh=cc8950f62400f91ef10184ef0c67b351&oe=59F48DC8
Thanks!
Post by: Bubbly_bluey on October 27, 2017, 07:22:08 pm
Hello!Hey! I'll give a crack at explaining the 3rd one. So on the diagram on one side the coil is wrapped around. Lets look at the left side first So you have 4 things going on:
Can someone help explain the 1st and 3rd point in the answers? I'm not used to seeing DC motors like this so I am very confused.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22851172_1362180510574135_1756100432_o.jpg?oh=c99e6aa9f3e2042b617df10595f4c77b&oe=59F488F0
A: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22906523_1362180517240801_1608008686_o.jpg?oh=63611363f57c0174f90fb5a1afcdc067&oe=59F56D36
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22851579_1362180543907465_554963234_o.jpg?oh=cc8950f62400f91ef10184ef0c67b351&oe=59F48DC8
Thanks!
1.The wire (from where it begins) starts to wrap vertically up. Using the right hand palm rule it goes down towards the pin direction. (not possible in motors)
2.Then the wire is wrapped horizontally at the top which current going in. Resultant force should be up, so thats possible
3. The current then wraps vertically down. Again observation (1) applies
4. The wire wraps horizontally at the bottom of the rotor so current going towards the pin. Resultant force would be down.
Now looking at 2 and 4, when a wire is wrapped just once, you have two opposing forces up and down. And so thats why I think its a mistake. The same goes for the wrapped coil on the right. Hope I'm making sense and someone corrent me if I'm wrong :D
Post by: Bubbly_bluey on October 27, 2017, 07:27:08 pm
Thanks :D
Post by: jaskirat on October 27, 2017, 07:45:51 pm
Post by: hinakamishiro on October 27, 2017, 08:13:33 pm
Post by: imda.beast on October 27, 2017, 08:21:07 pm
i don't quite understand how this question is done. can you help? thanx in advance
Post by: pikachu975 on October 27, 2017, 08:24:57 pm
hi,
i don't quite understand how this question is done. can you help? thanx in advance
Boron doping means it is p type so extra holes.
Positive flow goes left in the silicon block hence holes move left, so therefore electrons move right in the valence band as they causing the apparent hole movement, thus A.
Post by: Bubbly_bluey on October 27, 2017, 08:32:44 pm
Hey guys, need help with these 3 questions :)Hey! So q14 is converting the magnetic flux graph to emf. (My explanation is using maths so I'm sorry if it doesn't make sense but I'll try) These two are linked together with the E= -change flux/ change time so like a differentiation q (you have to differentiate magnetic flux to get emf. Solooking at the magnetic flux graph at the origin is a point of inflection (POI) with positive graident, meaning on the emf graph should start above the x axis. BUT, remember the formula has a negative sign so its the opposite and actually starts at the bottom. Leaving options C and D. However when differentiating a POI you get a stat point so it should be D.
To confirm you can look back at the magnetic flux graph at the next time and you have a stationary point, gradient =0. Therefore that is an indication that you should have an x-int next.
q16 Im a bit iffy on back emf. But i think maybe because has the DC motor gain speed until it reaches constant velocity when back emf=supply emf. So there would not be any extra work/ no net force done because the induced emf would balance with the supply emf?? dont quote me on this but hence why I would pick C?
q20 I would find the energy of the photon at the peakof the graph so maybe around 8-9micrometres in wavelength. Use E=hf to find that energy. Then convert it to eV by dividing it by (1.602x10^-19). So I worked that out to be 0.155 eV which is how much you need. So 0.17eV would be enough to provide this.
Hope this help :D
Post by: winstondarmawan on October 27, 2017, 08:38:45 pm
hey guys! if a q says to describe how g-forces affects of astronaunts in launch: what exactly do you have to write? They experience an increasing force due to acceleration but I don't know how to describe it. Is it refering to multistage rockets?I think it's best to talk about the Law of Conservation of Momentum as well as F=ma to prove that the spacecraft is accelerating. From there you talk about how g-forces are directly proportional to acceleration.
Thanks :D
Post by: statues on October 27, 2017, 08:43:42 pm
What I understand is that it's a DC source which means that instead of trying to minimise the Voltage as we would if it were an AC source in order to maximise the current - we are instead trying to maximise the voltage as this corresponds to an increase in emf which corresponds to an increase in current? I don't understand why in this case a high voltage in the primary coil corresponds to a high current.
This is from the 2015 hsc q18. The answer is A.
please help
Many Thanks :)
Post by: Bubbly_bluey on October 27, 2017, 09:08:02 pm
Hey guys i'm just having trouble understanding/ working out these questions and i would really appreciate if anyone could give me a hand! I know it looks like a lot, but they are all just multiple choice questions which didn't have any explanations that i could find. Cheers :)hey! q12: So it says centripetal force that means we have to apply F=mv^2/r. We have mass and orbital radius but not velocity. So note that its orbiting so we apply the orbital velocity remembering that mass (M) is the earth!. You sub it back into the centripetal force and you should get D.
8 ) maximum force would always be at 90degrees. b/c sin90=1. You can elimate B and D. But moving charges in magnetic fields only experience a force by F=qvbsin(theta). Hence C
7) For torque to be proportional to current, there must be a uniform change in magnetic flux. Hence a radial magnetic (B) because it would give maximum torque at every direction and so magnetic flux would always change to induce current.
8 ) Use F=nBilsin(theta). Working out the angle first that should be between the magnet and the wire (hence not 30 degrees). A trick is to hold a pen vertically down into the page for flux and another pen for the wire. you will see that it always makes 90 degrees with the magnetic field. To find length you sin rule. hence A.
3) Current moving down X-Y. From birdseye field lines are cutting the wire horizontally. Use Right hand rule and force is into the page so it must go clockwise.
20) I would find the energy of the photon at the peakof the graph so maybe around 8-9micrometres in wavelength. Use E=hf to find that energy. Then convert it to eV by dividing it by (1.602x10^-19). So I worked that out to be 0.155 eV which is how much you need. So 0.17eV would be enough to provide this.
19) idk too XD
17) DC is direct current so has it is turned on there is changes in flux>induced current>induced force> therefore torque. But DC motors have back emf. As it increases supply emf=back emf eventually. Hence the current decrease.But current is proportional to torque by T=NBIA cos, so torque decreases too.
20) So first convert the frequency of the 450nm wavelength to see if it surpasses any of the thresholds of X or Y. I got 6.66x10^14 Hz. So only x would emit. Intensity only introduces more photons so more electrons can be emmitted not increasing KE. Hence B
15) Angle for torque is the angle between the armature plane with the magnetic field hence 30. Angle for force on wire always the angle the wire (XW or YZ) makes with the magnetic field, 90. You can try use the pen trick I said earlier to visualise.
14) For projectile motion to happen you need two components and horizontal and vertical. Accelerating would not projuce a projectile curve so its not B or D.The horizontal from the pushing of the cue against the ball. Now to find the vertical start drawing vector coponents. You can see vertical is down. (from your view on the page) Now if you look the front of the train its to the left so try to imagine: what would happen if you turned left or right IF the ball was just stationary? If you turn right the ball will go opposite and go left (from the train driver's view) But to our view its equivalent to moving down as shown in the vector. Hence C
Post by: CyberScopes on October 27, 2017, 09:14:22 pm
Hey guys i'm just having trouble understanding/ working out these questions and i would really appreciate if anyone could give me a hand! I know it looks like a lot, but they are all just multiple choice questions which didn't have any explanations that i could find. Cheers :)
Ok ill see what I can do
1. Remembering that centripetal force is the same as gravitational force, so use Newtons Gravitation Law:
F = GMm/d2
= (6.67*10-11 * 6*1024 * 7.35*1022) / (3.83*108)2
= 2*1020N
= A
2. Using F = qvBsinθ:
-> For A, v = 0, therefore F = 0, so it cant be A.
- > For B, the particle is moving parallel to the magnetic field lines: sin0 = 0 which means F = 0, so cant be B.
-> For C, F = q(40)B*sin90
Since we're comparing, we can assume q and B are just 1, so: F = 40
-> For D, considering B and q as 1, F = 50sin45 = 35 < 40
Therefore answer must be C
3. A radial magnetic field assures that the plane of the coil is always flat at 0 degrees, and cos0 = 1, so this is what makes the torque directly proportional to the current (since T = nBIAcosθ and n, B, I and A are constant)
So answer must be B
4. F = BILsinθ
B = 0.05
F = 0.03
θ = 90 degrees (since its perpendicular to the magnetic field lines)
L = 0.4 (using trig to find the length of just the wire)
Plug in and rearrange for I and you get 1.5
Answer is A
5. This was answered previously in the thread a few posts back
6. Not sure
7. Not sure, remember having troubles with this one before
8. A does not make sense and B does not consider back EMF.
I think it may be D since it considers that after a while, back EMF impacts the torque and slow it down, but I am not completely sure.
9. X hits at around 4 and Y hits at around 9
First you need to check which ones actually can eject electrons:
It needs to be within the threshold frequency of the light:
f = c/λ = 3*108 / 450*10-9 = 6.67*1014Hz
X has around 4*1014Hz so its within the frequency to emit electrons, but Y is above it so it cannot eject electrons, so C and D are eliminated.
Knowing how the photoelectric effect occurs, more intensity only increases the photo-current that is ejected, not the energy of the electrons.
So the answer is B.
10. The angle for torque is the plane of the coil to the magnetic field lines, which is 30 degrees.
The angle for force, for example by taking line WX, is 90 degrees since its always perpendicular to the magnetic field lines (same as for YZ)
So answer is A
11. This one is sort of by using "common sense" or just sort of understanding physics environments. Imagine leaving a ball in a box that cannot move, how can u make it so the ball is touching the left side of the box by moving the box? This can only be done by rotating, or moving, the box to the right. You can't "shift" the train to the right but it can turn to the right.
Answer is C
I did most of these on the spot so I may have made mistakes, just let me know if I did.
Post by: pikachu975 on October 27, 2017, 09:14:43 pm
Could Someone please help explain this to me
What I understand is that it's a DC source which means that instead of trying to minimise the Voltage as we would if it were an AC source in order to maximise the current - we are instead trying to maximise the voltage as this corresponds to an increase in emf which corresponds to an increase in current? I don't understand why in this case a high voltage in the primary coil corresponds to a high current.
This is from the 2015 hsc q18. The answer is A.
please help
Many Thanks :)
This is out of HSC so it's dumb that they put it in, since we only work with AC in HSC
Post by: beau77bro on October 27, 2017, 09:20:56 pm
Post by: blasonduo on October 27, 2017, 09:21:58 pm
Hey guys i'm just having trouble understanding/ working out these questions and i would really appreciate if anyone could give me a hand! I know it looks like a lot, but they are all just multiple choice questions which didn't have any explanations that i could find. Cheers :)
Hey!!!!! Well we are in for a looooooong ride :))
12) For this question we must firstly find the orbital velocity:
=
8.) This whole question is going off the formula:
For A, the velocity is 0, so the force is 0, so that's wrong.
For B, the velocity is 60, but it is travelling Parallel to this magnetic field (ie sin0 = 0)
For C, everything works out, so the force is 40
For D, everything works out, the velocity is 50, but due to sinϑ = 45, the force is half, and thus 25
This means C will have the highest force :D
7) This is something that you really just need to know, In a galvanometer, this is the reason why the radial magnet is so vital, to get accurate measurements! B!
8.) Big note here, the wire is still PERPENDICULAR to the field, so sinϑ = 90.
To find the length, just use Pythagoras, and you should find length to be 0.4m
3) This was on the previous page ;) This is more applying your knowledge of the Motor Effect.
For this, we have to use our good ol' right hand!
We know the current is going down the page (thumb)
We know its a north pole and the magnetic field lines are going away from the magnet (fingers)
We then can see where our palm is facing, (which is into the page!) but as we move backwards our direction of magnetic field changes VERY slightly! (as we see in the diagram) It is this slight chance which causes the wire to rotate CLOCKWISE around the magnet ;) A
20 Bubbly_bluey Just made a very good explanation about this above! If you need more help, feel free to ask ;) B
19) This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why i'll be happy to!)
17) again, Bubbly_bluey pretty much explained this above! (Well done btw!!) The work done (which is torque) once hitting it's fixed maximum speed will be 0, so it must decrease! D
20) We here are given wavelength, but the graph shows frequency, so we must change the wavelength to frequency.
As this is before Line Y, this means NOTHING will happen to Y, so C and D are out!
The next part is all about the properties of photons, where INTENSITY means more electrons, not more energy, so B
15) We know that from the motors, the force is ALWAYS perpendicular to the coils, so it has to be A or C, but as torque varies with the angle, we need to know where torque = 1, (cos 0) which is where the coil is "flat" meaning that moving it 30 will make cos 30 and thus it is A :)
14) 14) For these questions especially, I'd like to picture myself in these situations and go from there. Imagine you are driving a car, speeding it up or slowing it down will NOT make you experience a force to the left or right (there is no circular motion!), so this eliminates both B and D. Now referring back to the car analogy, when you turn the car, you'd always experience a force opposite to the way you are turning. The car is beginning its turning motion while you continue in a straight line path. This is because of Newton's first law.
So back to the question, the ball began to roll down the page, and from what was said before, it must mean the train was turning up the page, or to the RIGHT, so it is C.
WHEWWWWWWWWW That WAS a ride!!!! I hope this helps! Ask away if any of this made no sense, I'll be happy to help!!!
Post by: imda.beast on October 27, 2017, 09:31:44 pm
Boron doping means it is p type so extra holes.
Positive flow goes left in the silicon block hence holes move left, so therefore electrons move right in the valence band as they causing the apparent hole movement, thus A.
hi, thank you for your help, but the correct answer was B from 2008 HSC Q15.
any idea why?
Post by: CyberScopes on October 27, 2017, 09:34:05 pm
could someone please explain T1 and T2 relaxation times simply, as well as in detail/explanation hard out. im seriously struggling with this and it seems both overly complicated and lackingly explained all at once. thanks and pliz.
As far as ive seen, every explanation either seems beyond the scope of the syllabus or too shallow to understand the concept, but i can try from what I know.
Assume that the radio waves have already been projected, and the nuclei of the atoms have already absorbed the energy, etc.
When the radio waves are switched off, different tissues with different densities release energy at different time periods.
When pulses of radio waves are quickly switched on and off, only tissues with short T1 values release their energy quickly and can be detected by the scanning procedure. These tissues are tissues like fat, liver, spleen, etc. This allows the MRI image to contrast between these tissues and watery tissues.
If the MRI wants to the opposite and instead highlight the watery tissues, it uses delayed radio pulses that are slow and only allow tissues with long T2 values are detected (like the brain), so the fatty tissues mentioned earlier will not appear as contrasted as the watery tissues.
This isnt the best explanation but its all I have :-\
Post by: Bubbly_bluey on October 27, 2017, 09:35:35 pm
Could Someone please help explain this to meI think the best way to work this out is to use a bit of maths. Using Np/Ns=Vp/Vs. So q ask for greatest deflect infering that we want the greatest volatage in secondary coil. Make Vs the subject and you get Vs= (Vp x Ns)/ Np. So if you decrease Np (number of coils in primary you have greater secondary volatage) :) Therefore A
What I understand is that it's a DC source which means that instead of trying to minimise the Voltage as we would if it were an AC source in order to maximise the current - we are instead trying to maximise the voltage as this corresponds to an increase in emf which corresponds to an increase in current? I don't understand why in this case a high voltage in the primary coil corresponds to a high current.
This is from the 2015 hsc q18. The answer is A.
please help
Many Thanks :)
Post by: blasonduo on October 27, 2017, 09:36:16 pm
hi, thank you for your help, but the correct answer was B from 2008 HSC Q15.
any idea why?
I think there might've been a mixup!
The way I interpreted this was just to follow the positive terminal, and you see current flows left to right but as electrons move in the opposite direction and thus RIGHT to LEFT (going left) so B
Post by: beau77bro on October 27, 2017, 09:41:17 pm
As far as ive seen, every explanation either seems beyond the scope of the syllabus or too shallow to understand the concept, but i can try from what I know.
Assume that the radio waves have already been projected, and the nuclei of the atoms have already absorbed the energy, etc.
When the radio waves are switched off, different tissues with different densities release energy at different time periods.
When pulses of radio waves are quickly switched on and off, only tissues with short T1 values release their energy quickly and can be detected by the scanning procedure. These tissues are tissues like fat, liver, spleen, etc. This allows the MRI image to contrast between these tissues and watery tissues.
If the MRI wants to the opposite and instead highlight the watery tissues, it uses delayed radio pulses that are slow and only allow tissues with long T2 values are detected (like the brain), so the fatty tissues mentioned earlier will not appear as contrasted as the watery tissues.
This isnt the best explanation but its all I have :-\
thanks cyber scope, the explanation made it a little clearer. but im struggling to understand how the time of the radio pulses allows them to measure relaxation time. like how are we measuring these and what is the distinct difference which we are observing and quantifying? all the explanations just seem inadequate. like so the magnetic field is in the transcerse plane when the hydrogen nuclei move to anti parallel - but then what, why does that matter and what does that have to do with the relation time? is it to do with that magnetic vector or the hydrogen atoms precessions im just very confused. im trying to work out analogies and stuff to simply and explain to my friends and class mates but im struggling
Post by: CyberScopes on October 27, 2017, 09:44:23 pm
thanks cyber scope, the explanation made it a little clearer. but im struggling to understand how the time of the radio pulses allows them to measure relaxation time. like how are we measuring these and what is the distinct difference which we are observing and quantifying? all the explanations just seem inadequate. like so the magnetic field is in the transcerse plane when the hydrogen nuclei move to anti parallel - but then what, why does that matter and what does that have to do with the relation time? is it to do with that magnetic vector or the hydrogen atoms precessions im just very confused. im trying to work out analogies and stuff to simply and explain to my friends and class mates but im struggling
I honestly wish I knew, I feel like this entire topic is lacking. The content that we need to know for a strong understanding isnt a part of the syllabus, and its difficult to wrap our heads around whats actually going on if we dont understand this missing content. If you ever find anything please let me know too, Im as lost as you are haha
Post by: beau77bro on October 27, 2017, 09:48:37 pm
I honestly wish I knew, I feel like this entire topic is lacking. The content that we need to know for a strong understanding isnt a part of the syllabus, and its difficult to wrap our heads around whats actually going on if we dont understand this missing content. If you ever find anything please let me know too, Im as lost as you are hahau will be the first to know - i have high hopes that atarGODS will save us in our time of need.
btw which avenger u reckon each atarGOD is...? lmao. thats a q for another time doe.
Post by: statues on October 27, 2017, 09:50:39 pm
I think there might've been a mixup!
The way I interpreted this was just to follow the positive terminal, and you see current flows left to right but as electrons move in the opposite direction and thus RIGHT to LEFT (going left) so B
Hey thanks for responding - It's attached to a galvanometer which measures current so a large current would induce a large deflection not as a direct result of voltage. However voltage apparently causes a large current which is what I don't understand.
Post by: blasonduo on October 27, 2017, 09:53:30 pm
Hey thanks for responding - It's attached to a galvanometer which measures current so a large current would induce a large deflection not as a direct result of voltage. However voltage apparently causes a large current which is what I don't understand.
Sorry, where's the galvanometer?
Post by: statues on October 27, 2017, 10:10:34 pm
Sorry, where's the galvanometer?
It's attached to the secondary coil
Post by: blasonduo on October 27, 2017, 10:17:18 pm
It's attached to the secondary coil
Oh I see! We were talking about different questions! My apologies!
Post by: statues on October 27, 2017, 10:23:06 pm
Oh I see! We were talking about different questions! My apologies!oh jeez I quoted the wrong person, my bad! It's evidently been a long day of study haha
Post by: statues on October 27, 2017, 10:25:12 pm
I think the best way to work this out is to use a bit of maths. Using Np/Ns=Vp/Vs. So q ask for greatest deflect infering that we want the greatest volatage in secondary coil. Make Vs the subject and you get Vs= (Vp x Ns)/ Np. So if you decrease Np (number of coils in primary you have greater secondary volatage) :) Therefore AHey thanks for responding - It's attached to a galvanometer which measures current so a large current would induce a large deflection not as a direct result of voltage. However voltage apparently causes a large current which is what I don't understand.
Post by: pikachu975 on October 27, 2017, 10:26:07 pm
hi, thank you for your help, but the correct answer was B from 2008 HSC Q15.
any idea why?
Oops my bad I thought the right one was positive for some reason!
Hey thanks for responding - It's attached to a galvanometer which measures current so a large current would induce a large deflection not as a direct result of voltage. However voltage apparently causes a large current which is what I don't understand.
As stated before, the reason why isn't in the HSC course. It has to do with the fact that the input voltage is DC which is why the formula np/ns = Is/Ip does not work. It's dumb that BOSTES put that question in.
If it was an AC input the answer should be what you said.
Post by: Bubbly_bluey on October 27, 2017, 10:30:18 pm
Hey thanks for responding - It's attached to a galvanometer which measures current so a large current would induce a large deflection not as a direct result of voltage. However voltage apparently causes a large current which is what I don't understand.oml lol sorry yeah galvanometers measure current. :D But the maths would still work in this case you would use Np/Ns=Is/Ip. rearrange to make Is subject. Is=(Np x Ip) / Ns. So yes it would be B. Sorry that was a really bad mistake XD. Hope this clears it up :)
Post by: Bubbly_bluey on October 27, 2017, 10:33:21 pm
Oops my bad I thought the right one was positive for some reason!whaat itsnot in the syllabus >: ( How did you know it was DC?
As stated before, the reason why isn't in the HSC course. It has to do with the fact that the input voltage is DC which is why the formula np/ns = Is/Ip does not work. It's dumb that BOSTES put that question in.
If it was an AC input the answer should be what you said.
Post by: blasonduo on October 27, 2017, 10:37:50 pm
whaat itsnot in the syllabus >: ( How did you know it was DC?
The battery pack indicates this DC pack. I'm curious though, because if we were to change the voltage to AC, would it change the answer?
Post by: pikachu975 on October 27, 2017, 10:39:04 pm
whaat itsnot in the syllabus >: ( How did you know it was DC?
I knew cause I read online that it's a bad question because we don't use DC transformers in HSC plus the input isn't AC.
The battery pack indicates this DC pack. I'm curious though, because if we were to change the voltage to AC, would it change the answer?
Yeah it'd be different
Post by: winstondarmawan on October 27, 2017, 10:39:35 pm
As stated before, the reason why isn't in the HSC course. It has to do with the fact that the input voltage is DC which is why the formula np/ns = Is/Ip does not work. It's dumb that BOSTES put that question in.
If it was an AC input the answer should be what you said.
Actually, it does work even if it is only for a little while. Because there is a change in flux as soon as the switch is closed.
Anyways, if anyone could give a quick rundown on the scientists and their contributions to the Manhattan Project for Quanta to Quarks that would be SWEETTT. I'm having trouble finding the exact scientists which contributed both directly and indirectly.
TIA.
Post by: pikachu975 on October 27, 2017, 10:41:55 pm
Actually, it does work even if it is only for a little while. Because there is a change in flux as soon as the switch is closed.
Anyways, if anyone could give a quick rundown on the scientists and their contributions to the Manhattan Project for Quanta to Quarks that would be SWEETTT. I'm having trouble finding the exact scientists which contributed both directly and indirectly.
TIA.
However we don't learn the specifics of how number of coils affects the output etc like we do for AC.
A quick one would be Einstein who suggested to start the project.
Post by: jaskirat on October 27, 2017, 10:55:32 pm
Hey! So q14 is converting the magnetic flux graph to emf. (My explanation is using maths so I'm sorry if it doesn't make sense but I'll try) These two are linked together with the E= -change flux/ change time so like a differentiation q (you have to differentiate magnetic flux to get emf. Solooking at the magnetic flux graph at the origin is a point of inflection (POI) with positive graident, meaning on the emf graph should start above the x axis. BUT, remember the formula has a negative sign so its the opposite and actually starts at the bottom. Leaving options C and D. However when differentiating a POI you get a stat point so it should be D.
To confirm you can look back at the magnetic flux graph at the next time and you have a stationary point, gradient =0. Therefore that is an indication that you should have an x-int next.
q16 Im a bit iffy on back emf. But i think maybe because has the DC motor gain speed until it reaches constant velocity when back emf=supply emf. So there would not be any extra work/ no net force done because the induced emf would balance with the supply emf?? dont quote me on this but hence why I would pick C?
q20 I would find the energy of the photon at the peakof the graph so maybe around 8-9micrometres in wavelength. Use E=hf to find that energy. Then convert it to eV by dividing it by (1.602x10^-19). So I worked that out to be 0.155 eV which is how much you need. So 0.17eV would be enough to provide this.
Hope this help :D
Post by: winstondarmawan on October 27, 2017, 11:19:31 pm
Also, in discussing the positive and negative impacts of the Manhattan Project to society, I have been told that stating 'Potentially saving the lives of many Ally soldiers' was not a good positive impact to put as it is quite controversial.
Thoughts?
Post by: statues on October 27, 2017, 11:21:34 pm
oml lol sorry yeah galvanometers measure current. :D But the maths would still work in this case you would use Np/Ns=Is/Ip. rearrange to make Is subject. Is=(Np x Ip) / Ns. So yes it would be B. Sorry that was a really bad mistake XD. Hope this clears it up :)
Hey thanks so much for the help. I totally agree with your work but the HSC answer is A. So I'm still a bit confused. If the questions just wrong was it omitted from the hsc?
Post by: statues on October 27, 2017, 11:28:53 pm
Quanta to Quarks
Also, in discussing the positive and negative impacts of the Manhattan Project to society, I have been told that stating 'Potentially saving the lives of many Ally soldiers' was not a good positive impact to put as it is quite controversial.
Thoughts?
I would tend to agree with that assessment - don't say that. There's loads of positive impacts of the manhattan project such as the development of nuclear power, radioisotopes , understanding of quantum physics , the end of ww2 etc. It's just not necessary to make that point - while it's potentially true it seems to undermine the slaughter of 80000 Japanese civilians - which is a negative point you should discuss too. Having both points would seem to form to some degree an incongruous argument. There's really nothing to be gained using this point- there are no doubt HSC markers who are going to be offended - like imagine if you got some Japanese dude marking your question, it's not going to go well for you.
Anyways best of luck I'm doing the same option! :)
Post by: winstondarmawan on October 28, 2017, 12:15:37 am
6. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22883926_1362403753885144_473053551_o.jpg?oh=51a20dd5d7c37a1fa4e92f54e4601332&oe=59F5ABED
9. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22883814_1362403813885138_1145465314_o.jpg?oh=2fb169176f21fc70c3a7d1ddc8aa51d3&oe=59F581D8
Thought the answer would be B, but the excel book has put the answer as C.
TIA.
Post by: CyberScopes on October 28, 2017, 12:34:52 am
Hey thanks so much for the help. I totally agree with your work but the HSC answer is A. So I'm still a bit confused. If the questions just wrong was it omitted from the hsc?
Heres how I think of it:
We know Vp/Vs = np/ns
We also know that C and D are obviously wrong.
So we're left with A and B.
Both options assume Vp is constant, so set Vp to 1 for ease of calculation.
Since we're trying to maximise Vs for maximum deflection (Remembering that V = IR so more V means more current for deflection), we flip both sides so it becomes:
Vs/1 = ns/np
So Vs = ns/np
Now from this we can either increase ns, or decrease np to increase Vs.
Since increasing ns is not an option, u can only decrease np to increase Vs
Therefore answer is A
Post by: jamonwindeyer on October 28, 2017, 12:51:57 am
Hello! Would appreciate help with these MCs:
6. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22883926_1362403753885144_473053551_o.jpg?oh=51a20dd5d7c37a1fa4e92f54e4601332&oe=59F5ABED
Hey! For your first one, a current is produced in the wire due to its motion relative to the earths magnetic field. Note that the wire has very low resistance. So, going through the options:
- The orientation of the wire makes a huge difference - It needs to cut through magnetic field lines for an induced current to flow. So D is out.
- The speed of rotation affects the rate of change of magnetic flux, so C is out.
- The length of the wire also affects the size of the induced current (think of it like the longer wire being able to cut through more flux lines). So A is out.
The answer is B - The wire has a very low resistance so increasing its thickness won't do much at all ;D
Quote
9. https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22883814_1362403813885138_1145465314_o.jpg?oh=2fb169176f21fc70c3a7d1ddc8aa51d3&oe=59F581D8
Thought the answer would be B, but the excel book has put the answer as C.
TIA.
The Excel book has simplified the scenario a tad - I know what you are thinking, but the eddy currents in both P and R are both negligible. P is an insulator so they don't flow, and the break in R has the same effect. So assuming both are negligible, they'll hit the bottom at the same time and then Q will follow ;D
Post by: jamonwindeyer on October 28, 2017, 12:57:41 am
If any questions got missed or we were stuck anywhere, repost/quote it below and I'll lend a hand if I can!! I've just finished submitting three assignments in 24 hours so I might not be much use at all, but hey, keen to do what I can to help you guys smash it out of the park on Monday ;D
Post by: Dante1091 on October 28, 2017, 01:37:11 am
Are we allowed to use an ampersand (&), in replacement of 'and' in science tests. Since I know we aren't allowed to use it in subjects like English, but I'm unsure about the sciences.
Post by: jamonwindeyer on October 28, 2017, 02:15:45 am
Hello guys,
Are we allowed to use an ampersand (&), in replacement of 'and' in science tests. Since I know we aren't allowed to use it in subjects like English, but I'm unsure about the sciences.
Absolutely - It probably would have been fine in English too but definitely fine here :)
Post by: bsdfjnlkasn on October 28, 2017, 08:13:05 am
Boron doping means it is p type so extra holes.
Positive flow goes left in the silicon block hence holes move left, so therefore electrons move right in the valence band as they causing the apparent hole movement, thus A.
Hey there,
Wouldn't the negative electrons move towards the positive terminal, so the left? Against conventional current i.e. positive hole flow?
Not sure, let me know if you can please clarify :)
Post by: blasonduo on October 28, 2017, 08:28:18 am
Hey there,
Wouldn't the negative electrons move towards the positive terminal, so the left? Against conventional current i.e. positive hole flow?
Not sure, let me know if you can please clarify :)
Yep! :)) He accidentally thought the positive terminal was on the right :P all good!
Post by: bsdfjnlkasn on October 28, 2017, 08:38:31 am
I'm not sure how to calculate this, legit done it 3 times and am still getting B? I believe I've calculated the period and mass and converted it... but still confused
With the M-M question, all the answers seem wrong to be honest
With the coil, doesn't the CW current produce a magnetic field going into the page between the two rings? So that causes the inner ring to experience an increase in flux there meaning it will produce a current such that the magnetic field is out of the page (to nullify the increase going into the page?) This current will have to be CW. Answer is D?
Thanks!
Post by: blasonduo on October 28, 2017, 08:55:46 am
Hey there,
I'm not sure how to calculate this, legit done it 3 times and am still getting B? I believe I've calculated the period and mass and converted it... but still confused
With the M-M question, all the answers seem wrong to be honest
With the coil, doesn't the CW current produce a magnetic field going into the page between the two rings? So that causes the inner ring to experience an increase in flux there meaning it will produce a current such that the magnetic field is out of the page (to nullify the increase going into the page?) This current will have to be CW. Answer is D?
Thanks!
Hey! Lets see what I can do :))
6) Straight out of the bat, we know that Michelson and Morley had a very valid experiment, so C and D are wrong. If we look at A, it says the experiment tested the theory of relativity, which is very wrong, the theory of relativity was not tested, however the test did strengthen that theory.
5) This question requires us to use the formula:
which is closest to D :)
11) Your theory is all right! However, if the second ring were to go clockwise the magnetic field in the middle of the rings would be into the page, and would help increase the magnetic field. If it were to go anti-clockwise, we would get that magentic field out of the page, opposing the change in flux, so it HAS to be anti-clockwise.
Hope this helps ;)
Post by: maevecouch on October 28, 2017, 08:58:38 am
thanks cyber scope, the explanation made it a little clearer. but im struggling to understand how the time of the radio pulses allows them to measure relaxation time. like how are we measuring these and what is the distinct difference which we are observing and quantifying? all the explanations just seem inadequate. like so the magnetic field is in the transcerse plane when the hydrogen nuclei move to anti parallel - but then what, why does that matter and what does that have to do with the relation time? is it to do with that magnetic vector or the hydrogen atoms precessions im just very confused. im trying to work out analogies and stuff to simply and explain to my friends and class mates but im strugglingOk so I'm doing medical physics too and I have to agree that the "Relaxtion" section of the syllabus is vague around the complexity. If I'm right (which I'm most likely not), I believe that when the radio waves are applied to nuclei, the parallel aligned nuclei will absorb the energy and move into their higher anti parallel alignments. When the radio waves being applied are switched off, in accordance to the Law of Conservation of Energy, the nuclei 'relax' or return from their anti parralel to parallel alignments and release the radio wave energy they absorbed. The time it takes to do such is measured by MRI scanners which pick up these radio waves to enhance tissue contrast (or ehance the areas where this relaxtion is occuring) and use Echo Delay Time or Repetition time depending on whether they want to measure T1 or T2 relaxtion.
Post by: winstondarmawan on October 28, 2017, 09:11:05 am
The Excel book has simplified the scenario a tad - I know what you are thinking, but the eddy currents in both P and R are both negligible. P is an insulator so they don't flow, and the break in R has the same effect. So assuming both are negligible, they'll hit the bottom at the same time and then Q will follow ;D
Thank you for your response!
I'm just wondering why the eddy currents on R are negligible. I've always learnt that they had an impact on these types of questions, because even thought they are significantly smaller than the looped current produced in Q, they are still there.
Post by: bsdfjnlkasn on October 28, 2017, 09:21:43 am
11) Your theory is all right! However, if the second ring were to go clockwise the magnetic field in the middle of the rings would be into the page, and would help increase the magnetic field. If it were to go anti-clockwise, we would get that magentic field out of the page, opposing the change in flux, so it HAS to be anti-clockwise.
Hope this helps ;)
Hey there,
Still not so sure about this last explanation..
If the current in the inner coil is going ACW, then between the two rings, there will be a very large B field going into the page instead of there being nothing, which the CW current in the innermost coil would achieve.
Maybe if you could draw the magnetic fields or something? Really struggling with visualising it.. Thank you! :D
Post by: blasonduo on October 28, 2017, 09:44:54 am
Hey there,
Still not so sure about this last explanation..
If the current in the inner coil is going ACW, then between the two rings, there will be a very large B field going into the page instead of there being nothing, which the CW current in the innermost coil would achieve.
Maybe if you could draw the magnetic fields or something? Really struggling with visualising it.. Thank you! :D
Hey I've attempted a drawing down below XD
I think there are 2 keys components here;
1) Clockwise produces a SOUTH pole, Anti-clockwise a NORTH pole
2) Look at the bigger picture, yeah, maybe the area between them might contradict, but look at what would have a bigger impact (which is the area in the middle) The overall net force here will decrease :)
Still unsure? Keep asking away! :))
Post by: beau77bro on October 28, 2017, 09:58:19 am
Ok so I'm doing medical physics too and I have to agree that the "Relaxtion" section of the syllabus is vague around the complexity. If I'm right (which I'm most likely not), I believe that when the radio waves are applied to nuclei, the parallel aligned nuclei will absorb the energy and move into their higher anti parallel alignments. When the radio waves being applied are switched off, in accordance to the Law of Conservation of Energy, the nuclei 'relax' or return from their anti parralel to parallel alignments and release the radio wave energy they absorbed. The time it takes to do such is measured by MRI scanners which pick up these radio waves to enhance tissue contrast (or ehance the areas where this relaxtion is occuring) and use Echo Delay Time or Repetition time depending on whether they want to measure T1 or T2 relaxtion.see i understood u right up until the end - how do they pick up this relaxtion time - i understand they simply measure the time to return, or for the magnetic vector to return, but how is that localised (is that with the gradient coil - but how does a big as radio coil which is measuring a large change identify just one localised area - ik this is the complicated part so ill ignore that) and also echo delay and repetition time - i feel uve said everything i need to know - i just need to grasp it. thankyou soo much! could u just explain that last bit?
Post by: Mymy409 on October 28, 2017, 10:27:23 am
Post by: CyberScopes on October 28, 2017, 10:45:23 am
TIA
Post by: bsdfjnlkasn on October 28, 2017, 10:47:48 am
Hey I've attempted a drawing down below XD
I think there are 2 keys components here;
1) Clockwise produces a SOUTH pole, Anti-clockwise a NORTH pole
2) Look at the bigger picture, yeah, maybe the area between them might contradict, but look at what would have a bigger impact (which is the area in the middle) The overall net force here will decrease :)
Still unsure? Keep asking away! :))
Thanks! Makes heaps of sense now - it was the bigger picture I should have been looking at :D
Post by: Zainbow on October 28, 2017, 10:55:38 am
Post by: Zainbow on October 28, 2017, 11:03:18 am
Sort of a general question, in many questions and answers ive seen that when the value of 'q' of an electron is taken, its taken as a positive rather than a negative as stated in the data sheet. My question is what does the negative actually mean, and why do we take a positive for the value of the charge?
TIA
Usually, the questions would require you to calculate the magnitude of a force and its direction, so to calculate the magnitude you only need the value of the charge. The negative is only there to indicate direction, or that it behaves opposite to a positively charged particle with the same value for charge.
Post by: maevecouch on October 28, 2017, 11:04:30 am
see i understood u right up until the end - how do they pick up this relaxtion time - i understand they simply measure the time to return, or for the magnetic vector to return, but how is that localised (is that with the gradient coil - but how does a big as radio coil which is measuring a large change identify just one localised area - ik this is the complicated part so ill ignore that) and also echo delay and repetition time - i feel uve said everything i need to know - i just need to grasp it. thankyou soo much! could u just explain that last bit?It's pretty convultued hey? By the last part I'm guessing you mean Repition Time and Echo Delay. For T1 Relaxtion they use a short Repition Time to match the short T1 they use. The repition time is the time elapsed between consecutive pulse of input radio waves. The relaxtion of the M vector happens very quickly, leaving some 'free' radio waves to be absorbed. So, the extra radio waves will make the image appear brighter as it is a more 'active' area. Whereas T2 use a long Echo Delay Time to match the long T2 in use. Echo time delay is the time delay betwee the emission of radio waves and measurement of those which return first. So in T2 images, the signals from the T2 relaxtion are measured after the initial radio waves are sent out which results in tissues with only long T2 contributing to the returning signals. Also, in T1 images, the Echo Delay Time is short so as to suppress any T2 and likewise, a long Repition Time is used in T2 images to suppress any T1. Hope that makes sense?
Post by: raymatar on October 28, 2017, 11:20:47 am
Post by: blasonduo on October 28, 2017, 11:22:06 am
Does anyone have a resource with answers for papers between 2001 and 2007?
https://tianjara.net/hsc/notes/Phys_Past_Paper_Sol.pdf
This is the best one I've found ;)
Post by: raymatar on October 28, 2017, 11:24:48 am
https://tianjara.net/hsc/notes/Phys_Past_Paper_Sol.pdfw
This is the best one I've found ;)
Yeah that's the same one I have but it's missing some answers and medical physics responses. Thanks for the help though.
Post by: beau77bro on October 28, 2017, 11:30:09 am
It's pretty convultued hey? By the last part I'm guessing you mean Repition Time and Echo Delay. For T1 Relaxtion they use a short Repition Time to match the short T1 they use. The repition time is the time elapsed between consecutive pulse of input radio waves. The relaxtion of the M vector happens very quickly, leaving some 'free' radio waves to be absorbed. So, the extra radio waves will make the image appear brighter as it is a more 'active' area. Whereas T2 use a long Echo Delay Time to match the long T2 in use. Echo time delay is the time delay betwee the emission of radio waves and measurement of those which return first. So in T2 images, the signals from the T2 relaxtion are measured after the initial radio waves are sent out which results in tissues with only long T2 contributing to the returning signals. Also, in T1 images, the Echo Delay Time is short so as to suppress any T2 and likewise, a long Repition Time is used in T2 images to suppress any T1. Hope that makes sense?
VERY NICELY PUT (I FEEL) - GONNA NEED TO READ OVER A BIT AND A COUPLE TIMES. THANKYOU. anyone else wants to add anything greatly appreciated.
Post by: Baylsskool on October 28, 2017, 12:15:58 pm
I mean with the ultraviolet catastrophe,why can't there be an infinite source of energy for an infinitly small wavelength, cause it checks out mathematically, we just havnt discovered any frequencies higher than gamma, of course it can't be infinite but surely there's more to discover so how can the peak wavelength be so?? Shouldn't it technically, if we did find the smallest wavelength possible at least hit the wall of the graph and decrease from there???
Or advise someone I can talk to about this?
Post by: mary123987 on October 28, 2017, 12:48:18 pm
also q 25 b) is it sufficient to use Vp/Vs = Is/Ip and say the info is incorrect as Vp/Vs ≠Is/Ip?
thankyou :)
Post by: winstondarmawan on October 28, 2017, 01:08:43 pm
Thank you for your response!
I'm just wondering why the eddy currents on R are negligible. I've always learnt that they had an impact on these types of questions, because even thought they are significantly smaller than the looped current produced in Q, they are still there.
Bumpp
Post by: mary123987 on October 28, 2017, 01:13:19 pm
Can anyone explain why black bodies actually have peaks, i mean theoretically it would work if the curve was exponentially decreasing but that's not what Planck discovered, why is there a section of energy missing at the start, why does the body not emit large amounts of energy due to the lower wavelengths being given off?? Is there an explanation for this missing gap?Hey so just to give you a quick explanation the initial curve as indicated by classical theorty suggested that as wavelength decreased radiance/intensity would increase to infinity now classical theory was wrong as firstly evidence showed otherwise and secondly thus increase in energy would violate the principle of consevation of energy as energy is supposedly being created without anthing causing it !
I mean with the ultraviolet catastrophe,why can't there be an infinite source of energy for an infinitly small wavelength, cause it checks out mathematically, we just havnt discovered any frequencies higher than gamma, of course it can't be infinite but surely there's more to discover so how can the peak wavelength be so?? Shouldn't it technically, if we did find the smallest wavelength possible at least hit the wall of the graph and decrease from there???
Or advise someone I can talk to about this?
Also from my understanding there is no actual section of energy missing in the beginning .
I get that your confused about why there can't be an infinite source of energy for an infinitly small wavelength but the reason is radiation emitted and absorbed by the walls of a black body cavity is quantised meaning the energy of photons is related to the frequency and thus in accordance to Einstein's contribution there is a threshold frquency , even if a light is very bright and carries large energy .
So whilst E=hf (which equals E= hc/λ)suggests that there can be an infinite source of energy for an infinitly small wavelength Einstein said nope actually there is a threshold frquency and this theory was discovered based on experimental data thus the reason for the peak .
Hope it makes sense
Post by: jamonwindeyer on October 28, 2017, 01:21:17 pm
VERY NICELY PUT (I FEEL) - GONNA NEED TO READ OVER A BIT AND A COUPLE TIMES. THANKYOU. anyone else wants to add anything greatly appreciated.
Keep in mind that the data can be localised because the magnetic field is slightly different in every voxel, as set up by the gradient coils - So we can use this to figure out where our readings are coming from ;D
I can't possibly comprehend how complex these algorithms must be to reconstruct these images, crazy stuff!
Thank you for your response!
I'm just wondering why the eddy currents on R are negligible. I've always learnt that they had an impact on these types of questions, because even thought they are significantly smaller than the looped current produced in Q, they are still there.
I mean you could just as easily say that there are eddy currents in the plastic - But again, they are significantly smaller than the looped current produced in Q ;D so I suppose we are comparing insignificants, and the Excel book has chosen to ignore the difference. There probably would be one, but it would be tiny - Both would behave essentially as they normally would under gravity ;D
Post by: mary123987 on October 28, 2017, 01:22:08 pm
BumppHey Although the charge in R experiences a force the break in the copper ring ,the emf established does not have a complete circuit so current cant flow so it lands at the same time as the plastic ring as nothing is opposing its downward force thus it accelerated down at the same rate as the pastic one
Post by: jamonwindeyer on October 28, 2017, 01:25:29 pm
Hey guys was just doing question 24 b and i am confused as to why the force isnt negative as when i subbed it in to F=qV/d i got -4.005 x 10 ^-14 (only thing that hints it out is part c) where you have to find velocity and considering eventually you have to squareroot it doesnt make sense to have a negative well its impossible!)
also q 25 b) is it sufficient to use Vp/Vs = Is/Ip and say the info is incorrect as Vp/Vs ≠Is/Ip?
thankyou :)
Hey! Which paper is this from? I think I know these questions though sooo:
- When you find force, you can ignore the negative in \(q\) because it only affects the direction. So instead of \(-0.00004nN\) (your answer), you just write \(0.00004nN\) down, or up, or however you define the negative direction. You ignore any negatives in the formula itself ;D
- Definitely okay, but it is better to examine it in terms of the conservation of energy. If you compare \(V_pI_p\) to \(V_sI_s\), you'll find that you've created energy in the secondary coils, seemingly from nothing -> Output power is higher than input! If my memory of this question serves me right, at least ;D
Post by: mary123987 on October 28, 2017, 01:30:56 pm
yes your right power in the first one is 0.12 whilst in the second 0.2 that makes sense thanks for that your a legend
Post by: winstondarmawan on October 28, 2017, 02:12:15 pm
Post by: jamonwindeyer on October 28, 2017, 02:36:57 pm
How would one go about assessing the reliability, validity and accuracy of second hand sources?
A reliable source will match with other sources when cross checked, a valid source will be free from bias (was it written by someone from the Flat Earth Society, for example?) and an accurate source will have data and information that matches known fact ;D
Post by: winstondarmawan on October 28, 2017, 03:24:13 pm
I mean you could just as easily say that there are eddy currents in the plastic - But again, they are significantly smaller than the looped current produced in Q ;D so I suppose we are comparing insignificants, and the Excel book has chosen to ignore the difference. There probably would be one, but it would be tiny - Both would behave essentially as they normally would under gravity ;D
Hey Although the charge in R experiences a force the break in the copper ring ,the emf established does not have a complete circuit so current cant flow so it lands at the same time as the plastic ring as nothing is opposing its downward force thus it accelerated down at the same rate as the pastic one
Thank you! What about the case of slits in a metal sheet, the eddy currents are somewhat significant there right?
So now it is my understanding that rings require the entirety of the ring for a significant current to flow whereas sheets and cubes (e.g. transformers) can have slits. Would this be correct?
Post by: 12carpim on October 28, 2017, 03:55:44 pm
Got a question why does a three-phase generator use electricity to generate electricity? Makes no sense if someone could help.
Thank you very much!
Post by: julies on October 28, 2017, 04:43:43 pm
Also would a really heavy magnet be able to hover above the superconductor, or is the opposing magnetic force unable to cancel out its weight force?
If somebody could please clarify, it would be greatly appreciated! : )
Post by: khadeeja_ on October 28, 2017, 04:54:59 pm
Post by: austv99 on October 28, 2017, 04:57:34 pm
Braggs question: I'm not too sure how the interference pattern produced in the experiment exactly led them to conclude the crystal lattic structure. Is it because it showed a geometric/periodic structure? Is there any significance with the contructive and deconstructive interferences produced?
Quanta to Quarks: Would I includes Chadwicks discovery of the neutron and the corresponding explanation by conservation laws, then fermis demonstration of nuclear fission, increasing the understanding of the ability to break up the atomic structure for energy production. Also, would i include that he used paulis proposal of the neutrino to comprehensively explain beta decay?
TIA
Post by: CyberScopes on October 28, 2017, 05:10:52 pm
Hey guys, are we supposed to explain how a magnet levitates over a superconductor by Lenz's Law? I've heard that this is technically incorrect?
Also would a really heavy magnet be able to hover above the superconductor, or is the opposing magnetic force unable to cancel out its weight force?
If somebody could please clarify, it would be greatly appreciated! : )
This depends on the question, if the question asks how the magnet levitates if the magnet is dropped on the superconductor then yes, the Lenz's Law method should be correct. But if the magnet is initially left stationary on the superconductor, and say helium was poured so it reached its critical temperature and attained its superconducting qualities, then Lenz's Law cannot be used since there is no relative motion.
For this case, it can be explained through one of the vital properties of superconductors: that magnetic fields cannot penetrate through them. Because of this, if a magnet is left on the superconductor, its magnetic field lines are cut off at the boundaries of the superconductor. Now, the magnetic field lines need to somehow reconnect back to the magnet (or just go somewhere for a matter of fact), but since the superconductor cannot be penetrated, this cannot happen. As a result, this creates a diamagnetic surface at the superconductor. In order for the magnetic field lines to enter the opposite pole of the magnet, the magnet is then pushed upwards, thus causing it to appear as if its levitating.
Pretty much: Superconductors dont take in magnetic field lines -> Magnets field lines have to re-enter the opposite pole of the magnet -> Forces magnet to lift up.
Post by: julia9102 on October 28, 2017, 05:44:01 pm
could anyone please explain what eddy currents are and how we can determine its direction?
Post by: hinakamishiro on October 28, 2017, 08:36:11 pm
BTW i was also hoping someone could explain the difference between magnetic flux and magnetic flux density?
Thanks! :)
Post by: Jayden Nguyen on October 28, 2017, 08:37:45 pm
(https://i.imgur.com/NLszvJZ.png)
Post by: Baylsskool on October 28, 2017, 08:51:52 pm
Hey so just to give you a quick explanation the initial curve as indicated by classical theorty suggested that as wavelength decreased radiance/intensity would increase to infinity now classical theory was wrong as firstly evidence showed otherwise and secondly thus increase in energy would violate the principle of consevation of energy as energy is supposedly being created without anthing causing it !
Also from my understanding there is no actual section of energy missing in the beginning .
I get that your confused about why there can't be an infinite source of energy for an infinitly small wavelength but the reason is radiation emitted and absorbed by the walls of a black body cavity is quantised meaning the energy of photons is related to the frequency and thus in accordance to Einstein's contribution there is a threshold frquency , even if a light is very bright and carries large energy .
So whilst E=hf (which equals E= hc/λ)suggests that there can be an infinite source of energy for an infinitly small wavelength Einstein said nope actually there is a threshold frquency and this theory was discovered based on experimental data thus the reason for the peak .
Hope it makes sense
Yeah I guess so, so it safe to say that it doesn't increase infinittly due to the threshold frequency of the sun, so there's a certain point of frequency where a hot star emits high frequency but then due to the threshold, it stops it from going higher ?
Post by: sidzeman on October 28, 2017, 09:56:09 pm
Post by: hobocop on October 28, 2017, 10:27:30 pm
For instance, the radius value in the orbital velocity equation v= (Gm/r)^0.5 because the solutions in this past paper did not account for radius of earth.
Post by: beau77bro on October 28, 2017, 10:39:15 pm
i get the electrons are emitted by the photoelectric effect and travel towards the n-type since it is positively charged by the diode's depletion zone. but why does that create a current throuhgout? its only migrating a short distance? or does it go around the whole circuit? confused
Post by: CyberScopes on October 28, 2017, 10:44:26 pm
Hey guys can I please have some help with these questions?
BTW i was also hoping someone could explain the difference between magnetic flux and magnetic flux density?
Thanks! :)
Ill do what I can:
19. An electron is in an electric field, so its attracted to the positive plate. Since electric fields go from +ve to -ve, the image indicates that the bottom plate is positive and the top is negative. Therefore, the electron is attracted to the bottom plate, and will have force (D) pulling it down.
20. If GPE is half its original value, then its distance between Earth and the satellite has decreased. Assuming this, you can say that d goes from 1 to 1/2. Now assuming the other variables are constant, you can make Newtons Gravitational Equation become F = 1/d2. If d = 1, then F = 1. Since d is now 1/2, F becomes 1/(1/2)2 = 4. So the answer is C
(Please check this not sure if i did it correct, there may be another method.)
20. Im not sure I might get back to you if i find out
15. A stronger magnetic field means that the circle will have a smaller radius since it will turn quicker, so it cannot be C or D. Knowing F = qvB, F is proportional to B, so the force should double. In this case, the radius should be halved, so its B. (Sorry theres probably a better way to explain this by equating centripetal force with magnetic force but im too lazy :D)
Correct me if im wrong on any of them
Post by: CyberScopes on October 28, 2017, 10:46:53 pm
Hi, do all the radius values in satellite orbit equations take into account the objects altitude above earth and radius of earth?
For instance, the radius value in the orbital velocity equation v= (Gm/r)^0.5 because the solutions in this past paper did not account for radius of earth.
They should account for the distance from the centre of both masses, which includes the radius of the Earth. Weird that you say one of the questions didnt, maybe the radius given includes the radius of the Earth aswell?
Post by: hobocop on October 28, 2017, 10:55:30 pm
They should account for the distance from the centre of both masses, which includes the radius of the Earth. Weird that you say one of the questions didnt, maybe the radius given includes the radius of the Earth aswell?
It might just be a mistake. This is from the sample answers for the 2015 HSC paper, question 26c).
Post by: beau77bro on October 28, 2017, 10:59:00 pm
It might just be a mistake. This is from the sample answers for the 2015 HSC paper, question 26c).are u sure it wasnt r + a?
Post by: blasonduo on October 28, 2017, 11:01:54 pm
Hey guys can I please have some help with these questions?
BTW i was also hoping someone could explain the difference between magnetic flux and magnetic flux density?
Thanks! :)
ooh! This place is backlogged with questions!
I shall help with the electromagnets ;)
The first step is to figure out the direction of the magnetic field (either left to right of right to left)
From the positive terminal you can see that the magnets are going to be ELECTROMAGNETS, you can see that it wraps around the LEFT block in clockwise way, You know it has a SOUTH facing pole (I use Nanti-Socks to remember :P )
To double check, we'll do the other side, from the NEGATIVE terminal, this time, it is travelling in an ANTI-clockwise way (remember to look at it standing in the middle) This means that the RIGHT block is the NORTH pole. So the magnetic field lines follow from RIGHT to LEFT.
The next part is the simple right hand palm rule, as you know the + and - terminals, follow the current lines until the right option works. In this case, B is the only one that rotates clockwise. (Left side coil current goes up (so thumb up!) and magnetic field right to left (so fingers to right) shows the palm upwards and thus it rotating clockwise.
Basically, this question is asking about your ability to figure out electromagnets, As B and D are almost the same, where D's RIGHT block is coiled the wrong way!
Post by: blasonduo on October 28, 2017, 11:07:52 pm
It might just be a mistake. This is from the sample answers for the 2015 HSC paper, question 26c).
Hmm, I just went over the question, and the answers do account for the radius of the earth, if you are unsure, would you like me to show the working?
Post by: hobocop on October 28, 2017, 11:47:36 pm
Hmm, I just went over the question, and the answers do account for the radius of the earth, if you are unsure, would you like me to show the working?
Hmm, I just went over the question, and the answers do account for the radius of the earth, if you are unsure, would you like me to show the working?
Is this not just accounting for altitude? (Not sure if my solutions are same as yours)
Post by: CyberScopes on October 29, 2017, 12:36:44 am
ooh! This place is backlogged with questions!
I shall help with the electromagnets ;)
The first step is to figure out the direction of the magnetic field (either left to right of right to left)
From the positive terminal you can see that the magnets are going to be ELECTROMAGNETS, you can see that it wraps around the LEFT block in clockwise way, You know it has a SOUTH facing pole (I use Nanti-Socks to remember :P )
To double check, we'll do the other side, from the NEGATIVE terminal, this time, it is travelling in an ANTI-clockwise way (remember to look at it standing in the middle) This means that the RIGHT block is the NORTH pole. So the magnetic field lines follow from RIGHT to LEFT.
The next part is the simple right hand palm rule, as you know the + and - terminals, follow the current lines until the right option works. In this case, B is the only one that rotates clockwise. (Left side coil current goes up (so thumb up!) and magnetic field right to left (so fingers to right) shows the palm upwards and thus it rotating clockwise.
Basically, this question is asking about your ability to figure out electromagnets, As B and D are almost the same, where D's RIGHT block is coiled the wrong way!
Can you please explain how you determined the poles of the electromagnets? Is there a rule to determine this?
Post by: jamonwindeyer on October 29, 2017, 01:00:37 am
Thank you! What about the case of slits in a metal sheet, the eddy currents are somewhat significant there right?
So now it is my understanding that rings require the entirety of the ring for a significant current to flow whereas sheets and cubes (e.g. transformers) can have slits. Would this be correct?
Somewhat, but lessened significantly. Whether you still want to say they exist is really up to how pedantic you want to be ;D
Hey there!
Got a question why does a three-phase generator use electricity to generate electricity? Makes no sense if someone could help.
Thank you very much!
I assume you mean the current we use setting up the magnetic field? Basically, we get out more electricity than we put in setting up that field ;D
Hey could someone explain how transistors work? don't get the whole current flowing through emitter and base and collector thing..
You actually don't need to explain how a transistor works in the HSC course! You just need to know what they are used for - An electrical switch, or an amplifier ;D
Post by: jamonwindeyer on October 29, 2017, 01:09:50 am
Hey there,
could anyone please explain what eddy currents are and how we can determine its direction?
Hi Julia! Eddy currents are loops of induced current formed in conductors in response to a changing magnetic field. Remember that induced eddy currents flow to create a new magnetic field, to oppose the original change in magnetic field. We determine their direction using the right hand grip rule.
If we hold our right hand in a thumbs up/grip, then the thumb represents where we need the north pole to be. Our fingers represent the direction of the eddy current loop ;D
For the PE effect of this question, I would discuss how Einsteins ability to explain the PE effect with the quantisation of energy which classical physics could not - evidence for quantum theory and dual wave particle model for light. However I'm not sure what I would discuss for the special relativity theory - what did it change about the model of light, besides the fact that it is constant in all frames of references?
You've nailed it, we originally thought that light travelled in a medium and that its speed would be affected by relative motion in that medium - Einstein's Special Theory establishes the fact that \(c\) is a constant. That's a change in the model ;D
can someone explain photovoltaic cells please? i dont get the directions and attractions -the photons strike the n-p junction and the electrons travel where and why? how does it create a current from this?
i get the electrons are emitted by the photoelectric effect and travel towards the n-type since it is positively charged by the diode's depletion zone. but why does that create a current throuhgout? its only migrating a short distance? or does it go around the whole circuit? confused
It goes around the whole circuit, you need to think of the depletion zone like a voltage source. It is setting up an electric field which 'pushes' electrons around the circuit, just like a voltage source would. The electric field is directed from N to P, so electrons will flow in the opposite direction around the whole circuit ;D
Post by: pikachu975 on October 29, 2017, 02:02:35 am
This question so hard. From qanta to qwarks can someone help me with an example of how they would answer it? TIA
(https://i.imgur.com/NLszvJZ.png)
Both fission and fusion result in less mass of products hence the extra mass is converted into energy (E=mc^2) and released
Post by: arunasva on October 29, 2017, 02:14:06 am
[b ]the picture might be too small check out question 12 here http://www.acehsc.net/wp-content/uploads/TrialPapers/2011/Physics/2011_Physics_-_James_Ruse_Trial_with_Solutions.pdf [/b]
Post by: jamonwindeyer on October 29, 2017, 02:30:55 am
Hey yo, any help for this question is appreciate. Guess this is the last time I'm askin for help here
[b ]the picture might be too small check out question 12 here http://www.acehsc.net/wp-content/uploads/TrialPapers/2011/Physics/2011_Physics_-_James_Ruse_Trial_with_Solutions.pdf [/b]
Hey! Induced emf (E) is proportional to rate of change (or gradient) of magnetic flux. Magnetic flux has two regions where it is changing, both linearly. Since both changes are linear, their rate of change is constant, so we'll get a constant induced emf. This narrows down to A or B.
The rate of change of flux at the tail end of the graph is greater - Steeper gradient. So, this will cause a larger induced emf, in the opposite direction to what we had initially. The answer is A ;D
Post by: arunasva on October 29, 2017, 03:10:25 am
Hey! Induced emf (E) is proportional to rate of change (or gradient) of magnetic flux. Magnetic flux has two regions where it is changing, both linearly. Since both changes are linear, their rate of change is constant, so we'll get a constant induced emf. This narrows down to A or B.Thanks man !
The rate of change of flux at the tail end of the graph is greater - Steeper gradient. So, this will cause a larger induced emf, in the opposite direction to what we had initially. The answer is A ;D
Post by: hobocop on October 29, 2017, 08:15:46 am
Thanks.
Post by: beau77bro on October 29, 2017, 08:36:38 am
Keep in mind that the data can be localised because the magnetic field is slightly different in every voxel, as set up by the gradient coils - So we can use this to figure out where our readings are coming from ;D
wait so does the mri measure the change in Magnetic vector or the re-emitted radio waves? how do they make the image? so the radio wave returns in the specific larmor frequency set up by the gradient coils and what information do they carry? oh wait is the magnetic vector just a way of explaining the time taken for the relaxation to happen?
Post by: austv99 on October 29, 2017, 09:45:56 am
Hey! Induced emf (E) is proportional to rate of change (or gradient) of magnetic flux. Magnetic flux has two regions where it is changing, both linearly. Since both changes are linear, their rate of change is constant, so we'll get a constant induced emf. This narrows down to A or B.Isnt the induced emf equal to negative change in flux over change in time, meaning the graph should be flipped?
The rate of change of flux at the tail end of the graph is greater - Steeper gradient. So, this will cause a larger induced emf, in the opposite direction to what we had initially. The answer is A ;D
Post by: austv99 on October 29, 2017, 09:49:06 am
Would appreciate if someone could help me clarify these questions.Bump :)
Braggs question: I'm not too sure how the interference pattern produced in the experiment exactly led them to conclude the crystal lattic structure. Is it because it showed a geometric/periodic structure? Is there any significance with the contructive and deconstructive interferences produced?
Quanta to Quarks: Would I includes Chadwicks discovery of the neutron and the corresponding explanation by conservation laws, then fermis demonstration of nuclear fission, increasing the understanding of the ability to break up the atomic structure for energy production. Also, would i include that he used paulis proposal of the neutrino to comprehensively explain beta decay?
TIA
Post by: Baylsskool on October 29, 2017, 09:53:07 am
Use this link for the photo of why I'm asking https://m.imgur.com/a/QL5F2
Post by: jamonwindeyer on October 29, 2017, 11:31:52 am
wait so does the mri measure the change in Magnetic vector or the re-emitted radio waves? how do they make the image? so the radio wave returns in the specific larmor frequency set up by the gradient coils and what information do they carry? oh wait is the magnetic vector just a way of explaining the time taken for the relaxation to happen?
The radio waves, but because the magnetic field is different everywhere, this affects the radio waves that are emitted in that location!
Hi, could I please get an explanation on why this is C?
Thanks.
The radial magnetic field removes the cosine term from the torque equation by ensuring the force on the conductor lengths is always perpendicular to the coil, but it actually doesn't do anything to the magnitude of the force! That is always just \(F=BIl\), it just isn't oriented optimally in a non-radial magnetic field ;D so force is always constant (swapping direction every half turn in a DC motor), which matches C :)
Isnt the induced emf equal to negative change in flux over change in time, meaning the graph should be flipped?
You are right, but the flipped version isn't presented! Which means we are just measuring EMF the opposite way, and that's okay ;D
Post by: jamonwindeyer on October 29, 2017, 11:36:08 am
So normally a moving conductor inside or being affected by a magnetic field,(key word moving), has induced current into it(not perfect), however in this case, a superconductor apparently only needs to be affected by a magnetic field to have a perfect induced current, am I reading it wrong, or is it that the superconductor doesn't need to move? And if so my question is why wouldn't it need to move to do this, as this contradicts Lenz law??
Use this link for the photo of why I'm asking https://m.imgur.com/a/QL5F2
This highlights the issue with explaining the Meisner Effect in this way, the textbook is incorrect! The Meisner Effect is distinct from the idea of eddy currents, so yes, a magnet would need to move for eddy currents to form! And the Meisner Effect is separate, it happens for other reasons that you don't really need to know ;D
Post by: eternalconflux on October 29, 2017, 12:04:25 pm
Here's the question.
When a zinc plate was exposed to ultraviolet light, electrons were released. One electron had kinetic energy of 3.0 x 10^-19 J
a) If a retarding voltage of 0.80V was applied, calculate the kinetic energy of the electron when it reached the anode.
b) Find the minimum retarding voltage that would inhibit the electron from reaching the anode.
Since my inability to solve this stems from my lack of understanding on the relationship between work function, retarding/stopping voltage and threshold frequency, an explanation would be much appreciated on how to solve these questions! :D I get the general gist of what photoelectric effect is (planck proposed light was packets of discrete energy, and Einstein, from Planck's observations, made further discoveries and derived that light was instead quantised into photons, and that there was a threshold frequency required before photoelectric effect could take place, and then intensity would take its part after threshold frequency was reached.
Post by: beau77bro on October 29, 2017, 12:39:01 pm
Post by: sidzeman on October 29, 2017, 12:55:47 pm
For part b) i - what would you discuss. I'm not quite sure why its given 3 marks
Post by: Baylsskool on October 29, 2017, 12:56:25 pm
This highlights the issue with explaining the Meisner Effect in this way, the textbook is incorrect! The Meisner Effect is distinct from the idea of eddy currents, so yes, a magnet would need to move for eddy currents to form! And the Meisner Effect is separate, it happens for other reasons that you don't really need to know ;DYeah alright cheers jamon
Could you explain the exclusion principle as electrons cannot be in the same place at the same time at the same spin, so what did Heisenberg then tell us when he said that the momentum and displacement of a certain electron could not be determined simultaneously ? That both combined meant, it's uncertain where electrons are but they are never in the same spot as each other basically??
Post by: aryak on October 29, 2017, 01:35:46 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf
Post by: winstondarmawan on October 29, 2017, 01:41:57 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/23022130_1363865803738939_1475372061_n.png?oh=4cf36567bde7eb0f93a759853295f976&oe=59F77BE6
Post by: beau77bro on October 29, 2017, 01:42:06 pm
Does anyone know where I can get sample answers for the option medical phys from hsc 2001 - 2007? The sample answers of Andrew Harvey only include the option Quanta to Quarks
For part b) i - what would you discuss. I'm not quite sure why its given 3 marks
for the sample answers - you most likely need to buy a success one or coroneos book.
in terms of reliability you can say:
- you checked if the author was credible, and whether their sources were reputable.
- you compared this info with other sources to see if it was consistent
- and you checked with your teacher (you can also say that the principles used clearly show the logic/reasoning to support the concepts - but hard to say with how complex mri is)
Post by: sidzeman on October 29, 2017, 01:49:15 pm
for the sample answers - you most likely need to buy a success one or coroneos book.Wouldn't the authors being credible relate to accuracy, and not reliability?
in terms of reliability you can say:
- you checked if the author was credible, and whether their sources were reputable.
- you compared this info with other sources to see if it was consistent
- and you checked with your teacher (you can also say that the principles used clearly show the logic/reasoning to support the concepts - but hard to say with how complex mri is)
Post by: beau77bro on October 29, 2017, 01:51:40 pm
Hi guys. I have a question related to photoelectric effect that i need help with.
Here's the question.
When a zinc plate was exposed to ultraviolet light, electrons were released. One electron had kinetic energy of 3.0 x 10^-19 J
a) If a retarding voltage of 0.80V was applied, calculate the kinetic energy of the electron when it reached the anode.
b) Find the minimum retarding voltage that would inhibit the electron from reaching the anode.
Since my inability to solve this stems from my lack of understanding on the relationship between work function, retarding/stopping voltage and threshold frequency, an explanation would be much appreciated on how to solve these questions! :D I get the general gist of what photoelectric effect is (planck proposed light was packets of discrete energy, and Einstein, from Planck's observations, made further discoveries and derived that light was instead quantised into photons, and that there was a threshold frequency required before photoelectric effect could take place, and then intensity would take its part after threshold frequency was reached.
If these answers are right I can explain it. I'm not sure though, so if they aren't I can't help and actually need help too hahah
(http://uploads.tapatalk-cdn.com/20171029/c1994ee812960b3a8df408c92721a49e.jpg)
Post by: beau77bro on October 29, 2017, 01:55:32 pm
Can someone explain to me 2014, question 26 (b). How do I find the x -intercept? Help would be appreciated.volts are equal to eV since the units are 1:1 (im pretty sure - jamon can help here), so what you can do is create a linear y = mx + b, with y = eV, x = frequency, b = - 4.1 eV and M = the gradient from the line. then u can sub in the eV = 1.2 and rearrange to find the frequency, or just do it by graphing the new line by making eV = 0 and finding the threshold frequency whilst keeping the gradient the same as the first graph.
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf
Post by: beau77bro on October 29, 2017, 02:01:13 pm
Wouldn't the authors being credible relate to accuracy, and not reliability?yes, but i also think being credible means their information is reliable (aka consistent or trustworthy - since it's hard to repeat a research task without just comparing this is a good measure of reliability). also reliability is about repeatability - so if the source isn't credible/is faulty then it's unlikely you will find a common answer from an esteemed professor or whomever.
Post by: sidzeman on October 29, 2017, 02:02:44 pm
yes, but i also think being credible means their information is reliable (aka consistent or trustworthy - since it's hard to repeat a research task without just comparing this is a good measure of reliability). also reliability is about repeatability - so if the source isn't credible/is faulty then it's unlikely you will find a common answer from an esteemed professor or whomever.Ahhh I see thank you very much!
Could someone also help me with part d of question 20 from the 2006 hsc?
i attempted it by multiplying the mass of the rod by 9.8 to give to force that should act on the balance. Then using a given current and the mass at that force i calculated the actual force that was acting on the balance (e.g. at I = 2.8, reading = 0.5485 - times that by 9.8 to give force on balance). Subtract the original weight from this = force due to parallel conductors = equate to lkII/d to find d - however this does not work
Post by: beau77bro on October 29, 2017, 02:13:27 pm
Ahhh I see thank you very much!
Could someone also help me with part d of question 20 from the 2006 hsc?
i attempted it by multiplying the mass of the rod by 9.8 to give to force that should act on the balance. Then using a given current and the mass at that force i calculated the actual force that was acting on the balance (e.g. at I = 2.8, reading = 0.5485 - times that by 9.8 to give force on balance). Subtract the original weight from this = force due to parallel conductors = equate to lkII/d to find d - however this does not work
ok so the force is actually the difference between the weight and a measurement taken, so make sure to go weight - measurement then x by 9.8. wait u did that. ill give it a go
Post by: beau77bro on October 29, 2017, 02:23:28 pm
I think ur (or my graph) may be off because when I graphed it I got 3.2 for 0.5485 not 2.8 - is that the answer?
Post by: mary123987 on October 29, 2017, 02:31:58 pm
Post by: banway on October 29, 2017, 02:35:43 pm
Post by: beau77bro on October 29, 2017, 02:43:26 pm
What sort of bottom raw mark do you reckon you'll need for a band 5 or 6 in physics?
depends on how hard/easy the paper is so who knows yet - this gives and indication tho http://rawmarks.info/wiki/Main_Page
Post by: aryak on October 29, 2017, 03:07:50 pm
volts are equal to eV since the units are 1:1 (im pretty sure - jamon can help here), so what you can do is create a linear y = mx + b, with y = eV, x = frequency, b = - 4.1 eV and M = the gradient from the line. then u can sub in the eV = 1.2 and rearrange to find the frequency, or just do it by graphing the new line by making eV = 0 and finding the threshold frequency whilst keeping the gradient the same as the first graph.[/quote]
How did you find the value of b?
Post by: sidzeman on October 29, 2017, 03:25:29 pm
(http://uploads.tapatalk-cdn.com/20171029/2a3fbc08d7b2fe35db7e3442e754d0b0.jpg)Yup around there I'm pretty sure the solutions were just wrong hahaha thank you! Also (this is probably a common question but) what is the difference between relativity and special relativity?
I think ur (or my graph) may be off because when I graphed it I got 3.2 for 0.5485 not 2.8 - is that the answer?
Post by: beau77bro on October 29, 2017, 03:29:04 pm
Yup around there I'm pretty sure the solutions were just wrong hahaha thank you! Also (this is probably a common question but) what is the difference between relativity and special relativity?well i think relativity can either just be in general physics where things are relative. or its the general theory of relativity which im pre sure includes non-inertial frames of reference (so like i think it relates alot to gravity like acceleration due to gravity - which apparently isn't real lmao), whilst special is just inertial.
How did you find the value of b?
oh because in the original graph there is an additional 4.1V helping them pass, or 4.1eV, so if u minus that it will give you the graph with a 0 voltages, or 0 eV, applied
Mod edit: Removed duplicate post. Do not double post unless necessary. Edit existing post instead
Post by: beau77bro on October 29, 2017, 03:35:29 pm
BUMPno their answer is correct, because the side AB is always perpendicular. the other side may not be, but AB will always be.
Post by: beau77bro on October 29, 2017, 03:38:25 pm
hey so why do we use positron emitters rather than just gamma emitters?BUMP
Post by: sidzeman on October 29, 2017, 03:58:28 pm
well i think relativity can either just be in general physics where things are relative. or its the general theory of relativity which im pre sure includes non-inertial frames of reference (so like i think it relates alot to gravity like acceleration due to gravity - which apparently isn't real lmao), whilst special is just inertial.Hmmm I still dont fully understand sorry. Also, could someone explain why the answer to this is B and not A?
Post by: beau77bro on October 29, 2017, 04:11:03 pm
Note: the electron needs more eV than what is required by the band gap. So it has to be less than the 0.155eV we get from our working. So it has to be A... ok I'm confused. Where's this from?
Post by: blasonduo on October 29, 2017, 04:15:53 pm
BUMP
My understanding was that positron was just as effective as gamma emitters, as, in the end, gamma radiation is always produced. For example, Technetium 99m is a gamma emitter, and it is very useful in the medical world!
Post by: pikachu975 on October 29, 2017, 04:26:50 pm
Since the products have higher mass, does this come from the fact that the products have less binding energy so the energy is converted to mass? Also how is energy released in nuclear reactions?
Thanks
Post by: khadeeja_ on October 29, 2017, 04:44:00 pm
Are there any good sources that summarise the impacts of medical physics on society, for x-rays, ct scans, ultrasound etc.. like in tables?
Mod edit: Removed duplicate posting. Please edit your previous post in future.
Post by: winstondarmawan on October 29, 2017, 04:58:54 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/23113597_1363981670394019_1360459847_n.png?oh=c3f94f6c6d29cbbf6ba0dc38a8752be5&oe=59F726B1
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894985_1363980263727493_1683153853_n.png?oh=820e57fbde8c1b68d7cea1ed59384be5&oe=59F82BC1
How many plots should their be until it is considered a valid relationship?
Post by: CyberScopes on October 29, 2017, 05:17:30 pm
From the 2016 HSC:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/23113597_1363981670394019_1360459847_n.png?oh=c3f94f6c6d29cbbf6ba0dc38a8752be5&oe=59F726B1
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894985_1363980263727493_1683153853_n.png?oh=820e57fbde8c1b68d7cea1ed59384be5&oe=59F82BC1
How many plots should their be until it is considered a valid relationship?
I dont think theres a certain amount, but I believe the question wanted to find a flaw/benefit of each data set, and the flaw of A would be that it didnt have as much data as that of B.
Post by: sidzeman on October 29, 2017, 05:18:59 pm
Ok so we estimate the wavelength is roughly 8um. Thats 8 x 10^-6m, we sub that into E = hf = hc/wavelength and that gives us our energy in joules. Now eV is in joules so we simply divide by the value given on the formula sheet - 1.602x10^-19. And that gives us our value of eV.The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well
Note: the electron needs more eV than what is required by the band gap. So it has to be less than the 0.155eV we get from our working. So it has to be A... ok I'm confused. Where's this from?
Post by: Checkmate123 on October 29, 2017, 05:20:27 pm
Can someone pls help with Q2Q reactions:Not quite sure about the reactions you are talking about. But, I can answer on why energy is released in nuclear reactions.
Since the products have higher mass, does this come from the fact that the products have less binding energy so the energy is converted to mass? Also how is energy released in nuclear reactions?
Thanks
If you look at the mass number vs average binding energy graph, you can see that average binding energy increases up until around Iron, then decreases.
In fission, you are splitting a nucleus into smaller nuclei, such as breaking down uranium into barium and krypton. From the graph, baryon and krypton have higher binding energy than uranium. Since binding energy is related to mass defect, the products of this reaction have higher mass defect than uranium. So, if you were to actually measure the masses of baryon, krypton and uranium, you would see that there appears to be a loss of mass. This 'missing mass' is released in the form of energy by energy-mass equivalence. Similarly, fusion of smaller nuclei to form a large nuclei can also release energy, provided the average binding energy of product is higher than reactants.
Note: Nuclear reactions always release energy provided the binding energy of products is greater than binding energy of reactants.
Post by: Mymy409 on October 29, 2017, 05:30:23 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf
Post by: blasonduo on October 29, 2017, 05:33:46 pm
The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well
Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A :))
Can someone please help me with Q12 from the 2014 HSC? I don't quite understand what's going on.
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf
Hey! This was answered a few pages back!
https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472
Here it is!
Post by: sidzeman on October 29, 2017, 05:48:37 pm
Sorry I don't have the book on me anymore, so I can't show you :((
Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A :))
Hey! This was answered a few pages back!
https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472
Here it is!
Post by: pikachu975 on October 29, 2017, 05:59:29 pm
The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well
Should be B! 0.155 eV doesn't mean the semiconductor wont work, as moving slightly away from the peak would give 0.155 eV but still have a pretty high intensity on the curve. Using the semiconductor from A, with like 0.03 eV, there are barely any photons with that energy hence it wouldn't be suitable.
Post by: sidzeman on October 29, 2017, 06:03:49 pm
Post by: blasonduo on October 29, 2017, 06:07:55 pm
Sorry I don't have the book on me anymore, so I can't show you :((
Just as pikachu said, but i'll use calculations
I say this because of this;
Post by: winstondarmawan on October 29, 2017, 06:10:10 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894854_1364012660390920_933578363_n.png?oh=b87e2f32349fb86ce6953e8bc9eaba08&oe=59F85C98
TIA!
Post by: Zainbow on October 29, 2017, 06:10:35 pm
Can anyone help me with 19 - should the electrical resistance be high for rapid heating, as high resistance = greater heat by resistive heating? Therefore answer should be D, but is in fact C
Hey! The pot gets heated up when eddy currents are formed in the base. The larger the eddy currents, the more resistance THEY will form and hence the more heat produced. The pot base itself doesn't require high resistivity - in fact, the more electrical resistance it has, the less likely eddy currents will form, knowing that the more electrically conductive the material is, the larger these eddy currents will be.
Post by: blasonduo on October 29, 2017, 06:12:40 pm
Can anyone help me with 19 - should the electrical resistance be high for rapid heating, as high resistance = greater heat by resistive heating? Therefore answer should be D, but is in fact C
For this we must take in account power loss, P = I^2 x R
Yes, resistance is there, but the current is a much larger factor than resistance for power loss, but how do we increase current?
V = IR
I = V/R
For current it be high, we need Voltage to be HIGH and resistance to be LOW
And that's where we find our answer! to optimise our heat loss, we need current to be at its highest, and it is at its highest when RESISTANCE is low ;)
EDIT
Would appreciate help with this question (2015 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894854_1364012660390920_933578363_n.png?oh=b87e2f32349fb86ce6953e8bc9eaba08&oe=59F85C98
TIA!
19) This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why i'll be happy to!)
Post by: khadeeja_ on October 29, 2017, 06:16:43 pm
Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A :))
Hey! This was answered a few pages back!
https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472
Here it is!
I checked the excel book to help you guys out..
so it says
A - Max. energy output from human = E=hf
=hc / wavelength
= (6.626x10^-34) x( 3x10^8) / (9x10^8)
= 0.138 eV
Therefore the band gap should be less than 0.138 eV. This makes HgCdTe the best choice.
Post by: Dante1091 on October 29, 2017, 06:22:00 pm
Post by: blasonduo on October 29, 2017, 06:27:43 pm
I checked the excel book to help you guys out..
so it says
A - Max. energy output from human = E=hf
=hc / wavelength
= (6.626x10^-34) x( 3x10^8) / (9x10^8)
= 0.138 eV
Therefore the band gap should be less than 0.138 eV. This makes HgCdTe the best choice.
Hey! thanks for that!
I still agree with pikachu here, A is such a silly answer for the Excel book to go for.
Theres no way A would be more suited than B here
Post by: winstondarmawan on October 29, 2017, 06:40:51 pm
EDIT
19) This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why i'll be happy to!)
Yes please. :)
Post by: blasonduo on October 29, 2017, 06:58:52 pm
Yes please. :)
Sure thing! This kinda is hard to understand, so I'll see if I can explain it properly
We know that acceleration = F/m
and that gravitational force is GmM/d^2
but as we have now defined force as F/m, we must also divide the gravitational force by m (the mass of the object)
So since we have defined acceleration as the objects force over their respective masses (the inversely proportional part of the answer!!)
And that force equalling the gravitational force, but without their respective masses. We here have explained how answer D is correct!
This is kinda tough to get your head around (and hence why I said this is best answered by removing the wrong answers) but I hope this makes sense.
If it didn't I'll happy to elaborate the confusion :)
Post by: sidzeman on October 29, 2017, 07:17:43 pm
Post by: jamonwindeyer on October 29, 2017, 07:42:12 pm
What stuff about Westinghouse and Edison do we have to know? Since I think I know too much irrelevant information about this dotpoint
If you know it already keep it in the noggin! No harm in knowing a little too much ;D
The principle of relativity states that you cannot conduct any mechanical experiments within an inertial FOR to reveal if you are moving at a constant velocity or stationary. However what I dont understand is how nonintertial FOR apply to relativity
Basically, non-inertial FOR does weird shit to relativity. And you don't need to understand anything beyond what they are, and that if you want to do anything with them you end up creating fictional forces to account for your observations. Anything beyond those two bits of knowledge is irrelevant ;D
Post by: sidzeman on October 29, 2017, 07:46:49 pm
Basically, non-inertial FOR does weird shit to relativity. And you don't need to understand anything beyond what they are, and that if you want to do anything with them you end up creating fictional forces to account for your observations. Anything beyond those two bits of knowledge is irrelevant ;DAhhh I see thank you. Also for med phys, can MRI be used to image bone? Some sources say that it image the spinal cord, but I dont see how this is possible when bone does not contain any hydrogen
Post by: jamonwindeyer on October 29, 2017, 07:58:00 pm
Ahhh I see thank you. Also for med phys, can MRI be used to image bone? Some sources say that it image the spinal cord, but I dont see how this is possible when bone does not contain any hydrogen
Spinal cord is a lovely bunch of neurons that send stuff to the rest of your body - It actually isn't a bone! There are bones around it of course ;D
That said I think you can take an MRI of bones, not sure exactly how it works but I think it can be done! Probably targets other types of nuclei? :)
Post by: Mymy409 on October 29, 2017, 08:10:07 pm
Hey! This was answered a few pages back!
https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472
Here it is!
Thanks! However, it's still a bit unclear to me. Could you explain how the force doesn't change?
Post by: jamonwindeyer on October 29, 2017, 08:15:37 pm
Thanks! However, it's still a bit unclear to me. Could you explain how the force doesn't change?
So the force on a current carrying conductor is \(F=BIL\sin{\theta}\), where \(\theta\) is the angle with the field. Even as the coil rotates, the wire is always perpendicular to the field. The coil isn't, that definitely changes, but the wire is always perpendicular (\(\theta=90\))! So the size of the force doesn't change ;D
Post by: beau77bro on October 29, 2017, 08:37:02 pm
https://educationstandards.nsw.edu.au/wps/wcm/connect/5c414559-9792-41e6-bd98-e43c9097168d/2016-hsc-physics.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-5c414559-9792-41e6-bd98-e43c9097168d-lHPDHJd
Post by: beau77bro on October 29, 2017, 08:42:15 pm
Spinal cord is a lovely bunch of neurons that send stuff to the rest of your body - It actually isn't a bone! There are bones around it of course ;D
That said I think you can take an MRI of bones, not sure exactly how it works but I think it can be done! Probably targets other types of nuclei? :)
your bones do contain some water - bone marrow and all isnt just dust and most organic compounds are hydrogen based anyway? right?
Post by: CyberScopes on October 29, 2017, 08:43:38 pm
hi, q 15 and 16 have me stumped for 2016 hsc... and also 1 - because i thought batteries need DC to reverse the reactions within the battery so that it can be used again... so how is it still AC - i get it is lowered because the phone cant take high voltages but still i just thought it would be DC.
https://educationstandards.nsw.edu.au/wps/wcm/connect/5c414559-9792-41e6-bd98-e43c9097168d/2016-hsc-physics.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-5c414559-9792-41e6-bd98-e43c9097168d-lHPDHJd
For 16, the speaker thing is a generator since mechanical energy (pushing the coil) is being converted into a current. According to lenzs law, the direction of the current has to oppose the cause of induction - i.e. the force has to push to the right. Therefore, check the direction of current by using the RHPR at any point - say for example the top of the coil - and you find that for the force to push the coil to the right, the current flows through the coil from y to x. But since we're tryna find the direction in terms of the conductor, if the current continues to flow with that as a reference point, it flows from the x side of the conductor to y. So the answer is
Edit: I meant A not B sorry
Post by: khadeeja_ on October 29, 2017, 08:45:38 pm
the answer is B
Post by: blasonduo on October 29, 2017, 08:48:26 pm
hi, q 15 and 16 have me stumped for 2016 hsc... and also 1 - because i thought batteries need DC to reverse the reactions within the battery so that it can be used again... so how is it still AC - i get it is lowered because the phone cant take high voltages but still i just thought it would be DC.
https://educationstandards.nsw.edu.au/wps/wcm/connect/5c414559-9792-41e6-bd98-e43c9097168d/2016-hsc-physics.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-5c414559-9792-41e6-bd98-e43c9097168d-lHPDHJd
Hello!
For 1) You are confusing a transformer for a rectifier A transformer only steps-up or steps-down a voltage. A rectifier changes AC to DC
15) B is outright wrong as a solar cell consists of an n-type and p-type "squished" together, light can't convert this
C is wrong because if this were true, the solar cell wouldn't work, the light adds energy for an electron to move around the external circuit.
D is wrong as its the n-type the light transfers energy to.
16) As we are moving the cone, it must be a generator. C and D are out.
Now due to Lenz's law when we push the solenoid, the solenoid will produce a NORTH pole to resist the change. From here we can figure out the direction of current (NANTI - SOCK)
and we see that it flows X to Y, so A :)
Post by: beau77bro on October 29, 2017, 08:50:49 pm
Hello!OHHH I GET 1 - IT SAID TRANSFORMER MY BAD
For 1) You are confusing a transformer for a rectifier A transformer only steps-up or steps-down a voltage. A rectifier changes AC to DC
15) B is outright wrong as a solar cell consists of an n-type and p-type "squished" together, light can't convert this
C is wrong because if this were true, the solar cell wouldn't work, the light adds energy for an electron to move around the external circuit.
D is wrong as its the n-type the light transfers energy to.
16) As we are moving the cone, it must be a generator. C and D are out.
Now due to Lenz's law when we push the solenoid, the solenoid will produce a NORTH pole to resist the change. From here we can figure out the direction of current (NANTI - SOCK)
and we see that it flows X to Y, so A :)
16 I READ WRONG OML.
15 - WHY IS C WRONG?
Post by: blasonduo on October 29, 2017, 09:00:19 pm
OHHH I GET 1 - IT SAID TRANSFORMER MY BAD
16 I READ WRONG OML.
15 - WHY IS C WRONG?
Remember, between this junction is a Depletion zone, and this depletion zone creates a potential difference between it. This is what stops the electrons from continuously moving from the n-type to the p-type. When an electron gains energy, it will move to a place that is "easiest" to move to. Jumping that depletion zone is a lot more work than moving through the external circuit, so the electrons will always move into the external circuit. PLUS if they were to jump the depletion zone, the solar cell wouldn't do its purpose!
Post by: CyberScopes on October 29, 2017, 09:07:25 pm
Hey could someone explain q20 from 2015
the answer is B
Δx = Uxt
so Ux = Δx/t = 70/3.5 = 20
To find Uy, assume Vy = 0 at max height, and take t as half (3.5/2 = 1.75):
Vy = Uy + ayt
0 = Uy + (-9.8 )(1.75)
Uy = 17.15
Now θ = tan-1(Uy/Ux) = tan-1(17.15/20)
= 40.61
The closest answer is 40 degrees, which is B.
Post by: beau77bro on October 29, 2017, 09:08:09 pm
It goes around the whole circuit, you need to think of the depletion zone like a voltage source. It is setting up an electric field which 'pushes' electrons around the circuit, just like a voltage source would. The electric field is directed from N to P, so electrons will flow in the opposite direction around the whole circuit ;D
im still struggling with this? so the Photon strikes the n-type, which produces a positive hole and a ejects an electron. now the positive hole is attracted to the negative portion of the depletion zone (the p-type) and the electron is attracted to the positive portion of the n-type. so how do they get there? does the electron just go straight to the positive or does it go all the way around the circuit. or it the positive hole that goes all the way around the circuit to the negative part, or does it just travel striaght to the negative through the junction?
i understand that it sets up a positve to negative electric field but which way. how do the electrons and holes move? WAIT LEGENDBLASONDUO JUST ANSWERED OMG THANK YOU!!!! but wait. isnt the electron just emitted in the N-type so it can go straight to the positive part of the depletion zone?
Post by: blasonduo on October 29, 2017, 09:15:11 pm
im still struggling with this? so the Photon strikes the n-type, which produces a positive hole and a ejects an electron. now the positive hole is attracted to the negative portion of the depletion zone (the p-type) and the electron is attracted to the positive portion of the n-type. so how do they get there? does the electron just go straight to the positive or does it go all the way around the circuit. or it the positive hole that goes all the way around the circuit to the negative part, or does it just travel striaght to the negative through the junction?
i understand that it sets up a positve to negative electric field but which way. how do the electrons and holes move? WAIT LEGENDBLASONDUO JUST ANSWERED OMG THANK YOU!!!! but wait. isnt the electron just emitted in the N-type so it can go straight to the positive part of the depletion zone?
It could, but why would it? There would be no purpose!
Post by: khadeeja_ on October 29, 2017, 09:34:19 pm
Post by: beau77bro on October 29, 2017, 09:34:35 pm
But isn't it the simple mode of travel - the n-type is the one that is exposed to sun, so its the one the produces the extra electron - why wouldn't it go straight to the positive with the positive hole going through the external to the negative part of the depletion zone?
Post by: blasonduo on October 29, 2017, 09:43:11 pm
(http://uploads.tapatalk-cdn.com/20171029/e835aacb0188d4e0f3b84179e005a211.jpg)
But isn't it the simple mode of travel - the n-type is the one that is exposed to sun, so its the one the produces the extra electron - why wouldn't it go straight to the positive with the positive hole going through the external to the negative part of the depletion zone?
Are you saying why the electrons can't go directly to the P-type? IF so, it is because it physically cannot. The potential difference is negative in the p-type, and electrons are repelled by that!
Post by: jamonwindeyer on October 29, 2017, 09:46:50 pm
2016 q14.. :o :o
Kepler's Law of Periods states:
If we quadruple the mass of earth on the RHS, then that factor of four needs to come to the LHS. If the radius is set, it must come from the orbital period. For the ratio \(\frac{r^3}{T^2}\) to get four times larger, the period needs to be cut in half:
So the answer is B :)
Post by: beau77bro on October 29, 2017, 09:49:48 pm
Are you saying why the electrons can't go directly to the P-type? IF so, it is because it physically cannot. The potential difference is negative in the p-type, and electrons are repelled by that!
no i'm saying they will just go straight to the N-type, and not pass through the external circuit or the depletion zone. is that what happens?
Post by: CyberScopes on October 29, 2017, 09:50:56 pm
2016 q14.. :o :o
So you have two satellites orbiting two planets, with the only difference being one has mass 4 times the other. Using Keplers Law of Periods:
r3/t2 = GM/4π2
We know r is the same so lets just set it to 1, and we dont actually care about G, π, etc, since its all just comparison, so lets simplify this equation:
1/t2 = M
Ok lets chuck in M = 1 for earth:
1/t2 = 1
t = 1
So lets say tEarth = 1
Now lets chuck M = 4:
1/t2 = 4
t2 = 1/4
tX = 1/2
So since were tryna get the value of T in terms of Earth values we can rewrite as such:
tX = 1/2
-> Sub tEarth = 1
tX = T/2
Your answer should be B :)
Post by: CyberScopes on October 29, 2017, 09:51:52 pm
Kepler's Law of Periods states:
If we quadruple the mass of earth on the RHS, then that factor of four needs to come to the LHS. If the radius is set, it must come from the orbital period. For the ratio \(\frac{r^3}{T^2}\) to get four times larger, the period needs to be cut in half:
So the answer is B :)
Dammit why are you so quick give us a chance ;D
Post by: khadeeja_ on October 29, 2017, 10:03:29 pm
So you have two satellites orbiting two planets, with the only difference being one has mass 4 times the other. Using Keplers Law of Periods:
r3/t2 = GM/4π2
We know r is the same so lets just set it to 1, and we dont actually care about G, π, etc, since its all just comparison, so lets simplify this equation:
1/t2 = M
Ok lets chuck in M = 1 for earth:
1/t2 = 1
t = 1
So lets say tEarth = 1
Now lets chuck M = 4:
1/t2 = 4
t2 = 1/4
tX = 1/2
So since were tryna get the value of T in terms of Earth values we can rewrite as such:
tX = 1/2
-> Sub tEarth = 1
tX = T/2
Your answer should be B :)
Thanks so much and to Jamon too :')
Post by: khadeeja_ on October 29, 2017, 10:04:25 pm
and 2014 q15 ( i never know how to solve these.. :-\)
Post by: CyberScopes on October 29, 2017, 10:10:31 pm
how about 2016 q18
The way I did this may be different to some people but here:
You need to find velocity so that the effects of gravity do not pull the bike down as it does a spin. The bike is moving in a circle so thats centripetal force. So equate weight force and centripetal force to find the min velocity required to maintain its circular motion:
F=mg
F=mv2/r
mg = mv2/r
g = v2/r
v2 = gr
v = square root of g*r = root 9.8*3.6
= 5.93...
= C
Post by: khadeeja_ on October 29, 2017, 10:15:37 pm
The way I did this may be different to some people but here:Thanks omg, you made it so much simpler :D
You need to find velocity so that the effects of gravity do not pull the bike down as it does a spin. The bike is moving in a circle so thats centripetal force. So equate weight force and centripetal force to find the min velocity required to maintain its circular motion:
F=mg
F=mv2/r
mg = mv2/r
g = v2/r
v2 = gr
v = square root of g*r = root 9.8*3.6
= 5.93...
= C
Post by: CyberScopes on October 29, 2017, 10:16:24 pm
how about 2016 q18
and 2014 q15 ( i never know how to solve these.. :-\)
For your other question:
Whenever I come across these questions, it just requires some algebra and assuming. Some people do it a more 'common sense' sort of way but when I use algebra it just makes much more sense in my head.
For this, use F = GMm/d2
For comparison, just assume all irrelevant variables (G, M and m) are 1, it makes the whole thing easier to understand:
F = 1/d2
Now say at the start d = 1:
F = 1/1 = 1
For the new distance, say d is doubled, so d = 2:
F = 1/22 = 1/4
But since F should actually be 12 when d = 1, times the new answer by 12:
F = 12/4 = 3N
Answer should be A
Post by: khadeeja_ on October 29, 2017, 10:24:52 pm
For your other question:i kind of just messed around with stuff and i got 3 as well, is this way alright too?
Whenever I come across these questions, it just requires some algebra and assuming. Some people do it a more 'common sense' sort of way but when I use algebra it just makes much more sense in my head.
For this, use F = GMm/d2
For comparison, just assume all irrelevant variables (G, M and m) are 1, it makes the whole thing easier to understand:
F = 1/d2
Now say at the start d = 1:
F = 1/1 = 1
For the new distance, say d is doubled, so d = 2:
F = 1/22 = 1/4
But since F should actually be 12 when d = 1, times the new answer by 12:
F = 12/4 = 3N
Answer should be A
1/d^2 is proportional to F
so when d is doubled --> 1/2d^2 = 12/2 (since d is inversely proportional, multiplying d with 2 would mean dividing 12 with 2)
and then d^2 = 3
Post by: jamonwindeyer on October 29, 2017, 10:35:21 pm
Dammit why are you so quick give us a chance ;D
Your method was completely different to mine, so I'm glad you chimed in as well! ;D
Post by: winstondarmawan on October 29, 2017, 10:39:21 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894992_1364158777042975_445018130_n.png?oh=f95370301f82d11379b817756f62a61e&oe=59F7EA1B
It was similar to the one that I asked you about why the copper slit ring falls at the same time as the plastic ring, however this question is suggesting that it is indeed different.
Post by: CyberScopes on October 29, 2017, 10:40:07 pm
i kind of just messed around with stuff and i got 3 as well, is this way alright too?
1/d^2 is proportional to F
so when d is doubled --> 1/2d^2 = 12/2 (since d is inversely proportional, multiplying d with 2 would mean dividing 12 with 2)
and then d^2 = 3
Hmm not sure since the answer gives d^2 = 3 not F = 3 or 1/d2 = 3, but I can see where youre going, maybe you can try something like this?
F∝1/d2
For when F = 12:
12 = 1/d2
Now double d:
12 = 1/(2d)2
12 = 1/4d2
3 = 1/d2
Force becomes 3 once d is doubled
Post by: sidzeman on October 29, 2017, 10:41:34 pm
Some sources say Einstein believed science was not removed - had a direct impact on the nations goals and so should not be used to advance dangerous agenda's e.g Nazi Party
However other sources say that it was Planck who believed that science was removed - that advancements in science should be persued regardless of the possible use to commit atrocities.
Both arguments are correct, but which is the one that BOSTES expects?
Post by: jamonwindeyer on October 29, 2017, 10:41:57 pm
Hey Jamon, I found a question that is troubling me.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894992_1364158777042975_445018130_n.png?oh=f95370301f82d11379b817756f62a61e&oe=59F7EA1B
It was similar to the one that I asked you about why the copper slit ring falls at the same time as the plastic ring, however this question is suggesting that it is indeed different.
Sure, and in my previous answers I've always said it is definitely different. It's just that the difference is tiny, and the eddy currents involved in both those situations are tiny anyway, so it is your call as to whether you define it as significant or not :)
Essentially it's like, yes, 1cm is different to 5cm. But in the context of kilometres, who really cares? Aha ;D
Post by: jamonwindeyer on October 29, 2017, 10:48:58 pm
I'm a bit confused about the Einstein and Planck "is science removed from social and poltiical (SAC) factors" dotpoint
Some sources say Einstein believed science was not removed - had a direct impact on the nations goals and so should not be used to advance dangerous agenda's e.g Nazi Party
However other sources say that it was Planck who believed that science was removed - that advancements in science should be persued regardless of the possible use to commit atrocities.
Both arguments are correct, but which is the one that BOSTES expects?
You've correctly defined their arguments, it is just about what you define as 'removed.' I frame it as:
- Einstein believed they should be removed, that science shouldn't play a role or be weaponised in a political sense. He wasn't a patriot, didn't believe science should be used for war. Etc.
- Planck believed scientists had a duty to serve their country, and so believed science was not removed from politics.
Provided you explain your argument you'll be paid regardless :)
Post by: CyberScopes on October 29, 2017, 10:50:49 pm
Make sure you:
- Read the question carefully
- Remember your SI units and converting
- Remember to use the data sheet if your lost
- Add in relevant verbs for better answers
- Manage your time well
- And most importantly dont stress out and be confident!
Oh and jamon can you give me a quick shout on your snapchat you know just sneak it in somewhere k thanks bye ;D
Post by: khadeeja_ on October 29, 2017, 11:00:19 pm
Hmm not sure since the answer gives d^2 = 3 not F = 3 or 1/d2 = 3, but I can see where youre going, maybe you can try something like this?yupp thanks :D
F∝1/d2
For when F = 12:
12 = 1/d2
Now double d:
12 = 1/(2d)2
12 = 1/4d2
3 = 1/d2
Force becomes 3 once d is doubled
Post by: khadeeja_ on October 29, 2017, 11:02:54 pm
Post by: jamonwindeyer on October 29, 2017, 11:14:23 pm
i dont understand why 2010 q4 is B
We know that the kinetic energy will go down and up as it rises and falls (slows down then speeds up). The question is whether it hits zero, and the answer to that is no it doesn't, because it always moving horizontally! SO it always has some kinetic energy ;D
Post by: khadeeja_ on October 29, 2017, 11:21:06 pm
We know that the kinetic energy will go down and up as it rises and falls (slows down then speeds up). The question is whether it hits zero, and the answer to that is no it doesn't, because it always moving horizontally! SO it always has some kinetic energy ;DThank youuuu!
Post by: winstondarmawan on October 29, 2017, 11:38:51 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22908566_1364204840371702_510182205_o.png?oh=2e30420e7fbcb7cdb524e9f75b0c4cd3&oe=59F7D808
TIA!
Post by: sidzeman on October 29, 2017, 11:46:51 pm
Post by: CyberScopes on October 30, 2017, 12:08:50 am
What can MRI be used to diagnose, besides cancerous growths, and how?
It can be used to track and determine how blood flow is, since blood is full of water content and thus makes it easy to detect with T2 relaxation times. This can be used to diagnose things like stenosis, blood clots, etc.
It can also be used to distinguish between white and grey matter in the brain and spinal cord since their compositions have varying T1 and T2 times, making it easy to differentiate between the two. This often helps with determining brain tumors, strokes, etc. This is also especially helpful for diagnosing multiple sclerosis, a demyelination disease where nerve covers are damaged, thus exposing the internal white matter with the grey matter. If images show grey matter and white matter sort of merging into each other in an MRI scan, this could be an indication of multiple sclerosis.
This is from the top of my head, someone please correct me if im not accurate
Post by: beau77bro on October 30, 2017, 08:09:27 am
(http://uploads.tapatalk-cdn.com/20171029/e835aacb0188d4e0f3b84179e005a211.jpg)BUMP - please help this is killing me. which direction do they go? electrons from N to the positive part of the positive part of the depletion zone in the n-type, but does it go around the circuit or straight to the positive? - see image from the post.
But isn't it the simple mode of travel - the n-type is the one that is exposed to sun, so its the one the produces the extra electron - why wouldn't it go straight to the positive portion of the depletion zone with the positive hole going through the external to the negative part of the depletion zone?
Post by: jamonwindeyer on October 30, 2017, 09:27:45 am
BUMP - please help this is killing me. which direction do they go? electrons from N to the positive part of the positive part of the depletion zone in the n-type, but does it go around the circuit or straight to the positive? - see image from the post.
Around the circuit!
Post by: RoJoHo on November 01, 2017, 06:17:57 pm
Hey Chloe! Happy Physics Land, you did an awesome job answering the question, spot on!
The important thing to remember is that the tension in the rope is a force which corresponds directly to the force that the rope is applying to the crate. In this way, it is a reactive force.
To confirm the answer, Chloe, your reasoning that an acceleration of the crate would result in greater tension in the rope. But Happy has correctly identified that, when moving at a constant speed, there is no acceleration in the crate. Therefore, the tension is identical to if it was stationary. This is assuming the crate isn't on the ground, in which case, there would be no tension in the rope.
For the second part however, I think my interpretation is different. Consider the crate travelling upward at a constant speed, with some level of tension. To accelerate it upwards would require an additional upwards force, provided by the rope, and thus tension would increase. If the crate then accelerated downward (still moving upwards, remember), this downwards acceleration would in fact be caused by gravity, no tension involved. I would therefore say that tension is higher in the rope when accelerating upwards.
Your second question is very interesting indeed. The key is in this phrase:
The car and caravan travel along with equal velocities and speed up at the same rate as they are connected with a rope.
What this tells us is that acceleration is identical for the car and trailer. So, if acceleration is equal, but the masses are different, then the forces must also be different. The resultant calculation is an application of Newton's 2nd Law:
Total Force = Car Force + Trailer Force
Car Force = 500a
Trailer Force = 200a
So Total Force = 700a
18000 = 700a
a = 25.7...
So Car Force = 12850, Trailer Force = 5140 (there is a rounding error there), and so the car uses more force pulling itself than the trailer! :D
I hope this explanation helps Chloe, and Happy Physics Land, thanks heaps for tackling that doozy of a first question! I hope it is also helping you on your HSC journey, it is certainly helping a lot of other people ;D
Hey Jamon, im just starting the HSC physics course and was wondering - could these kinds of prelim moving about questions be asked in the HSC exam directly?
Thanks!
Post by: CyberScopes on November 01, 2017, 07:58:14 pm
Hey Jamon, im just starting the HSC physics course and was wondering - could these kinds of prelim moving about questions be asked in the HSC exam directly?
Thanks!
Im not Jamon obviously lol, but from the past papers ive seen its highly unlikely moving about questions are asked. A pure question only involving moving about is impossible since its not a part of the syllabus and they can only test the syllabus, but there are some universal parts - such as Newtons Laws and a general understanding of velocity, acceleration, etc - that may be asked in conjunction with another HSC topic (e.g. projectile motion and other parts of space).
Post by: mxrylyn on November 12, 2017, 10:20:26 pm
I'm having trouble with this HSC question.
I tried tore-arrange the formula and got root (r cubed x 4 x pie squared) / (G x M) = T
But I used the mass of Ba for M, but I didn't get any of the choices of answers.
I also confused because I thought planet Ba was orbiting a star, so wouldn't the mass need to be the mass of the star? Or do I have that wrong too?
Is there any other way to get the answer to that question that I am missing?
Post by: mxrylyn on November 12, 2017, 10:36:58 pm
Hello.
I'm having trouble with this HSC question.
I tried tore-arrange the formula and got root (r cubed x 4 x pie squared) / (G x M) = T
But I used the mass of Ba for M, but I didn't get any of the choices of answers.
I also confused because I thought planet Ba was orbiting a star, so wouldn't the mass need to be the mass of the star? Or do I have that wrong too?
Is there any other way to get the answer to that question that I am missing?
Nvm, I totally overlooked the rule for two bodies orbiting the same body.
Just double checking that the answer is d?
Post by: blasonduo on November 12, 2017, 10:59:59 pm
Nvm, I totally overlooked the rule for two bodies orbiting the same body.
Just double checking that the answer is d?
Yep! That's what I got! :))
That rule is very important in HSC physics ;) remember it!
Post by: justwannawish on November 13, 2017, 06:56:36 pm
Post by: KiNSKi01 on November 13, 2017, 07:38:53 pm
I'm not fully aware of all the content covered in HSC physics but I am doing VCE 1/2 physics
Generally, if you can use equations related to energy you can get the questions done quicker (as Falling objects convert all their to energy to KE from GPE)
Also, always clearly split up the questions into two parts :horizontal motion and vertical motion. You only really need to know speed=distance/time for horizontal motion and xuvat equations for vertical motion ;D
More specific tip, remember that the time taken to reach ground from max height is 1/2 the total time spent in the air (unless it is a question involving a cliff/ object is thrown onto a lower level)
Hope I helped :P
Post by: blasonduo on November 13, 2017, 08:36:03 pm
Before answering, clearly identify what values you have and don't have.
Now figure out if any of the unknown values could equal 0, this is used in many questions. (ie rolling a ball off a table has the initial vertical velocity as 0)
Relate these "knowns" to the formula sheet to see if any equations only leave one "unknown" Even if it isn't the one the question asked for, it may be the step to figure it out. All equations you need are on the formula sheet, don't try to make it complicated!
Post by: justwannawish on November 14, 2017, 10:07:52 am
Yo justwannawish!
I'm not fully aware of all the content covered in HSC physics but I am doing VCE 1/2 physics
Generally, if you can use equations related to energy you can get the questions done quicker (as Falling objects convert all their to energy to KE from GPE)
Also, always clearly split up the questions into two parts :horizontal motion and vertical motion. You only really need to know speed=distance/time for horizontal motion and xuvat equations for vertical motion ;D
More specific tip, remember that the time taken to reach ground from max height is 1/2 the total time spent in the air (unless it is a question involving a cliff/ object is thrown onto a lower level)
Hope I helped :P
Great stuff!! Just adding a few things just in case you really struggle with projectile motion.
Before answering, clearly identify what values you have and don't have.
Now figure out if any of the unknown values could equal 0, this is used in many questions. (ie rolling a ball off a table has the initial vertical velocity as 0)
Relate these "knowns" to the formula sheet to see if any equations only leave one "unknown" Even if it isn't the one the question asked for, it may be the step to figure it out. All equations you need are on the formula sheet, don't try to make it complicated!
Thank you! I'm alright with the equations themselves, but I just need a list or something of all the tricks and rules and stuff (e.g. vy = 0 at maximum height) so anything like that will be really useful!
Post by: CyberScopes on November 14, 2017, 10:21:56 am
Thank you! I'm alright with the equations themselves, but I just need a list or something of all the tricks and rules and stuff (e.g. vy = 0 at maximum height) so anything like that will be really useful!
Here are a couple general tips:
- Max horizontal range occurs when the angle is 45 degrees
- Vy = 0 at max height
- Ux = Vx -> Horizontal motion has no acceleration
- When calculating 't' using the equations from the 'y' axis of motion, it is only the time to reach half the total trajectory time (i.e. when Vy = 0), but the equation Δx = Uxt considers the value of t for the entire trajectory.
- If a projectile is launched horizontally (from the max height):
-> Uy = 0
-> U = Ux
-> Δy should be taken as a negative value generally in these cases
Post by: owidjaja on November 14, 2017, 07:57:17 pm
So I just received a notification for my physics exam and noted that it was a practical exam and could be assessed on either projectile motion, pendulum or centripetal force. Since in my last practical exam my partner and I were just blindly going through the practical component of the exam (and mostly flailing and shrugging since we weren't sure what we were doing), are there any tips as to how to know blindly do the experiment and know what you're doing (not sure if I worded that correctly)?
Thanks!
Post by: blasonduo on November 14, 2017, 08:15:34 pm
Hey guys,
So I just received a notification for my physics exam and noted that it was a practical exam and could be assessed on either projectile motion, pendulum or centripetal force. Since in my last practical exam my partner and I were just blindly going through the practical component of the exam (and mostly flailing and shrugging since we weren't sure what we were doing), are there any tips as to how to know blindly do the experiment and know what you're doing (not sure if I worded that correctly)?
Thanks!
As you know the experiment is going to be one of those 3 experiments, go over them.
It's totally fine to be confused about conducting the experiment the first time around, but you should now have the general gist of it afterwards. Go over the sheets you received, (google it if needed) and explain to yourself the reason a particular step was needed. Make sure you understand what you are trying to figure out and why the steps taken allow this.
For example in projectile motion.
The height of the table was recorded
Why? So we are able to figure out the time taken.
The horizontal distance travelled by the ball was recorded
Why? To calculate the initial velocity of the ball.
The experiment was repeated numerous times
Why? To increase reliability.
If you have any specific questions relating to any of the experiments, feel free to ask here :))
Post by: justwannawish on November 14, 2017, 09:14:58 pm
Here are a couple general tips:
- Max horizontal range occurs when the angle is 45 degrees
- Vy = 0 at max height
- Ux = Vx -> Horizontal motion has no acceleration
- When calculating 't' using the equations from the 'y' axis of motion, it is only the time to reach half the total trajectory time (i.e. when Vy = 0), but the equation Δx = Uxt considers the value of t for the entire trajectory.
- If a projectile is launched horizontally (from the max height):
-> Uy = 0
-> U = Ux
-> Δy should be taken as a negative value generally in these cases
thank you so much!
Could someone please advise me what to cut down for the following exam question? I feel like I talked for too long :(
Discuss the contribution of von Braun to the development of space exploration (4 marks) .
Spoiler
Wernher von Braun, a German aerospace engineer, has made many significant contributions to space exploration, regarding liquid-fuelled propelled rockets, the Jupiter C, Saturn 5 and V2 rocket.
Working with Hermann Oberth, von Braun designed liquid-fuel propelled rockets, which can be combined with greater efficiency and, thus, greater power than the solid-fuel engines. His work controlled the g-forces experienced by astronauts and varied the thrust produced by the rocket, with these liquid fuels now used in modern satellites and in his own Saturn V thrust rocket that powered the Apollo missions.
After defecting, he developed advanced rockets for space exploration, including the Jupiter-C, a modified Redstone ballistic missile that launched the US’s first satellite, Explorer 1, into orbit. Shaping new research and developing space travel, his contributions have enhanced existing knowledge of space exploration. As NASA’s Marshall space flight centre’s director, he later led and constructed the Saturn 5 Rocket, the only space launch vehicle to launch missions that carried humans beyond low Earth orbit. His project also was the booster that launched Apollo 11 into outer space, sending man to the moon and ultimately developing rocket design, fabrication and space travel operation.
However, his role as the director and designer of the V2 rocket, the first intercontinental ballistic missile, had a controversial effect on space exploration. Demonstrating the use of gyroscopes to stabilise large rockets, the V2 was launched 80km above Earth in a 190km trajectory. Its liquid ethanol-oxygen propellent and bullet structure was structured to attack London during World War II, a detriment to rocketry, but was later mimicked in the rockets of the US and Soviet space exploration to launch their space programs .
Working with Hermann Oberth, von Braun designed liquid-fuel propelled rockets, which can be combined with greater efficiency and, thus, greater power than the solid-fuel engines. His work controlled the g-forces experienced by astronauts and varied the thrust produced by the rocket, with these liquid fuels now used in modern satellites and in his own Saturn V thrust rocket that powered the Apollo missions.
After defecting, he developed advanced rockets for space exploration, including the Jupiter-C, a modified Redstone ballistic missile that launched the US’s first satellite, Explorer 1, into orbit. Shaping new research and developing space travel, his contributions have enhanced existing knowledge of space exploration. As NASA’s Marshall space flight centre’s director, he later led and constructed the Saturn 5 Rocket, the only space launch vehicle to launch missions that carried humans beyond low Earth orbit. His project also was the booster that launched Apollo 11 into outer space, sending man to the moon and ultimately developing rocket design, fabrication and space travel operation.
However, his role as the director and designer of the V2 rocket, the first intercontinental ballistic missile, had a controversial effect on space exploration. Demonstrating the use of gyroscopes to stabilise large rockets, the V2 was launched 80km above Earth in a 190km trajectory. Its liquid ethanol-oxygen propellent and bullet structure was structured to attack London during World War II, a detriment to rocketry, but was later mimicked in the rockets of the US and Soviet space exploration to launch their space programs .
Post by: CyberScopes on November 16, 2017, 05:40:26 pm
thank you so much!
Could someone please advise me what to cut down for the following exam question? I feel like I talked for too long :(
Discuss the contribution of von Braun to the development of space exploration (4 marks) .SpoilerWernher von Braun, a German aerospace engineer, has made many significant contributions to space exploration, regarding liquid-fuelled propelled rockets, the Jupiter C, Saturn 5 and V2 rocket.
Working with Hermann Oberth, von Braun designed liquid-fuel propelled rockets, which can be combined with greater efficiency and, thus, greater power than the solid-fuel engines. His work controlled the g-forces experienced by astronauts and varied the thrust produced by the rocket, with these liquid fuels now used in modern satellites and in his own Saturn V thrust rocket that powered the Apollo missions.
After defecting, he developed advanced rockets for space exploration, including the Jupiter-C, a modified Redstone ballistic missile that launched the US’s first satellite, Explorer 1, into orbit. Shaping new research and developing space travel, his contributions have enhanced existing knowledge of space exploration. As NASA’s Marshall space flight centre’s director, he later led and constructed the Saturn 5 Rocket, the only space launch vehicle to launch missions that carried humans beyond low Earth orbit. His project also was the booster that launched Apollo 11 into outer space, sending man to the moon and ultimately developing rocket design, fabrication and space travel operation.
However, his role as the director and designer of the V2 rocket, the first intercontinental ballistic missile, had a controversial effect on space exploration. Demonstrating the use of gyroscopes to stabilise large rockets, the V2 was launched 80km above Earth in a 190km trajectory. Its liquid ethanol-oxygen propellent and bullet structure was structured to attack London during World War II, a detriment to rocketry, but was later mimicked in the rockets of the US and Soviet space exploration to launch their space programs .
Its a very well put answer. Personally, I would remove some of the specific details that doesnt exactly add much value to the question, especially since u have a lot of examples in the answers and history (such as in the last paragraph)
Post by: justwannawish on November 17, 2017, 10:24:53 pm
Its a very well put answer. Personally, I would remove some of the specific details that doesnt exactly add much value to the question, especially since u have a lot of examples in the answers and history (such as in the last paragraph)
Thank you! I'll try to reduce it, but wanted to discuss both sides of the issue so that's why I included the history
Could you also help me with finding details of projectile motion with air resistance taken into factor? Why do angles larger than 45 produce longer ranges?
Sorry for asking so many questions!
Post by: blasonduo on November 17, 2017, 10:39:48 pm
Thank you! I'll try to reduce it, but wanted to discuss both sides of the issue so that's why I included the history
Could you also help me with finding details of projectile motion with air resistance taken into factor? Why do angles larger than 45 produce longer ranges?
Sorry for asking so many questions!
In the HSC physics course, we do not need to factor in air resistance into the equations, so don't worry about that.
Also, 45 degrees will always give the furthest range, unless given more velocity. There is no way a projectile angles at 89 degrees to the horizontal will travel further than one 45 degrees to the horizontal ;)
Post by: justwannawish on November 17, 2017, 11:08:32 pm
In the HSC physics course, we do not need to factor in air resistance into the equations, so don't worry about that.
Also, 45 degrees will always give the furthest range, unless given more velocity. There is no way a projectile angles at 89 degrees to the horizontal will travel further than one 45 degrees to the horizontal ;)
I was just wondering about air resistance, since the values are experiment and the angle with maximum range was like 50 with a concave down parabola on a graoh of the results. We have to research it and present information for the secondary experiment, and I was wondering how air resistancce affected the theoretical answer
Post by: jamonwindeyer on November 18, 2017, 12:41:09 am
I was just wondering about air resistance, since the values are experiment and the angle with maximum range was like 50 with a concave down parabola on a graoh of the results. We have to research it and present information for the secondary experiment, and I was wondering how air resistancce affected the theoretical answer
This is actually a fairly interesting problem. When we factor in air resistance, we get an additional acceleration in the horizontal direction, opposing the motion. However, do we consider this as a constant acceleration (drag), or is it proportional to the speed of the object itself? It is standard to make the retarding acceleration proportional to the square of the speed, but sometimes we use other functions.
All of this will effect the value of optimum launch angle. Typically we'd obtain something less than 45 degrees, as we need to put a bit more of our velocity into the horizontal component to compensate for the horizontal de-acceleration. If you think about it intuitively, this should make sense. If something is now pushing back against us, it makes sense that we'd want to shoot a bit lower. As another way to think of it, this will reduce the time of flight and thus reduce the amount of time air resistance can slow down the projectile :)
Definitely not impossible to have 50 degrees as an optimum angle though - It depends on how drag affects the object :)
Post by: justwannawish on November 18, 2017, 06:41:50 pm
This is actually a fairly interesting problem. When we factor in air resistance, we get an additional acceleration in the horizontal direction, opposing the motion. However, do we consider this as a constant acceleration (drag), or is it proportional to the speed of the object itself? It is standard to make the retarding acceleration proportional to the square of the speed, but sometimes we use other functions.
All of this will effect the value of optimum launch angle. Typically we'd obtain something less than 45 degrees, as we need to put a bit more of our velocity into the horizontal component to compensate for the horizontal de-acceleration. If you think about it intuitively, this should make sense. If something is now pushing back against us, it makes sense that we'd want to shoot a bit lower. As another way to think of it, this will reduce the time of flight and thus reduce the amount of time air resistance can slow down the projectile :)
Definitely not impossible to have 50 degrees as an optimum angle though - It depends on how drag affects the object :)
Then do you think i should say it opposed the theory of 45 degrees being the optimal launch angle?
Also, if you have three trials and get values of say 50, 52, 56, is the 56 an outlier? Should it be excluded when averaging the results?
Thank you once again
Post by: jamonwindeyer on November 19, 2017, 01:20:43 am
Then do you think i should say it opposed the theory of 45 degrees being the optimal launch angle?
Also, if you have three trials and get values of say 50, 52, 56, is the 56 an outlier? Should it be excluded when averaging the results?
Thank you once again
I would say it "appears to contradict," then give an explanation as to why. Ultimately the textbook theory is still valid when air resistance is minimal ;D
Mathematically, the 56 isn't an outlier, especially with only three data points. You'd want to include it unless there is a good reason not to (EG - a significant experimental flaw with that trial) :)
Post by: justwannawish on November 20, 2017, 12:29:38 pm
I would say it "appears to contradict," then give an explanation as to why. Ultimately the textbook theory is still valid when air resistance is minimal ;D
Mathematically, the 56 isn't an outlier, especially with only three data points. You'd want to include it unless there is a good reason not to (EG - a significant experimental flaw with that trial) :)
Thank you! Could you also help me compare the theory (and theoretical value) with the actual value of 50 degrees? A scaffold or anything would be great
Also if the qualitative trend is a parabola, what is the quantitative trend?
Post by: justwannawish on November 25, 2017, 10:18:07 am
Thank you! Could you also help me compare the theory (and theoretical value) with the actual value of 50 degrees? A scaffold or anything would be great
Also if the qualitative trend is a parabola, what is the quantitative trend?
Hey guys could anyone help me with this question, I'm so sorry for being a bother!
Post by: jamonwindeyer on November 25, 2017, 05:22:14 pm
Thank you! Could you also help me compare the theory (and theoretical value) with the actual value of 50 degrees? A scaffold or anything would be great
Also if the qualitative trend is a parabola, what is the quantitative trend?
Hey guys could anyone help me with this question, I'm so sorry for being a bother!
No bother! I don't know how I missed it the first time ;D
To compare the two, I think you'd basically just give the two values and explain why they are different. How does the theoretical value come to be, and what does the theory ignore that causes the difference? No scaffold for how or in what order, but that is vaguely what you'd cover.
Qualitative data you really can't link to just a graph, because you can't graph a qualitative trend. Maybe even just one variable goes up while the other goes up, or one goes down while the other goes up? :)
Post by: justwannawish on November 25, 2017, 09:08:04 pm
No bother! I don't know how I missed it the first time ;D
To compare the two, I think you'd basically just give the two values and explain why they are different. How does the theoretical value come to be, and what does the theory ignore that causes the difference? No scaffold for how or in what order, but that is vaguely what you'd cover.
Qualitative data you really can't link to just a graph, because you can't graph a qualitative trend. Maybe even just one variable goes up while the other goes up, or one goes down while the other goes up? :)
Okay thank you!
Also, does performing an experiment outside on different days make it invalid if the results are still consistent? When factoring wind speeds, humidity and stuff like that? e.g. with the pendulum experiment, wouldn't vibrations affect the period of the pendulum, making it invalid?
Post by: Mate2425 on November 26, 2017, 10:16:29 pm
Thank you.
Post by: jamonwindeyer on November 27, 2017, 12:59:01 am
Okay thank you!
Also, does performing an experiment outside on different days make it invalid if the results are still consistent? When factoring wind speeds, humidity and stuff like that? e.g. with the pendulum experiment, wouldn't vibrations affect the period of the pendulum, making it invalid?
It lessens the validity if those variables would be a factor in the experiment, yes! But it is your judgement as to how much :)
Hi could someone please help me with "what are the features of the Kennedy Space Center which makes it an effective launch site".
Thank you.
Primarily (would love to hear other ideas) it is located about as close to the equator as it can be within mainland America! Being close to the equator allows the earths rotational motion to be maximised to its fullest extent ;D
Post by: Mate2425 on December 02, 2017, 09:05:24 pm
It lessens the validity if those variables would be a factor in the experiment, yes! But it is your judgement as to how much :)
Primarily (would love to hear other ideas) it is located about as close to the equator as it can be within mainland America! Being close to the equator allows the earths rotational motion to be maximised to its fullest extent ;D
Post by: mxrylyn on December 05, 2017, 07:37:13 pm
These are the results I found from a first hand investigation
Post by: jakesilove on December 05, 2017, 08:24:33 pm
Is there a linear relationship between the Period and Length of the pendulum, and a parabolic relationship between the period squared and the length of a pendulum?
These are the results I found from a first hand investigation
Hey!
The mathematical relationship describing the pendulum's motion is
Where T is the period, l is the length of the string and g is the acceleration due to gravity.
So, you shouldn't be getting a linear relationship when plotting T against l; the equation is more complication than that. However, if you square both sides;
and plot T squared against l, then you should get a linear plot. So, is it possible you've just written your results the wrong way around?
Jake
Post by: mxrylyn on December 05, 2017, 09:01:51 pm
Hey!
The mathematical relationship describing the pendulum's motion is
Where T is the period, l is the length of the string and g is the acceleration due to gravity.
So, you shouldn't be getting a linear relationship when plotting T against l; the equation is more complication than that. However, if you square both sides;
and plot T squared against l, then you should get a linear plot. So, is it possible you've just written your results the wrong way around?
Jake
YES! THANK YOU!
Post by: itssona on December 10, 2017, 11:15:00 am
An astronaut travelling at 0.5c takes 10 hours ship time to reach her destination. Calculate how much time has passed on Earth as measured by:
(a) an observer on earth
(b) the astronaut
so I did part a) since I figured I need to find out Tv and my answer was 11.55 hours which is correct buuuuut how to do b? the answer is 8.7 hours but how :(
thank you :)
Post by: KiNSKi01 on December 10, 2017, 01:47:07 pm
so I did part a) since I figured I need to find out Tv and my answer was 11.55 hours which is correct buuuuut how to do b? the answer is 8.7 hours but how :(
thank you :)
The astronaut would observe time to be running slow for earth so all you have to do is divide 10 hours by what you used for gamma in the previous question ;)
Post by: itssona on December 10, 2017, 03:29:02 pm
The astronaut would observe time to be running slow for earth so all you have to do is divide 10 hours by what you used for gamma in the previous question ;)Gamma??
Post by: RuiAce on December 10, 2017, 03:35:14 pm
Gamma??
Having said that
The astronaut would observe time to be running slow for earth so all you have to do is divide 10 hours by what you used for gamma in the previous question ;)Please keep in mind that in the HSC, they aren't taught the specific terminology of the Lorenz factor. The formulas involving relativity automatically substitute \(\frac{1}{\sqrt{1-v^2/c^2}} \) straight in
Post by: itssona on December 10, 2017, 04:10:44 pm
ohhhh thank you both
Having said thatPlease keep in mind that in the HSC, they aren't taught the specific terminology of the Lorenz factor. The formulas involving relativity automatically substitute \(\frac{1}{\sqrt{1-v^2/c^2}} \) straight in
Post by: itssona on December 10, 2017, 04:16:52 pm
Alpha centauri is 4.5 light years away from earth. Imagine a spaceship able to fly at 0.9c. How long would it take to get there as observed by its pilot, and by mission control on Earth?
so I got that the time for astronaut is Tv which is 2.16 years but how do I find how long it takes as observed by mission control on earth?
my values are v = T_v= 2.16 T_0=0.5c
kinda confused
thank you :)
Post by: yum.z2 on December 10, 2017, 05:06:45 pm
for a physics assignment the questions are : the scientist is VON BRAUN
-Outline the contribution of the chosen scientist towards rocket development and link how the disocveries are important in making rocket development possible
-Describe the impact of their discovery to advancement of modern rocketry
need helppp
Post by: Mate2425 on December 15, 2017, 04:42:41 pm
Thanks.
Post by: justwannawish on December 16, 2017, 12:59:20 pm
If a bus accelerates to the right, which way would a person have to lean to prevent falling? I think it's right? but don't get why
which way would the direction of gravity seem to be?
And if you dropped something, which direction would it go?
Post by: Natasha.97 on December 16, 2017, 01:15:47 pm
Hey guys, some questions about inertial frames of reference
If a bus accelerates to the right, which way would a person have to lean to prevent falling? I think it's right? but don't get why
which way would the direction of gravity seem to be?
And if you dropped something, which direction would it go?
Hi!
Disclaimer: Don’t do Physics so I’m not 100% sure if this is correct:
The way I think about it is the person is still “stuck” in the original position and wants to remain there until they purposely move. Hence, the person leans to the right to avoid falling towards their original position (which is towards the left). If they dropped something, the same logic applies that it will fall towards the left. Think about it this way: if the bus you were on was accelerating and suddenly stopped at a red traffic light, you’ll lean backwards to prevent falling forwards.
Hope this helps :)
Post by: jamonwindeyer on December 17, 2017, 03:28:54 pm
Hi can i get some help with this question please... What is the acceleration due to gravity on a moon of mass 1.67×10^18 Kg and has a diameter of 4765km.
Thanks.
Hey! Link \(F=mg\) to universal law of gravitation, and you'll see stuff cancel:
You have these values! \(M\) is the mass of the moon and \(d\) is the radius of the moon (cut the diameter in half) - Remember to convert to SI units! ;D
Post by: Mate2425 on December 19, 2017, 12:18:58 am
Hey! Link \(F=mg\) to universal law of gravitation, and you'll see stuff cancel:
You have these values! \(M\) is the mass of the moon and \(d\) is the radius of the moon (cut the diameter in half) - Remember to convert to SI units! ;D
Post by: Aaron12038488 on December 20, 2017, 10:08:22 am
Post by: jamonwindeyer on December 20, 2017, 10:14:16 am
i have a question- say for example you are required to use escape velocity formula to answer a question. Do you have to derive it as I've heard formulas that are not on the formula sheet must be derived?
I think it depends a little bit on the mark allocation, but almost always you should be fine to just use the formula. Or, they will structure the question to specifically ask for the derivation ;D
Post by: mxrylyn on December 22, 2017, 01:49:30 pm
mean that the work done on and object is equal to the potential energy?
And if it is true, is it true only of potential eneergy? or all energies ie. heat energy ect.
Post by: blasonduo on December 22, 2017, 02:39:30 pm
does the sentence "When work is done on an object, there is a corresponding change in the kinetic and/or potential energy of the object"
mean that the work done on and object is equal to the potential energy?
And if it is true, is it true only of potential eneergy? or all energies ie. heat energy ect.
Hello! Yes, this is true!
Just recall the law of conservation of energy; where no energy can be created nor destroyed. Imagine I drop a ball 20km above the earth's surface, and as that ball begins to fall, its potential energy DECREASES, so to obey this law, kinetic energy INCREASES. Work done normally means changing the energy from one form to another, so applying a certain amount of work to an object will change a certain amount of energy :)
Remember,
For your second question (I hope I interpreted it correctly), energy being converted to heat energy is considered a non-conservative force as it is making energy leave the system (on the ball) and hence slowing down its kinetic energy. If the ball did not have any of these non-conservative forces and just used kinetic and potential energy, then all energy would be would be retained. In short, I don't think so, if I dropped my ball on a windier day, there would be more heat energy produced, even if dropped at the same height.
Hope this helps :)
Post by: swordkillz on January 04, 2018, 09:21:04 pm
Thank you so much!!
Post by: blasonduo on January 05, 2018, 03:34:40 pm
Would someone be able to explain why space and time become relative if c is constant and justify this statement (3 marks). Most of the information I find online state this fact without justifying with evidence why such is so. Do I link back to the 2 postulates proposed in special relativity? Do I used scientific examples like muons or atomic clocks?
Thank you so much!!
Hello!
Yes! That is the approach I would go for, and I would specifically mention the thought experiment that uses a mirror. Where if the train was moving, the light would have to travel slightly further than when it was stationary and hence would need time to go slower to accommodate for this. (If this experiment is not ringing any bells, I'd be happy to fully explain it) I'd draw a diagram to fully show this.
I'd justify this exactly like you did, with examples, just make sure you include the results of them (ie the atomic clock that was moving appeared to show time passed slower)
Hope this helps! :))
Post by: Aaron12038488 on January 08, 2018, 01:56:25 pm
Post by: blasonduo on January 08, 2018, 03:00:25 pm
hey! i need help in gathering info on einsteins thought experiment on time dilation. And the equations involved : 2L /c?
Hello! For time dilation, the only equation involved for that is:
Remember that:
to = the time elapsed for someone who is stationary, observing the moving object
tv = the time elapsed for someone travelling in the object at a certain speed.
v = the speed of the object compared to the speed of light (c).
When you say you need help gathering information, are you saying that you are finding it difficult to search for secondary sources on the topic? If so, here are a few:
https://science.howstuffworks.com/science-vs-myth/everyday-myths/relativity10.htm
http://www.phy.olemiss.edu/HEP/QuarkNet/time.html
https://futurism.com/einsteins-weird-world-of-relativistic-time-dilation/
For secondary sources, remember you still need to look into valid, reliable and accurate sources, and to identify if these links fall into these categories.
a Valid source is one that relates back to the question asked
a Reliable source is when it comes from a reputable source (such as peer-reviewed journals or .gov sites)
an Accurate source is when it can be cross-referenced by a reliable source, and contains consistent information.
If this was not what you were asking, and wanted help on the idea of time dilation itself, sorry!! Just send in another post asking :))
Hope this helps :))
Post by: RuiAce on January 08, 2018, 03:08:10 pm
Hello! For time dilation, the only equation involved for that is:Just thought I'd pop in on the first bit. When he spoke about the 2L/c bit I think he meant this:
Hint: Use the underscore to open a subscript :)
(https://i.imgur.com/iGtwhYx.png)
(https://i.imgur.com/qy1qZlH.png)
(I don't know how any of that works anymore though so if you want to clarify it for him please do)
Although, having said that, pretty sure 2L/c is related to how time dilation actually works and not an equation we actually use for computations
Post by: blasonduo on January 08, 2018, 03:29:21 pm
Just thought I'd pop in on the first bit. When he spoke about the 2L/c bit I think he meant this:
(https://i.imgur.com/iGtwhYx.png)
(https://i.imgur.com/qy1qZlH.png)
(I don't know how any of that works anymore though so if you want to clarify it for him please do)
Although, having said that, pretty sure 2L/c is related to how time dilation actually works and not an equation we actually use for computations
Ahh! That makes more sense! Thanks! :P but yeah, Rui is right.
That equation you asked about is just showing the time taken for the light ray to return to it's original position as
Then it uses that to prove that that time taken must be longer when its moving, as the hypotenuse is always the largest side :)
Its a proof of how time dilation actually works :P
Hint: Use the underscore to open a subscript :)
No clue what to do :P
Post by: RuiAce on January 08, 2018, 03:41:31 pm
No clue what to do :PYou'll pick up on it ;) can always practice it under the guide
Post by: owidjaja on January 16, 2018, 09:50:58 pm
Could someone please explain the Slingshot effect in relation to Newton's Laws of Motion and Universal Gravitation? I'm struggling how to formulate my response and integrate both concepts.
Thanks in advance :)
Post by: fireives1967 on January 17, 2018, 01:13:44 am
What percentage or ratio do you guys recommend of how I should balance my Physics study?
80% Calculations 20% Content?
65-35?
Post by: blasonduo on January 17, 2018, 10:02:22 am
Hey guys!
Could someone please explain the Slingshot effect in relation to Newton's Laws of Motion and Universal Gravitation? I'm struggling how to formulate my response and integrate both concepts.
Thanks in advance :)
Hey!!
As known, we use the slingshot effect to increase the velocity and change direction of the spacecraft to conserve fuel. The spacecraft is able to gain more kinetic energy through an elastic collision with the planet. Due to the law of conservation of momentum, which states that the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision and thus:
Through this, we are able to see that if the velocity of the spacecraft increases, it's overall momentum increases and thus momentum from the planet must decrease. The planet's velocity decreases and because the mass of the planet is extremely large, the planet's velocity decrease is only very minute.
As the slingshot effect utilises the planet's gravitational field, the stronger this gravitational field strength, the more force can be placed on the spacecraft and the more velocity it can gain. Looking at the formula:
We can see that the smaller the distance between the objects and the larger the planet affects the force between the two objects and thus utilising a larger planet can give the spacecraft higher velocities :))
I hope this helps! Feel free to ask more!
Hey guys, these holidays I've been studying for physics though I've been pretty worried that I'm using my time inefficiently. Since Physics is definitely more maths than content (or maybe I'm wrong) I feel like I should be practising calculations more rather than writing my syllabus notes.
What percentage or ratio do you guys recommend of how I should balance my Physics study?
80% Calculations 20% Content?
65-35?
Hey!
This is a very difficult question to answer, as it really depends on how difficult you find the math components. However, most of the physics calculations in exams are just pulled directly out of the formula sheet and it becomes a simple substitute. Sometimes you'll need to "swap out" ie changing
F to ma. (you'll see these questions when practising, and you'll get used to it)
80% calculations, in my opinion, is way too much, some physics concepts do take a while to get down packed! If you are struggling that much on calculations where you need 80% work on it, just keep asking questions until it sinks in :)
For me personally, I spent around 35% on calculations and 65% on content, which slowly decreased to about 15% calculations and 85% content right before the HSC exams.
As I said, it is different for everybody, and the main thing to do is work on the stuff you struggle with, and if that's calculations, then go for it. :)
Post by: Aaron12038488 on January 18, 2018, 10:27:34 am
Post by: blasonduo on January 18, 2018, 11:05:49 am
i need help understanding the right hand grip rule, i find the right palm rule easy tho. Do we have to know how to use the right hand grip for M and G?
Hey!!
If you understand the palm rule, the grip rule will be a breeze! As we know, a current carrying conductor will produce a magnetic field, and the grip rule shows us the direction of the magnetic field. It also explains why 2 wires carrying current in the same direction will attract. (I shall attach an image to show this)
This rule is fantastic at finding the north and south poles of a solenoid very quickly, as wrapping around your fingers in the same direction as current flow allows you to find the north side of the solenoid.
My personal experience is that we were required to use the right-hand palm rule a lot more times than the right-hand grip rule, but it is still very important!
Hope this helps! :))
Post by: Abi21 on January 22, 2018, 02:30:10 pm
1) 0.5 kg of ice at 0 degrees is mixed with 0.1kg of steam at 100 degrees. What will be the final temperature?
2) A student attempts to identify metal by measuring the specific heat capacity. 100g of the metal is heated to 75 degrees and then transferred to a 70g copper calorimeter containing 200g of water at 20 degrees. The temperature of the final mixture is 25 degrees? What metal is the student testing? ( given a bunch of metals with their approximate specific heat capacity)
Thanks in advance!
Post by: blasonduo on January 22, 2018, 02:33:01 pm
Hi, I'm struggling with two physics questions, hoping for a little bit of help please!
1) 0.5 kg of ice at 0 degrees is mixed with 0.1kg of steam at 100 degrees. What will be the final temperature?
2) A student attempts to identify metal by measuring the specific heat capacity. 100g of the metal is heated to 75 degrees and then transferred to a 70g copper calorimeter containing 200g of water at 20 degrees. The temperature of the final mixture is 25 degrees? What metal is the student testing? ( given a bunch of metals with their approximate specific heat capacity)
Thanks in advance!
Hey! This content isn't in the HSC physics syllabus!
Maybe you were thinking of another thread?
Post by: Abi21 on January 22, 2018, 02:35:01 pm
Post by: clovvy on January 25, 2018, 11:44:12 pm
A daredevil tries to jump a canyon of width 10m. To do so, he drives his motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed is necessary to clear the canyon?
Post by: blasonduo on January 26, 2018, 12:01:19 pm
Spoiler
WHEWWW!!! That was a biggie! That being said, this is wayyy too hard for HSC physics, never expect this type of difficulty in your exams :))
Post by: jamonwindeyer on January 26, 2018, 01:43:12 pm
I am stuck with this projectile question:
A daredevil tries to jump a canyon of width 10m. To do so, he drives his motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed is necessary to clear the canyon?
We can approach this a little more simply than above if we're tricky. We know that the peak of motion will be at the halfway point of the parabola, meaning, when the range is five metres. However, we also know that the vertical velocity is zero at this point, so:
This needs to be, as mentioned, at a horizontal distance of 5 metres. So use the horizontal range formula:
Absolutely agree though, this is way up there as a projectiles question. If it did appear in a HSC exam, it would be right near the end, and probably in parts ;D
Post by: blasonduo on January 26, 2018, 02:11:19 pm
We can approach this a little more simply than above if we're tricky.
Ah! I was wondering if there was an easier method! It still uses the substitution to solve it too!
Thanks for the help :))
Post by: clovvy on January 26, 2018, 08:30:37 pm
We can approach this a little more simply than above if we're tricky. We know that the peak of motion will be at the halfway point of the parabola, meaning, when the range is five metres. However, we also know that the vertical velocity is zero at this point, so:
This needs to be, as mentioned, at a horizontal distance of 5 metres. So use the horizontal range formula:
Absolutely agree though, this is way up there as a projectiles question. If it did appear in a HSC exam, it would be right near the end, and probably in parts ;D
Hey guys thanks a lot for the assist because I did end up with a sec^2 at some point so I want to know a simpler way to do it.. by the way, why did you put the distance as 5m Jamon?
Post by: jamonwindeyer on January 26, 2018, 11:54:15 pm
Hey guys thanks a lot for the assist because I did end up with a sec^2 at some point so I want to know a simpler way to do it.. by the way, why did you put the distance as 5m Jamon?
Glad to help! This is because I wanted to analyse the point at the peak of the motion, when vertical velocity is zero. This always happens halfway through the motion, so if we need to travel a 10 metre distance total, that means we are analysing the variables at 5 metres - The halfway point ;D
Post by: Mate2425 on January 29, 2018, 09:52:10 pm
Thanks,
Kind Regards,
Mate2425
Post by: blasonduo on January 29, 2018, 10:14:51 pm
Hi, could someone please help me with the topic of projectile motion. I am finding it hard to understand the setting out for questions and when and when not to use certain formulas. I was wondering if anyone knew of any great resources or questions and answers that could help with this section of the syllabus.
Thanks,
Kind Regards,
Mate2425
Hey!!
If you struggle with trying to figure out how to complete a question, here are some tips: :)
1) On the right-hand side, write down every variable. (ie range = ..... , time = ....)
2) Label the variable you are trying to find with a ?
3) Find if any variables = 0 (this is very prevalent in questions that drop an object, or is needed to calculate it's maximum height reached.
4) Look at the formula sheet and see if any equations have only your ? variable, and if so substitute and rearrange.
5) If not, try to solve the other variables, the more variables you can find, the easier it will be to find your known.
In my opinion, the questions that trick most people are the ones where you need to assume a variable as 0, and people just forget, remember to keep looking over this!
http://nsb.wikidot.com/pl-9-2-2-1 is a very good site to help you with this! (used it myself)
http://www.itute.com/2010/12/25/free-download-physics-worksheet-projectile-motion/ This is also a good source of questions!
If you need further help with anything specific, Feel free to ask! Always happy to help :))
Post by: Mate2425 on January 30, 2018, 09:41:24 pm
Hey!!
If you struggle with trying to figure out how to complete a question, here are some tips: :)
1) On the right-hand side, write down every variable. (ie range = ..... , time = ....)
2) Label the variable you are trying to find with a ?
3) Find if any variables = 0 (this is very prevalent in questions that drop an object, or is needed to calculate it's maximum height reached.
4) Look at the formula sheet and see if any equations have only your ? variable, and if so substitute and rearrange.
5) If not, try to solve the other variables, the more variables you can find, the easier it will be to find your known.
In my opinion, the questions that trick most people are the ones where you need to assume a variable as 0, and people just forget, remember to keep looking over this!
http://nsb.wikidot.com/pl-9-2-2-1 is a very good site to help you with this! (used it myself)
http://www.itute.com/2010/12/25/free-download-physics-worksheet-projectile-motion/ This is also a good source of questions!
If you need further help with anything specific, Feel free to ask! Always happy to help :))
Post by: itssona on February 05, 2018, 07:05:38 pm
(a) State ONE such prediction.
(b) Describe an experiment to test this prediction.
(c) Explain how technological advances since 1905 have made it possible to carry out this experiment. (HSC 2005)
So I said for part a about time dilation, and then I talked about the atomic clock experiment for (b) but what do I say for part C? kinda stuck
thanks guys :)
Post by: not a mystery mark on February 05, 2018, 07:22:36 pm
I have a pretty easy physics question, anyway, here it goes.
A car accelerates at 2.5m/s^2 East for 16s. After this time the car is moving at 10m/s West. Calculate it's initial velocity.
The answer I got was -30m/s. I just can't picture in my head somebody travelling West then all of sudden East.
ALSO: is it -30m/s East or West (or something completely different)
Thank you <3
Post by: clovvy on February 05, 2018, 07:34:35 pm
Einstein’s 1905 theory of special relativity made several predictions that could not be verified for many years.
(a) State ONE such prediction.
(b) Describe an experiment to test this prediction.
(c) Explain how technological advances since 1905 have made it possible to carry out this experiment. (HSC 2005)
So I said for part a about time dilation, and then I talked about the atomic clock experiment for (b) but what do I say for part C? kinda stuck
thanks guys :)
a) One prediction from special relativity was that time was relative and time dilation would occur when observing an object travelling at relativistic speeds. This would result in the time recorded by stationary observer of fast moving object being longer than the time for the same event measured in the moving frame of reference
b) A test for time dilation could involve measuring the time taken for an unstable particle (like muon) to decay when observed at rest compared to when travelling at very high speeds. This is possible as muons can be produced on earth in particle accelerators when high energy cosmic radiation encounters astmopheric gases about 100km above the surface. This shows that at rest the decay rate is on average about 2.2 micro-meters, while the info is collected by detectors has shown that when travelling at very high speed, the decay rate measures muchlonger as predicted by special relativity
c) The technological advances in 1905 that have made this experiment possible are development of various technologies utilised in particle accelerators including detectors used to study sub-atomic particles, along with development of atomic clock to allow very small times be accurately measured, making it possible to assess time dilation. In particle accelerators it became possible to produce carefully controlled magnetic fields to control the paths of matter particles and accelerate them to relativistic speed. The study of collision of particles in accelerators (particularly electrons), allowed the muon to be found and with the advances in producing complicated magnetic fields and use of atomic clocks, they could be intensively studied at rest on surface. When technology allow sensors to detect muons being produced by cosmic radiation, and these muons were studied at different altitudes, the evidence for the relative nature of time and time dilation was strongly supported, with these very high speed muons decaying at a much slower rate when compared to those studied at rest on surface...
I hope this makes sense
Post by: blasonduo on February 05, 2018, 07:36:21 pm
Einstein’s 1905 theory of special relativity made several predictions that could not be verified for many years.
(a) State ONE such prediction.
(b) Describe an experiment to test this prediction.
(c) Explain how technological advances since 1905 have made it possible to carry out this experiment. (HSC 2005)
So I said for part a about time dilation, and then I talked about the atomic clock experiment for (b) but what do I say for part C? kinda stuck
thanks guys :)
Hey! Clovvy has done a great job! but I'll try to answer with the example you gave. Both are excellent examples :) Hopefully, I can guide you through this :))
This question is asking you what technologies were needed to actually do the experiment when talking about time dilation, the 2 that come to mind is the atomic clock and the plane.
Remember, for the time dilation effect to be easily observed, the faster the object has to go, so the slower it goes, the more difficult it is to identify any change in time, this is why technologies like planes are needed, as trying to observe this phenomenon would be near impossible walking down the street.
Even though planes are fast, they are nowhere near as fast as the speed of light, meaning the time dilation effect is still very minuscule, so a VERY accurate measuring device is needed, such as an atomic clock. Atomic clocks are very precise and can identify up to the nanosecond, which means that even if a nanosecond difference is detected between the 2 atomic clocks, we are able to observe it.
It is both of these technologies that allow this experiment to be conducted, and for results to be observed, so advancements in technologies are a big deal when it comes to science :)
Hope this helps :))
Post by: fireives1967 on February 05, 2018, 10:27:03 pm
Edit: sorry, the bottom one jumping upward
Edit 2: final answer, the top one jumping upward
Post by: jamonwindeyer on February 05, 2018, 11:15:37 pm
Heyyy.
I have a pretty easy physics question, anyway, here it goes.
A car accelerates at 2.5m/s^2 East for 16s. After this time the car is moving at 10m/s West. Calculate it's initial velocity.
The answer I got was -30m/s. I just can't picture in my head somebody travelling West then all of sudden East.
ALSO: is it -30m/s East or West (or something completely different)
Thank you <3
Hey! No stress, happy to help ;D
I think, failing all else, use the formula. Let East be positive and let West be negative:
So indeed, you were correct! Since we assigned negative to be West to begin with, it is 30m/s West! Remember, accelerating East doesn't necessarily mean moving East - In this context it is actually still moving West, just more and more slowly as it accelerates in the opposite direction (deaccelerates) ;D
Post by: blasonduo on February 06, 2018, 08:16:43 am
not everyone will make it, though.
seeing you are determined, let me give you a little tip:
a little birdie told me you're searching for the number "15th"
not of the newest, but the oldest.
i'm not sure what this means though, and it is up to you to find the next clue
pull up this post and your next step will be waiting.
i wish you the best of luck.
Post by: not a mystery mark on February 07, 2018, 10:15:09 am
Hey! No stress, happy to help ;D
I think, failing all else, use the formula. Let East be positive and let West be negative:
So indeed, you were correct! Since we assigned negative to be West to begin with, it is 30m/s West! Remember, accelerating East doesn't necessarily mean moving East - In this context it is actually still moving West, just more and more slowly as it accelerates in the opposite direction (deaccelerates) ;D
Thank you so so so much lord Jamon. Everything is hyper clear now haha.
Post by: philgee on February 12, 2018, 06:46:53 pm
Post by: blasonduo on February 12, 2018, 07:36:09 pm
Hello, I was wondering how i could write a band 6 response for this question. Would you mind writing one? Thank you.
Hey! This question is basically an information dump! I shall give you the key points I'd use in this question.
Launch:
- Law of conservation - the change of momentum of a system is equal to 0. ie P (rocket) + P (exhaust gas) = 0
- F = ma, The force the exhaust gases exert are always constant, and when the gas gets "used" up, the mass will decrease, meaning that acceleration MUST increase!!
- Escape velocity, the initial velocity required to completely escape the gravitational field of a planet, so the kinetic energy must be equal to the potential gravitational energy. State the formula.
- Newton's first law, that an object will stay in motion until an external force is applied to it, this is when the spacecraft is travelling in space.
Slingshot:
- Conservation of momentum again, mv = mv, so if the spacecraft's velocity increases, the planet's velocity must decrease, but due to its mass, it will be insignificant. Used to conserve fuel due to velocity boost.
- Utilises the planets gravitational acceleration, and by the formula, the bigger the planet, the larger the gravitational force and the more force on the spacecraft.
Orbit:
- Utilises the gravitational force to keep it in a stable orbit. and state orbital velocity, how centripetal force must equal gravitational force for it to remain stable.
There's probably more you could say, but this is what I'd say :) Just use these in a logical and sophisticated order, and you've got the marks, Love to see you give it a try :)
Hope this helps! :))
Post by: philgee on February 12, 2018, 08:11:23 pm
Post by: clovvy on February 16, 2018, 12:25:55 am
Post by: blasonduo on February 16, 2018, 11:06:07 am
Hi, my class have started motors only recently and I have read ahead, all of a sudden the teacher introduced F=qvBsinx and I realised it was in the Ideas to Implementation syllabus, are they really connected to motor effects? How close would that be to the motors syllabus?
Hello! You are right about this little bit being in Ideas to Implementation, but this can be used in motors and generators, because the formula relates to the force on a charged particle, and current is the flow of electrons (which are negatively charged) so we can use this to calculate the force applied to them in motors and generators :))
Hope this helps :)
Post by: jamonwindeyer on February 16, 2018, 06:45:18 pm
Hello! You are right about this little bit being in Ideas to Implementation, but this can be used in motors and generators, because the formula relates to the force on a charged particle, and current is the flow of electrons (which are negatively charged) so we can use this to calculate the force applied to them in motors and generators :))
Hope this helps :)
To expand on this idea, consider \(F=BIL\) vs \(F=Bqv\) (ignoring the sine parts). A current is just a flow of charge per unit time, so:
But \(\frac{L}{t}=v\), that's a speed (distance over time)!
Bazinga ;D
Post by: jasn9776 on February 17, 2018, 09:36:42 pm
Is it because the two forces cancel out?
(https://s.wordpress.com/latex.php?latex=%5Ctau%20%3DnBIA%5Ccos%20%5Ctheta&bg=FFFFFF&fg=000000&s=4)
see diagram attached:
Post by: blasonduo on February 18, 2018, 10:02:40 am
how would you explain the reason why a motor with a single coil loses torque when turns. Is it because of F=BILsintheta? I dont think so because The coil should still experience the same force since its angle to the magnetic field isn't changed.
Is it because the two forces cancel out?
(https://s.wordpress.com/latex.php?latex=%5Ctau%20%3DnBIA%5Ccos%20%5Ctheta&bg=FFFFFF&fg=000000&s=4)
see diagram attached:
Hello! For this, we are going to focus on the current "I" in the torque formula :))
Recall when a motor begins to spin, there is a change in magnetic flux which causes a back EMF to be induced in the wire, the faster it goes, the more EMF that gets induced, until the supply voltage roughly equals the back EMF, and this is where the motor cannot spin any faster.
Because the voltages are opposite in direction, the means the difference between the supply voltage and back EMF will be the new "supply voltage". Meaning the faster the motor spins, the less overall voltage the wire carries.
As current is equal to (I = V/R), this means as the overall voltage in the wire decreases, so does the overall current and from the torque formula, as current decreases, so does the torque! :))
Hope this helps! :)
Post by: jamonwindeyer on February 18, 2018, 11:14:01 am
how would you explain the reason why a motor with a single coil loses torque when turns. Is it because of F=BILsintheta? I dont think so because The coil should still experience the same force since its angle to the magnetic field isn't changed.
Is it because the two forces cancel out?
(https://s.wordpress.com/latex.php?latex=%5Ctau%20%3DnBIA%5Ccos%20%5Ctheta&bg=FFFFFF&fg=000000&s=4)
see diagram attached:
Hey! In addition to above, do you mean when the coil is vertical that no torque is present? As implied in your diagram, it's basically because the forces aren't in a direction that would create torque. For torque you need some component of force that is tangential to the direction of rotation, something to actually push it sideways. At the vertical position the forces aren't doing this, they are acting perpendicular to the rotation direction (if that makes sense) which won't produce any torque ;D
There's a better, mathematical reason involving vectors that the new syllabus sort of gives - But for us, try to consider it intuitively ;D
Post by: jess-steenbeeke on February 19, 2018, 06:41:58 pm
Einstein’s 1905 theory of special relativity made several predictions that could not be verified for many years.
(a) State ONE such prediction.
(b) Describe an experiment to test this prediction.
(c) Explain how technological advances since 1905 have made it possible to carry out this experiment. (HSC 2005)
So I said for part a about time dilation, and then I talked about the atomic clock experiment for (b) but what do I say for part C? kinda stuck
thanks guys :)
okay so basically the atomic clocks weren't around in his day- so we didn't have way to measure time that accurately (because effects of time dilation would be very very small) also you need to get somewhat fast to be able to measure a difference anyway... the atomic clock experiment was performed in high speed jets. Hope this helps :)
Post by: justwannawish on February 21, 2018, 06:05:45 pm
I just have a question about how to determine the velocity/current direction for AC generators. I know the direction of the magnetic field is from north to south and I'm trying to use the right hand palm rule, but i cannot determine which is the negative/positive end?
Any help will be appreciated
Post by: beatroot on February 21, 2018, 07:07:02 pm
Knowing the magnetic field already, we need either the current or force to find the other. The diagram shows a clockwise direction, so we’ll know that the force on the wire MN is down.
Using the righthand palm rule, we can determine that the current runs from N to M. (because of induction)
We also know that current runs from the positive terminal to the negative terminal, so following the current will reveal the negative terminal!
From this, we can see that the top terminal is positive and the bottom is negative.
I hope I’m right :)
I’ll be happy to explain anything that doesn’t make sense.
Hope this helps! :)
Post by: blasonduo on February 21, 2018, 07:20:01 pm
Beat me to it!
Post by: Aaron12038488 on February 21, 2018, 07:46:26 pm
What is the mass of Jupiter?
i get 1.89 x 10^27 kg but the answer is 1.51 x 10^28 kg.
Post by: RuiAce on February 22, 2018, 10:49:35 am
Europa, a moon of Jupiter, has an orbital diameter of 1.34 x 10^9m, and a period of 3.55 days.
What is the mass of Jupiter?
i get 1.89 x 10^27 kg but the answer is 1.51 x 10^28 kg.
\begin{align*}\frac{(1.34\times 10^9)^3}{(3.55 \times 24\times 3600)^2}&= \frac{6.67\times 10^{-11} \times M}{4\pi^2}\\ M &\approx 1.51\times 10^{28}\, \text{kg}\end{align*}
Make sure you've converted everything to SI units. The period is given in days, but we need to work in seconds.
Post by: Aaron12038488 on February 22, 2018, 05:29:26 pm
shouldn't divide diameter by 2? to get radius?
\begin{align*}\frac{(1.34\times 10^9)^3}{(3.55 \times 24\times 3600)^2}&= \frac{6.67\times 10^{-11} \times M}{4\pi^2}\\ M &\approx 1.51\times 10^{28}\, \text{kg}\end{align*}
Make sure you've converted everything to SI units. The period is given in days, but we need to work in seconds.
Also is it better to do HSC space questions, or from the THSC half-yearly bank? which is more likely to be similar for my half-yearly?
Post by: RuiAce on February 22, 2018, 05:35:02 pm
shouldn't divide diameter by 2? to get radius?Oh, very good point. Thanks for the pick-up. In that case you should certainly divide by 2; the given answers are wrong
Post by: jamonwindeyer on February 22, 2018, 05:45:42 pm
Also is it better to do HSC space questions, or from the THSC half-yearly bank? which is more likely to be similar for my half-yearly?
Either! Half Yearlies all try and replicate the HSC to an extent so pretty much anything you practice that is similar will be helpful ;D
Post by: justwannawish on February 22, 2018, 07:04:33 pm
I still don't understand where the current runs, sorry. Could you please explain how you used induction to find it flows from N to M?
Hey!
Knowing the magnetic field already, we need either the current or force to find the other. The diagram shows a clockwise direction, so we’ll know that the force on the wire MN is down.
Using the righthand palm rule, we can determine that the current runs from N to M. (because of induction)
We also know that current runs from the positive terminal to the negative terminal, so following the current will reveal the negative terminal!
From this, we can see that the top terminal is positive and the bottom is negative.
I hope I’m right :)
I’ll be happy to explain anything that doesn’t make sense.
Hope this helps! :)
Post by: beatroot on February 22, 2018, 07:37:55 pm
hey, thank you for your answer.
I still don't understand where the current runs, sorry. Could you please explain how you used induction to find it flows from N to M?
Hey there!
Assuming you know the right-hand palm rule, You'd see that it would flow from M to N, however, by lenz's law, current will always be induced in a way to oppose the change in flux, so it must flow from N to M. (kinda think of it like if there was current flowing through the wire as with a motor. The induced current will always oppose the supply current, but in this case, there is no supply current, so the "back emf" is the current in generator.)
Hope this makes sense! :)
Post by: justwannawish on February 22, 2018, 10:24:21 pm
Hey there!
Assuming you know the right-hand palm rule, You'd see that it would flow from M to N, however, by lenz's law, current will always be induced in a way to oppose the change in flux, so it must flow from N to M. (kinda think of it like if there was current flowing through the wire as with a motor. The induced current will always oppose the supply current, but in this case, there is no supply current, so the "back emf" is the current in generator.)
Hope this makes sense! :)
yep, that makes sense, but if there wasn't a torque direction, would it be possible to find the direction of the current flow?
Also, is motor effect the flow of positively charged particles or negatively?
Post by: beatroot on February 22, 2018, 10:41:11 pm
yep, that makes sense, but if there wasn't a torque direction, would it be possible to find the direction of the current flow?
Also, is motor effect the flow of positively charged particles or negatively?
Hey!
That would be impossible, as we would have 2 unknowns, and because of this, we could never really know the direction of current (because if the force on MN was up, then the current would flow in the opposite direction)
Remember, current is the flow of electrons, and electrons are negatively charged :)
Post by: justwannawish on February 23, 2018, 04:35:15 pm
Hey!
That would be impossible, as we would have 2 unknowns, and because of this, we could never really know the direction of current (because if the force on MN was up, then the current would flow in the opposite direction)
Remember, current is the flow of electrons, and electrons are negatively charged :)
Hey, thank you. It's just that we learned it in class as positively charged particles as my teacher said it was in terms of conventional current rather than electric flow so I was a bit confused about that
Post by: Bri MT on February 23, 2018, 05:35:47 pm
Hey, thank you. It's just that we learned it in class as positively charged particles as my teacher said it was in terms of conventional current rather than electric flow so I was a bit confused about that
Conventional current which is shown on diagram could be described as the theoretical flow of positively charged particles; however, actual current is related to electrons (and thus negative charge).
So it can be useful when learning right hand rules etc. to do so in terms of conveniently current but physical phenomenon should be decided in terms of current actually is.
(I studied VCE so feel free to correct me if the HSC course says differently )
Post by: itssona on February 26, 2018, 02:08:21 pm
I used the grip rule and got anticlockwise. How do I use the right hand palm rule?
Post by: Dragomistress on February 26, 2018, 05:44:12 pm
Post by: jamonwindeyer on February 26, 2018, 08:16:29 pm
hi could someone please tell me why its not anticlockwise, and instead, its clockwise?
I used the grip rule and got anticlockwise. How do I use the right hand palm rule?
Grip rule gives me clockwise! We are introducing a field out of the page, so we want a field into the page to compensate. Point your right thumb into the page, and your right fingers should be wrapping clockwise! ;D
Hey, what would be an appropriate response to, "Assess the role of split ring commutator in allowing DC motors to spin continuously in one direction"
Hey! In a sentence, "It reverses current direction every half turn to maintain a constant direction of torque." Essentially, if we didn't have the split ring commutator, the current direction would not switch. This means the forces on each wire would stay in the same direction, which means after every half turn, they'd start pushing the coil back in the other direction. The split ring commutator makes that switch to keep the torque generated in the same direction ;D
Post by: clovvy on March 07, 2018, 11:24:32 pm
I was reading through ATAR Notes physics ((Jamon's notes)... One of the syllabus heading says 'conduct an investigation to verify the effects of distance, magnetic field strength, and relative motion, on electromagnetic induction in a coil by a moving magnet'... I liked the explanations on this but I could not find this anywhere in the syllabus... Can anyone explain? Thanks
Post by: blasonduo on March 08, 2018, 11:45:33 am
Hi,
I was reading through ATAR Notes physics ((Jamon's notes)... One of the syllabus heading says 'conduct an investigation to verify the effects of distance, magnetic field strength, and relative motion, on electromagnetic induction in a coil by a moving magnet'... I liked the explanations on this but I could not find this anywhere in the syllabus... Can anyone explain? Thanks
Hello! He has just simplified this dot point:
plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when:
– the distance between the coil and magnet is varied
– the strength of the magnet is varied
– the relative motion between the coil and the magnet is varied
:)
Post by: radnan11 on March 10, 2018, 05:31:39 pm
Do the brushes in a DC motor necessarily need to be made out of graphite(carbon) or can they be other materials eg. copper?
Thanks :)
Post by: jamonwindeyer on March 11, 2018, 09:53:52 am
Hi, AN
Do the brushes in a DC motor necessarily need to be made out of graphite(carbon) or can they be other materials eg. copper?
Thanks :)
Hey there radnan11! Can definitely be made of anything that conducts electricity - And it depends on the application as well. Copper brushes tend to have better conductance, so sometimes they'll be used. Usually though, as you say, it's a carbon brush (usually with some copper powder in there as well, to make it slightly less resistive) ;D
Lots of reasons for that preference. The physical composition of these is better for brushes as it is a little softer (will wear away itself instead of wearing away the commutator), has a higher melting point (the arcs that occur in brushed motors get hoooooot, need to make sure yo stuff don't melt), and tends not to have as much friction at the contact. The slightly higher resistance of these brushes is also good for dealing with the large currents that exist while the motor is starting ;D
^ None of that is assessable knowledge, but since you are asking about material which is already sort of beyond the syllabus, thought I'd give you some background ;D
Post by: stels on March 13, 2018, 12:35:35 am
"Plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when the distance between the coil and magnet is varied, the strength of the magnet is varied and the relative motion between the coil and the magnet is varied"
For the hypothesis, I need 3 predictions with 3 correct explanations, so basically 3 sentences.
The theory is:
At smaller distances, the magnetic field is denser, meaning more flux lines will be cut and so a greater EMF will be induced.
So can I just add the words "It is predicted" in front of the statement above?
"It is predicted that at smaller distances, the magnetic field is denser, meaning more flux lines will be cut and so a greater EMF will be induced."
...or is that too limited for a hypothesis since I didn't say anything about larger distances, should I edit anything?
For the other two hypothesis:
It is predicted that through increasing the number of identical magnets, a stronger magnetic field will be created, meaning greater amount of flux is cut (is there another word for this since I don't entirely get what it means when a flux is cut), and because of this, it induces a greater EMF.
It is predicted that as the relative motion becomes greater, there will be a greater induced EMF because there is a faster change in flux.
ALSO, I need 2 correctly identified risks and two minimisation strategies. I feel like this experiment is fairly safe, no safety glasses were needed or anything.
Post by: Fizzycyst on March 13, 2018, 11:12:38 am
I have always taught that hypotheses are just a possible answer to the aim, it is a statement, no need to write it is predicted, but different teachers do different things. Just say "as the distance between the magnet and the coil decreases..." tbh, you shouldnt even need to mention about rate of change of flux as it is not an explanation, it is legit just "what would happen", that is that the current should increase.
Relative motion refers to both speed and direction, you have identified speed, but what if the direction of relative motion was changed?
Without knowing how you will be conducting the prac, its pree impossible to help with hazard minimisation -- how do you intend on undertaking the prac? It's not an easy prac to do well! You need to keep in mind appropriate strategies to control variables which you do not want to change, for example for one set of results you want to change strength of magnet (pretty easy), but how will you keep the other things (distance and relative motion) controlled for this?
Post by: kaustubh.patel on March 13, 2018, 01:40:20 pm
Post by: jamonwindeyer on March 13, 2018, 08:50:03 pm
tbh, you shouldnt even need to mention about rate of change of flux as it is not an explanation, it is legit just "what would happen", that is that the current should increase.
I was taught the same for a hypothesis, that it doesn't need to explain what you will observe, just to go through the observations themselves :) but roll with your teacher since they will mark!
As some ideas for risks/prevention strategies:
- You will be using electricity in this experiment, which presents the risk of electrocution. Ensure that apparatus is not connected directly to mains power, and if possible, has circuit breakers/protection devices built in (otherwise, ensure the mains supply has these). Ensure there is no water near the experiment.
- You will be moving magnets quickly in this experiment, which means you might need to protect yourself against projectiles if you slip. Safety goggles :)
Post by: jamonwindeyer on March 13, 2018, 11:00:42 pm
Hwy a question about AC induction motor. If the main reason for the induction motor to work is due to the induced eddy currents creating a magnetic which interacts with the rotating magnetic field then why are there laminated discs in the squirrel cage. To enhance the eddy currents there shouldn't be laminated discs so the magnetic field those large eddy current produce will have a greater strength and thus have greater force output.
Hey! Great question - The HSC simplifies the construction of a squirrel cage rotor a bit which is why this gets a little tough. So:
- A squirrel cage rotor has a core made up of laminations (the bit you are talking about), but the actual squirrel cage itself is made of conductive bars that aren't laminated.
- The purpose of that core is to direct the magnetic field efficiently through the conductive bars of the squirrel cage, thus inducing a current and... You know the rest.
- We want all of that energy (or as much as possible) going to the conductive bars to be useful, meaning we need to minimise any induced currents in the core. Exactly the same as a transformer. This is why the rotor has laminations.
tl;dr - The laminations are only in specific places on the squirrel cage rotor that stop currents flowing unless they will directly assist with rotation :)
Post by: justwannawish on March 15, 2018, 06:49:04 pm
Was just wondering what angle the theta in torque=NBIAcos theta was referring to? I know torque is max when theta=0 but when does that happen? And is it the opposite angle to F=BILsintheta?
Also I'm a bit confused with how to interpret coil diagrams in terms of torque? E.g. Questions like the attached photo don't make sense!!
Post by: Fizzycyst on March 15, 2018, 06:59:43 pm
The theta in the force equation, is the angle between the 'current carrying conductor' or flow of charged particles and the external magnetic field. In the diagram, the conductor itself is going into / out of the page and the field is going across the page, these two directions are perpendicular and hence theta = 90. When you do your right hand palm rule it is the angle made between your thumb and your index finger.
Basically, as a rule of thumb (Yes, pun intended), in motors the angle in the force equation is 90 degrees all the time through the rotation of the coil in the field.
Post by: jamonwindeyer on March 15, 2018, 08:19:08 pm
In the torque equation theta refers to the angle between the plane of the coil and the magnetic field, which would be the theta as shown in the question.
Yep! But careful justwannawish, that \(\theta\) angle in the image isn't what we need. It's the other part of that 90 degree angle, so the angle formed between the coil and the magnetic field, as I've marked in green down here ;D
(Love your work, Fizzycyst ;D)
(https://i.imgur.com/daXeA1s.png)
Post by: Fizzycyst on March 15, 2018, 08:45:16 pm
Yep! But careful justwannawish, that \(\theta\) angle in the image isn't what we need. It's the other part of that 90 degree angle, so the angle formed between the coil and the magnetic field, as I've marked in green down here ;D
(Love your work, Fizzycyst ;D)
(https://i.imgur.com/daXeA1s.png)
^^ this
TY for correcting!! :D
Post by: not a mystery mark on March 17, 2018, 07:20:58 pm
Two aeroplanes flying on a collision course at 10.0km apart.
Aeroplane A is flying at 500km/hr on a heading of North 30° East.
Aeroplane B is flying at 600km/hr on a heading of West 30° North.
If neither deviated from its original course - Calculate the time it would take before they collided.
I literally have no idea how to approach this. Thank you for checking it out. Blesu desu. <3
Post by: jamonwindeyer on March 18, 2018, 01:27:27 pm
I have another question... please send help.
Two aeroplanes flying on a collision course at 10.0km apart.
Aeroplane A is flying at 500km/hr on a heading of North 30° East.
Aeroplane B is flying at 600km/hr on a heading of West 30° North.
If neither deviated from its original course - Calculate the time it would take before they collided.
I literally have no idea how to approach this. Thank you for checking it out. Blesu desu. <3
Hey hey! I assume this is a Prelim question? If you are a current Year 12 student, you won't need to tackle something like this ;D
That said, draw yourself a triangle showing the situation. The bottom side should be horizontal, 10km long. Then draw two lines coming up from that at the angles described in the question. So, Plane A's line should come up from the left at 30 degrees from the vertical, then Plane B's from the right at 30 degrees from the horizontal.
The two sides you have drawn both have a length that depends on the time that has passed. Plane A's will be \(500t\) and Plane B's will be \(600t\), where \(t\) is in hours!
So we have a triangle with three sides: 10, 500t and 600t. We also know the angle on the right, 30 degrees. We can tie all that together with the cosine rule, which hopefully you've seen before?
This gives you a quadratic equation that you can solve to find \(t\)! Does this help? :)
Post by: itssona on March 18, 2018, 02:57:35 pm
A train conductor sets the clocks on all City-rail stations to the same time. He synchronised them. A passenger is on a train somewhere on its journey from Central to Redfern. According to this passenger, when the clock strikes noon at Central, what time is it at Redfern? Noon, before noon, or afternoon .
Post by: 8Dadeedo on March 18, 2018, 03:28:09 pm
Hope this is the right place haha
Thank ya
Post by: blasonduo on March 19, 2018, 08:41:34 am
hi any help with this relativity simultaneity question please :)
A train conductor sets the clocks on all City-rail stations to the same time. He synchronised them. A passenger is on a train somewhere on its journey from Central to Redfern. According to this passenger, when the clock strikes noon at Central, what time is it at Redfern? Noon, before noon, or afternoon .
Hey! I would say Afternoon!
As he is travelling to Redfern, by simultaneity, he should observe the clock in Redfern to be slightly ahead. This is because light from Redfern will get to the passenger before the light from central (as he is moving away from central and closer to Redfern) This change in time would be minuscule though!
Post by: jamonwindeyer on March 19, 2018, 08:49:59 pm
Hey, I was hoping to get some help with the following question :)
Hope this is the right place haha
Thank ya
Welcome to the forums! It is indeed :)
So what we are looking for is \(F_\text{moon}=F_\text{earth}\). Let's define \(r\) as the distance from the centre of the earth, making the distance from the centre of the moon \(d=3.85\times10^8-r\). We also know that \(M_e=81.4M_m\). Put all this together with Newton's Law of Gravitation, where \(m\) is the mass of the spacecraft:
After all the cancellations, just solve for \(r\) from there!!
Post by: talitha_h on March 20, 2018, 07:23:35 pm
This is worth 1 mark, the answer is that craft is decelerating as it approaches the planet x. But I don't understand why this is so ??
thanks in advance
Post by: blasonduo on March 20, 2018, 08:43:15 pm
Q: Near the end of the rocket journey to planet X the astronaut strands on the scales every hour and records the readings 82 kg, 85, 111, 128. Make an inference about the motion of the rocket during this time.
This is worth 1 mark, the answer is that craft is decelerating as it approaches the planet x. But I don't understand why this is so ??
thanks in advance
Hello! ^~^
I think it is really useful to put yourself into the situation, like an elevator. Imagine you are currently going down to the ground floor (ie the planet in the scenario) and as you approach the floor, you'll have to slow down, and thus decelerate. When this happens, you start to feel much heavier (due to Newton's law of inertia, where our bodies want to continue with the same speed, so we apply more force to the floor and more force onto the scales)
Hope this helps! :))
If anything isn't clear, let me know, I'd love to explain :D
Post by: jasn9776 on March 22, 2018, 12:37:26 pm
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.
However, i am confused as there are explanations involving a tall mountain and firing horizontally.
Thanks
Post by: RuiAce on March 22, 2018, 12:47:27 pm
What is escape velocity?Whilst what you've said is true, it's missing that extra piece of information as you've stated below.
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.
However, i am confused as there are explanations involving a tall mountain and firing horizontally.
Thanks
Our analysis of escape velocity is entirely dependent on firing horizontally over a tall mountain. What this means is that the projectile cannot be fired from an angle. Most of the time when throwing a ball a long distance or something, you'd fire it at an angle. If you wanted to maximise its distance, you'd fire it at 45 degrees, and if you wanted to maximise its distance, you'd fire it vertically upwards.
Here, the whole aim of firing it horizontally (i.e. 0 degrees) is to actually bring the effect of Earth's gravity into play. Normally, if you fire something horizontally (at say, chest height), it wouldn't go up, because the Earth's gravity would bring it down. However if you fired it quickly enough (as though you launched a rocket from your chest), it would probably keep going and eventually clash with some trees in front of you.
So instead we fire it from a high mountain to ensure that there's no obstacles in our way. And the idea is that the more velocity we fire it with, the more the distance it'll cover before it lands.
(That's quite important. Eventually, we'll fire it so quickly that it doesn't land, and in fact the projectile ends up orbiting around the Earth. If we keep going up, then eventually it will manage to escape. In fact, it's escaping without us firing it at an angle; what's so powerful is that it manages to escape despite the fact that we've fired it horizontally.)
Post by: blasonduo on March 22, 2018, 12:51:12 pm
What is escape velocity?EDIT just adding on because typing is slow >_<
Is my definition correct?:
escape velocity is the minimum initial velocity at the surface of a body required for an object to escape from the influence of the body's gravitational field and not return.
However, I am confused as there are explanations involving a tall mountain and firing horizontally.
Thanks
Hello! You are absolutely right!
This throwing stuff comes from the theorization of how escape velocity came to be, kind of Newtons "Thought experiment"
His theory was a progression, such that if he were to throw an object horizontally out, it would make a parabolic shape and hit the ground.
If he threw it with a larger horizontal speed, it would still be parabolic, but it would hit the ground further away.
If he threw it fast enough so that the speed of the object and its rate of falling exactly matched the curvature of the earth, the object would be in orbit.
And this is where escape velocity comes in; At even faster speeds, he thought he could eventually escape the earth's gravitational pull entirely (11.2km/s straight up :D )
So this whole throwing a stone thing was Newton's way of thinking of the motions of objects due to earth's gravitational pull :))
Hope this helps! ^~^
Post by: jasn9776 on March 22, 2018, 06:25:17 pm
Here, the whole aim of firing it horizontally (i.e. 0 degrees) is to actually bring the effect of Earth's gravity into play. Normally, if you fire something horizontally (at say, chest height), it wouldn't go up, because the Earth's gravity would bring it down. However if you fired it quickly enough (as though you launched a rocket from your chest), it would probably keep going and eventually clash with some trees in front of you.
So instead we fire it from a high mountain to ensure that there's no obstacles in our way. And the idea is that the more velocity we fire it with, the more the distance it'll cover before it lands.
(That's quite important. Eventually, we'll fire it so quickly that it doesn't land, and in fact the projectile ends up orbiting around the Earth. If we keep going up, then eventually it will manage to escape. In fact, it's escaping without us firing it at an angle; what's so powerful is that it manages to escape despite the fact that we've fired it horizontally.)
So what your saying is that the height of the mountain doesn't matter as long as there are no obstacles in the way. But doesn't gravity decrease the further you are from the centre of the mass. So wouldn't a very tall mountain require less escape velocity than a short mountain? v=sqrt(2GM/r) so as r increases the escape velocity decreases even if minutely?
Also i don't get why firing it horizontally is the only way to 'bring the effect of Earth's gravity into play'. I would still be fully affected by gravity if i fired it vertically upwards from the surface right? So is the escape velocity the same if i fire it from the surface vertically upwards compared to horizontally ???
Anyways thanks for posting so fast. Its amazing what you can get free answers so quick.
Post by: RuiAce on March 22, 2018, 09:45:38 pm
So what your saying is that the height of the mountain doesn't matter as long as there are no obstacles in the way. But doesn't gravity decrease the further you are from the centre of the mass. So wouldn't a very tall mountain require less escape velocity than a short mountain? v=sqrt(2GM/r) so as r increases the escape velocity decreases even if minutely?It matters once you consider the magnitude of the velocity - I was pretty much just illustrating the concept. As per the formula, indeed theoretically speaking the higher you are on the mountain (or, tbh, just ON a higher mountain), the formula \( v = \sqrt{\frac{2GM}{r}} \) suggests that there will be a minute decrease in the velocity for some minute increase in \(r\).
Also i don't get why firing it horizontally is the only way to 'bring the effect of Earth's gravity into play'. I would still be fully affected by gravity if i fired it vertically upwards from the surface right? So is the escape velocity the same if i fire it from the surface vertically upwards compared to horizontally ???
Anyways thanks for posting so fast. Its amazing what you can get free answers so quick.
(Essentially, because \(r\) actually measures the distance you are from the centre of the Earth. But again, I was really just illustrating a concept, because that's the more important bit.)
Perhaps, the word 'only' is a mistake. We should replace it with something like "fully" instead. Because a projection upwards would mean that you somewhat had a thrust in the upwards direction. Whereas when you project horizontally, the only force in the vertical direction is gravity.
Post by: sadfd45678 on March 22, 2018, 09:46:12 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf (Q19)
B was my answer because if the astronaut was in orbit, it wouldnt "fall" down, it would just go around and around the planet.
Thanks so much
Post by: blasonduo on March 22, 2018, 10:39:35 pm
Hi guys i have a couple HSC questions where I have trouble wrapping my head around:
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf (Q19)
B was my answer because if the astronaut was in orbit, it wouldnt "fall" down, it would just go around and around the planet.
Thanks so much
Hello!!
This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why, i'll be happy to!)
Most people do pick B for that reason, they don't really understand C or D, so go with B, because it is technically correct but it doesn't actually explain anything relevant to the question
I hope this helps! :)
Post by: jasn9776 on March 23, 2018, 05:47:32 pm
why is this complete rubbish (I found it in the marking criteria)?
Many students wrote that the momentum of the rocket must be constant, and since p = mv and
mass is decreasing as fuel is burnt, velocity must increase to keep the momentum of the rocket
constant. This is COMPLETE RUBBISH!. The rocket accelerates (increases its velocity) because there
is a non-zero net force acting on the rocket in the forward direction! The rate of acceleration
increases for the reasons outlined above.
looks pretty similar to the sample answer from the 2016 physics q28 guidelines:
Since the mass of the rocket is decreasing while the force acting on it due to the gases is
constant, the acceleration of the rocket increases as it ascends.
So i guess the mathematical explanation ignores the physics even though it is technically correct but doesn't actually explain it? is momentum is not constant!? Why is that a flawed explanation? because it didn't use F=ma? OF COURSE IT IS NOT CONSTANT! It is constantly changing. it is change in momentum that is equal between gasses and rocket
Momentum of the rocket:
Δp rocket = -Δp fuel (Newton’s 3rd Law) and the rocket is continually gaining momentum as Δp rocket is positive
(upwards).
Post by: Fizzycyst on March 24, 2018, 08:51:53 pm
Hi guys i have a couple HSC questions where I have trouble wrapping my head around:
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf (Q19)
B was my answer because if the astronaut was in orbit, it wouldnt "fall" down, it would just go around and around the planet.
Thanks so much
One of the worst (if not THE worst) MCQ in HSC history.
I remember presenting at HSC Meet The Markers for that years HSC Physics paper. That MCQ caused a massive uproar amongst the 100+ Physics teachers, academics and university professors in both sessions and myself and the other 2 presenters really could not defend that question.
The audience was legit fuming, until I started mentioning the question which I marked for the HSC that year (the semiconducting modelling question) and how my favourite response was a drawing of an elephant. LoL.
From what I have heard it was a last minute replacement question which replaced another question (could it have been worse?) and it didn’t go through the regular checks that the questions usually do.
It really does not have a good, correct answer. The given response, (d), as it reads implies that they would certainly experience different accelerations due to their vastly different masses.
gg HSC.
Post by: Aaron12038488 on March 28, 2018, 02:40:17 pm
Post by: sadfd45678 on March 28, 2018, 08:17:27 pm
Hello!!
This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.
We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why, i'll be happy to!)
Most people do pick B for that reason, they don't really understand C or D, so go with B, because it is technically correct but it doesn't actually explain anything relevant to the question
I hope this helps! :)
Hi Can ou explain why it is D then. Also, is my understanding correct: Anything in orbit is constantly freefalling due to the same acceleration?
Post by: Fizzycyst on March 29, 2018, 08:27:23 am
not sure if it is just me or for everyone else, but I cant seem to access the old physics stage 6 syllabus . It keeps redirecting me to the NESA website, and all I cn find is the new syllabus instead of the old...
https://educationstandards.nsw.edu.au/wps/wcm/connect/0f85cdff-baa0-402e-b066-57b684cea56a/physics-st6-syl+Physics.pdf?MOD=AJPERES&CVID=
Post by: jamonwindeyer on March 29, 2018, 10:47:19 pm
Hi Can ou explain why it is D then. Also, is my understanding correct: Anything in orbit is constantly freefalling due to the same acceleration?
Anything in Orbit is constantly freefalling due to the same acceleration: I'd say this is fair! I'd say it's a tad better to say that an object is in constant freefall because the curvature of its flight path matches the curvature of the earth (or whatever planet). Only because I associate those two things a bit more closely. But you're right - Anything in orbit is undergoing (roughly) uniform circular motion, which means a constant magnitude of acceleration :)
Post by: cthulu on March 30, 2018, 03:50:47 pm
(https://i.imgur.com/KLX1Ngt.png)
(https://i.imgur.com/bpT8S46.png)
Post by: Fizzycyst on March 30, 2018, 10:37:58 pm
Hey, just joined this forums a few weeks ago and have been lurking the entire time, got my half-yearlies next week and I am unable to ask my teacher for help so I came here! I found this question recently in a Girraween past paper and I am unable to solve it, would be appreciated if anyone can help! Thanks!
(https://i.imgur.com/KLX1Ngt.png)
(https://i.imgur.com/bpT8S46.png)
Using RHG rule, it is noted that in region II, both currents give rise to a magnetic field in the same direction (into the page), therefore this is where the field in strongest as the fields combine.
Part (d), you have field lines into the page, proton moving to the right — so you can use the RHP Rule to determine the direction of force on the proton being up the page or towards the wire carrying 10A right. Therefore you could say the proton accelerates or deflects upwards.
I hope that helps!
Post by: itssona on April 01, 2018, 05:50:02 pm
"There is constant back emf induced in transmission wires as eddy currents are produced in wires, resulting in current loss."
thank you!
Post by: dermite on April 02, 2018, 09:30:47 am
Thanks in advance,
Post by: clovvy on April 02, 2018, 07:46:28 pm
Its maximum height and the velocity it is projected
Post by: blasonduo on April 02, 2018, 09:05:36 pm
how does back emf mean current loss in regards to AC Generators?
"There is constant back emf induced in transmission wires as eddy currents are produced in wires, resulting in current loss."
thank you!
Hey! Back EMF is only produced where there is a change in magnetic flux, so theoretically, if anything is near-by that is able to conduct electricity, eddy currents will be induced (even if its minuscule!). These eddy currents do not actually get transferred with the rest of the current, so there will be a current loss.
Hope this answers your question, if not, please ask away!! :))
Hi, I would like some help with this question.
Thanks in advance,
Hey! I'll try to guide you through it!
This whole question relates to the production of back emf. As we know, back emf opposes the flow of the supply current, so it decreases the net flow of current. The slower the motor, the less back emf and thus more supply current. This increase in current can destroy the wires due to heat. Hopefully this will be able to guide you through the question! :)) If not, please ask!!
A projectile has time of flight of 7.5s and a range of 1200 m. Calculate:
Its maximum height and the velocity it is projected
Although it does not state where it was launched from the ground, nor landed on ground level, I am going to assume this.
viy = initial velocity (vertical)
vfy = final velocity (vertical)
others should be self-explanatory.
Finding initial horizontal velocity;
Pretty easy....
Finding initial vertical velocity;
For height
Post by: Dragomistress on April 04, 2018, 07:14:54 am
It says that we need one AC electric motor design so, can I use the motor which is highly similar to the DC motor or do I also have to do the AC induction motor?
Post by: Fizzycyst on April 04, 2018, 02:11:13 pm
For the syllabus point, "Describe the main features of an AC electric motor".
It says that we need one AC electric motor design so, can I use the motor which is highly similar to the DC motor or do I also have to do the AC induction motor?
Tbh, I would go with both. Questions are always asked about both.
If you have time also look over the universal motor and synchronous motors, just in case.
The universal motor did come up in the 2016 HSC (MC Q20), synchronous motors have never been asked anywhere, but doesn't mean it would happen, it is a type of AC motor and it is in most textbooks, which is something the exam committee takes into account. They are quite similar in construction to an AC induction, so its not learning that much extra.
Do also keep in mind that 'simple' AC motors aren't really around -- as they would need to be rotating at the same frequency as the AC input
Post by: jamonwindeyer on April 04, 2018, 07:52:28 pm
For the syllabus point, "Describe the main features of an AC electric motor".
It says that we need one AC electric motor design so, can I use the motor which is highly similar to the DC motor or do I also have to do the AC induction motor?
^ To add to above, if you are doing your notes induction motors do have their own dot point later in the course!! When I did mine I just put a few quick notes on universal and 'basic' AC motors ;D
Post by: itssona on April 06, 2018, 07:36:52 am
Post by: kiwiberry on April 06, 2018, 11:51:14 am
any help pls? (quick if u can since exam soon)
On the moon, W=224 and g=1.6. Using W=mg, we can find the astronaut’s mass to be 140kg. Using W=mg again with m=140 and g=3.72 instead, we can calculate their weight on Mars to be 520.8N :)
Post by: Fizzycyst on April 06, 2018, 01:06:22 pm
On the moon, W=224 and g=1.6. Using W=mg, we can find the astronaut’s mass to be 140kg. Using W=mg again with m=140 and g=3.72 instead, we can calculate their weight on Mars to be 520.8N :)
Careful! The question says mass. So it will be 140kg :)
Post by: kiwiberry on April 06, 2018, 01:08:20 pm
Careful! The question says mass. So it will be 140kg :)
Yes!!! Sorry, my bad!
Post by: sadfd45678 on April 06, 2018, 08:12:32 pm
When a satellite is orbiting, it is constantly falling under gravity. So why wouldn't it acceleration be 9.8 or whatever it is after it is re-arranged due to radius the satellite is orbiting at, rather than a=v^2/r?
Thanks
Post by: jamonwindeyer on April 06, 2018, 10:32:49 pm
Hi guys,
When a satellite is orbiting, it is constantly falling under gravity. So why wouldn't it acceleration be 9.8 or whatever it is after it is re-arranged due to radius the satellite is orbiting at, rather than a=v^2/r?
Thanks
Hey! So the \(g\) you mean in the first bit is from Newton's Law right?
But in an orbit, gravity provides a centripetal acceleration:
Essentially, you are right - It is 9.8 (or a slightly lower value anyway, based on altitude). And it is also \(\frac{v^2}{r}\). The two are the same, and when you put them together you get a way to derive the formula for orbital velocity ;D
Post by: Dragomistress on April 08, 2018, 02:51:04 pm
^ To add to above, if you are doing your notes induction motors do have their own dot point later in the course!! When I did mine I just put a few quick notes on universal and 'basic' AC motors ;D
Just curious, what is the dot point for the induction motor?
Post by: Fizzycyst on April 08, 2018, 03:15:32 pm
Just curious, what is the dot point for the induction motor?
The induction motor doesn’t have a theory dot point as such, just the practical dot point where you need to perform a practical to demonstrate the principle behind it.
But, in order to show how it demonstrates the principle, you need to know how it works.
I have seen Questions worth up to 5 marks in HYE and trials basically asking you to explain how an AC induction motor works.
Post by: jamonwindeyer on April 08, 2018, 04:21:03 pm
Just curious, what is the dot point for the induction motor?
Yep as above, now that I've looked at the syllabus again it isn't actually that much later at all - It is literally directly beside the "features of AC motors" one ;D just a structural thing, I preferred to stick my notes on induction motors with the notes for the prac, important to structure your notes how best makes sense to you though :)
Post by: 8Dadeedo on April 10, 2018, 12:32:55 pm
Thank ya :D
Post by: Fizzycyst on April 10, 2018, 03:13:54 pm
Hey can I please get some help with the following?
Thank ya :D
This seems about right!
Don't judge my drawings >:(
Post by: sadfd45678 on April 13, 2018, 09:09:56 pm
How do i prepare for physics MCQ. No matter how much I do there is always a couple that stumps me like the HSC 2017 Q17. Like there's always something that you havent thought about. If anyone has anytips on how to do well at it, please let me know. Thanks
Post by: Fizzycyst on April 14, 2018, 08:32:02 pm
Past papers are a good way to consolidate understanding and look for different types of questions, but undoubtedly, there will be questions which ask you to bend your knowledge a bit to an unfamiliar context — this is what High Band 5 / Band 6 students are able to do. See a scenario, think of how their existing knowledge applies to the new scenario and then apply it.
You can’t know everything beforehand, you need to be able to APPLY what you know to the unfamiliar scenarios. No textbook and/or teacher and/or tutor teaches every possible type of question for every possible dot point, even if they think they do.
Teachers equip you with the skills you need, but you need to be able to APPLY.
Engineers aren’t like “Bruvv, never learnt this exact scenario in my 5 yr degree, must be impossibru” Rather, they twist what they are familiar with to apply to the given scenario.
This is where you need to be to address some of the harder MCQ. There is no “trick”, there is no magic NZT-48 for this, just hard work.
Hi Guys
How do i prepare for physics MCQ. No matter how much I do there is always a couple that stumps me like the HSC 2017 Q17. Like there's always something that you havent thought about. If anyone has anytips on how to do well at it, please let me know. Thanks
Post by: clovvy on April 18, 2018, 03:55:36 pm
Post by: jamonwindeyer on April 18, 2018, 05:51:24 pm
A motor is connected in series with an ammeter and a 12V power supply. The internal resistance of the motor is 1.5 Ohms. When the motor is running at full speed, the ammeter reads 3.0 A. Account for this.
Hey! So this is a Back EMF thing - If you use Ohm's Law with \(V=12\) and \(R=1.5\) -> You get \(I=8\)A! But when the motor is running, you get Back EMF induced, which will subtract from the supply current - In this case, it has to be \(I_\text{Back}=5\)A, since that is what the ammeter measures as the total current when it is running ;D
There's a bit of explanation on Back EMF here, in an old guide I wrote many a year ago ;D
Post by: clovvy on April 18, 2018, 06:44:52 pm
Hey! So this is a Back EMF thing - If you use Ohm's Law with \(V=12\) and \(R=1.5\) -> You get \(I=8\)A! But when the motor is running, you get Back EMF induced, which will subtract from the supply current - In this case, it has to be \(I_\text{Back}=5\)A, since that is what the ammeter measures as the total current when it is running ;D
There's a bit of explanation on Back EMF here, in an old guide I wrote many a year ago ;D
That's weird, the answer for this is 7.5 and where did 5A come from?
Post by: Fizzycyst on April 18, 2018, 06:59:20 pm
5A came from the expected current (12/1.5) minus the measured current (3A).
It’s worthwile to note the formula EMFnet = EMFsupply - EMFback and note this can be used also with current, which is essentially what Jamon did.
That's weird, the answer for this is 7.5 and where did 5A come from?
Post by: jamonwindeyer on April 18, 2018, 07:43:58 pm
That's weird, the answer for this is 7.5 and where did 5A come from?
Weird indeed - Not sure where they'd get 7.5A from:
Edit: Ahh, they want the EMF not the current - Fair enough!! It's a written response so they'd be happy with either or there, they are more after your account for why it is happening in the first place ;D
Post by: clovvy on April 18, 2018, 07:48:00 pm
Weird indeed - Not sure where they'd get 7.5A from:
Edit: Ahh, they want the EMF not the current - Fair enough!! It's a written response so they'd be happy with either or there, they are more after your account for why it is happening in the first place ;D
It's in volts not amps
Post by: jamonwindeyer on April 18, 2018, 07:49:39 pm
It's in volts not amps
Yep, clicked as soon as I saw Fizzycyst's response, see my edit above - Make sure you can explain it that's the main thing they'll be looking for! Since it is an account question they want the reasons, the maths should only be there as a support ;D
Post by: jasn9776 on April 20, 2018, 10:08:27 pm
So the exam question asked: Why when the load of DC motor is increased, it is forced to slow down, the torque it applies onto the load increases.
Is it enough to say the back emf is reduced because of slowed turning, therefore slow change in magnetic flux, thus lower emf produced. Vnet=Vsupply - Vback. V=IR as Vnet goes up I goes up. torque=BIAcostheta.
But the marking guidelines said the reversal of current causes back emf every-time. As the DC commutator reverses direction of current, back-emf is produced. When under load, spins slower therefore reversal of current direction is less frequent, hence actual voltage is greater.
Which one is better?
Post by: danaodish on April 21, 2018, 05:08:35 pm
Just quickly, my teacher said that we'd be given the 2019 Data sheet for this years (2018) HSC... I don't think that's true but does anyone know if it is? :-\
Post by: RuiAce on April 21, 2018, 05:18:07 pm
Hi!That seems absurd since it literally says that it's the data sheet for 2019, but here's the link.
Just quickly, my teacher said that we'd be given the 2019 Data sheet for this years (2018) HSC... I don't think that's true but does anyone know if it is? :-\
(Pending someone more qualified to re-answer your question though.)
Post by: jamonwindeyer on April 21, 2018, 11:24:34 pm
Hi!
Just quickly, my teacher said that we'd be given the 2019 Data sheet for this years (2018) HSC... I don't think that's true but does anyone know if it is? :-\
I am also skeptical that this is true! Either way, it has all the values you'll need (and more). You'll definitely get the current formula sheet since there are formulas on the old one that aren't on the new one :)
But yeah, I'd guess there's been a misunderstanding somewhere along the line :)
Post by: jamonwindeyer on April 21, 2018, 11:28:56 pm
Is it true that when the current stops suddenly or changes direction, a back-emf is induced?
So the exam question asked: Why when the load of DC motor is increased, it is forced to slow down, the torque it applies onto the load increases.
Is it enough to say the back emf is reduced because of slowed turning, therefore slow change in magnetic flux, thus lower emf produced. Vnet=Vsupply - Vback. V=IR as Vnet goes up I goes up. torque=BIAcostheta.
But the marking guidelines said the reversal of current causes back emf every-time. As the DC commutator reverses direction of current, back-emf is produced. When under load, spins slower therefore reversal of current direction is less frequent, hence actual voltage is greater.
Which one is better?
Hiya! I like the first version with the formulas waaaay better. A current changing direction doesn't really generate back emf. There would be induced current flow in conductive parts of the motor because of that switch, yes. But like...
Yeah, I'm clutching at straws here trying to give that answer some credit, but I don't really get it. The first one seems far better to me :) where's this marking criteria from? :)
Post by: jasn9776 on April 22, 2018, 11:35:16 am
Hiya! I like the first version with the formulas waaaay better. A current changing direction doesn't really generate back emf. There would be induced current flow in conductive parts of the motor because of that switch, yes. But like...From QATs 2015
Yeah, I'm clutching at straws here trying to give that answer some credit, but I don't really get it. The first one seems far better to me :) where's this marking criteria from? :)
Describe the change in current direction through the coils of a DC motor
Relates this to reduced operating voltage or current due to back-emf.
Relates this slowing down of the motor to increased current or torque.
Part of sample answer
"The reversal of direction of current through the coil causes back-emf every time, meaning that the operating voltage is always less than the applied voltage. When a load forces the motor to slow down the reversal of current direction through the coils is less frequent, hence the actual voltage - and current - is greater, hence torque is increases."
Also is a universal motor using AC unable to increase torque? The next question says the high rate of current reversal when AC is applied means that the slower rate of coils turning has little effect on back-emf experienced in the coils.
Post by: jamonwindeyer on April 22, 2018, 11:55:50 pm
From QATs 2015
Describe the change in current direction through the coils of a DC motor
Relates this to reduced operating voltage or current due to back-emf.
Relates this slowing down of the motor to increased current or torque.
Part of sample answer
"The reversal of direction of current through the coil causes back-emf every time, meaning that the operating voltage is always less than the applied voltage. When a load forces the motor to slow down the reversal of current direction through the coils is less frequent, hence the actual voltage - and current - is greater, hence torque is increases."
Interesting! Yeah, as I said, not a fan - Maybe I'm just not clicking with what it means but I'd be putting red crosses on that if I were marking it, especially next to the other answer you gave which I think makes more sense (and is actually correct, aha).
Quote
Also is a universal motor using AC unable to increase torque? The next question says the high rate of current reversal when AC is applied means that the slower rate of coils turning has little effect on back-emf experienced in the coils.
Sorry my friend, I've literally got no idea what the heck this exam is going on about :P
(Universal motors can vary their torque, but I'd say you'd be hard pressed to find that asked anywhere near a HSC exam) :)
Post by: itssona on April 24, 2018, 07:17:02 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-physics.pdf
Post by: Floatzel98 on April 24, 2018, 07:41:07 pm
can anyone explain why in 2009 Q21 (a) the answer is X?? how do we know which way to current is going in order to see that the force on X is up? thanks :)Look at the orientation of the battery.
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-physics.pdf
Post by: jamonwindeyer on April 24, 2018, 09:46:40 pm
can anyone explain why in 2009 Q21 (a) the answer is X?? how do we know which way to current is going in order to see that the force on X is up? thanks :)
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-physics.pdf
Hey! As above, the battery is labelled just below the coil (though in a really non obvious spot tbh). Even without the labels, when a battery is represented with the two lines like that, the longer line is always the positive terminal. Conventional current flows from positive to negative, so out of the page on Side X :)
From there, right hand slap rule. Current (thumb) out of the page, magnetic field to the right, force is therefore upwards :)
Post by: itssona on April 25, 2018, 12:32:32 pm
The answer says that Earth & Sun exert a gravitational field which creates a centripetal force (?) that keeps JWsT in orbit but how does this answer the question :/
thanks :)
Post by: jamonwindeyer on April 25, 2018, 03:17:31 pm
can someone help with 2010 q24 please :(
The answer says that Earth & Sun exert a gravitational field which creates a centripetal force (?) that keeps JWsT in orbit but how does this answer the question :/
thanks :)
Hey! Are you referencing the NESA sample answer? I think it answers the question:
Both the Sun and the Earth exert a force on the JWST which is towards the Sun. Therefore the
centripetal force on the JWST is greater than that produced by the Sun alone. This means that
the orbital speed of the JWST around the Sun is greater than it would be if the Sun alone were
producing the centripetal force on the JWST and therefore the period is less than would be
predicted from Kepler’s law of periods, given that its orbital radius is greater than that of the
Earth.
To expand on the sample answer - We understand that the combined gravitational attraction of the sun and the earth produces a larger centripetal force than what the earth experiences from the sun alone. According to:
An increase in force will correspond to an increase in velocity. As F goes up, v goes up. This allows it to contradict Kepler's Law of Periods. Even though its orbital radius is slightly greater than that of earth, it can still orbit with the same orbital period as the earth, since it is moving faster than it would normally be moving at that orbital radius ;D
Post by: itssona on April 26, 2018, 12:43:13 am
Hey! Are you referencing the NESA sample answer? I think it answers the question:shh makes sense!! thank you jamon :)
Both the Sun and the Earth exert a force on the JWST which is towards the Sun. Therefore the
centripetal force on the JWST is greater than that produced by the Sun alone. This means that
the orbital speed of the JWST around the Sun is greater than it would be if the Sun alone were
producing the centripetal force on the JWST and therefore the period is less than would be
predicted from Kepler’s law of periods, given that its orbital radius is greater than that of the
Earth.
To expand on the sample answer - We understand that the combined gravitational attraction of the sun and the earth produces a larger centripetal force than what the earth experiences from the sun alone. According to:
An increase in force will correspond to an increase in velocity. As F goes up, v goes up. This allows it to contradict Kepler's Law of Periods. Even though its orbital radius is slightly greater than that of earth, it can still orbit with the same orbital period as the earth, since it is moving faster than it would normally be moving at that orbital radius ;D
Post by: itssona on April 29, 2018, 01:07:18 pm
Post by: jamonwindeyer on April 29, 2018, 01:20:19 pm
kinda confused - what does the rotor do in an induction motor? I know it isn't connected to the power source and instead it creates current but what is its role? thanks :)
It is designed as a place for the induced currents to flow in response to the changing magnetic field generated by the stator! The rotor then spins due to the motor effect to do some sort of mechanical work. It's the exact same as any other motor, except the source of current is electromagnetic induction, rather than a direct electrical supply ;D
Post by: philgee on May 11, 2018, 07:57:04 pm
"Superconductors are the future of communication technology. Either we begin to use them every day, or we fade into technological stasis."
Assess this statement with reference to at least 2 specific applications of superconductors and the Advantages and Disadvantages of their use. (7M)
Post by: jamonwindeyer on May 12, 2018, 12:03:05 am
Hello, I am unsure how to plan out my answer for this question.
"Superconductors are the future of communication technology. Either we begin to use them every day, or we fade into technological stasis."
Assess this statement with reference to at least 2 specific applications of superconductors and the Advantages and Disadvantages of their use. (7M)
Hey philgee! I like this question. So I'd start by filling in the specific things it requests:
- Application #1
- Application #2
- Advantages
- Disadvantages
The only other real thing it wants is an assessment/judgement, and that would probably go at the end. So in general the response looks like:
Superconductors are _____. They are currently used in EXPLAIN APPLICATION ONE. They are also used in EXPLAIN APPLICATION TWO. Generally, superconductors are advantageous because ________. Their limits include _______. Overall, superconductors are MAKE YOUR JUDGEMENT.
From there it is honing, and perhaps doing things like tables/dot points, decide how best to present it ;D is that helpful? Is all the content you need for this question sitting okay for you? :)
Post by: philgee on May 12, 2018, 03:10:20 pm
Thank you sooooo much!!! I clearly see what the set up should be, and the mark allocation. I appreciate it.
Post by: owidjaja on May 22, 2018, 07:53:02 pm
I was going through the applications of superconductors and was reading on its use in superconducting quantum interference device (SQUID) and how it's able to detect tiny magnetic fields with an insulator in between two superconducting materials. But how are they able to detect magnetic fields through the insulator?
Post by: jamonwindeyer on May 22, 2018, 09:53:37 pm
Hey guys,
I was going through the applications of superconductors and was reading on its use in superconducting quantum interference device (SQUID) and how it's able to detect tiny magnetic fields with an insulator in between two superconducting materials. But how are they able to detect magnetic fields through the insulator?
Hey!
Syllabus Answer: Not assessable, so don't worry!!
Actual Answer, very much rough because this is way out of my league: What those superconductors and the insulator form is called a Josephson Junction. These junctions can have a current flowing through them without any voltage applied, due to the quantum tunnelling of cooper pairs from one superconductor to the other. A magnetic field applied to this junction (in a specific arrangement, I think there is multiple in a SQUID?) will change the current flowing through it, and this allows super precise measurement of the field. Last sentence super waffly, I know, but that's my rough understanding of it. Try Googling the Josephson effect if you want to learn more!
But yeah, waaaaay beyond the syllabus, so don't stress ;D
Post by: talitha_h on May 25, 2018, 08:50:23 pm
Is it simply because it lacks a fluctuating field that comes from AC power?
Thank you
Post by: jamonwindeyer on May 25, 2018, 09:47:42 pm
Is it simply because it lacks a fluctuating field that comes from AC power?
Thank you
Hey! Exactly, AA/AAA (or whatever) batteries are sources of DC power. No changing field, no induction, so no transformer ;D
Post by: talitha_h on May 26, 2018, 04:46:55 pm
Post by: Jane20 on May 26, 2018, 10:42:33 pm
I am so confused.
For example suppose in a very long train that is moving at relativistic speed and fireworks are launched at both ends of the train at the same time. Which observer will see that the launching simultaneous ? The one inside the train and stands in the middle or the observer from outside ?
Thanks alot guys !!
Post by: blasonduo on May 26, 2018, 11:31:13 pm
So for my assessment we are doing a transformer prac. I don't specifically know what we will be doing but we will be asked on possible sources of error and suggestions for improvement for the prac. Any help is appreciated, thanks.
Hey! Assuming you measure the output voltage compared to the input voltage, I'll give you the two main ones I can think of. :))
1) We know practically, transformers are not 100% efficient, and that through flux leakage and eddy currents, the output voltage will be lower than the theoretical. To improve this, using better core materials such as iron and making sure it is laminated etc...
2) The process in which we are measuring, are we using an analogue voltmeter or a digital voltmeter? In what ways would recording a value from an analogue voltmeter differ from a digital one?
Hope this helps.
Hi guys can anyone pls explain the relative simultaneity for me :(
I am so confused.
For example suppose in a very long train that is moving at relativistic speed and fireworks are launched at both ends of the train at the same time. Which observer will see that the launching simultaneous ? The one inside the train and stands in the middle or the observer from outside ?
Thanks alot guys !!
Hello!! :))
The outside observer would see the fireworks explode at the same time.
To figure this out, imagine being on the train, when the fireworks explode, we are travelling towards one firework and away from the other. As light is constant for all observers, the light from the firework in front of us will get to us slightly faster. Meaning we will see one firework before the other.
Another way to which could (?) make it easier to understand...
Imagine a 100-meter track, there are two identical runners, one at the finish line and one at the starting line; they will be running towards each other. You, however, stand exactly 50meters between them. When the race starts, you begin to run towards the finish line. From this, we know we are running towards the guy at the finish line, and hence we will get to him first.
Hope this helps :))
Post by: owidjaja on June 01, 2018, 07:41:17 pm
I have an assignment coming up where I have to build a model- I've decided to make a magnetic induction/induction motor (as you can see in my screenshot). I know I'll be needing: string (not sure which type), a stand of some sort to hold up my string, a pie base, a strong magnet and something that can rotate really fast.
Would anyone here happen to know where I can find a strong magnet? And what could I use that can rotate really fast? My Excel textbook suggested a drill chuck but I kinda find that a bit too much, and my physics teacher suggested a motor, but I have no clue how to make a motor work...
Thanks in advance :)
Post by: blasonduo on June 01, 2018, 09:58:19 pm
Hey guys,
I have an assignment coming up where I have to build a model- I've decided to make a magnetic induction/induction motor (as you can see in my screenshot). I know I'll be needing: string (not sure which type), a stand of some sort to hold up my string, a pie base, a strong magnet and something that can rotate really fast.
Would anyone here happen to know where I can find a strong magnet? And what could I use that can rotate really fast? My Excel textbook suggested a drill chuck, but I kinda find that a bit too much, and my physics teacher suggested a motor, but I have no clue how to make a motor work...
Thanks in advance :)
Hey! I know that Jaycar sells Neodymium rare Earth magnets which are very strong and would be suitable for your experiment, but they are pretty pricey (i.e. $10-20)
When I made my generator in the HSC, I made my own handcrank, which utilised mechanical advantages, such as a big wheel for me to crank while the axle spinning the actual magnets pretty small. Cogs are the easy way to go (but I used rubberbands ;) ) Also make sure you have a solid base on which you are cranking. If it is unsteady, your "spin potential" dramatically decreases! I guess it's really up to you to find what works for you (and what's easy to construct ;) )
All in all, with the strong magnets, you might not need a fast spinning thingo, the induction was actually quite obvious even with weak magnets!
If you need any other specific question, ask away! I'll try my best!
I'd love to see your progress on this!
Post by: owidjaja on June 01, 2018, 10:23:48 pm
Hey! I know that Jaycar sells Neodymium rare Earth magnets which are very strong and would be suitable for your experiment, but they are pretty pricey (i.e. $10-20)Hmm, that rubber band is a great idea! I was thinking of having the magnet glued to a bunch of craft sticks and have a rubber band somehow attached to the craft sticks so it will rotate (kinda like those rubber band cars). But I'm trying to figure out where exactly the rubber band would be located if I were to do so. Any ideas?
When I made my generator in the HSC, I made my own handcrank, which utilised mechanical advantages, such as a big wheel for me to crank while the axle spinning the actual magnets pretty small. Cogs are the easy way to go (but I used rubberbands ;) ) Also make sure you have a solid base on which you are cranking. If it is unsteady, your "spin potential" dramatically decreases! I guess it's really up to you to find what works for you (and what's easy to construct ;) )
All in all, with the strong magnets, you might not need a fast spinning thingo, the induction was actually quite obvious even with weak magnets!
If you need any other specific question, ask away! I'll try my best!
I'd love to see your progress on this!
Post by: not a mystery mark on June 03, 2018, 05:12:11 pm
Curious to how to approach this question.
"A toy car at rest of mass 250g is hit by a toy truck and travels horizontally for 50.0m in 2.0s. Calculate the instantaneous speed just after being hit."
I just keep getting 0m/s :')
Tyy for checking this out 🌷🌷🌷🌸🌸
Post by: S200 on June 03, 2018, 05:48:04 pm
Howdy again.Are we including resistive forces? If not, isn't it just 25 m/s?
Curious to how to approach this question.
"A toy car at rest of mass 250g is hit by a toy truck and travels horizontally for 50.0m in 2.0s. Calculate the instantaneous speed just after being hit."
I just keep getting 0m/s :')
Tyy for checking this out 🌷🌷🌷🌸🌸
Post by: owidjaja on June 06, 2018, 10:36:18 pm
Just a general question on superconductors: is there any benefit in using liquid helium instead of liquid nitrogen?
Post by: RuiAce on June 06, 2018, 10:42:19 pm
Hey guys,I think it's just the issue of temperature and melting points. By the time helium has finally condensed into a liquid, nitrogen has already solidified.
Just a general question on superconductors: is there any benefit in using liquid helium instead of liquid nitrogen?
(So liquid helium, which is just colder by nature, should be able to cool things down even further.)
Post by: Fizzycyst on June 12, 2018, 07:20:10 pm
Hey guys,
Just a general question on superconductors: is there any benefit in using liquid helium instead of liquid nitrogen?
Liquid He is used to cool down the Low Tc superconductors (Type I Metals and Type II Metals and Metal Alloys) as Liquid Nitrogen is not cold enough to get below the Tc of these materials, whereas Liquid Nitrogen is used to cool the High Tc superconductors (Type II Ceramics, cuprates etc..) as their Tc are above the boiling point of Nitrogen. You could use Liquid He to cool High Tc superconductors, it would kind of defeat the purpose though.
One thing which would arise is that by making the superconductor cooler, it can withstand a higher external magnetic field -- so it could carry larger currents
Post by: justwannawish on June 15, 2018, 11:32:27 pm
Our assessment at school is a skills task on the photoelectric effect and Thomson's experiment. What are the types of things we should expect? I've done the past HSC questions on the two experiments and they are basically largely around descibing the experiments, and calcuating work function/threshold frequency for photoelectric. Also could anyone help me out with the accuracy reliability and validity aspects of those experiments? Our teacher said this would definitely be included in the exam :/
Post by: justwannawish on June 17, 2018, 07:17:35 pm
Post by: jamonwindeyer on June 18, 2018, 12:17:59 am
Hi guys, kinda in hand with my other q, does increasing the amount of frequencies measured for photoelectric improve the reliability or accuracy of it? any other qays to improve accuracy, validity and reliability of data?
It depends on what you are measuring! But if you are measuring the work function of your metal, definitely improves the reliability of the experiment, since it is additional repetition. Each experiment for every frequency will yield a value for the work function, if they are all close then you can deem your experiment to be more reliable since it can get the same result (or similar) with repetition. Accuracy comes from how close the thing is to the expected result, so your repetitions might not necessarily improve how accurate it is. For example, if your experiment is yielding a work function ten times greater than the actual value, then it doesn't matter how many times you get that value, it is still inaccurate ;D
Validity is controlling variables. Think of sources of error in the experiments you are reading about/studying and propose ways they could be minimised (better equipment, different method, etc) :)
Post by: envisagator on June 24, 2018, 12:11:57 pm
Post by: S200 on June 24, 2018, 06:57:58 pm
I would just expound on the greatness of electricity, not discarding the fundamentalists, Tesla and Edison.
Post by: blasonduo on June 25, 2018, 05:24:23 pm
How do I go about assessing the validity of a statement e.g. "Electricity, as a form of energy, is good for killing people and little else" assess the vaidity of this statment??
A statement is deemed valid when the conclusion the student draws upon follows current works or theories.
From the given statement, I see three distinct conclusions that the text draws upon;
1) Electricity is a form of energy
2) It is good for killing people
3) Its use is very limited
So in your response, you need to address each point individually and then as an overall conclusion.
Energy is the capacity to do work, electrical energy is the movement of electrons and thus contains kinetic energy. So the first statement is valid (although saying "electricity" isn't the best wording, but I'll assume that's fine :P )
It indeed has the ability to easily kill people, as little as 0.1 amps can stop the heart and has proven fatal. (This, however, is not assessed in HSC physics, so just acknowledging the fact that this statement holds some truth should be sufficient)
The last statement is probably the most critical aspect that cracks the validity, you should already know through the motors and generators topic, that electrical energy has huge uses. (globalisation, advanced medical equipment, more leisure through mechanical applications, etc...) use points such as these to prove that this point is indeed invalid.
Sum everything up, show that the statement has some truths, but overall does not have accurate conclusions and is thus invalid.
Hope this helps :)
Post by: envisagator on June 25, 2018, 09:01:59 pm
A statement is deemed valid when the conclusion the student draws upon follows current works or theories.this is fantastic, thank you very much!!!
From the given statement, I see three distinct conclusions that the text draws upon;
1) Electricity is a form of energy
2) It is good for killing people
3) Its use is very limited
So in your response, you need to address each point individually and then as an overall conclusion.
Energy is the capacity to do work, electrical energy is the movement of electrons and thus contains kinetic energy. So the first statement is valid (although saying "electricity" isn't the best wording, but I'll assume that's fine :P )
It indeed has the ability to easily kill people, as little as 0.1 amps can stop the heart and has proven fatal. (This, however, is not assessed in HSC physics, so just acknowledging the fact that this statement holds some truth should be sufficient)
The last statement is probably the most critical aspect that cracks the validity, you should already know through the motors and generators topic, that electrical energy has huge uses. (globalisation, advanced medical equipment, more leisure through mechanical applications, etc...) use points such as these to prove that this point is indeed invalid.
Sum everything up, show that the statement has some truths, but overall does not have accurate conclusions and is thus invalid.
Hope this helps :)
Post by: moq418 on June 28, 2018, 08:26:08 pm
Post by: envisagator on June 30, 2018, 12:26:39 pm
For the hsc and Trials physics exam are there any questions that are going to be always ask if so please tell me what they are thanks>!!A good start is going through the syllabus and get confident with how the syllabus points are phrased, you should be/get comfortable with the ones that asks you to assess, discuss, justify, explain, analyse.... as the questions in exams basically come from these dot points with a few twists which require you to combine knowledge from each section. Some common questions are like: Assess the contribution of Hertz, Einstein, Plank to quantum theory, competition between Westinghouse and Edison, and those development of these or that to society/ environment. Hope this helps!!!
Post by: envisagator on June 30, 2018, 12:33:58 pm
I'm having trouble phrasing the answer to this question, this is what my answer is sort off leading to: Faradays law of electromagnetic induction, then talking about how current in a circuit takes a brief moment to reach maximum or zero value resulting in a changing magnetic field (hence mag.flux) inducing an emf, is this correct?? or what should I include??
Post by: blasonduo on June 30, 2018, 12:53:34 pm
The question is: Explain why magnetic flux is important when generating emf.
I'm having trouble phrasing the answer to this question, this is what my answer is sort off leading to: Faradays law of electromagnetic induction, then talking about how current in a circuit takes a brief moment to reach maximum or zero value resulting in a changing magnetic field (hence mag.flux) inducing an emf, is this correct?? or what should I include??
This question is emphasising on Faraday's Law, but I would argue that it isn't the magnetic flux that is important, but more the change of magnetic flux. I would state Faraday's law and include the formula; I think applying the formula to the question really shows your knowledge on the topic. I would then explain how the change of magnetic flux can optimise the EMF, showing how important it really is.
I wouldn't think this would be a high mark question.
Hope this helps! Ask away if you need any clarification!
Post by: envisagator on June 30, 2018, 01:06:13 pm
This question is emphasising on Faraday's Law, but I would argue that it isn't the magnetic flux that is important, but more the change of magnetic flux. I would state Faraday's law and include the formula; I think applying the formula to the question really shows your knowledge on the topic. I would then explain how the change of magnetic flux can optimise the EMF, showing how important it really is.Its a 3 mark question. The problem i had was whether explaining faradays law in terms of the requirments for generating emf would give me the marks, I was thinking you had to explictly say explaing how magnetic flux generates an emf which I'm not exactly sure about. Just a bit confused!!
I wouldn't think this would be a high mark question.
Hope this helps! Ask away if you need any clarification!
Post by: blasonduo on June 30, 2018, 01:45:09 pm
Its a 3 mark question. The problem i had was whether explaining faradays law in terms of the requirments for generating emf would give me the marks, I was thinking you had to explicitly say explaing how magnetic flux generates an emf which I'm not exactly sure about. Just a bit confused!!
Yep! Definitely include that in! Normally, I would coincide that with Faraday's law, but that is definitely an important part of the question!
Post by: Mate2425 on July 03, 2018, 09:44:25 pm
Thanks
Post by: blasonduo on July 07, 2018, 05:30:47 pm
Hi could someone please help me with understanding Forward bias and Reverse bias in p-n junction with simple terminology and if possible the correct way to draw diagrams.
Thanks
Hey! To my knowledge, this content is beyond the scope of this course!
Post by: jamonwindeyer on July 09, 2018, 12:39:19 am
Hi could someone please help me with understanding Forward bias and Reverse bias in p-n junction with simple terminology and if possible the correct way to draw diagrams.
Thanks
Hey! To my knowledge, this content is beyond the scope of this course!
Indeed it is right on the edge of the course ;D do you understand how/why the PN junction forms? If you do, that's pretty much what you need!
Remember also that reverse bias increases the size of the depletion zone in the junction, thus making it nearly impossible for current to flow through. Forward bias allows current to flow through quite easily, because the size of that depletion zone is reduced (beyond a threshold forward-bias voltage, say about 0.7V, you can have pretty much as much current as you want flowing through without any additional voltage) :)
Happy to elaborate on anything you need though - But yeah, blasonduo is right in that this is a little beyond what you need for the course ;D
Post by: Mate2425 on July 09, 2018, 12:56:57 am
Indeed it is right on the edge of the course ;D do you understand how/why the PN junction forms? If you do, that's pretty much what you need!
Remember also that reverse bias increases the size of the depletion zone in the junction, thus making it nearly impossible for current to flow through. Forward bias allows current to flow through quite easily, because the size of that depletion zone is reduced (beyond a threshold forward-bias voltage, say about 0.7V, you can have pretty much as much current as you want flowing through without any additional voltage) :)
Happy to elaborate on anything you need though - But yeah, blasonduo is right in that this is a little beyond what you need for the course ;D
Hi, thank you everyone for you help!! If i may clarify with you Jamon, is the reason for the P/N junction forming in relation to the doping of semiconductors with Group 3 and 5 elements?
Thank you in advance!
Post by: jamonwindeyer on July 09, 2018, 12:57:26 am
Hi, thank you everyone for you help!! If i may clarify with you Jamon, is the reason for the P/N junction forming in relation to the doping of semiconductors with Group 3 and 5 elements?
Thank you in advance!
Yep that's it! And the subsequent forming of covalent bonds when you bring the two together in a junction :)
Post by: key to success on July 17, 2018, 07:43:09 pm
If someone can help me with the attached multiple choice, I'll much appreciate it! :) :) :)
Thanks!
Post by: envisagator on July 17, 2018, 08:08:31 pm
Hi guys,This question has 2 components:
If someone can help me with the attached multiple choice, I'll much appreciate it! :) :) :)
Thanks!
First calculate the weight force of the hanging mass: F=mg, which gives 1.96N
You know that the question refers to satellite in orbit around Earth, so you use F=mv^2 / r . where F is the centripetal force which in this case is the hanging mass which 'pulls' the stopper to the centre. So F= 1.96N (found above)
We know that v = 8.37m/s . therefore, v^2 = approx 70 m/s
m = the mass of the stopper
r= 0.6m (given in Q)
Substituting into centripetal force equation, and rearranging to solve for m we get m= 0.01678 kg, which is approx 17g, therefore B is the answer.
Hope this helps, note how we didnt need to use time, its one of those questions which gives more than you need. :)
Post by: key to success on July 17, 2018, 09:24:16 pm
This question has 2 components:
First calculate the weight force of the hanging mass: F=mg, which gives 1.96N
You know that the question refers to satellite in orbit around Earth, so you use F=mv^2 / r . where F is the centripetal force which in this case is the hanging mass which 'pulls' the stopper to the centre. So F= 1.96N (found above)
We know that v = 8.37m/s . therefore, v^2 = approx 70 m/s
m = the mass of the stopper
r= 0.6m (given in Q)
Substituting into centripetal force equation, and rearranging to solve for m we get m= 0.01678 kg, which is approx 17g, therefore B is the answer.
Hope this helps, note how we didnt need to use time, its one of those questions which gives more than you need. :)
That helps heaps! Thanks mate!
Post by: The Slow Hare on July 20, 2018, 07:48:15 am
*Askin this question because I have to draw two sperate diagrams for them but I don’t know what the difference is.
P.S. first time posting a question, so if I did anything wrong -> pls don’t hate
Post by: S200 on July 20, 2018, 07:57:47 am
What’s the difference between a solar cell and a photocell. From my understanding solar cell uses light hits the semiconductor complex which induces a current in the external circuit. But isn’t the photocell the exact same.Although I couldn't help you with the diagram, a quick google search gives this...
*Askin this question because I have to draw two sperate diagrams for them but I don’t know what the difference is.
P.S. first time posting a question, so if I did anything wrong -> pls don’t hate
Quote
A solar cell produces power for an electrical circuit while a photocell is a light-activated control switch. ... The only commonality between the two is that light is needed for them to work. Solar cells can be found everywhere, while photocells are hidden.
Post by: envisagator on July 20, 2018, 10:49:06 am
What’s the difference between a solar cell and a photocell. From my understanding solar cell uses light hits the semiconductor complex which induces a current in the external circuit. But isn’t the photocell the exact same.A solar cell is a device that converts sunlight into electrical energy using a p-n junction. Basic working principle is: When light (photons) of a sufficiently high frequency hits the cell, each photon of light frees an electron from the junction between the layers, thus creating holes and free electrons. The free electrons flow to the n-type layer and the holes to the p-type layer. This creates an electric potential between the layers. The electrons flow via the external circuit to the p-type layer, releasing energy to the load.
*Askin this question because I have to draw two sperate diagrams for them but I don’t know what the difference is.
A photocell on the other hand is a device that converts light energy into electrical energy by the photoelectric effect. Basic principle: A light source is directed towards a semi-conductive metal plate, causing electrons to be emmited by the photoelectric effect; the metal plate is curve so that the photoelectrons emitted are focused towards a collecting anode and subsequently pass through an external circuit as current. Used in motion detectors.
Hope this helps :)
Post by: key to success on July 20, 2018, 03:42:28 pm
Although I couldn't help you with the diagram, a quick google search gives this...
Although I couldn't help you with the diagram, a quick google search gives this...
The answers here are pretty thorough but very simply put;
solar cells- need light hence photocurrent to make devices work (power produced is used)
photocells- need an absence of light- photocurrent to trigger a switch to make things work
Post by: Dragomistress on July 22, 2018, 09:25:49 am
I am wonder why this is the case. A piece of silicon is doped with an element from group V of the periodic table and it has no overall charge? Why does it have no overall charge?
Post by: Dragomistress on July 22, 2018, 09:28:24 am
In an experiment to demonstrate the photoelectric effect, it was found that when light with a frequency f and an intensity I was shone on the cathode, the voltage needed to completely stop the photoelectric current was V volts. The effect of increasing the incident light intensity shining on the cathode is that the voltage V needed to be:
Answer: kept the same.
Post by: envisagator on July 22, 2018, 10:47:58 am
Hey,In the most basic way I could explain this is the fact that the impurity atoms (the dopant) is already in a neutral state, so when it is added to a neutral solid (semiconductor) the overall charge remains zero.
I am wonder why this is the case. A piece of silicon is doped with an element from group V of the periodic table and it has no overall charge? Why does it have no overall charge?
Hope this helps :)
Post by: jamonwindeyer on July 22, 2018, 04:40:09 pm
Also, I am wondering why this is the case.
In an experiment to demonstrate the photoelectric effect, it was found that when light with a frequency f and an intensity I was shone on the cathode, the voltage needed to completely stop the photoelectric current was V volts. The effect of increasing the incident light intensity shining on the cathode is that the voltage V needed to be:
Answer: kept the same.
Increasing the intensity of the light means that you have more photons striking the metal. This means more released photoelectrons, so, more current! However, each individual electron still has the same energy, since this is determined by frequency, not intensity. Energy per electron is what voltage depends on, so the voltage is the same even though the current is larger ;D
Post by: Duarashid on July 22, 2018, 10:29:57 pm
Thanks :)
Post by: jamonwindeyer on July 22, 2018, 10:36:03 pm
Hi! Im not sure what to label my Y axis for the black body emission spectrum graph (incase we have to draw it). I've seen a few images, with 'Spectral energy density' and others with 'intensity'. I know for the X-axis it is wavelength. Should I write the units too?
Thanks :)
Hey! Intensity is totally fine on the Y-Axis. If you wanted units, you could put W/m^2, but I doubt they'll stress that much in a diagram of the graph ;D spectral radiance is similar to intensity, but it has a more specific and more complex definition, so just roll with intensity ;D
Post by: moq418 on July 25, 2018, 11:32:47 pm
Post by: mxrylyn on July 26, 2018, 12:56:36 am
We have started the term with only 4 weeks to fully learn the option module before trials.
At the beginning of the year we were told that we would be doing from Quanta to quarks. So anticipating this 4 week crunch , I spent a lot of time last term pre learning and writting out Sylabus dot points for Quanta to quarks.
However we have a new teacher this term and they have decided that we will be changing to astrophysics. I really want to still do quantum quarks as I already know it extremely well and spent numerous hours on it last term and I was wondering if it is possible for me to do Quanta to quarks in the HSC while everyone else in the class does astrophysics.
I'm still happy to do the class assessments as they are on astrophysics.
Note: there is only two of us in the class
Post by: S200 on July 26, 2018, 12:59:28 am
-snip-
Do all students in a school have to do the same option?
We have started the term with only 4 weeks to fully learn the option module before trials.
At the beginning of the year we were told that we would be doing from Quanta to quarks. So anticipating this 4 week crunch , I spent a lot of time last term pre learning and writting out Sylabus dot points for Quanta to quarks.
However we have a new teacher this term and they have decided that we will be changing to astrophysics. I really want to still do quantum quarks as I already know it extremely well and spent numerous hours on it last term and I was wondering if it is possible for me to do Quanta to quarks in the HSC while everyone else in the class does astrophysics.
I'm still happy to do the class assessments as they are on astrophysics.
Note: there is only two of us in the class
If HSC is anything like VCE, and you are referring to individual module analysis, I would say no.
Bear in mind that I am not a HSC specialist.
Post by: jamonwindeyer on July 26, 2018, 01:00:38 am
How to draw a transformer diagram
Hey! The key elements would be the two coils wrapped around a core, all labelled (including designating which is the primary core). Optionally you could show things like where the magnetic flux and current are flowing! Pretty much any diagram you've seen in a textbook or similar will do the trick ;D
Do all students in a school have to do the same option?
We have started the term with only 4 weeks to fully learn the option module before trials.
At the beginning of the year we were told that we would be doing from Quanta to quarks. So anticipating this 4 week crunch , I spent a lot of time last term pre learning and writting out Sylabus dot points for Quanta to quarks.
However we have a new teacher this term and they have decided that we will be changing to astrophysics. I really want to still do quantum quarks as I already know it extremely well and spent numerous hours on it last term and I was wondering if it is possible for me to do Quanta to quarks in the HSC while everyone else in the class does astrophysics.
I'm still happy to do the class assessments as they are on astrophysics.
Note: there is only two of us in the class
Hey! Definitely can do whatever option you like in the external exam - If you prefer Q2Q, get through your internal assessments on Astro then swap over for the HSC exam! Make sure you can handle learning it properly without any direct help from your teacher ;D
Post by: Dragomistress on July 27, 2018, 02:59:29 pm
Post by: S200 on July 27, 2018, 04:59:00 pm
Why are superconductors used on a maglev train?Heat dissapation?
Imagine how hot the electromagnets would get!
I know that the Synchrotron uses water-cooled copper pipes in most of their magnets, but even they still need superconductors for some operations...
Post by: Jane20 on July 28, 2018, 12:41:02 am
Can anyone please help me with this question!
How do you determine the force between 2 parallel current carrying conductors Experimentally ?
Thank you :)
Post by: blasonduo on July 28, 2018, 10:40:39 am
Why are superconductors used on a maglev train?
Ordinary trains need to use wheels and thus are susceptible to static friction. The friction removes energy from the system which will eventually cause the train to stop. As maglev trains do not, they require less energy to run, making it both more energy efficient and faster. It also removes the constant screeching from the tracks and reduced noise pollution for nearby residents.
Hope this helps! :))
Post by: dermite on July 28, 2018, 11:25:07 am
Hi guys
Can anyone please help me with this question!
How do you determine the force between 2 parallel current carrying conductors Experimentally ?
Thank you :)
use
F = Kl I1I2/d2
Post by: jamonwindeyer on July 28, 2018, 12:07:10 pm
Hi guys
Can anyone please help me with this question!
How do you determine the force between 2 parallel current carrying conductors Experimentally ?
Thank you :)
Hey! To expand on dermite's answer, if you pass a known current through two wires of known length, a known distance apart, then the force is the only unknown in the formula! That can be used for verification. You could also place one wire on a scale, and measure how its weight changes when another wire carrying the same current is hung just above the first wire! This second method is a direct measurement and so probably more like what they are after, potentially? :)
Post by: Jane20 on July 28, 2018, 03:14:39 pm
Hey! To expand on dermite's answer, if you pass a known current through two wires of known length, a known distance apart, then the force is the only unknown in the formula! That can be used for verification. You could also place one wire on a scale, and measure how its weight changes when another wire carrying the same current is hung just above the first wire! This second method is a direct measurement and so probably more like what they are after, potentially? :)
I think the 2nd method would work since the question asks for measurement from experiment.
Thank you ! :)
Post by: Jasoon on August 02, 2018, 09:58:29 pm
Could you please help me with this dilemma. So I've researched some stuff on induction cooktops and in some notes they say the heat is generated more efficiently due to a higher resistance in the pot because of resistive heating between atoms and eddy currents? But in the 2012 HSC for multiple choice (question 19) they say the heat is better generated due to a lower resistance as P=I^2R and if R is small then I will be large as V=IR and thus power will be high leading to a greater heat?
Post by: envisagator on August 02, 2018, 10:39:30 pm
Hi All,Welcome to the forums Jasoon!!!!
Could you please help me with this dilemma. So I've researched some stuff on induction cooktops and in some notes they say the heat is generated more efficiently due to a higher resistance in the pot because of resistive heating between atoms and eddy currents? But in the 2012 HSC for multiple choice (question 19) they say the heat is better generated due to a lower resistance as P=I^2R and if R is small then I will be large as V=IR and thus power will be high leading to a greater heat?
So the first part of what you said is correct, that is, heat is generated more efficiently with increased resistive heating from eddy currents. The tricky part of this MC is to consider: V=IR, where 'I' is eddy current which you want to make very large so that heat generated by resistive heating is high. But if the resistance is high then the current will be low (inversely proportional), therefore, the resistance must be low for the current to be high to create a more efficient cooktop. Hence, answer is C
Hope this helps :)
Post by: jamonwindeyer on August 03, 2018, 07:04:08 am
Post by: Fergus6748 on August 03, 2018, 11:18:45 am
"Discuss the relationship between thought and reality"
Thanks a bunch
Post by: jamonwindeyer on August 03, 2018, 06:39:18 pm
Hey guys, I'm having a bit of trouble with this, how would you approach answering this question:
"Discuss the relationship between thought and reality"
Thanks a bunch
Hey! This is the Physics Q+A not Philosophy Q+A ;)
... Kidding, what a weird question! I think it is a play at the Einstein's thought experiment stuff. My response would be:
- Thought experiments can be used to formulate aspects of our reality long before experimental evidence exists.
- Rattle off a bunch of evidence based on Einstein's thought experiments, and the later experimental proof of those theories
- Conclude
;D
Post by: S200 on August 03, 2018, 08:06:40 pm
Hey! This is the Physics Q+A not Philosophy Q+A ;)Further on this, I would particularly use the trains TE that kinda well describes length contraction and everything else.
... Kidding, what a weird question! I think it is a play at the Einstein's thought experiment stuff. My response would be:
- Thought experiments can be used to formulate aspects of our reality long before experimental evidence exists.
- Rattle off a bunch of evidence based on Einstein's thought experiments, and the later experimental proof of those theories
- Conclude
;D
Post by: Jasoon on August 03, 2018, 08:39:22 pm
Thanks for your reply. I think the actual resistance that's used is a moderate resistance but I can nowsee why a low resistance is preferred over a high.
I also have another question :) . Does the meisnner effect involve any eddy currents (creating a magnetic field that opposes the external magnetic field) at all, and how the magnet levitates? Or is it just the total exclusion of a magnetic field and no eddy currents are induced from an emf, because to have eddy currents there must be a change in magnetic flux, and if the magnet is levitating there isn't any. I've read up on some theory and experiments like quantum pinning (which exceeds the HSC I believe) and how if a magnetic was placed on a superconductor that was cooled down to it's critical temp it would levitate, showing that although there is no change in magnetic flux the magnet still levitates.
Thanks
Post by: jamonwindeyer on August 03, 2018, 10:56:05 pm
Hey, envisagator and Jamon
Thanks for your reply. I think the actual resistance that's used is a moderate resistance but I can nowsee why a low resistance is preferred over a high.
I also have another question :) . Does the meisnner effect involve any eddy currents (creating a magnetic field that opposes the external magnetic field) at all, and how the magnet levitates? Or is it just the total exclusion of a magnetic field and no eddy currents are induced from an emf, because to have eddy currents there must be a change in magnetic flux, and if the magnet is levitating there isn't any. I've read up on some theory and experiments like quantum pinning (which exceeds the HSC I believe) and how if a magnetic was placed on a superconductor that was cooled down to it's critical temp it would levitate, showing that although there is no change in magnetic flux the magnet still levitates.
Thanks
Very welcome! :)
So as you say it is a totally separate effect, nothing to do with eddy currents. You don't need to know why the flux is excluded, just that it happens, and that we call it Meisner Effect :) and yep, all that quantum pinning stuff is beyond the course as well, the syllabus requirements here are actually pretty simplistic ;D
Post by: Jane20 on August 05, 2018, 10:43:59 pm
Given an electron with its quantum wavelength is 4.75x10^(-9). Calculate its velocity.
For this i use de Broglie's equation
But then part b) it says
Find the quantum frequency of the electron.
For this question, why do we have to use light velocity c=fx(wavelength) instead of the actual velocity of the electron (calculated from part a) ?
Thank you :)
Post by: Dragomistress on August 12, 2018, 08:58:19 pm
May someone help me with these three questions?
Thanks!
Post by: jamonwindeyer on August 12, 2018, 10:48:28 pm
Hey!
May someone help me with these three questions?
Thanks!
I'll take the first one! It is A - Max Planck absolutely believed that science should play into politics, and he certainly practiced in Germany. That leaves A as the only possible answer ;D
Post by: justwannawish on August 13, 2018, 09:15:39 am
Could someone explain this to me? The answer's A :)
Post by: Fergus6748 on August 13, 2018, 01:06:23 pm
Hi!Hi so basically because of Lenz's law, a south pole would form at the end of the solenoid to oppose the movement of the magnet, so that means on the other side of the solenoid would form a North pole. Thus that means the magnetic field at P will flow from left to right, and using right hand rule, pointing your thumb towards the north end you can find that the current is travelling anticlockwise so from X to Y. Hope this helps!!
Could someone explain this to me? The answer's A :)
Post by: elvis810 on August 13, 2018, 05:41:40 pm
Post by: justwannawish on August 13, 2018, 06:07:45 pm
Hi so basically because of Lenz's law, a south pole would form at the end of the solenoid to oppose the movement of the magnet, so that means on the other side of the solenoid would form a North pole. Thus that means the magnetic field at P will flow from left to right, and using right hand rule, pointing your thumb towards the north end you can find that the current is travelling anticlockwise so from X to Y. Hope this helps!!
Hey,
Thank you so much for answering! I get the field bit and that the current travels anticlockwise, but I don't get why it can't go from Y to X?
Post by: blasonduo on August 13, 2018, 08:22:51 pm
Hey!
May someone help me with these three questions?
Thanks!
Hey! I'll do the 2nd one!! :))
This gives me the answer of C
Additional note: Now, what is really weird (but freaking awesome) is that you notice that the distance is not dependant in the equation, yet it is work, and work always coincides with how far it was moved (work is force times distance for crying out loud!). Even if I had the plates 100m apart, and this particle moved 50m, the amount of work done would be the same, and this is all because of the Force the particle experiences. An example would be throwing a ball, you throw one on earth with a certain amount of energy, it will go up a certain height. Go to the moon and use the same energy, the ball will go higher because there is less gravitational force, even thought you put in the same amount of work. The fact that this just kinda cancels out is super gobsmacking in my opinion, I love physics :))
Post by: jamonwindeyer on August 13, 2018, 10:52:39 pm
Could someone help me with this question? The answer is D. Please explain why it moves up. Thanks!
Welcome to the forums Elvis! ;D
I'm thinking the answers mean the ring is what jumps, but the logic is the same regardless. When we flick the switch, we have a current flowing into the coil. This introduces a magnetic field - Now it is constant once the current is flowing, but while it ramps up, it is changing. That changing field induces eddy currents in the ring.
Now the induced current will act to oppose the change which created it (Lenz's Law). The ring wasn't in a magnetic field, now it is, so the eddy currents will act to create forces to get the ring away from the field. Rotating won't do this, not moving won't do this - Jumping up will do this! Hence the answer!
This is the intuitive explanation without an exact analysis of forces, but this does the trick for this question! ;D
Post by: owidjaja on August 13, 2018, 11:16:57 pm
So tomorrow's my physics exam and I'm still unsure on the quanta to quarks stuff. Is it better to cram for that or just do more past papers? I love exam clashes
Post by: jamonwindeyer on August 13, 2018, 11:21:48 pm
Hey guys,
So tomorrow's my physics exam and I'm still unsure on the quanta to quarks stuff. Is it better to cram for that or just do more past papers? I love exam clashes
Normally practice is the way to go, but you gotta make sure you know your content, at least a little. So I'd say cram quanta for an hour or so, just download a set of notes and read, try and remember the main points. Then get some sleep!! ;D
Post by: elvis810 on August 14, 2018, 08:54:39 am
Welcome to the forums Elvis! ;D
I'm thinking the answers mean the ring is what jumps, but the logic is the same regardless. When we flick the switch, we have a current flowing into the coil. This introduces a magnetic field - Now it is constant once the current is flowing, but while it ramps up, it is changing. That changing field induces eddy currents in the ring.
Now the induced current will act to oppose the change which created it (Lenz's Law). The ring wasn't in a magnetic field, now it is, so the eddy currents will act to create forces to get the ring away from the field. Rotating won't do this, not moving won't do this - Jumping up will do this! Hence the answer!
This is the intuitive explanation without an exact analysis of forces, but this does the trick for this question! ;D
Thank you!
Post by: dermite on August 16, 2018, 06:31:49 pm
I'd appreciate help on the subject.
thanks
Post by: elvis810 on August 19, 2018, 06:22:57 pm
An 80 kg daredevil stunt person jumps from a helium-filled balloon at an altitude of 20 km. When at an altitude of 10 km, the stunt person is falling at 250 ms-1 through the atmosphere. The amount of heat and sound energy produced by the friction while the stuntperson falls through this 10 km distance is closest to:
(A) 2500 kJ
(B) 4560 kJ
(C) 5350 kJ
(D) 7850 kJ
Thanks again for all your help!
Hi there, im struggling with understanding the operation of a galvanometer and loudspeaker (in simple terms) for the respective dot point of the motor effect.
I'd appreciate help on the subject.
thanks
A galvanometer consists of 2 curved magnets, a spiral spring and a pointer connected to a coil which is wrapped around an iron core. When the galvanometer is connected to an external circuit, current flows through the coil. The coil will experience a force as it is inside a magnetic field (motor effect) and starts to rotate. As the coil rotates, the spring is stretched and the opposing torque exerted by the spring increases. When the opposing torque of the spring is equivalent to the forward torque of the coil, the coil will stop rotating. A scale is calibrated such that the pointer reads the current through the coil.
A loudspeaker is a device that transforms electrical energy into sound energy. A coil of wire (known as the voice coil) sits in the space between the pole pieces of the magnet. Electrical signal inputs to the loudspeaker are in the form of alternating currents. When the current flows through the coil, the coil experiences a force as a consequence of the motor effect. This force will cause the coil to vibrate back or forth depending on the direction of the AC. The voice coil is connected to a paper speaker cone that creates sound waves in the air as it vibrates.
Mod edit: Merged double posts --Calebark
Post by: S200 on August 19, 2018, 06:56:24 pm
I'm not sure where to start.Hey there!
An 80 kg daredevil stunt person jumps from a helium-filled balloon at an altitude of 20 km. When at an altitude of 10 km, the stunt person is falling at 250 ms-1 through the atmosphere. The amount of heat and sound energy produced by the friction while the stuntperson falls through this 10 km distance is closest to:
(A) 2500 kJ
(B) 4560 kJ
(C) 5350 kJ
(D) 7850 kJ
Thanks again for all your help!
SO, for this, use Gravitational Potential energy vs Kinetic, and the energy loss must be heat/sound.
So;
Post by: Mate2425 on August 29, 2018, 09:28:05 pm
Thankyoou!
Post by: S200 on August 30, 2018, 05:23:09 am
Hi guys, could someone please help me in getting a basic understanding with examples of rotational and transitional velocity - Astrophysics.Although this is more down to earth than astrophysics, this YouTube video does explain the concept of concurrent translational and rotational velocity...
Thankyoou!
Hope it helps
Post by: dermite on August 31, 2018, 04:54:56 pm
While drilling into a tough material, the DC motor in an electric drill is slowed significantly. This causes its coils to overheat. Explain why this occurs with reference to physics principles (4mks)
thanks in adv!
Post by: S200 on August 31, 2018, 10:58:49 pm
Hi there, i need some help with this question:Hmmm.
While drilling into a tough material, the DC motor in an electric drill is slowed significantly. This causes its coils to overheat. Explain why this occurs with reference to physics principles (4mks)
thanks in adv!
Well the motor would be under stress, and going slower. If its slower, would that mean that the slip rings are in contact longer and the current causes the overheating?
Post by: jamonwindeyer on September 01, 2018, 12:07:44 am
Hi there, i need some help with this question:
While drilling into a tough material, the DC motor in an electric drill is slowed significantly. This causes its coils to overheat. Explain why this occurs with reference to physics principles (4mks)
thanks in adv!
Hey! To extend on above, it is back emf. In regular operation, back emf acts against the supply current to reduce the current flowing in the coils of the motor. We design motors expecting this reduced current flow. When we force the motor to slow down, we reduce the amount of back emf, since the coil experiences a lower rate of change of magnetic flux. This increases the current in the coils beyond what was designed to be tolerable for extended periods, causing the coils to overheat! ;D
Key principles to mention in your answer would be induction and back emf :)
Post by: clovvy on September 05, 2018, 03:43:05 pm
Post by: blasonduo on September 05, 2018, 06:59:43 pm
Hey, I have something that has bugged me for a while regarding the solutions for the physics topic test gravity test 2 q8 solution... I am not getting the results provided in the solution when I plug them in my calculator so I am suspiscious that there may be an error made.... Finding Ep was fine, but when finding the velocity where it became an issue....
Hey Hey! I just did the calculations, and they worked for me, may I see your working so I can see the problem? :))
An important note here though is how to define potential energy, especially as if you should make the surface of the earth as 0, or the centre of the earth as 0. This tripped me up a tonne, but the primary rule of thumb that occurred to me was that if they say radius, they refer to the centre of the earth. If they say altitude, they refer to the surface :)) Nonetheless, this question is looking out for the change in potential energy. Maybe this will help in your new calculations :))
Post by: clovvy on September 07, 2018, 04:38:47 pm
Hey Hey! I just did the calculations, and they worked for me, may I see your working so I can see the problem? :))Unfortunately I was unable to get around that so I gave up..
An important note here though is how to define potential energy, especially as if you should make the surface of the earth as 0, or the centre of the earth as 0. This tripped me up a tonne, but the primary rule of thumb that occurred to me was that if they say radius, they refer to the centre of the earth. If they say altitude, they refer to the surface :)) Nonetheless, this question is looking out for the change in potential energy. Maybe this will help in your new calculations :))
My next question is "assess the effectiveness of solid state devices compared to their termionic equivalence"... This question did not have an answer provided to me, and instinct alone tell me to list the advantages of solid states instead.. How should I answer this question (this is also from Jamon's physics topic test)..
Post by: radnan11 on September 09, 2018, 10:22:19 pm
Post by: S200 on September 09, 2018, 10:24:43 pm
And for Magnitude, shouldn't it just be that \(F=nBiL\)?
So \(20\times 0.005 \times 0.025 \times 0.15\)? So \(\therefore \quad F=0.0375\) upwards??
The angle should only affect the flux produced right?
Post by: Fergus6748 on September 10, 2018, 12:29:04 pm
Isn't it just the right hand slap rule, so force is straightThat's what I got as well. Except the distance is in metres for SI units. So:downup?
And for Magnitude, shouldn't it just be that \(F=nBiL\)?
So \(20\times 0.005 \times 0.025 \times 15\)? So \(\therefore \quad F=0.0375\) upwards??
The angle should only affect the flux produced right?
15cm = 0.15m
Post by: iimooncabbageii on September 12, 2018, 09:18:05 pm
Why is the answer A? Wouldn't the bright spot move sideways?
(https://snag.gy/JC1ifB.jpg)
Post by: jamonwindeyer on September 12, 2018, 10:44:48 pm
Hi, bit confused with this question.
Why is the answer A? Wouldn't the bright spot move sideways?
Hey there!! Not quite - We have to remember we are talking about a magnetic field not an electric field! For those we use the right hand slap rule, or a similar rule.
Magnetic field to the left of the page, charges moving out of the page. So, fingers to the left of the page, thumb pointed out of the page - This points our palm downwards. However, electrons are negative charges, so we flip our answer - The electrons will experience a force upwards towards A :)
Post by: clovvy on September 15, 2018, 11:32:00 pm
Post by: jamonwindeyer on September 15, 2018, 11:48:39 pm
Hey, now that I am paying close attention to Jake's lecture slide, he mentioned that the BCS theory is wrong so I decided to google it and I still don't quite understand... can anyone explain?
You don't need to! The BCS theory is what you learn in this course - With contemporary knowledge, we know it doesn't fully explain superconductivity, but that isn't something you need to worry about. For HSC Physics, treat it as gospel ;D
Post by: Jane20 on September 21, 2018, 07:55:42 am
How can I explain the Meissner effect? I read different sources and they all say different things, induced current, exclusion of magnetic field. I dont know what should I write in the exam?
Thank you !!
Post by: talitha_h on September 22, 2018, 01:33:18 pm
Hi guys
How can I explain the Meissner effect? I read different sources and they all say different things, induced current, exclusion of magnetic field. I dont know what should I write in the exam?
Thank you !!
Hope this helps, anyone can correct me if I'm wrong.
The Meissner effect is the phenomenon that a superconductor is able to totally exclude external magnetic fields below its critical temperature, therefore its internal magnetic field is always zero.
When an external magnetic field attempts to enter a superconductor, it induces a perfect eddy current to circulate in the superconductor, as a result of zero resistance. This ‘perfect’ current flows in such a direction that the magnetic field it produces is just as strong, but in the opposite direction to the external magnetic field (Lenz’s Law). This leads to a total cancellation of this external magnetic field and allows none of it to penetrate through the superconductor.
This idea can also be used to explain why a small magnet is able to hover over a piece of superconductor. The perfect flow of induced current in the superconductor will allow it to set up magnetic poles that are strong enough to repel the small magnet forcefully enough to overcome its weight force. Superconductors: - Have 0 electrical resistance. - Demonstrate the Meissner effect.
Post by: jamonwindeyer on September 22, 2018, 01:56:18 pm
Hope this helps, anyone can correct me if I'm wrong.
The Meissner effect is the phenomenon that a superconductor is able to totally exclude external magnetic fields below its critical temperature, therefore its internal magnetic field is always zero.
When an external magnetic field attempts to enter a superconductor, it induces a perfect eddy current to circulate in the superconductor, as a result of zero resistance. This ‘perfect’ current flows in such a direction that the magnetic field it produces is just as strong, but in the opposite direction to the external magnetic field (Lenz’s Law). This leads to a total cancellation of this external magnetic field and allows none of it to penetrate through the superconductor.
This idea can also be used to explain why a small magnet is able to hover over a piece of superconductor. The perfect flow of induced current in the superconductor will allow it to set up magnetic poles that are strong enough to repel the small magnet forcefully enough to overcome its weight force. Superconductors: - Have 0 electrical resistance. - Demonstrate the Meissner effect.
Almost perfect Talitha - I'll just fix one thing if that's okay.
The Meissner Effect cannot be attributed to eddy currents. We know this because we can hold a magnet above a metal, cool it below critical temperature, and then release the magnet, and it will hover. The magnet wasn't moving, so there was no change in flux, so there can't have been eddy currents. Yet, the magnetic flux is still excluded.
I actually think this paragraph from Wikipedia does a fantastic job explaining it:
Any perfect conductor will prevent any change to magnetic flux passing through its surface due to ordinary electromagnetic induction at zero resistance. The Meissner effect is distinct from this: when an ordinary conductor is cooled so that it makes the transition to a superconducting state in the presence of a constant applied magnetic field, the magnetic flux is expelled during the transition. This effect cannot be explained by infinite conductivity. The placement and subsequent levitation of a magnet above an already superconducting material does not demonstrate the Meissner effect, while an initially stationary magnet later being repelled by a superconductor as it is cooled through its critical temperature does.
As to actually explaining why the flux is excluded, that is beyond the HSC syllabus. You just need to know what the Meissner effect is ;D
Post by: Dragomistress on October 01, 2018, 11:37:38 am
Why is the answer D?
Post by: jamonwindeyer on October 01, 2018, 11:47:36 am
Hi,
Why is the answer D?
Hey! Because for a line to be in gradient-form, it needs to be in the form \(y=mx\), not \(y=m\sqrt{x}\). So squaring both sides:
In this form, \(y=T^2\), and \(x=l\), which leaves the gradient as the expression in D ;D
Note: This is a poorly worded question, so I don't blame you for getting a bit confused ;D
Post by: parallaxd on October 02, 2018, 04:29:57 pm
Post by: clovvy on October 02, 2018, 05:42:29 pm
For the striation patterns part of ideas to implementation syllabus I'm confused as to whether high pressure means a low amount of gas particles or a high amount???High pressure should definitely mean high amount of gas particles... Because within a closed system, the more gas particles present, the pressure will build up for sure... And generally this is always the case..
Unless my understanding is wrong, then please correct me before I lose marks in HSC...
Post by: jamonwindeyer on October 02, 2018, 08:12:34 pm
For the striation patterns part of ideas to implementation syllabus I'm confused as to whether high pressure means a low amount of gas particles or a high amount???
High pressure should definitely mean high amount of had particles... Because within a close system, the more gas particles present, the pressure will build up for sure... And generally this is always the case..
Unless my understanding is wrong, then please correct me before I lose marks in HSC...
You're all good clovvy, high pressure means more particles (meaning more collisions with electrons) ;D
Post by: Mate2425 on October 02, 2018, 08:15:32 pm
is there some sort of standard process or helpful trick i can employ to ensure i am on the right track or know that i have the right sort of answers for these sort of questions?
Thank you!!
Post by: clovvy on October 02, 2018, 09:37:36 pm
Hi guys having real trouble trying to understand the graphs for motors and generators in HSC multiple choice questions e.g HSC 2014 https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdfI don't see a graph question as MC but I do see graphs provided for answer, are you refering to Q12 or Q14?
is there some sort of standard process or helpful trick i can employ to ensure i am on the right track or know that i have the right sort of answers for these sort of questions?
Thank you!!
Post by: Mate2425 on October 02, 2018, 09:56:27 pm
I don't see a graph question as MC but I do see graphs provided for answer, are you refering to Q14?
Yes thank you that's the sort of questions i am struggling with! :)
Post by: clovvy on October 02, 2018, 10:06:46 pm
Yes thank you that's the sort of questions i am struggling with! :)I'll try to explain both as best as I can with my limited knowledge anyway..
So for Q12, the force acting on the wire PQ is given by F=BILsinθ (θ is the angle of the wire in relation to the field), and so in both cases the angle is 90°. As the coil rotates through 360° the current supplied is constant, the length PQ does not change, and the magnetic field is constant in the case of the parallel field, and may be considered uniform in the case of the radial field. So the answer should be C..
From the diagram provided in Q14, when the switch is open the resistance in the outside circuit is infinite, so the potential difference generated produces 0 current (V=IR, when R is infinity). However, when the switch is closed a cycle of AC potential difference causes a cycle of current to flow, hence the answer is C..
This is as best as I can explain it, if anyone sees any incorrect/incomplete info or want to add something to my explanation, that would be appreaciated...
Post by: jamonwindeyer on October 02, 2018, 11:19:10 pm
Yes thank you that's the sort of questions i am struggling with! :)
So for Q12, the force acting on the wire PQ is given by F=BILsinθ (θ is the angle of the wire in relation to the field), and so in both cases the angle is 90°. As the coil rotates through 360° the current supplied is constant, the length PQ does not change, and the magnetic field is constant in the case of the parallel field, and may be considered uniform in the case of the radial field. So the answer should be C..
Love your answer for Q14, but the one for Q12 is just a tiny bit off - Let me help :)
For Q12, we remember that \(\tau=nBIA\cos{\theta}\), the torque is dependent on the angle of the coil with the field. What the radial field does is eliminates this angle - As you hint at, it is considered to always be \(\theta=90^\circ\). So, the magnitude of the torque for the radial motor is constant. For the parallel field motor, it varies sinusoidally with the \(\cos{\theta}\) term. In both, we need the torque to swap magnitude every half turn.
The graph that matches is actually B - The radial field has constant magnitude of torque (just swapping direction), and the parallel field has sinusoidal variation as required ;D
As for your question on how to tackle graphs, Mate, unfortunately there aren't any shortcuts. You just have to know the content. Use process of elimination to knock out obviously incorrect answers to give yourself the best chances of a correct answer. But the fact you are struggling with these is normal - They are tough questions ;D
Post by: Bruh01 on October 03, 2018, 06:44:11 pm
Post by: clovvy on October 03, 2018, 10:23:11 pm
Love your answer for Q14, but the one for Q12 is just a tiny bit off - Let me help :)
For Q12, we remember that \(\tau=nBIA\cos{\theta}\), the torque is dependent on the angle of the coil with the field. What the radial field does is eliminates this angle - As you hint at, it is considered to always be \(\theta=90^\circ\). So, the magnitude of the torque for the radial motor is constant. For the parallel field motor, it varies sinusoidally with the \(\cos{\theta}\) term. In both, we need the torque to swap magnitude every half turn.
The graph that matches is actually B - The radial field has constant magnitude of torque (just swapping direction), and the parallel field has sinusoidal variation as required ;D
As for your question on how to tackle graphs, Mate, unfortunately there aren't any shortcuts. You just have to know the content. Use process of elimination to knock out obviously incorrect answers to give yourself the best chances of a correct answer. But the fact you are struggling with these is normal - They are tough questions ;D
Oddly enough BOSTES answers confirms it to be C... Unless it is one of those years where they actually screw up the answer.. After I look at the graphs more closely I am sort of leaning towards your answer but idk why BOSTES pick C as the correct answer
Post by: jamonwindeyer on October 04, 2018, 08:46:28 am
Oddly enough BOSTES answers confirms it to be C... Unless it is one of those years where they actually screw up the answer.. After I look at the graphs more closely I am sort of leaning towards your answer but idk why BOSTES pick C as the correct answer
Ahh no you were right! I did TORQUE, the question wanted FORCE, I didn’t read the question properly
sorry!!
Post by: clovvy on October 04, 2018, 09:20:07 am
Ahh no you were right! I did TORQUE, the question wanted FORCE, I didn’t read the question properlyIn return I deserve a +1 upvote as compensation for making me twist my head over this question overnight :P :P :P
Post by: Dragomistress on October 04, 2018, 09:50:08 pm
Why is 20 A and 14 D?
Post by: Mate2425 on October 05, 2018, 12:03:14 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/d9a632e8-e0a0-439a-9869-3ce2795be292/physics-hsc-exam-2014.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-d9a632e8-e0a0-439a-9869-3ce2795be292-lG96GS9
Thankkkkyyyyooouu!!
Post by: clovvy on October 05, 2018, 12:39:40 pm
Hey guys would appreciate it if someone could give me a solid explanation on HSC 2014 Q17. MCHi,
http://educationstandards.nsw.edu.au/wps/wcm/connect/d9a632e8-e0a0-439a-9869-3ce2795be292/physics-hsc-exam-2014.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-d9a632e8-e0a0-439a-9869-3ce2795be292-lG96GS9
Thankkkkyyyyooouu!!
So for this question, because the alpha particle is twice that of a proton, the force applied to it within the electric field is twice as great. However, since the mass of he alpha particle is 4x that of proton, and taking into account F=ma, its acceleration is half that of the proton... Hence the correct answer should be B
Post by: Mate2425 on October 05, 2018, 01:39:32 pm
Hi,
So for this question, because the alpha particle is twice that of a proton, the force applied to it within the electric field is twice as great. However, since the mass of he alpha particle is 4x that of proton, and taking into account F=ma, its acceleration is half that of the proton... Hence the correct answer should be B
Thanks Clovvy 8)
Post by: Mate2425 on October 06, 2018, 11:59:40 am
Thanks, :D :D
Post by: clovvy on October 06, 2018, 12:30:05 pm
Hey, how do i do the process for HSC 2015 Q24C. using the elctrical energy and kinetic energy method.Hi, this is my solution
Thanks, :D :D
a=F/m=4×10^14/9.1×10^-11=4.4×10^16m/s/s..
Then you use v^2=uy^2+2ay
So v=(2×4.4×10^16×0.02)^0.5=4.19×10^7m/s..
Sorry for some reason LaTeX is not working properly so I use a rough way of writing instead but hopefully that helps..
Post by: Mate2425 on October 06, 2018, 12:39:39 pm
And also by any chance would you happen to know how to do the electrical energy and kinetic energy method.
Post by: Dragomistress on October 06, 2018, 05:49:31 pm
I always take quite some time in converting integers to using powers. Do you get penalised for not using powers of 10 for something like 0.00000000251? Or do you have to write 2.51x10^-9.
Can I also get some help on the attached questions?
Post by: clovvy on October 06, 2018, 07:38:18 pm
Hi, could someone explain the effect of the magnetic field on the proton (as in whether it goes clockwise or anti) and how the resultant vector points to the perpendicular direction. Thankshey, sorry for the delay I missed this question..
So the magnetic field will cause the proton in a circular path which moves out of the page initially, continues in an anti-clockwise direction as viewed from the right. At the same time, the electric field will cause the proton to move to the left with increasing speed. So the resultant motion will be basically be the sum of the vectors based on the direction of the motion of proton, which will look like its stretched towards the left. Also the radius shown will be decreasing as the proton loses energy as it radiates electromagnetic radiation as it is an accelerating charge.
Post by: clovvy on October 06, 2018, 07:51:51 pm
Thanks Clovvy, why do we use the projectile motion formula?hey sorry I didn't respond quick enough... It is asking for velocity and this is probably the best way to do it imo.., because the initial velocity (hence the KE) of the electron on the cathode is considered to be effectively zero. The acceleration of an electron within a uniform field can easily be found with F=ma since you know the force and the mass (force calculated from part ii assuming you've done it).. with regard to using projectile, it works because the only force acting on the electron is the electric field as it moves (similar to the behaviour of projectiles)- although I am hoping that someone can explain that better than myself...
And also by any chance would you happen to know how to do the electrical energy and kinetic energy method.
and I don't know how to use electrical/kinetic energy method sorry.., if Jamon or someone expert could lend a hand that would be a great help for me too
Post by: clovvy on October 06, 2018, 08:53:20 pm
Heyyyy!hey man,
I always take quite some time in converting integers to using powers. Do you get penalised for not using powers of 10 for something like 0.00000000251? Or do you have to write 2.51x10^-9.
Can I also get some help on the attached questions?
So for the Michelson and Morley question, I remember that one as I did do the ruse trials and tbh it is confusing, like I am 100% certain the the aether is not a solid so I do pick D when I first did it online..., still doesn't make sense to me imo so I need someone to be able to explain that to me.. I also suspect that the answer provided is a typo..
So for the motors question...
This is more like an elimination of wrong answers... eliminate A as it is obviously wrong so eliminate that... and given where A and B is, both the force and the torque will be at maximum. therefore both F and the torque will reverse direction after rotating 90 degrees... so we would expect sudden change in force so D is eliminated... obviously the torque will not be at stationary hence the most correct should be B (hope my explanation make sense)..
For the last one, just plug in sin30 because it enters the magnetic field at an angle of 30, and using the rhsr, the force should go into page...
I hope I have helped you a bit..
Post by: clovvy on October 06, 2018, 09:27:13 pm
Hey,For Q14, as side Y enters the magnetic field a voltage will be induced. As X enters it will also produce an equal voltage, as it is moving through the field in the same direction. This leads to NO difference in voltage between X and Y while they are both within the field. As they leave the field on the other side, the voltage is again induced but in opposite direction to the original voltage when it enters the field. So out of the 4 option, only A shows this..
Why is 20 A and 14 D?
For Q20, if the gravitational force weakens, the period it took to complete one whole orbit will reduce, which eliminates option A and B straight away.. To pick between 0.28 or 0.5, I am not sure how to do it either so I will need help with that if anyone is kind enough to help...
Post by: Mate2425 on October 06, 2018, 11:21:26 pm
For HSC 2011 Q21a, i am strugling to understand why they didn't connect line of best fit from point of (T=12) and (T=36) and also how do they know precisely where to position the line between the array of points provided. I followed my method and determined a resistance of 0.1265 ohms which was out by 0.0002 ohms would they penalise this answer and the graph or one or the other?
Many Thanks! 😁
Post by: clovvy on October 07, 2018, 08:34:12 am
Hey Jamon,Hey man,
For HSC 2011 Q21a, i am strugling to understand why they didn't connect line of best fit from point of (T=12) and (T=36) and also how do they know precisely where to position the line between the array of points provided. I followed my method and determined a resistance of 0.1265 ohms which was out by 0.0002 ohms would they penalise this answer and the graph or one or the other?
Many Thanks! 😁
For these type of question, you are looking at trends shown with the given values and you try to make an estimate... In this case, the relationship looks linear and you want to 'average out' your results which is why the line is drawn 'near' it but did not cross it as it is a 'line of best fit'...
I hope that make sense.... regarding the value of resistance at 24 degrees, your answer is close enough so I don't think they will penalise you too much I am sure range between 0.1265-127 is acceptable...
hey man,Getting back to the Michelson and Morley question, the correct answer is indeed D so I was correct... The solution provided for Q7 was a typo so the answer is wrong because the aether is not a solid (I was positive I was correct, and I double checked with a ruse physics teacher which I knew in person)...
So for the Michelson and Morley question, I remember that one as I did do the ruse trials and tbh it is confusing, like I am 100% certain the the aether is not a solid so I do pick D when I first did it online..., still doesn't make sense to me imo so I need someone to be able to explain that to me.. I also suspect that the answer provided is a typo..
So for the motors question...
This is more like an elimination of wrong answers... eliminate A as it is obviously wrong so eliminate that... and given where A and B is, both the force and the torque will be at maximum. therefore both F and the torque will reverse direction after rotating 90 degrees... so we would expect sudden change in force so D is eliminated... obviously the torque will not be at stationary hence the most correct should be B (hope my explanation make sense)..
For the last one, just plug in sin30 because it enters the magnetic field at an angle of 30, and using the rhsr, the force should go into page...
I hope I have helped you a bit..
Post by: kaustubh.patel on October 07, 2018, 01:30:12 pm
Post by: clovvy on October 07, 2018, 03:11:29 pm
hey guys need some help with 2012 physics MC q19 and 20, i think 19 is D but the answer is C and 20 i just dont know but its answer is B.for question 19, electric power raises the temperature of the pot. The pot base has a low resistance. P=I^2R. The more rapid the change in flux, the greater the induced emf. This creates a larger current and more rapid heating saucepan. Therefore it is C
for 20, Max Enerhy output=E=hf=hc/λ =(6.626x10^-34x3x10^8)/(9x10^6)=0.138eV. Therefore the band gap should be less than 0.138eV. This makes HgCdTe the best choice so A.
Post by: Mate2425 on October 07, 2018, 07:00:05 pm
Also Q20 HSC 2014 could someone please provide me a thorough working out, step by step to this question.
Thank you!!
Post by: clovvy on October 07, 2018, 08:07:11 pm
Hey guys so with HSC 2012 Q18. how do you get the figures cause i measured the distance from centre of Earth to point X to be 2.5 and Y to be 3.5 cm respectively?Hi again!!
Also Q20 HSC 2014 could someone please provide me a thorough working out, step by step to this question.
Thank you!!
So for Q18 this is how I do it...
So the answer is C
For Q20 of HSC 2014, this is my solution
Post by: Jane20 on October 08, 2018, 01:44:17 pm
Thank you
Post by: clovvy on October 08, 2018, 06:24:35 pm
Hi can anyone please show me how to work out the graph question 26b HSC 2014 paper :'(Before drawing the graph you need to find the frequency first:
Thank you
Mod Edit: Tidied the LaTex, thanks clovvy!
Post by: Pokechimp7 on October 08, 2018, 07:14:49 pm
I need some help with this projectile motion question!
An object is launched at a velocity of 20 m/s in a direction of 25 degrees upward from the horizontal.
a) What's the maximum height reached by the object?
b) What's the total flight time?
c) What's the horizontal range?
d) What's the magnitude of the velocity of the object just before it hits the ground?
Any help is greatly appreciated!
Post by: S200 on October 08, 2018, 07:27:09 pm
Hi everyone,a.) \(v^2=u^2+2as\), where \(v=0\) and \(u=20\sin{25}\)
I need some help with this projectile motion question!
An object is launched at a velocity of 20 m/s in a direction of 25 degrees upward from the horizontal.
a) What's the maximum height reached by the object?
b) What's the total flight time?
c) What's the horizontal range?
d) What's the magnitude of the velocity of the object just before it hits the ground?
Any help is greatly appreciated!
b.) \(2\times [v=u+at]\), same conditions as above.
c.) Answer from b, \(\times 20\cos{25}\)
d.) Same as launch velocity. \(20ms^{-1}\) at 25 degrees to the negative horizontal.
Image
(https://i.ytimg.com/vi/ET3UMc7F64w/maxresdefault.jpg)
Post by: Jane20 on October 08, 2018, 09:32:59 pm
Before drawing the graph you need to find the frequency first:
Mod Edit: Tidied the LaTex, thanks clovvy!
Thanks Clovvy ! :)
Post by: not a mystery mark on October 16, 2018, 09:04:12 pm
A 600 kg car is merging onto a major highway via a curve banked at 7 degrees from the horizontal.
The radius of the curve is 240 m.
- What is the maximum safe speed for this car around the curve? -
The radius of the curve is 240 m.
- What is the maximum safe speed for this car around the curve? -
Post by: blasonduo on October 16, 2018, 09:37:46 pm
Howdyy, I have another question. The teacher today showed me her answer but I couldn't really understand the reasoning behind it. You guys are smart so thank you!A 600 kg car is merging onto a major highway via a curve banked at 7 degrees from the horizontal.
The radius of the curve is 240 m.
- What is the maximum safe speed for this car around the curve? -
Hey! A question from the new syllabus? How exciting!!
An object on a banked track does not use friction, but the velocity of the object to maintain its circular motion. If the velocity is too slow of too fast, the car will either "slip" and fall into the centre of the banked track or go over the top of the banked track. So I'm going to assume that the "safe speed" is the speed for the car to not move up or down the slope.
We know that with any object moving in a circle uniformly, the object travels in a circle at a constant speed due to a force accelerating it inwards, which is called centripetal force. We also know that the force of gravity acts on this car; I've made this diagram to visually help with this.
(https://i.imgur.com/Ssntlvj.png)
As you can see, both the gravitational force and the centripetal force are perpendicular, and the Normal force to any object is always perpendicular to the surface, we so know that the angle is 7 in this.
By then equating Tan (opposite over adjacent), we derive a formula (which does not have mass in it amazingly!)
Subbing that in, I get 16.9938 m/s
= 17m/s.
Let me know if i'm right, because "safe" is a broad term. Hope this helps! :))
Post by: not a mystery mark on October 16, 2018, 10:06:19 pm
An object on a banked track does not use friction, but the velocity of the object to maintain its circular motion. If the velocity is too slow of too fast, the car will either "slip" and fall into the centre of the banked track or go over the top of the banked track. So I'm going to assume that the "safe speed" is the speed for the car to not move up or down the slope.
As you can see, both the gravitational force and the centripetal force are perpendicular, and the Normal force to any object is always perpendicular to the surface, we so know that the angle is 7 in this.
By then equating Tan (opposite over adjacent), we derive a formula (which does not have mass in it amazingly!)
Subbing that in, I get 16.9938 m/s
= 17m/s.
Let me know if i'm right, because "safe" is a broad term. Hope this helps! :))
OH MY GOD YOU'RE MAGIC! Thank youu!! That was the answer she got, but this time I've walked away with understanding! Thank you heaps friend. I shall be back with more new syllabus questions. ahaha
Post by: owidjaja on October 19, 2018, 09:42:19 pm
I need help with the question attached. I've asked my teacher but I still didn't understand why the answer is C.
Post by: jasn9776 on October 19, 2018, 09:53:30 pm
Hey guys,ehh its hard to explain but basically, the magnetic field is going into the page from the diagram of the two people.
I need help with the question attached. I've asked my teacher but I still didn't understand why the answer is C.
Using lenzs law you want the current to go from P to Q. So at A, it will generate a current to oppose the change in flux, i.e. a current from Q to P. At C, the current will generate a force upwards. with right hand palm rule the current will go from P to Q.
Post by: radnan11 on October 21, 2018, 10:24:09 am
Post by: blasonduo on October 21, 2018, 11:02:36 am
Quick question: are we allowed to shorten low earth orbits as LEO in our answers. Thanks
If you need to refer to it many times, it'll be best to first write out Low Earth Orbit (LEO) and then continue using LEO afterwards. Just to make sure the marker understands what you mean by it :) As long as you do that, you'll be sweet!
Post by: Oscar322 on October 26, 2018, 12:24:01 am
Post by: Blissisignorance on October 26, 2018, 10:23:35 pm
At Q28) the graph in the sample answers has velocity remaining constant for most of AB before falling just before B all the way to just after C. But, I can't understand why velocity isn't dropping during AB too since AB=BC. My graph had velocity falling from A to B, then falling once again to C before being constant.
Post by: dermite on October 27, 2018, 09:35:53 am
Post by: jamonwindeyer on October 27, 2018, 03:51:53 pm
In the 2015 HSC,
At Q28) the graph in the sample answers has velocity remaining constant for most of AB before falling just before B all the way to just after C. But, I can't understand why velocity isn't dropping during AB too since AB=BC. My graph had velocity falling from A to B, then falling once again to C before being constant.
I agree with your answer Bliss!! You are right - The magnet would be opposing the motion towards it just as much as it would oppose the motion away from it (in terms of the eddy currents generated). I think the sample answer is wrong :)
hi, i need some help as to how to do this qn. The answer given is 3.13ms-1
Hey! The centripetal force on the rotating mass carrier is provided by the weight force of the mass carrier below. So:
You substitute and rearrange to find \(v\) ;D
Post by: moq418 on October 27, 2018, 04:19:18 pm
Describe quantitatively the force acting on a charge moving through a magnetic field:
F = qvB
Post by: Fergus6748 on October 27, 2018, 05:52:20 pm
How do you :Heya, are there any values with the question?
Describe quantitatively the force acting on a charge moving through a magnetic field:
F = qvB
Post by: dazza2020 on October 28, 2018, 02:43:08 pm
Just wondering what markers think of integrating option topic knowledge into core answers?
E.g. Describe the impact Planck's theories regarding Black Body Radiation had on Scientific Thinking? Beyond the obvious, I read somewhere that De-Broglie explained his wavelength's of electron orbits after learning that waves could be quantised and reading Planck's 1905 papers.
Post by: Fergus6748 on October 28, 2018, 04:56:26 pm
Hi All,Hey, I would assume that they would have no problem with it as they can't take marks or give you marks for info that doesn't relate to the question. If it the info is relevant than by all means go ahead and put it in. The only thing with that is that, even if it is relevant, if the info doesn't really answer or at least forward your answer than all you are doing is spending precious time writing info that may not help you get marks. Hope this helps!!
Just wondering what markers think of integrating option topic knowledge into core answers?
E.g. Describe the impact Planck's theories regarding Black Body Radiation had on Scientific Thinking? Beyond the obvious, I read somewhere that De-Broglie explained his wavelength's of electron orbits after learning that waves could be quantised and reading Planck's 1905 papers.
Post by: Mate2425 on October 30, 2018, 08:55:54 pm
Thank you!
Post by: sweetiepi on October 30, 2018, 08:57:55 pm
Hey guys, so walking into exam and during the exam should my calc be in RAD or in just degrees?For the sciences, you should always have your calculator in degrees. :)
Thank you!
Post by: phunky on October 31, 2018, 02:03:49 am
Post by: blasonduo on October 31, 2018, 08:56:54 am
Hi, I'm a bit confused about this question - how do you work out the poles of the magnets?
The first step is to figure out the direction of the magnetic field (either left to right of right to left)
From the positive terminal you can see that the magnets are going to be electromagnets, you can see that it wraps around the left block in clockwise way, You know it has a SOUTH facing pole (I use the term Nanti-Socks to remember if clockwise is South or North :P )
To double check, we'll do the other side, from the negative terminal, this time, it is travelling in an ANTI-clockwise way (remember to look at it standing in the middle) This means that the RIGHT block is the NORTH pole. So the magnetic field lines follow from right to left.
The next part is the simple right hand palm rule, as you know the + and - terminals, follow the current lines until the right option works. In this case, B is the only one that rotates clockwise. (Left side coil current goes up (so thumb up!) and magnetic field right to left (so fingers to right) shows the palm upwards and thus it rotating clockwise.
Basically, this question is asking about your ability to figure out electromagnets, As B and D are almost the same, where D's right block is coiled the wrong way!
Hope this helps!
Post by: Mate2425 on October 31, 2018, 02:07:21 pm
Thank you.
Post by: Fergus6748 on October 31, 2018, 02:16:36 pm
Hey could someone please explain why the answer is A in HSC 2016 Q16a?Heya, so basically the loudspeaker is being used in reverse, as the cone of the speaker is being pushed and that is moving the coil. So therefore it is acting as a generator, because it is transfering mechanical energy to electrical energy. The movement of the coil is creating current in the circuit, as according to Lenz's Law. Then you use right hand grip rule to find that current travels from X to Y. Hope this helps!!
Thank you.
Post by: Fergus6748 on October 31, 2018, 02:18:20 pm
The first step is to figure out the direction of the magnetic field (either left to right of right to left)Nice Mnemonic!!
From the positive terminal you can see that the magnets are going to be electromagnets, you can see that it wraps around the left block in clockwise way, You know it has a SOUTH facing pole (I use the term Nanti-Socks to remember if clockwise is South or North :P )
To double check, we'll do the other side, from the negative terminal, this time, it is travelling in an ANTI-clockwise way (remember to look at it standing in the middle) This means that the RIGHT block is the NORTH pole. So the magnetic field lines follow from right to left.
The next part is the simple right hand palm rule, as you know the + and - terminals, follow the current lines until the right option works. In this case, B is the only one that rotates clockwise. (Left side coil current goes up (so thumb up!) and magnetic field right to left (so fingers to right) shows the palm upwards and thus it rotating clockwise.
Basically, this question is asking about your ability to figure out electromagnets, As B and D are almost the same, where D's right block is coiled the wrong way!
Hope this helps!
Post by: Blissisignorance on October 31, 2018, 05:47:04 pm
Heya, so basically the loudspeaker is being used in reverse, as the cone of the speaker is being pushed and that is moving the coil. So therefore it is acting as a generator, because it is transfering mechanical energy to electrical energy. The movement of the coil is creating current in the circuit, as according to Lenz's Law. Then you use right hand grip rule to find that current travels from X to Y. Hope this helps!!But since the current is induced to oppose motion of speaker, shouldn't it be from Y to X then?
Post by: Ralopsi on October 31, 2018, 06:41:26 pm
Thanks for your help
Post by: Fergus6748 on October 31, 2018, 07:15:03 pm
What is the necessary explanation for why doping changes the electrical properties of semiconductors? Do i need to talk about donor/acceptor levels or is that unecessary?Heya, so it would depend on the question. If it's a 4-6 marker, you would need to describe what doping is and what p-type and n-type semiconductors are and how that it affects the electrical properties of semiconductors. You should also talk a bit about Undoped semiconductors as a comparison. You will need to talk about valence electrons and how that creates 'holes' or adds electrons which boosts the movement of electrical charge. Hope this helps!!
Thanks for your help
Post by: fkkiwi on October 31, 2018, 08:05:09 pm
Post by: Divayth Fyr on October 31, 2018, 10:09:13 pm
Is there a good explanation for this? The HSC solutions don't really help
We have to look at the effects of the two fields on the proton, and then look at how they will combine.
Electric Field
Remember that the electric field lines show the direction that a positive charge will move in them. As the particle is a proton, it will move in the direction of the field (to the left).
Magnetic Field
This part is a little bit more difficult. We need to use the right hand palm rule to see how it will interact with the field (right because it is positive). Using this, we can see that it will move out of the page, meaning that it is accelerating out of the page. This means that it will have a velocity out of the page. BUT, it already has a velocity going up the page. This means that the total velocity will be a combination of these two vectors, meaning it will be sort of diagonal. However, this is a charged particle moving in a magnetic field. So, this diagonal velocity will have a force applied on it causing it to gain an even greater diagonal velocity. No matter where it is, the force is pointed towards the center; it's a centripetal force. This means that the particle will undergo uniform circular motion (a fancy way of saying it will go in a circle).
It's kind of difficult to explain in a paragraph, so you should look at this picture: https://files.mtstatic.com/site_4539/8298/0?Expires=1540985526&Signature=Lwcy7Hau7aSkOnD1W1Pqon~3uyalcn4MNngMGnmRi~SNUc~HvHA41Z9mHW-JfC48UanNzky5D7rFMEHIkQ4pwxAANbACbIKCtMrHWivfUt8nXDH7s9-UlIYEgdfEkvlqk-rpzC78EUSF7qvrul-fTlrpmekHTEXnq22~pfHNopg_&Key-Pair-Id=APKAJ5Y6AV4GI7A555NA
Keep in mind that the picture shows an electron in the magnetic field, so you need to use the left hand palm rule in order to find the force on it.
Both Of Them Together
So, we have a particle that is moving to the left and going around in circles (starting with going out of the page). The velocity going to the left has no effect on the circular motion, as that is parallel to the magnetic field:
Once again, it is difficult to explain in a paragraph, so you should look at this animation of it: https://www.youtube.com/watch?v=a2_wUDBl-g8
I hope this helps.
Post by: phunky on October 31, 2018, 11:15:23 pm
I have no idea how to do this question??
Post by: Divayth Fyr on November 01, 2018, 08:16:55 am
haha thanks for the nansocks analogy!!
I have no idea how to do this question??
The solar cell uses the photoelectric effect, so you're looking to find the energy value for which electrons will be able to jump the gap and complete the circuit. This means you're using:
Remember that as intensity increases, more electrons will be able to bridge the gap. So, you want to pick an intensity around a wavelength of 10 micrometers. We need the frequency of that, so we use:
Put frequency into the equation, and then divide by \( 1.602\times10^{-19} \) and you end up with a value of \( 0.12 \) . However, we chose a value that was close to the maximum intensity. So this means that our answer is slightly off. Thus, we chose the closest one so the answer is B.
Post by: not a mystery mark on November 09, 2018, 06:57:51 pm
"A projectile is fired from the top of a 120m high cliff at 25m/s. It lands on the ground 6.4s after firing."
a) find the Intial Vertical Velocity (uy)
So my working is as follows:a) find the Intial Vertical Velocity (uy)
...Which is not correct. Infact - that's the final vertical velocity... what the heck.
Even weirder. When I do this the answer is "correct"?!?!?
My mind is McBlown. It seems that 'v' and 'u' have switched? Pls explain what is up. Thank you, I love you.
Post by: jamonwindeyer on November 09, 2018, 07:05:27 pm
(the reason this swap happens is interesting, and I can discuss it if you like!
Post by: not a mystery mark on November 09, 2018, 07:15:27 pm
Hello my friend ;D you are firing from a 120m high cliff, and landing on the ground. Your vertical displacement is downwards - It is negative. I think if you make \(s=-120\) in both those equations, you'll be MindUnBlown ;)
(the reason this swap happens is interesting, and I can discuss it if you like!
I hate that it was that simple ahahah thank you heaps. Also a discussion would be hella neato and interesting :D
Post by: jamonwindeyer on November 09, 2018, 11:37:54 pm
I hate that it was that simple ahahah thank you heaps. Also a discussion would be hella neato and interesting :D
Sweet! Okay, so it's pretty much about symmetry. Picture the path as defined in the question - It spends a certain amount of time (call it \(T_1\)) being accelerated downwards on its way up, which sets the velocity to 0 at the peak of motion. Then it spends another amount of time (call it \(T_2\)) being accelerated downwards to reach a final velocity.
In your mistake, you swapped the scenario - You started at the bottom, and you wanted to land on the cliff. To do that, you would be tracing your path in reverse. But to make sure you get to the same peak, you'd need to start with the vertical velocity you finished with in the above scenario, since you know that is how much will be taken away on the ascent. You spend \(T_2\) time getting accelerated the same as before, but now it acts against you - So to end up at \(v_y=0\), you need to start with that higher value of velocity. Then it is similar on the way down - Before, you were stopped after \(T_1\) seconds, so now doing it in reverse, you end up with the same velocity you started with in the first scenario.
Okay, that was actually tough to explain without a diagram - Maybe it made some sense? Maybe not... Haha ;D if anyone wants to jump in and clarify feel free :)
Post by: jamonwindeyer on November 11, 2018, 03:37:14 pm
Keen to see this thread be just as much of a collaborative resource for the new course as it was for the old :)
Post by: david.wang28 on November 20, 2018, 06:21:59 pm
I'm having trouble with a kinematics question which I have posted in the link. Can I get a detailed reply please? Thanks :)
Post by: fun_jirachi on November 20, 2018, 08:16:39 pm
b) Since we're given that the driver comes to a stop a few millimetres away from the hazard, the assumption is that this is negligible, and so we subtract our answer from part a) from 50 since that's the distance he spends braking after his reaction. So we get 43.5m
c)
Sub in your known values for s, v and u (which should be 43.5, 0 and 75/3.6 respectively) to get your t, which should be 4.2s.
d) You can use either s=ut+(at^2)/2 or v=u+at, since acceleration is constant, but use the second one, its so much easier. You should get -5.0m/s/s.
Hope this helps :)
Post by: david.wang28 on November 20, 2018, 08:54:29 pm
5. a) Remembering that s=ut, and converting 75 km/h to SI units, you get that he travelledThanks for the answer! Much appreciated :)
b) Since we're given that the driver comes to a stop a few millimetres away from the hazard, the assumption is that this is negligible, and so we subtract our answer from part a) from 50 since that's the distance he spends braking after his reaction. So we get 43.5m
c)
Sub in your known values for s, v and u (which should be 43.5, 0 and 75/3.6 respectively) to get your t, which should be 4.2s.
d) You can use either s=ut+(at^2)/2 or v=u+at, since acceleration is constant, but use the second one, its so much easier. You should get -5.0m/s/s.
Hope this helps :)
Post by: david.wang28 on November 23, 2018, 09:46:18 pm
I'm having trouble with a kinematics question which I have posted in the link. Can I get a detailed reply please? Thanks :)
Post by: fun_jirachi on November 23, 2018, 10:47:56 pm
You have the starting height 25m, acceleration 9.8m/s/s and u=0.
Ignore what I did before please, that was so stupid :)
So uh time taken for the whole thing is from s=ut+1/2(at^2). Subbing in the above values tells you it takes the pelican 2.26s (roughly) to complete the dive. For it to get to the 5m mark, sub in s= 20 instead.The difference in time will tell you how long it takes the pelican to dive the last 5m.
Spoiler
I got 0.23s, might be wrong but should be right. Therefore, since its bigger than 0.1s, the pelican goes hungry
Hope this helps :)
Post by: S200 on November 23, 2018, 11:16:00 pm
So, the pelican has "fallen" 20 metres, under the force of gravity. The fish starts to move .10 of a second after seeing it, so the pelican needs to cover the next 5 metres in under 0.10 seconds.
A velocity vs time graph could show it, or maybe a displacement vs time graph of the last 5 metres? (with a negative gradient).
calculations
\(s_1=u_1t_1+\frac{at_1^2}{2}\), so \(-20=0+(\frac{-9.8}{2} )\times (t_1)^2\)
This means that \(t_1\), the time taken before the fish spots it, is equal to \(\sqrt{\frac {20}{4.9}} \quad \therefore \quad t=2.02\)
\(v_1=u_1+at_1\), so \(v_1= 0+9.8\times t_1\), meaning that \(v_1=19.7989899\) or \(19.80\)
\(v_1\) now becomes your \(u_2\)...
\((v_2)^2=(u_2)^2+2as_2\) so \(v=\sqrt {19.80^2 +19.6\times 5}\)
So \(v= ~7\sqrt {10}\)
Now, back into \(v=u+at\), we get that \(t= \frac {7\sqrt {10} - 19.80} {9.8} \quad \therefore \quad t=0.23s\), which is greater than \(0.1s\), so the Pelican goes hungry...
Jirachi got in before me, but I thought I had better keep mine up there...
Nvm. 0.23 same same
Post by: david.wang28 on November 24, 2018, 12:41:28 pm
Post by: david.wang28 on November 26, 2018, 07:53:49 pm
I'm having trouble with 2 kinematics questions which I have posted in the link. Can I get a detailed reply please? Thanks
Post by: fun_jirachi on November 26, 2018, 08:27:36 pm
(https://i.imgur.com/FL9tt0G.jpg)
This is the diagram you should have gotten for the first question.
By the cosine rule, find the distance of the third side of the triangle. Then use that new information and the sine rule to find the angle. Answer in the spoiler.
Spoiler
Answer: 7.4km, N (It's actually about 0.4 of a degree to the west of north, but we ignore that assuming its to the nearest degree. If you want the exact stuff, its 24 minutes 3.11 seconds)
I'll do the other one when I have more time :)
Hope this helps :)
Post by: S200 on November 26, 2018, 10:05:55 pm
He only runs one leg... i.e; 25km.
Post by: david.wang28 on November 26, 2018, 10:13:05 pm
Isn't 6.a a trick question?That's why I'm asking :-[
He only runs one leg... i.e; 25km.
Post by: Bri MT on November 26, 2018, 10:25:53 pm
That's why I'm asking :-[
The wording would mean just the 25, even though it seems more likely that they would want the distance travelled since you need that for part c.
I really wouldn't stress over it - as long as you understand, that's the main thing
Post by: david.wang28 on November 28, 2018, 10:54:45 am
Sorry for being a pest, but I'm having trouble with the kinematics questions which I have posted in the link. Can I get a detailed reply please? Thanks :)
Post by: vox nihili on November 28, 2018, 11:24:01 am
Hi guys,
Sorry for being a pest, but I'm having trouble with the kinematics questions which I have posted in the link. Can I get a detailed reply please? Thanks :)
You'd really get a lot more out of this if you showed us what you've done so far on these questions. You learn really little from having people do the questions for you :)
Post by: david.wang28 on November 28, 2018, 08:20:45 pm
Post by: david.wang28 on November 28, 2018, 08:21:39 pm
Post by: david.wang28 on November 30, 2018, 05:55:12 pm
I'm having trouble with section 10 Q5 and all of section 11 which I have posted in the link. Can I get a detailed reply please? Thanks
Post by: blasonduo on November 30, 2018, 11:23:44 pm
Hi guys,
Sorry for being a pest, but I'm having trouble with the kinematics questions which I have posted in the link. Can I get a detailed reply please? Thanks :)
Hey! Sorry for the late reply!
5) I get the same displacement as you to 2 s.f, which is fantastic! However, our angle in which this displacement occurs from is different (Mine is around N0.4W). By trialling your angle, it seems to me that you claim that the boat has moved a net distance of 3.13km in the x-axis. Which from intuition, seems way to large (since the boat basically only went upwards as it travelled East then West.) Try this part again, and let me know if you have any trouble with it. (For all we know, it could've been a mistake on my end ;) )
6) Follows the same method. It seems to me that your complicate your working more than I, which is fine. I tend to draw individual right-angled triangles for each "event". You get the same answer in displacement as I do for the x-direction. There is a little in the y-direction (0.0386km North) but since you round to 2 s.f, it makes you assume that it is directly west, when in fact I get N78W; which is pretty west, but does show the 12 degree error, which I'm not a fan of, but your answer is still good! c) looks great!
5(ball) seems correct, follows exactly what I would do, might've been an error I don't see, but it's all good!
Your relative velocity seems perfect! Great job!
For section 10, we need to point to completely negate the velocity of the stream and have an x-axis of exactly 10m/s (the distance of the river over time) use these velocities to find the net velocity and thus the angle in which it needs to travel at, make your own coordinate system.
For section 11, this is somewhat tricky!
a) m*g*sin(theta) = m*a
b) same as a) include the force of friction, use projectile motion formula.
c) m*g*sin(theta) needs to equal frictional force (ie net force is 0)
d) same as above
e/f) bit tricker as you need to find the net force of the two systems as a whole system. Remember that one of the forces will change with the angle while the other will not, once you find the net force, use methods used for Diagram A.
Come back with some working out with these, and I'll happily help out where needed! This should give you a slight boost to the questions :))
If you notice a problem with any of my comments, feel free to ask!
Post by: david.wang28 on December 02, 2018, 04:54:39 pm
Hey! Sorry for the late reply!Here is my working out for section 10 and section 11, I have no idea what to do for Q5 and Q6 for section 11 (maybe it's because I haven't learnt the dynamics module yet). Can you at least check if my working out is correct? Thanks :)
5) I get the same displacement as you to 2 s.f, which is fantastic! However, our angle in which this displacement occurs from is different (Mine is around N0.4W). By trialling your angle, it seems to me that you claim that the boat has moved a net distance of 3.13km in the x-axis. Which from intuition, seems way to large (since the boat basically only went upwards as it travelled East then West.) Try this part again, and let me know if you have any trouble with it. (For all we know, it could've been a mistake on my end ;) )
6) Follows the same method. It seems to me that your complicate your working more than I, which is fine. I tend to draw individual right-angled triangles for each "event". You get the same answer in displacement as I do for the x-direction. There is a little in the y-direction (0.0386km North) but since you round to 2 s.f, it makes you assume that it is directly west, when in fact I get N78W; which is pretty west, but does show the 12 degree error, which I'm not a fan of, but your answer is still good! c) looks great!
5(ball) seems correct, follows exactly what I would do, might've been an error I don't see, but it's all good!
Your relative velocity seems perfect! Great job!
For section 10, we need to point to completely negate the velocity of the stream and have an x-axis of exactly 10m/s (the distance of the river over time) use these velocities to find the net velocity and thus the angle in which it needs to travel at, make your own coordinate system.
For section 11, this is somewhat tricky!
a) m*g*sin(theta) = m*a
b) same as a) include the force of friction, use projectile motion formula.
c) m*g*sin(theta) needs to equal frictional force (ie net force is 0)
d) same as above
e/f) bit tricker as you need to find the net force of the two systems as a whole system. Remember that one of the forces will change with the angle while the other will not, once you find the net force, use methods used for Diagram A.
Come back with some working out with these, and I'll happily help out where needed! This should give you a slight boost to the questions :))
If you notice a problem with any of my comments, feel free to ask!
Post by: worldno1 on December 25, 2018, 04:36:54 pm
can someone help me for this circular motion question? i'm quite lost:
1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.
Post by: redpanda83 on December 25, 2018, 06:18:17 pm
hi guys!first get the centripetal acceleration. 12.5 g = 122.5 m/s2. With a radius of 8.84m, Use a = v2 /r, to calculate the 32.91 m/s as the velocity.
can someone help me for this circular motion question? i'm quite lost:
1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.
Second step
use angular velocity (defined as rate of change of angle), which can be expressed as w = v / r
r1=8.84, r2 depends on if the person is inside or outside of the machine, so r2=6.84 if inside or 10.84 if outside
So v1 / r1 = v2 / r2
Should work
actually r2=6.84, as the max acceleration was 12.5g. So the person must be inside the machine obviosly. if you solve with r2=10.84, a2=15.32g, which is greater than 12.5g
hi guys!
can someone help me for this circular motion question? i'm quite lost:
1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.
Derivation of a_1_/r_1_ = a_2_/r_2_
Since tangential velocity v is given by
v = ω r
we can show that the equation for acceleration you have already mentioned becomes
a = ω^2 r
solving for the value we know is constant, ω , we get
a/r = ω^2
As ω is constant, then ω^2 is a constant as well.
Now that we have our equation solved for a constant, we can connect the two radii like so
a_1_/r_1_ = a_2_/r_2_
Mod Edit: Merged posts. Please refrain from double posting.
Post by: 006896 on January 05, 2019, 07:26:32 pm
I am wondering whether for HSC sciences, is the Prelim content able to be tested by NESA in the HSC exam? I know for Maths, apparently 30% of Prelim content will be tested in the HSC exam. Is this true for Physics (and Chemistry)? As in, should I be studying Prelim topics during these holidays?
Thanks
Post by: RuiAce on January 05, 2019, 08:08:48 pm
Hi,To my knowledge that is only for mathematics and not the sciences. It was how it was with the old science syllabuses but I don't see anything that hints at "prelim content is suddenly examinable for the new science syllabuses" either
I am wondering whether for HSC sciences, is the Prelim content able to be tested by NESA in the HSC exam? I know for Maths, apparently 30% of Prelim content will be tested in the HSC exam. Is this true for Physics (and Chemistry)? As in, should I be studying Prelim topics during these holidays?
Thanks
Post by: r1ckworthy on January 09, 2019, 07:31:21 pm
Currently, I have been summarising the information under each dot point, but I'm not too sure whether it is as effective as answering questions. What do you guys think?
PS. the dot point workbook is by Brian Shadwick
Post by: fun_jirachi on January 09, 2019, 09:06:41 pm
My school prescribes the Dot Point book, so I guess you could say I have experience using it. I think it definitely helps; you're not going to be rattling off your knowledge to the examiner for the majority of the exam. Even if the questions are redundant slightly in that they are pretty easy (in the fact that the use of concepts you learn is pretty straightforward, but dont really prep you for the much harder questions), it's still practice to make sure you learn everything and get question answering practice - definitely better than just summarisation. There a few stupid questions here and there though that my class tends to skip, but for the most part it thoroughly covers all the dot points (hence the name ;) ). Also, be mindful of the answers, they're not always right! If you've thoroughly done a concept, and you do a question, and the answer is different, do it again, and if the answer is different still, you're probably right.
That's about Dot Point and how it could help, but in terms of having it 'completely as my study resource' is not a great idea; doing timed tests (resources may be few right now, but its worth doing whatever you can find), and of course you do need to rehash content every now and then. It's pretty good, but only when used in conjunction with a few other things.
Hope this helps :)
Post by: louisaaa01 on January 10, 2019, 04:39:20 pm
What parts of my practical investigations do I need to include in my study notes?
Post by: jamonwindeyer on January 10, 2019, 07:27:24 pm
In the HSC/Trials, if a question were to ask you to outline a first-hand investigation you conducted on a certain topic/area, what information would you need to include in your response in order to get full marks?
What parts of my practical investigations do I need to include in my study notes?
Hey louisaaa! To outline means to "Sketch in general terms; indicate the main features of", that's according to NESA. So I'd be probably including:
- The aim (what was being measured/investigated?)
- Rough description of method
- Key result/trend identified
Would obviously depend a bit on the context of the question - For my pracs, I would just write a super brief dot point experimental report for each thing I did! So that's aim, method, results, discussion in super brief dot points for each practical ;D
Post by: david.wang28 on January 11, 2019, 12:20:55 pm
I'm having trouble with question 2.2 and 2.3 that are in the attachments. Some I have done working out, some not. Can anyone help me with those questions please? Thanks :)
Post by: jamonwindeyer on January 11, 2019, 07:37:17 pm
Hello,
I'm having trouble with question 2.2 and 2.3 that are in the attachments. Some I have done working out, some not. Can anyone help me with those questions please? Thanks :)
Hey David! You are on the right track for 2.2 - You just need to do a bit more work to get your voltage (\(V\)). That 120V isn't directly applied to the bar, the bar is in parallel with that second resistor. You need to do a bit of circuit analysis - Draw a version of the circuit with the bar replaced with a 10 ohm resistor, and try and figure out the voltage across it (Hint - The two parallel resistors can be reduced to one resistor using a formula, which makes the analysis easier!). After that, you do use the current you get from Ohm's Law like you did:
Only other thing is I think the 2.6N is supposed to be 2.6kg :)
2.3 is actually really tricky! Your Part (a) looks about right - To do Part (b) onwards, I'm fairly sure you need to make an assumption that the radius of curvature of that curved motion is large enough that you can assume the magnetic force is always directed to the left. Then, it becomes a projectile motion question - With the horizontal acceleration taking the place of gravity, meaning the horizontal direction actually becomes the one with acceleration!! Turn the image 90 degrees to the left, and you'll see what I mean ;D
Once it leaves the magnetic field, it's moving at a constant speed (another assumption, but an appropriate one), and it becomes \(v=\frac{d}{t}\) and a bit of vector arithmetic! I'll write out a proper solution if you need it, but it might need to wait until after lecture weekend! Or better yet, come chat to me about it in the Physics lecture tomorrow ;D tough question!!
Post by: david.wang28 on January 11, 2019, 10:35:14 pm
Hey David! You are on the right track for 2.2 - You just need to do a bit more work to get your voltage (\(V\)). That 120V isn't directly applied to the bar, the bar is in parallel with that second resistor. You need to do a bit of circuit analysis - Draw a version of the circuit with the bar replaced with a 10 ohm resistor, and try and figure out the voltage across it (Hint - The two parallel resistors can be reduced to one resistor using a formula, which makes the analysis easier!). After that, you do use the current you get from Ohm's Law like you did:Shame I can't come to the lectures tomorrow- I'm very busy with my homework. I would have definitely registered to listen to your top-quality lecture. Anyways, thanks for the help Jamon! :)
Only other thing is I think the 2.6N is supposed to be 2.6kg :)
2.3 is actually really tricky! Your Part (a) looks about right - To do Part (b) onwards, I'm fairly sure you need to make an assumption that the radius of curvature of that curved motion is large enough that you can assume the magnetic force is always directed to the left. Then, it becomes a projectile motion question - With the horizontal acceleration taking the place of gravity, meaning the horizontal direction actually becomes the one with acceleration!! Turn the image 90 degrees to the left, and you'll see what I mean ;D
Once it leaves the magnetic field, it's moving at a constant speed (another assumption, but an appropriate one), and it becomes \(v=\frac{d}{t}\) and a bit of vector arithmetic! I'll write out a proper solution if you need it, but it might need to wait until after lecture weekend! Or better yet, come chat to me about it in the Physics lecture tomorrow ;D tough question!!
Post by: jamonwindeyer on January 12, 2019, 02:23:09 pm
Post by: kahliacopland on January 14, 2019, 02:18:41 pm
"A proton is undergoing circular motion in a magnetic field with an angular speed of 1500 rad/s. If the net force acting on the proton is 3.8x10^-22N, what is the radius of its path?"
Could someone help me with this?? thanks :)
Post by: blasonduo on January 14, 2019, 02:55:24 pm
Hi, I'm having trouble with this circular motion question-
"A proton is undergoing circular motion in a magnetic field with an angular speed of 1500 rad/s. If the net force acting on the proton is 3.8x10^-22N, what is the radius of its path?"
Could someone help me with this?? thanks :)
Hello!
Firstly, we have the angular velocity, where the formula is given by :
\[ v = r \omega\]
This formula can be derived from the formula sheet, and is pretty easy to remember. (If you'd like, I can derive it for you :) )
We also know Centripetal force:
\[ F = \frac{mv^2}{r}\]
We have an expression for velocity so let's sub in and cancel...
\[ F = m\omega^2 r \]
Rearranging...
\[ r = \frac{F}{m\omega^2} \]
We have all those values! When I sub in I get 10cm as the radius.
Hope this helps!
Post by: Crystine_16 on January 16, 2019, 12:19:04 am
Post by: Yertle the Turtle on January 16, 2019, 01:10:26 am
Can someone please explain how transformers work like why does the voltage increase as the number of turns in the coil increase and vice versa. I get the mathematical ratio formula but I'm not sure on the explanation. Thanks in advance :)Hi, I didn't do HSC, but we cover this in VCE. Basically you have to think about the coils as being part of a solonoid. When you think about a solonoid, you use Faraday's Law: emf=-N((B*A)/t) As the N term is relating to the number of turns in the wire, the greater the number of turns the greater the induced emf. Conversely, by putting a certain voltage through the primary coil of a transformer, you generate a certain magnetic field (the B*A part of the equation) through the metal core. If the number of turns on the secondary coil is lower than that on the primary coil, then the voltage output will be smaller than the voltage input.
It is all to do with thinking in terms of Faraday's Law: The N and the emf terms are the ones that change, and when the number of turns decreases, so does the emf induced, while as it increases so does the number of turns. The change in either one can change the other one, and thus you can have both step-down and step-up transformers.
Hope this helps :D
Post by: not a mystery mark on January 17, 2019, 08:57:23 pm
Low-Density Stars have narrower spectral lines as there as fewer collisions between particles in the star’s photosphere.
But I'm still confused about how that works? How do the collisions actually affect the width of the spectral lines.
Does anybody have any ideas?
Post by: jamonwindeyer on January 17, 2019, 11:52:23 pm
Heyy! Currently studying the Spectra of Stars for Module 7 and I've found that:
Low-Density Stars have narrower spectral lines as there as fewer collisions between particles in the star’s photosphere.
But I'm still confused about how that works? How do the collisions actually affect the width of the spectral lines.
Does anybody have any ideas?
Hey hey! I don't necessarily think you need to know this for the HSC course, I think it's a bit more depth than necessary, but I do remember reading up on this back when I wrote the HSC Physics Notes for the new course. My understanding is:
- Spectral line width is a consequence of (among other things) the Heisenberg Uncertainty Principle, we can't know the exact energy (and the emitted photons we are detecting vary a bit in energy anyway). There is a quantum limit to the precision with which we can know the energy and time differences involved in these transitions.
- In high pressure environments, there are far more interactions between atoms, meaning these transitions happen more quickly. If there are less collisions, they happen more slowly. As time in the state goes up, energy variance goes down. This is the uncertainty principle. Formulaically, this is the cause:
- Since energy is related to frequency by \(E=hf\), energy variance going down also means less frequency spread, and therefore narrower lines ;D
(hopefully this helps! If I have said something you don't quite get let me know, as I said you shouldn't need this much depth in the HSC) :)
Post by: not a mystery mark on January 20, 2019, 12:01:02 am
- Spectral line width is a consequence of (among other things) the Heisenberg Uncertainty Principle, we can't know the exact energy (and the emitted photons we are detecting vary a bit in energy anyway). There is a quantum limit to the precision with which we can know the energy and time differences involved in these transitions.
- In high pressure environments, there are far more interactions between atoms, meaning these transitions happen more quickly. If there are less collisions, they happen more slowly. As time in the state goes up, energy variance goes down. This is the uncertainty principle. Formulaically, this is the cause:
- Since energy is related to frequency by \(E=hf\), energy variance going down also means less frequency spread, and therefore narrower lines ;D
Ahh thanks heaps! I think I understand now. Haha, just trying to learn all the nitty-gritty of the syllabus for greater understanding. You're amazing - again, thanks heaps!
Post by: jamonwindeyer on January 20, 2019, 12:07:13 am
Ahh thanks heaps! I think I understand now. Haha, just trying to learn all the nitty-gritty of the syllabus for greater understanding. You're amazing - again, thanks heaps!
Very welcome! Good on you for pushing beyond (you'll be able to teach me a thing or two before this year is done I reckon ;)) :)
Post by: r1ckworthy on January 23, 2019, 04:40:38 pm
Can't seem to figure out the following problem:
What is the orbital period of an Earth satellite having an orbital radius half that of the moon?
It's from HSC 2012, Q13 multiple choice.
I've done the question so that the period of the satellite would be equal to 1/8 of the period of the moon (using Kepler's laws). The problem is that I need to find the period of the moon, which I can't seem to do. Either I am on the correct way or am totally wrong. I am sure of the latter. Anyways, pls help!! :'( :'(
Post by: blasonduo on January 23, 2019, 05:05:45 pm
Hey all,
Can't seem to figure out the following problem:
What is the orbital period of an Earth satellite having an orbital radius half that of the moon?
It's from HSC 2012, Q13 multiple choice.
I've done the question so that the period of the satellite would be equal to 1/8 of the period of the moon (using Kepler's laws). The problem is that I need to find the period of the moon, which I can't seem to do. Either I am on the correct way or am totally wrong. I am sure of the latter. Anyways, pls help!! :'( :'(
Hey! There are multiple ways to do this, the method I went with is as follows:
We know the formula for orbital velocity:
\[ V = \sqrt{\frac{GM}{r}}\]
We want our orbital radius to be halfway, so r is halved.
\[ V = \sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{1.915\times10^{8}}}\]
\[V = 1445.62 m/s \]
We know period is given by:
\[ T = {\frac{2\pi r}{v}}\]
\[ T = {\frac{2\pi\times1.915\times10^{8}}{1445.62}}\]
\[T = 8.3\times10^{5} s\] which is B :)
You could also use Kepler's equation as well in this case, either works!
Hope this helps!
Post by: fun_jirachi on January 23, 2019, 05:20:33 pm
There is a data table above which tells you all the information you need to know! You can do it with two approaches: doing it as you've done (though note that radius cubed is proportional to time squared, so the period of the satellite would be 1/sqrt8, instead of 1/8) or by finding the radius of satellite then subbing into Kepler's Third Law, since all the data is on the reference sheet. Either way, you should use one piece of data in the table (in the former case, orbital period of the Moon around Earth or in the latter case, the radius of the moon.) Either way you should get B (that's what I got)
You are totally correct, just remember that if it's not on the reference sheet, and it's not in the syllabus, they have to give it to you! Exams are cruel but they won't just ask you say 'what is the force due to gravity on Uranus' without giving you the corresponding information; perhaps not even a single person in NSW sitting the HSC would know that off the top of their head. It's totally unfair. Just remember to note the information given!
Hope this helps :)
EDIT: blasonduo already posted, but gonna post anyway :)
Post by: r1ckworthy on January 27, 2019, 07:22:39 pm
I just wanted to ask what type of answer one would give to this question: An object was placed in the centre of the Earth. What would it's mass and weight be? Justify your answer.
For this, I explained why weight is zero mathematically ( I showed r=0 leading to g=0 and so W=0), and wrote that mass cannot change.
The textbook however explains that the gravitational pull on the mass due to the Earth would be equal (radially outwards) in all directions, so the net force on it would be zero.
The question I wanted to ask is: which approach is better, and which one will gain more marks?? In other words, is a quantitative or qualitative approach to questions better? Obviously, both are right, but I feel that the textbook's answer is better. I would like to know so that I can employ the specific type of reasoning in future questions.
BTW. The question was from the Dot point textbook (5.5.3.6)
Post by: Yertle the Turtle on January 27, 2019, 11:13:19 pm
Hello everyone,Technically both answers are correct, but I personally would instinctively use your answer, as often questions (at least in VCE Physics) tend to regard objects as point masses. Therefore I would tend to use your answer, though the other answer is possibly better.
I just wanted to ask what type of answer one would give to this question: An object was placed in the centre of the Earth. What would it's mass and weight be? Justify your answer.
For this, I explained why weight is zero mathematically ( I showed r=0 leading to g=0 and so W=0), and wrote that mass cannot change.
The textbook however explains that the gravitational pull on the mass due to the Earth would be equal (radially outwards) in all directions, so the net force on it would be zero.
The question I wanted to ask is: which approach is better, and which one will gain more marks?? In other words, is a quantitative or qualitative approach to questions better? Obviously, both are right, but I feel that the textbook's answer is better. I would like to know so that I can employ the specific type of reasoning in future questions.
BTW. The question was from the Dot point textbook (5.5.3.6)
Post by: jamonwindeyer on January 28, 2019, 12:30:01 pm
Hello everyone,
I just wanted to ask what type of answer one would give to this question: An object was placed in the centre of the Earth. What would it's mass and weight be? Justify your answer.
For this, I explained why weight is zero mathematically ( I showed r=0 leading to g=0 and so W=0), and wrote that mass cannot change.
The textbook however explains that the gravitational pull on the mass due to the Earth would be equal (radially outwards) in all directions, so the net force on it would be zero.
The question I wanted to ask is: which approach is better, and which one will gain more marks?? In other words, is a quantitative or qualitative approach to questions better? Obviously, both are right, but I feel that the textbook's answer is better. I would like to know so that I can employ the specific type of reasoning in future questions.
BTW. The question was from the Dot point textbook (5.5.3.6)
Hey! To give a HSC perspective, if this was, say, a 2 mark question - Both answers would score 2/2. One uses formula to prove the lack of a net force, one uses logic to prove the lack of a net force. In both cases, the approach is valid and gives the correct outcome! ;D
For your understanding though, I'd encourage you to always consider the qualitative reasons behind your formulaic analysis. It's important to understand why the numbers are doing what they are doing. In giving answers, I actually prefer your answer as well, but for your own study it's good to go deeper than formula ;D
Post by: david.wang28 on January 29, 2019, 05:56:45 pm
I'm having trouble with question 5 b) in the link below. The answer is 22 m/s, but I got 82 m/s, as shown in my working out. Can anyone please help me with this question please? Thanks :)
Post by: jamonwindeyer on January 29, 2019, 07:08:46 pm
Hello,
I'm having trouble with question 5 b) in the link below. The answer is 22 m/s, but I got 82 m/s, as shown in my working out. Can anyone please help me with this question please? Thanks :)
Howdy! So first, just to set it up how you did in the image, the impulse of \(I=3\text{kgms}^{-1}\) you found is also equal to the change in momentum, so:
Obviously mass is constant, so it's just the velocity change:
However, this is applied in the opposite direction to the initial velocity! The final direction of travel is opposite to the initial. So the final velocity is actually given by:
But because it is speed, we ignore direction, so the negative doesn't matter ;D that's all you missed, you had a bit of a sign error with how the final speed was related to the initial speed ;D
Post by: david.wang28 on January 29, 2019, 08:07:00 pm
Howdy! So first, just to set it up how you did in the image, the impulse of \(I=3\text{kgms}^{-1}\) you found is also equal to the change in momentum, so:Yeah, the process was there, but the understanding wasn't. Thanks Jamon! :)
Obviously mass is constant, so it's just the velocity change:
However, this is applied in the opposite direction to the initial velocity! The final direction of travel is opposite to the initial. So the final velocity is actually given by:
But because it is speed, we ignore direction, so the negative doesn't matter ;D that's all you missed, you had a bit of a sign error with how the final speed was related to the initial speed ;D
Post by: david.wang28 on January 31, 2019, 06:37:03 pm
I have trouble with question 3 and 4 in the link below. I have posted my working out, but somehow the answer for a) was 2.1 m/s and b) was 78%. Can anyone please help me with this question? Thanks :)
Post by: fun_jirachi on January 31, 2019, 06:58:40 pm
First part is perfect, got that right! The major flaw in the next part is that the question asks for what fraction of Laurence's initial kinetic energy is converted into other forms of energy. What you've done there is include Marcus's mass in the final calculation of kinetic energy, causing your value to be heavily skewed. Make sure you just use 39 x 1/2 x 2.1^2, and not 84 x 1/2 x 2.1^2 for the final.
Hope this helps :D
Post by: david.wang28 on January 31, 2019, 08:07:09 pm
Hey there!Yes this really helps. Thanks for the help! :)
First part is perfect, got that right! The major flaw in the next part is that the question asks for what fraction of Laurence's initial kinetic energy is converted into other forms of energy. What you've done there is include Marcus's mass in the final calculation of kinetic energy, causing your value to be heavily skewed. Make sure you just use 39 x 1/2 x 2.1^2, and not 84 x 1/2 x 2.1^2 for the final.
Hope this helps :D
Post by: r1ckworthy on February 02, 2019, 05:37:37 pm
How would you respond to the following question: Explain, in terms of the principles of physics involved, why gravitational potential energy is a negative quantity?
I kind of know the answer already, but I just don't know how to format and 'write' the answer itself. My answer right now is all over the place and isn't quite formal or logical.
Post by: jamonwindeyer on February 02, 2019, 11:46:43 pm
Hey all,
How would you respond to the following question: Explain, in terms of the principles of physics involved, why gravitational potential energy is a negative quantity?
I kind of know the answer already, but I just don't know how to format and 'write' the answer itself. My answer right now is all over the place and isn't quite formal or logical.
I reckon let's start with your answer! Can you share it? Chances are even if it isn't quite logical yet we can get it there with just a bit of moving things around ;D
Post by: r1ckworthy on February 03, 2019, 09:23:04 am
I reckon let's start with your answer! Can you share it? Chances are even if it isn't quite logical yet we can get it there with just a bit of moving things around ;D
Alright, here is my answer:
At a distance of infinity away from the gravitational field of a central mass, the gravitational potential energy of an object will be zero. However, when work is done to raise the object a distance against the gravitational field, gravitational potential energy must increase. Therefore, gravitational potential energy must be negative so it can match both descriptions.
I'm fine with most of my answer, except the last part.
Post by: jamonwindeyer on February 03, 2019, 12:28:29 pm
Alright, here is my answer:
At a distance of infinity away from the gravitational field of a central mass, the gravitational potential energy of an object will be zero. However, when work is done to raise the object a distance against the gravitational field, gravitational potential energy must increase. Therefore, gravitational potential energy must be negative so it can match both descriptions.
I'm fine with most of my answer, except the last part.
Cool! Yeah, I like your answer. It pretty much lines up with the one in my head! I like your last statement, you could even say it as, "GPE must therefore be negative for both of these principles to be true." or something like that. Logically though, your answer works fine, nice work ;D
Post by: Jefferson on February 04, 2019, 08:06:36 pm
Primary Coil:
- 2000W
- 1000V
- 2A
Secondary Coil
- 2000W
- 200V
- 10A
Question:
If this transformer is 80% efficient, calculate the output current.
Is it as simple as 0.8 × 10A?
If so, then why does efficiency effect current and not voltage? Shouldn't it affect power overall, (P=IV).
i.e. 0.8P = 0.8(IV).
Post by: jamonwindeyer on February 04, 2019, 11:57:27 pm
If so, then why does efficiency effect current and not voltage? Shouldn't it affect power overall, (P=IV).
i.e. 0.8P = 0.8(IV).
Hey! Really good question - The efficiency of transformers is actually a fairly complicated topic, the HSC glosses over it just a little bit. Essentially, the exact effect of efficiency on output voltage and current depends a bit on where the power loss occurs. Is it due to resistance in the windings? Eddy currents in the core? There are different models for transformer efficiency and different ways of accounting for this. The HSC tends to not worry about this and give you extra values to help you hone in on the answer.
Problem here is that you aren't given that additional direction - Like your induction question, I don't think this is worded in the best way. So you've got to make some assumptions to answer this. I think your assumption is fair - It says, "Okay, my output voltage is locked, so all of my power loss occurs due to some of the current being used elsewhere in the transformer." Incidentally, this would occur due to the current needed to magnetise the core, and you can also account for eddy current loss in this way. You've assumed there is no resistance in either primary or secondary coil, and that there is no leakage flux.
Are you expected to know all of that in your assumption? Of course not. You just assumed the output current was what was affected by efficiency, which given the lack of information suggesting otherwise, is a fair assumption to make :)
Post by: Jefferson on February 05, 2019, 07:22:46 am
Hey! Really good question - The efficiency of transformers is actually a fairly complicated topic, the HSC glosses over it just a little bit. Essentially, the exact effect of efficiency on output voltage and current depends a bit on where the power loss occurs. Is it due to resistance in the windings? Eddy currents in the core? There are different models for transformer efficiency and different ways of accounting for this. The HSC tends to not worry about this and give you extra values to help you hone in on the answer.
Problem here is that you aren't given that additional direction - Like your induction question, I don't think this is worded in the best way. So you've got to make some assumptions to answer this. I think your assumption is fair - It says, "Okay, my output voltage is locked, so all of my power loss occurs due to some of the current being used elsewhere in the transformer." Incidentally, this would occur due to the current needed to magnetise the core, and you can also account for eddy current loss in this way. You've assumed there is no resistance in either primary or secondary coil, and that there is no leakage flux.
Are you expected to know all of that in your assumption? Of course not. You just assumed the output current was what was affected by efficiency, which given the lack of information suggesting otherwise, is a fair assumption to make :)
Wow!
Thank you so much for a detailed explanation yet again.
That really helped. <3.
Post by: david.wang28 on February 05, 2019, 05:33:48 pm
I have trouble with the question 18 in the link below. Can anyone please help me with this question (I have no idea how to do it)? Thanks :)
Post by: fun_jirachi on February 05, 2019, 06:23:22 pm
In an elastic collision, both KE and momentum are preserved. In an inelastic collision, the masses 'smoosh together' and move off as one mass.
Because m1 = 3m2, you can just sub in that whenever you would otherwise sub in m1.
Because of the nature of an inelastic collision, this is a much easier calculation.
Hope this made sense! :)
Post by: david.wang28 on February 05, 2019, 08:10:01 pm
Hey there!Yes this made sense. Thanks a lot! :)
In an elastic collision, both KE and momentum are preserved. In an inelastic collision, the masses 'smoosh together' and move off as one mass.
Because m1 = 3m2, you can just sub in that whenever you would otherwise sub in m1.
Because of the nature of an inelastic collision, this is a much easier calculation.
Hope this made sense! :)
Post by: david.wang28 on February 08, 2019, 08:01:34 pm
I'm having trouble with Q 2.5 in the link below (I have no idea how to do it). Can anyone please help me with these questions? Thanks! :)
Post by: Bri MT on February 08, 2019, 08:37:22 pm
https://puzzel.org/en/jigsaw/play?p=-MSuld_tea2uknHFln2-
All of this work will pay off.
Post by: david.wang28 on February 08, 2019, 10:18:01 pm
The wheel has 22 cuts on it. We also know the radius of the wheel to be 0.250 metres. From the radius, we can find the diameter of the wheel, and hence the distance between 2 cuts (assume they are equally spaced).Not sure which formulas to use?
This value can then be used for part b
In c, wencan consider the time taken for the light to travel to the mirror and back & use that it needs to pass through a slit both times to find the period
Let us know how you go after these hints and where/if you get tripped up :)
Post by: jamonwindeyer on February 09, 2019, 09:18:43 am
Not sure which formulas to use?
The first one just needs your formula for the circumference of a circle, \(C=2\pi R\). The second one is the tangential velocity formula:
And finally, the last one will also use the tangential velocity formula, as well as the basic \(v=\frac{d}{t}\). What you are essentially doing is figuring out how fast the wheel needed to be spinning to allow light travelling at \(3.15\times10^8\) to pass back through on the adjacent slit. Of course, we know the light wasn't travelling that fast, but that was the measurement so we do the maths based on that :)
(a) and (b) are fairly easy, but C is tricky! Have a crack at it though and if not I can step you through it :)
Post by: david.wang28 on February 09, 2019, 10:30:08 am
The first one just needs your formula for the circumference of a circle, \(C=2\pi R\). The second one is the tangential velocity formula:For c), I used v = d/t and subbed in the values to get 6.35*10^-5 secs. I then multiplied that value by 22 to get 1.4*10^-3 secs (there are 22 cuts in the wheel for the light to travel through). Is this the right answer?
And finally, the last one will also use the tangential velocity formula, as well as the basic \(v=\frac{d}{t}\). What you are essentially doing is figuring out how fast the wheel needed to be spinning to allow light travelling at \(3.15\times10^8\) to pass back through on the adjacent slit. Of course, we know the light wasn't travelling that fast, but that was the measurement so we do the maths based on that :)
(a) and (b) are fairly easy, but C is tricky! Have a crack at it though and if not I can step you through it :)
Post by: jamonwindeyer on February 09, 2019, 04:14:59 pm
For c), I used v = d/t and subbed in the values to get 6.35*10^-5 secs. I then multiplied that value by 22 to get 1.4*10^-3 secs (there are 22 cuts in the wheel for the light to travel through). Is this the right answer?
Unless I'm mistaken, I think that is bang on ;D nice work!
Post by: david.wang28 on February 09, 2019, 09:13:32 pm
Unless I'm mistaken, I think that is bang on ;D nice work!Thanks Jamon! Means a lot coming from you :)
Post by: david.wang28 on February 14, 2019, 07:56:27 pm
I'm having trouble answering this question in the link below. Can anyone help me please? Thanks :)
Post by: r1ckworthy on February 15, 2019, 06:51:41 pm
This is question 19 from HSC 2017, and it is really stuffing me up. I'm not quite sure how the right hand palm/ slap rule works in this case, so I hope you guys could explain it to me.
Post by: zhudiac on February 15, 2019, 08:32:18 pm
At point A and C, the wire is being swung downwards which moves the conductor through equipotential lines; i.e. the B-field strength changes as the conductor is swung at points A and C, as can be seen in the Earth diagram. This is not the case for B and D, where the wire moves along equipotential lines (into/out of the page) and thus moves such that the B-field strength does not change. Thus, we can immediately eliminate B and D, as current is induced when there is a change in flux over change in time; which only occurs during A and C, by Faraday's Law.
To determine between A and C, we must use Lenz's law. By Lenz's law, the induced current must oppose the change in B-field which caused it; or to oppose the direction of motion. Thus, by applying right hand palm rule, we can see that by applying a current from P to Q at C (which interacts with the external Earth's B-field), a force is generated so as to oppose the direction of motion (which is down the page). Thus, it must be C and not A that produces a positive current from P to Q. Thus the answer is C.
Post by: jamonwindeyer on February 15, 2019, 08:55:30 pm
Hello,
I'm having trouble answering this question in the link below. Can anyone help me please? Thanks :)
Hey! Gleaming context from the question, essentially what this refers to is the absorption spectrum of the star! The frequencies of the lines represent energy differences in the atoms of the star, which can help characterise their elemental composition. Further, the density of the star can be gleamed from the width of the spectral lines (I explain this more here, but it is probably unnecessary to understand) :)
Post by: Crystine_16 on February 17, 2019, 04:40:14 pm
Thanks in advance :)
Post by: fun_jirachi on February 17, 2019, 05:23:51 pm
Basically, current induces a magnetic field. An AC motor produces an alternating current, so it produces a constantly shifting magnetic field by RHGR. Then because the magnetic field is constantly moving, it induces a change in flux over a period of time (because the field is always changing! By the definition of flux, which is magnetic field permeating an area, if the field is changing while the area that gets permeated is the same, the flux changes!). By Faraday's Law, we get an induced EMF, which again induces a potential difference. This potential difference forms a current proportional to the resistance of whatever it flows through, and then you get an induced current flowing from the current.
Hope this helps :)
Post by: jamonwindeyer on February 18, 2019, 12:26:13 am
Can someone please explain to me how AC induction motors work. I've spent nearly five hours on youtube and websites trying to understand it but I srsly can't :(
Thanks in advance :)
Hey! Jirachi's response above is great, it basically highlights a bit of a process. You need to understand each step before you move to the next:
1. Three phase AC input to stator creates rotating magnetic field
2. Rotating magnetic field creates changing magnetic flux for rotor
3. EMF (and so current) is generated in the rotor via electromagnetic induction
4. Force produced due to this current interacting with the field (motor effect)
If you were having trouble with a specific step here, we might be able to hone in a bit more :)
Post by: david.wang28 on February 18, 2019, 09:16:47 pm
I wanted to ask why Young's experiment had two slits instead of one. Can someone clarify this please? Thanks :)
Post by: jamonwindeyer on February 18, 2019, 10:15:52 pm
Hello,
I wanted to ask why Young's experiment had two slits instead of one. Can someone clarify this please? Thanks :)
Hello! Young's experiment was aiming to demonstrate the interference and diffraction of light waves. For that to happen, you need two light 'sources.' That's what the two slits approximate - Two light sources which then interfere with each other to produce the interference patterns Young observed in the experiment! :)
Post by: david.wang28 on February 18, 2019, 10:35:44 pm
Hello! Young's experiment was aiming to demonstrate the interference and diffraction of light waves. For that to happen, you need two light 'sources.' That's what the two slits approximate - Two light sources which then interfere with each other to produce the interference patterns Young observed in the experiment! :)Is it to increase the intensity as well?
Post by: jamonwindeyer on February 18, 2019, 10:46:55 pm
Is it to increase the intensity as well?
That's not really the reason!! We only get the pattern of maxima/minima when we use two slits. A single slit produces variation in intensity, but it isn't the same thing as the double slit. The double slit is the whole point of making the experiment work :)
(https://qph.fs.quoracdn.net/main-qimg-90fc1737cca4df643bca1b25ae41c6e8)
Post by: david.wang28 on February 19, 2019, 04:32:32 pm
That's not really the reason!! We only get the pattern of maxima/minima when we use two slits. A single slit produces variation in intensity, but it isn't the same thing as the double slit. The double slit is the whole point of making the experiment work :)Thank you! :)
(https://qph.fs.quoracdn.net/main-qimg-90fc1737cca4df643bca1b25ae41c6e8)
Post by: louisaaa01 on February 20, 2019, 02:45:40 pm
Thank you.
Post by: e2503 on February 20, 2019, 09:11:14 pm
I`ve been assigned a first hand investigation on kinematics to design an experiment that models the horizontal and vertical components of velocity on the time taken for the flight of a projectile. Whilst i think that the given topic is straight forward, i am unsure about what experiment to conduct that will satisfy the aim. :'( :'(
If anyone has any suggestions, please send them through :)
Post by: jamonwindeyer on February 20, 2019, 11:27:39 pm
Can someone please help with this question?
Thank you.
Hello! This one can basically be broken down as follows:
- Spinning magnet creates changing magnetic flux
- Changing flux results in eddy currents flowing in the disc due to electromagnetic induction
- Lenz's Law says these induced currents will flow to oppose the change that created them. In this context, this means they'll flow to try and attract the magnet back to the same relative position. That won't work, because the magnet is presumably moving in a fixed fashion (assuming the inertia of the magnet is large, basically). So instead, the disc will spin to try and keep up with the magnet!
tl;dr - > Induced eddy currents due to the changing flux cause the disc to spin and 'chase' the magnet to minimise the resultant change in flux ;D
Some reasonably tricky ideas in here, let me know if I can clarify ;D
Post by: jamonwindeyer on February 20, 2019, 11:34:16 pm
Hi :) :)
I`ve been assigned a first hand investigation on kinematics to design an experiment that models the horizontal and vertical components of velocity on the time taken for the flight of a projectile. Whilst i think that the given topic is straight forward, i am unsure about what experiment to conduct that will satisfy the aim. :'( :'(
If anyone has any suggestions, please send them through :)
Hello! For horizontal, you want to launch something horizontally off a raised platform and look at the impact on range. A way to do that is to use a ramp (think hot wheels!) that leads down and launches horizontally off the edge of a table - Changing the height of that ramp above the table will change your horizontal launching velocity :)
Vertical is tougher! Would be keen to hear other people's ideas here? I'd be launching vertically at different velocities and measuring effect on maximum height, but I can't immediately think of a way to predictably vary the vertical velocity? :) I bet there are lots of ideas out there on this!
Post by: e2503 on February 21, 2019, 07:30:10 am
Hello! For horizontal, you want to launch something horizontally off a raised platform and look at the impact on range. A way to do that is to use a ramp (think hot wheels!) that leads down and launches horizontally off the edge of a table - Changing the height of that ramp above the table will change your horizontal launching velocity :)
Vertical is tougher! Would be keen to hear other people's ideas here? I'd be launching vertically at different velocities and measuring effect on maximum height, but I can't immediately think of a way to predictably vary the vertical velocity? :) I bet there are lots of ideas out there on this!
Thank you so much Jamon :D
Post by: david.wang28 on February 21, 2019, 02:44:54 pm
I'm stuck with Q 2.2, Q 2.4 and Q 2.5 in the link below. Can anyone please help me with these questions? Thanks ")
Post by: _Himani_ on February 23, 2019, 05:23:20 pm
Can someone please explain why in some cases we write/use the formula to measure electric field strength as E=-delta v/d and others it is simply E=v/d. I understand that delta V is probably the 'more correct' form as it is the change in potential, but I don't understand the negative sign.
Thanks!
Post by: jamonwindeyer on February 23, 2019, 09:20:18 pm
Hello,
I'm stuck with Q 2.2, Q 2.4 and Q 2.5 in the link below. Can anyone please help me with these questions? Thanks ")
Hello!
- For 2.2, I can't see the referenced diagram! It is almost certainly just using Speed=Distance/Time? The distance probably comes from multiples of the wavelength and you are given the speed \(v\), so your answer will be something like ?A/v. Maybe?
- For 2.4(b), use \(d\sin{\theta}=m\lambda\), since that is effectively what this situation is. By increasing the frequency, the wavelength is reduced, which means the associated \(\theta\) for minimum disturbance lines is decreased. This is a great simulator for this exact situation ;D
- For 2.4c), places where the wavelets constructively interfere correspond to maxima. Where they destructively interfere corresponds to minima (least disturbance). The diagram will look really similar (identical) to that one I included in an answer a couple of days ago!
Make sure you understand the parallels between 2.4 and the double slit experiment, it's really important to have that conceptualisation set ;D
For 2.5, it's a bit weird to ask for diagrams, I'm guessing they want some similar to that in this video? https://www.youtube.com/watch?v=tlBgvHzppKk
Essentially, Newton thought light was a particle (a corpuscle), and that it reflected by the particle literally bouncing off a surface. Huygen thought it was a wave, reflecting essentially as a consequence of Huygen's Principle (that each point on a wavefront can be thought of as a source of wavelets, and that the leading edge of these wavelets is the actual wavefront). To my interpretation of the syllabus though, that isn't something you need to understand in a whole lot of depth ;D
Post by: jamonwindeyer on February 23, 2019, 09:44:04 pm
Hiya,
Can someone please explain why in some cases we write/use the formula to measure electric field strength as E=-delta v/d and others it is simply E=v/d. I understand that delta V is probably the 'more correct' form as it is the change in potential, but I don't understand the negative sign.
Thanks!
Hello! Great question. The \(\frac{-\Delta V}{d}\) form essentially comes from the proper mathematical link between electric field and voltage (meaning it's not something you need to worry about). Essentially, it's because the definition of \(E=\frac{V}{d}\) comes from the work done with a force over a distance, \(W=Fd\). Since \(W=Vq\) and \(F=Eq\):
However, \(W=Fd\) is for work done against a field. In our case, we are considering the work done moving a particle _with_ the field, so we actually use \(W=-Fd\) if we are doing it formally. Carry that through, and that's where the negative comes from.
This is a bit of a bad explanation in itself, but hopefully it gives you some intuition. Don't worry about this in normal working, just trust that it's a consequence of more complicated mathematics and use the formula as you know how to use it ;D
Post by: david.wang28 on February 24, 2019, 10:09:28 pm
Hello!Thank you Jamon! :)
- For 2.2, I can't see the referenced diagram! It is almost certainly just using Speed=Distance/Time? The distance probably comes from multiples of the wavelength and you are given the speed \(v\), so your answer will be something like ?A/v. Maybe?
- For 2.4(b), use \(d\sin{\theta}=m\lambda\), since that is effectively what this situation is. By increasing the frequency, the wavelength is reduced, which means the associated \(\theta\) for minimum disturbance lines is decreased. This is a great simulator for this exact situation ;D
- For 2.4c), places where the wavelets constructively interfere correspond to maxima. Where they destructively interfere corresponds to minima (least disturbance). The diagram will look really similar (identical) to that one I included in an answer a couple of days ago!
Make sure you understand the parallels between 2.4 and the double slit experiment, it's really important to have that conceptualisation set ;D
For 2.5, it's a bit weird to ask for diagrams, I'm guessing they want some similar to that in this video? https://www.youtube.com/watch?v=tlBgvHzppKk
Essentially, Newton thought light was a particle (a corpuscle), and that it reflected by the particle literally bouncing off a surface. Huygen thought it was a wave, reflecting essentially as a consequence of Huygen's Principle (that each point on a wavefront can be thought of as a source of wavelets, and that the leading edge of these wavelets is the actual wavefront). To my interpretation of the syllabus though, that isn't something you need to understand in a whole lot of depth ;D
Post by: carlasilvia on February 26, 2019, 12:25:25 pm
Post by: cruzj01 on February 26, 2019, 01:57:03 pm
"Two long, parallel, current-carrying conductors attract each other with a force of F newtons. The current in each, and the distance between them is tripled. Predict the new force between the wires"
heLP
Post by: fun_jirachi on February 26, 2019, 04:45:09 pm
Find the altitude of a satellite in orbit around Earth if its orbital speed is 5.0km/s.
Hey there :)
Okay but this question has me slightly confused,
"Two long, parallel, current-carrying conductors attract each other with a force of F newtons. The current in each, and the distance between them is tripled. Predict the new force between the wires"
heLP
Hey there :)
For some distance between them r, and for some current I1 and I2, the force per unit length is F. If you triple the current in one wire, the force triples, since we can see force and current are directly proportional. Similarly, if you triple the distance between them, the force is reduced by a factor of three since we can see force and current are inversely proportional. After applying all the changes specified in the question, the new force should be 3F.
Post by: blasonduo on February 26, 2019, 05:32:01 pm
Find the altitude of a satellite in orbit around Earth if its orbital speed is 5.0km/s.
Hey there :)
Just to add in here, if we want the orbital altitude make sure to subract Earth's radius from your answer from above :)
Post by: e2503 on February 27, 2019, 09:18:14 pm
How do i make calculations of projectile motion with the consideration of air resistance/ drag forces? And would the calculations including air resistance differ immensely compared to theoretical caclucations with no air resistance?
Post by: jamonwindeyer on February 27, 2019, 10:39:43 pm
Hi there :D
How do i make calculations of projectile motion with the consideration of air resistance/ drag forces? And would the calculations including air resistance differ immensely compared to theoretical caclucations with no air resistance?
Hello! So if you do want to do that (don't need to in this course!) there are a few ways you can do it, depending on how accurate you want to be with things ;D
Air resistance is generally modelled as being proportional to the square of velocity. The constant of proportionality is the drag coefficient, and is different depending on the object, its inclination, all sorts of stuff. I've seen \(C=0.5\) used a fair bit, I think it might be the standard one for spherical objects?
So the idea is you calculate the air resistance by taking the velocity \(V\) (the vector sum of horizontal and vertical velocities), calculating the resistive acceleration as \(a_\text{res}=0.5V^2\), and then redistributing that to the two directions.
Where \(\theta\) is the velocity angle at that point in time! As you can see, it gets complicated quickly! The other thing you can do is just add a constant horizontal deceleration to the object, which effectively turns your horizontal calculations into ones with acceleration just like the vertical. And there are more - It depends on how close you want to be to reality ;D
Edit: How much will it differ? Again, depends. Try and calculate the motion of an airplane without air resistance and things break pretty quickly! For most everyday situations with aerodynamic objects, you are fine. For example, I did an experiment with my Tutesmart class to predict the landing position of a hot wheels car launched from a ramp. The theoretically predicted position was exactly where the car landed ;D
Post by: e2503 on February 28, 2019, 06:12:37 pm
Hello! So if you do want to do that (don't need to in this course!) there are a few ways you can do it, depending on how accurate you want to be with things ;D
Air resistance is generally modelled as being proportional to the square of velocity. The constant of proportionality is the drag coefficient, and is different depending on the object, its inclination, all sorts of stuff. I've seen \(C=0.5\) used a fair bit, I think it might be the standard one for spherical objects?
So the idea is you calculate the air resistance by taking the velocity \(V\) (the vector sum of horizontal and vertical velocities), calculating the resistive acceleration as \(a_\text{res}=0.5V^2\), and then redistributing that to the two directions.
Where \(\theta\) is the velocity angle at that point in time! As you can see, it gets complicated quickly! The other thing you can do is just add a constant horizontal deceleration to the object, which effectively turns your horizontal calculations into ones with acceleration just like the vertical. And there are more - It depends on how close you want to be to reality ;D
Edit: How much will it differ? Again, depends. Try and calculate the motion of an airplane without air resistance and things break pretty quickly! For most everyday situations with aerodynamic objects, you are fine. For example, I did an experiment with my Tutesmart class to predict the landing position of a hot wheels car launched from a ramp. The theoretically predicted position was exactly where the car landed ;D
Thank you so much for your help :D :D :)
I will be sure to remeber your tips
Post by: sarrahbarodawala on March 03, 2019, 01:50:14 pm
Have a question regarding magnetic fields
"What is the minimum magnitude of a magnetic field necessary to apply a force of 1 x 10-12 to an electron moving at a speed of 500 kms-1? What electric field is necessary to apply this magnitude force?
There are so many formulas regarding magnetic fields that I just got so confused in working it out. Please help :)
Thank you!
btw answers are a) magnetic field 13T and b) electric field 6.3 x 106 NC-1
Post by: fun_jirachi on March 03, 2019, 04:14:23 pm
If there seem to be too many formulas, the quickfix is just to figure out what you have, and what you need to find. Then basically the formula with the most corresponding stuff is the one you need, otherwise you do a little substitution of one formula into another to make a derived formula you can use. In this case, it's totally unnecessary.
Post by: david.wang28 on March 05, 2019, 06:21:34 pm
I am stuck on all of the multiple choice questions in the attachment below. Can anyone please help me out? Thanks :)
Post by: jmbelger on March 05, 2019, 06:40:03 pm
So we have a prac investigation assessment coming up on the connection between electricity and magnetism (electromagnetism) and we have to discuss how electricity and magnetism are linked.
I've been looking everywhere for a simple explanation without just stating that a current-carrying wire interacts with a magnetic field.
BUT all I can find are people explaining Faraday, Lenz or Maxwell's laws as well as basic principles of magnetism and electrostatics.
Is there a simple explanation for how the two are linked - using basic terminology?
Any help is greatly appreciated!
Jem :D
Post by: jamonwindeyer on March 05, 2019, 07:00:01 pm
Hey,
So we have a prac investigation assessment coming up on the connection between electricity and magnetism (electromagnetism) and we have to discuss how electricity and magnetism are linked.
I've been looking everywhere for a simple explanation without just stating that a current-carrying wire interacts with a magnetic field.
BUT all I can find are people explaining Faraday, Lenz or Maxwell's laws as well as basic principles of magnetism and electrostatics.
Is there a simple explanation for how the two are linked - using basic terminology?
Any help is greatly appreciated!
Jem :D
Welcome to the forums Jem!! ;D
Electricity and magnetism are two tangible manifestations of the same electromagnetic force. Two faces, same person, sort of thing! In fact, we even have a name for the combined effects of electric/magnetic forces on an object - The Lorentz Force :)
If you do consider them separately, then:
- Moving electrical charges produce magnetic fields
- Changing magnetic fields produce electric fields
Those are the links between the two! :) to actually explain why they are the same is a fair bit above what a high school Physics student would be expected to know, so this should be plenty - The way I read that question, it wants you to explore the ways the two are connected in the physical world rather than theoretically ;D
Post by: jamonwindeyer on March 05, 2019, 07:08:31 pm
Hello,
I am stuck on all of the multiple choice questions in the attachment below. Can anyone please help me out? Thanks :)
Hey David! I'll give you a quick run down, but many of these are really just about knowing the theory for this particular part of the course rather than any sort of problem solving - You might need to revisit your textbook to understand these properly!!
- Q1.1 is C, the colour of a star is related to its surface temperature. This can then be linked to the characteristic wavelength by Wien's Displacement Law.
- Q1.2 is B (as I interpret it), the dominant wavelength of a star is related to the surface temperature by Wien's Displacement Law. However, that law applies only to actual black bodies, of which stars are only an approximate. Therefore, the actual temperature is likely to be slightly different. Wien's Law is also less accurate for shorter wavelengths.
- Q1.3 is B, you just need to know this one :)
- Q1.4 is D, by process of elimination. Photons have different energies (clearly), the amount of energy carried by a photon is inversely proportional to wavelength, and since colour and frequency are linked, C is also untrue. If you calculate the energy per photon for blue and green light, you'll realise that you need less photons of blue light than green light to make 1 joule :)
Post by: jmbelger on March 05, 2019, 07:11:01 pm
Welcome to the forums Jem!! ;D
Electricity and magnetism are two tangible manifestations of the same electromagnetic force. Two faces, same person, sort of thing! In fact, we even have a name for the combined effects of electric/magnetic forces on an object - The Lorentz Force :)
If you do consider them separately, then:
- Moving electrical charges produce magnetic fields
- Changing magnetic fields produce electric fields
Those are the links between the two! :) to actually explain why they are the same is a fair bit above what a high school Physics student would be expected to know, so this should be plenty - The way I read that question, it wants you to explore the ways the two are connected in the physical world rather than theoretically ;D
Thank you so much! This is exactly what I needed to know - minus the convoluted textbook mumble!
The prac is meant to examine an investigation which demonstrates the relationship between the two forces so I think I'll start with considering the two forces separately then together.
Once again - thank you!
Jem ;D
Post by: david.wang28 on March 05, 2019, 08:03:56 pm
Hey David! I'll give you a quick run down, but many of these are really just about knowing the theory for this particular part of the course rather than any sort of problem solving - You might need to revisit your textbook to understand these properly!!Thank you Jamon! :)
- Q1.1 is C, the colour of a star is related to its surface temperature. This can then be linked to the characteristic wavelength by Wien's Displacement Law.
- Q1.2 is B (as I interpret it), the dominant wavelength of a star is related to the surface temperature by Wien's Displacement Law. However, that law applies only to actual black bodies, of which stars are only an approximate. Therefore, the actual temperature is likely to be slightly different. Wien's Law is also less accurate for shorter wavelengths.
- Q1.3 is B, you just need to know this one :)
- Q1.4 is D, by process of elimination. Photons have different energies (clearly), the amount of energy carried by a photon is inversely proportional to wavelength, and since colour and frequency are linked, C is also untrue. If you calculate the energy per photon for blue and green light, you'll realise that you need less photons of blue light than green light to make 1 joule :)
Post by: louisaaa01 on March 07, 2019, 09:02:09 pm
Can you please help with this question (from AC generators)?
Sketch graphs to show the relationship between:
(a) The flux through the coil and the induced
voltage.
(b) The torque acting on the coil and the induced voltage.
The answers show a sine relationship for (a), and a cosine relationship for (b) - but I can't seem to reason as to why. I thought that, for (a) - as flux increases to some maximum value, induced emf would decrease - and for (b) some sort of direct (not necessarily directly proportional) relationship would occur. Also, I'm a bit confused as to how both flux and torque can be on the x-axis (in the suggested answers) when, in a generator, they don't increase indefinitely?
Thank you in advance for your assistance!
Post by: Coolmate on March 09, 2019, 09:25:21 pm
I am having trouble understanding the topic area of distance and displacement on horizontal planes, in particular the maths side of questions and formulas. Any guidance/ links to resources will be greatly appreciated.
Kind Regards,
Coolmate :D
Post by: jamonwindeyer on March 10, 2019, 11:25:46 am
Hi!
Can you please help with this question (from AC generators)?
Sketch graphs to show the relationship between:
(a) The flux through the coil and the induced
voltage.
(b) The torque acting on the coil and the induced voltage.
The answers show a sine relationship for (a), and a cosine relationship for (b) - but I can't seem to reason as to why. I thought that, for (a) - as flux increases to some maximum value, induced emf would decrease - and for (b) some sort of direct (not necessarily directly proportional) relationship would occur. Also, I'm a bit confused as to how both flux and torque can be on the x-axis (in the suggested answers) when, in a generator, they don't increase indefinitely?
Thank you in advance for your assistance!
Hello! I think there might be some issue with your question here, because you certainly shouldn't have a sinusoidal relationship between flux and voltage. Really, drawing a graph linking flux and induced emf is a faulty exercise, because induced emf depends on rate of change of flux. So you can't draw any sort of direct relationship between flux and emf.
Could you perhaps attach a picture of the graphs? Or a Dropbox link to an image or similar? Maybe I'm interpreting it incorrectly, but I'd tend to agree more with your points ;D
Post by: jamonwindeyer on March 10, 2019, 11:30:29 am
Hi I'm in Year 11,
I am having trouble understanding the topic area of distance and displacement on horizontal planes, in particular the maths side of questions and formulas. Any guidance/ links to resources will be greatly appreciated.
Kind Regards,
Coolmate :D
Hey Coolmate!! Tough stuff for sure! :) the general approach to any of these questions is roughly the same:
- Break/resolve any given vector into two perpendicular pieces (ie - horizontal/vertical, north-south/east-west, etc)
- Perform your kinematics calculations on each direction separately, yielding one 'result' in each direction
- Recombine your two results into a single vector using trigonometry
If you attach an example of a question you are having trouble with, we'd be happy to step through it for you ;D
Post by: Jefferson on March 10, 2019, 01:40:42 pm
Could someone show me how to solve this relative velocity problem using relative velocity vectors.
Angelica is inside train A moving at 80km/h South and sees Bob inside car B moving at 100 km/h at 300° T.
What is Bob's velocity relative to the ground?
Answers are 92km/h and 251° T.
Attachment below is what I did.
Post by: louisaaa01 on March 16, 2019, 01:47:26 pm
Hello! I think there might be some issue with your question here, because you certainly shouldn't have a sinusoidal relationship between flux and voltage. Really, drawing a graph linking flux and induced emf is a faulty exercise, because induced emf depends on rate of change of flux. So you can't draw any sort of direct relationship between flux and emf.
Could you perhaps attach a picture of the graphs? Or a Dropbox link to an image or similar? Maybe I'm interpreting it incorrectly, but I'd tend to agree more with your points ;D
I've attached the question and provided answer :)
Post by: david.wang28 on March 18, 2019, 05:11:40 pm
Hi all,Your working out is fine. Don't need to worry about relative velocity vectors.
Could someone show me how to solve this relative velocity problem using relative velocity vectors.
Angelica is inside train A moving at 80km/h South and sees Bob inside car B moving at 100 km/h at 300° T.
What is Bob's velocity relative to the ground?
Answers are 92km/h and 251° T.
Attachment below is what I did.
Post by: david.wang28 on March 18, 2019, 05:13:10 pm
I am stuck on all of the multiple choice questions and Q 2.1 b) in the link below. Can anyone please help me out? Thanks :)
Post by: _Himani_ on March 18, 2019, 07:48:46 pm
I was just wondering how I should prepare for a Processing and Data Analysis In-Class Assessment on modules 5 and 6. Where can I find questions to attempt? Would the general examples from the In Focus textbook be enough?
As always,
Thank You Very Much.
Post by: jamonwindeyer on March 18, 2019, 09:05:44 pm
Hello,
I am stuck on all of the multiple choice questions and Q 2.1 b) in the link below. Can anyone please help me out? Thanks :)
Sure thing David!
- 1.1 is (D), because all the others have nonsense words with nothing to do with anything ;)
- 1.2 you are right with (D)
- 1.3 is (B), the energy of an emitted photoelectron is \(E=hf-\phi\). It has nothing to do with intensity, only the frequency of light and the work function of the metal.
- 1.3 is C), because that's just how it was ;D
As for the last one, remember that the energy of emitted photoelectrons is always \(E=hf-\phi\). Again, intensity has nothing to do with that - Intensity only releases MORE photoelectrons, it doesn't change their individual energies. Therefore, if the light doesn't have a high enough frequency, you won't get a photocurrent no matter how intense your light is ;D
Post by: david.wang28 on March 18, 2019, 10:45:48 pm
Hey,You can buy the topic tests from atarnotes of course :)
I was just wondering how I should prepare for a Processing and Data Analysis In-Class Assessment on modules 5 and 6. Where can I find questions to attempt? Would the general examples from the In Focus textbook be enough?
As always,
Thank You Very Much.
Post by: david.wang28 on March 18, 2019, 10:46:51 pm
Sure thing David!Thanks Jamon! :)
- 1.1 is (D), because all the others have nonsense words with nothing to do with anything ;)
- 1.2 you are right with (D)
- 1.3 is (B), the energy of an emitted photoelectron is \(E=hf-\phi\). It has nothing to do with intensity, only the frequency of light and the work function of the metal.
- 1.3 is C), because that's just how it was ;D
As for the last one, remember that the energy of emitted photoelectrons is always \(E=hf-\phi\). Again, intensity has nothing to do with that - Intensity only releases MORE photoelectrons, it doesn't change their individual energies. Therefore, if the light doesn't have a high enough frequency, you won't get a photocurrent no matter how intense your light is ;D
Post by: jamonwindeyer on March 18, 2019, 10:52:52 pm
Hey,
I was just wondering how I should prepare for a Processing and Data Analysis In-Class Assessment on modules 5 and 6. Where can I find questions to attempt? Would the general examples from the In Focus textbook be enough?
As always,
Thank You Very Much.
These are hard to prepare for!! It's really just about building your general understanding of the topics. As David so kindly mentions above, you can buy the AN Topic Tests for that purpose ;) but of course, Physics In Focus or anything you already have will do just fine as well!!
On top of that, make sure you practice experiment skills. Validity/accuracy/reliability, graphing skills, variables, sources of error, that sort of stuff :) if you needed any of that clarified let us know!
Post by: jamonwindeyer on March 18, 2019, 10:55:32 pm
Thanks Jamon! :)
You're welcome! Happy to help :)
Post by: david.wang28 on March 20, 2019, 04:16:38 pm
I have trouble with all of Q 2.3 in the attachment below. Can anyone please help me out? Thanks ;)
Post by: Dillan on March 28, 2019, 10:04:24 pm
Got a physics depth study due soon on motors in particular DC ones.
Was just after a few ideas of an inquiry question to base my depth study on.
Also if anyone has any tips and tricks on writing writing a band 6 depth study, would be greatly appreciated if you can send it my way
Cheers
Post by: jamonwindeyer on March 28, 2019, 10:57:25 pm
Hello,
I have trouble with all of Q 2.3 in the attachment below. Can anyone please help me out? Thanks ;)
Hey David! How did you end up going with this? If you are struggling with an entire worksheet of questions on a topic it is probably worth going back to the notes/textbook on the topic, watching some YouTube videos, etc. It will be better for you than just having the answers served to you by someone else!! I'd also wager that wanting 10 questions worth of answers is probably why you didn't get much of a response to this post - You'll get help much faster if you work with us a bit and show us what understanding you do have and where your confusions lie ;D
Post by: jamonwindeyer on March 28, 2019, 11:01:41 pm
Hey everyone,
Got a physics depth study due soon on motors in particular DC ones.
Was just after a few ideas of an inquiry question to base my depth study on.
Also if anyone has any tips and tricks on writing writing a band 6 depth study, would be greatly appreciated if you can send it my way
Cheers
Hey there!
There are lots of inquiry questions you could ask about DC motors. Some that spring to mind:
- Something on efficiency (how does changing X, Y and Z effect the efficiency of the motor)
- Modifications to the motor structure (different materials) and how they effect the motor/its operation
- How the speed of the motor can be controlled
All of these go a bit beyond the syllabus, which is the whole point! ;D
As for tips on a B6 depth study, that's tougher because no one has ever gotten a B6 for a Physics Depth study yet (no one's done one ;)) ;D I think it's about demonstrating that you've gone beyond the syllabus, primarily! If you do the bare minimum, you've defeated the purpose of a depth study. Show that you've done your research. Additionally, make sure any experimental reports are done well with proper discussion of error, accuracy, reliability, all that usual stuff ;D
Post by: david.wang28 on March 29, 2019, 06:00:33 pm
Hey David! How did you end up going with this? If you are struggling with an entire worksheet of questions on a topic it is probably worth going back to the notes/textbook on the topic, watching some YouTube videos, etc. It will be better for you than just having the answers served to you by someone else!! I'd also wager that wanting 10 questions worth of answers is probably why you didn't get much of a response to this post - You'll get help much faster if you work with us a bit and show us what understanding you do have and where your confusions lie ;DI ended up doing fine. Thanks for asking :)
Post by: david.wang28 on March 29, 2019, 06:02:22 pm
I am stuck on two question, Q 2.2 b) and Q 2.5 a) in the attachments below. Can anyone please help me out? Thanks :)
Post by: DrDusk on March 29, 2019, 08:47:48 pm
Hello,First one:
I am stuck on two question, Q 2.2 b) and Q 2.5 a) in the attachments below. Can anyone please help me out? Thanks :)
Consider any hand on the clock, the minute or second hand. Now consider a point on the end of the minute hand. This point traces a circular path where the distance of this path is l = 2*pi*r where r is the length of the minute hand(Note this is using the arc length formula l = r* theta).
Now if you derive an expression for the period of rotation T, it will be T = (2*pi*r)/v which is just Time = distance/speed. Now if you make v the subject you get v = 2*pi*r/T - Eqn 1
Now remember the bolt in the clock always supplies the same centripetal force on the minute hand where F = m*v^2/r, but according to out equation 1 above, the time is dilating which means the period T will be different, but if our centripetal force is always constant how do we account for this difference?? Well simple length contraction!!!, the length of the minute hand r will decrease.
EDIT:
Note: The mass M will also increase so to keep F constant its a tug of war between M, v^2 and r. Since v^2 decreases much faster than m can increase, we need r to decrease which will end up balancing v^2 with m
P.S please do not write this in an exam, its way too long haha. I just wrote it so you can understand
Post by: jamonwindeyer on March 29, 2019, 08:53:27 pm
Hello,
I am stuck on two question, Q 2.2 b) and Q 2.5 a) in the attachments below. Can anyone please help me out? Thanks :)
I really like DrDusk's explanation above! I've never seen it before, might steal that as another way to explain it in the future ;D
2.5(a) is a trick question - Or at least I think it is! The concept of mass dilation only really makes sense for an object moving relative to the observer. If the observer is in the same frame of reference, they'll measure the mass the same as if the object is at rest - That's the whole point (and this confusing point is why we normally talk about relativistic momentum, not mass).
Neglecting this, however, the way the question is set up I'm thinking it is just a weird wording - The answer is that the rest mass is much, much smaller than the relativistic mass ;D
Post by: david.wang28 on March 29, 2019, 09:04:09 pm
First one:This is quite a creative way of explaining this question; I like it! Thanks a lot! :)
Consider any hand on the clock, the minute or second hand. Now consider a point on the end of the minute hand. This point traces a circular path where the distance of this path is l = 2*pi*r where r is the length of the minute hand(Note this is using the arc length formula l = r* theta).
Now if you derive an expression for the period of rotation T, it will be T = (2*pi*r)/v which is just Time = distance/speed. Now if you make v the subject you get v = 2*pi*r/T - Eqn 1
Now remember the bolt in the clock always supplies the same centripetal force on the minute hand where F = m*v^2/r, but according to out equation 1 above, the time is dilating which means the period T will be different, but if our centripetal force is always constant how do we account for this difference?? Well simple length contraction!!!, the length of the minute hand r will decrease.
P.S please do not write this in an exam, its way too long haha. I just wrote it so you can understand
Post by: david.wang28 on March 29, 2019, 09:07:28 pm
I really like DrDusk's explanation above! I've never seen it before, might steal that as another way to explain it in the future ;DI mean, rest mass is smaller than the relativistic mass, according to my calculations for the next part (not shown). Thanks Jamon! :)
2.5(a) is a trick question - Or at least I think it is! The concept of mass dilation only really makes sense for an object moving relative to the observer. If the observer is in the same frame of reference, they'll measure the mass the same as if the object is at rest - That's the whole point (and this confusing point is why we normally talk about relativistic momentum, not mass).
Neglecting this, however, the way the question is set up I'm thinking it is just a weird wording - The answer is that the rest mass is much, much smaller than the relativistic mass ;D
Post by: _Himani_ on March 30, 2019, 05:57:28 pm
Just a quick question from the charged particles, Conductors and ELectric and Magnetic Fields Topic Test from the ATAR Notes book.
Question 3: There are two parallel plates set up half a metre apart. The voltage can be used to accelerate particles from rest, from one plate to another. a) AN electron is used in this setup with a voltage of 10,000V. Calculate the speed of the electron when it strikes the top plate.
Okay, so I calculated my electric field (20,000) and the acceleration of the particle (3.52 x 10^15). Then I went on to use the formula v^2=u^2+2as which would be v^2=o+2(3.52 x 10^15)(0.5). The worked solution uses v^2=o+2(3.52 x 10^15) with, I'm assuming s=1. Why is this? The diagram with the question has also marked the distance between the plates as one metre and not half a metre. Is this just a typo or am I missing something in the calculation?
BTW: This book is a total lifesaver! Thanks!
Post by: DrDusk on March 30, 2019, 06:38:04 pm
Just building on that you can also use Work done to calculate its velocity where W = (m*v^2)/2 = F * s.
Post by: jamonwindeyer on March 30, 2019, 06:55:05 pm
Post by: sidzeman on March 31, 2019, 11:20:31 am
a) How long will the cat be in the air before she lands on the board?
b) At what distance must the skateboard be when the cat steps off the branch?
c) What is the combined horizontal speed of the skateboard and the cat after the cat lands on the board?
Please see below for my working out for part A and B - I was stuck on part C
Thanks :)
Post by: louisaaa01 on March 31, 2019, 06:58:10 pm
Post by: DrDusk on April 01, 2019, 12:11:26 am
Can someone please help with this question from the 2006 HSC Physics paper? Thanks!Well firstly you have relative motion between the disk and a magnet, since the disk is conductive it can harbor the production of a current. At the bottom near Y the disk is moving into the page and the magnetic field is to the right.
Now we know that since the disk is moving into the page, the produced current must produce a force that opposes this motion. Hence a force must be generated pointing out of the page. So using right hand palm rule, your palm points out of the page, fingers to the right, and you thumb then must point downwards as your thumb must be perpendicular to your fingers.
The direction of your thumb is the direction of your current and since the thumb points downwards from Y, we can say the current goes from Y to X. Thus B.
Oh and its not alternating because the disk only ever moves one direction relative to the magnetic field.
Post by: jamonwindeyer on April 01, 2019, 09:33:40 am
A 3.0kg cat jumps from a branch situated 2.0m above a point where a 1.8kg skateboard will pass moving at 4.2m/s. The cat must land on the skateboard.
a) How long will the cat be in the air before she lands on the board?
b) At what distance must the skateboard be when the cat steps off the branch?
c) What is the combined horizontal speed of the skateboard and the cat after the cat lands on the board?
Please see below for my working out for part A and B - I was stuck on part C
Thanks :)
Hey! Nice work on Parts A-B! So for Part C, we need to use the conservation of momentum in the horizontal direction. The skateboard initially possesses a horizontal momentum of \(\rho_i=1.8\times4.2=7.56\text{kgms}^{-1}\). This momentum must be the same after the mass of the cat is added (the cat doesn't contribute any extra momentum in the collision because it possesses only vertical momentum after the jump, which we assume is transferred to the ground and lost). So we are solving for:
Post by: Dillan on April 03, 2019, 09:53:29 pm
Just after a basic run-down on the pros and cons of an AC and DC motor, and what would be ideal for an appliance like a vacuum cleaner. (i.e. Brushless or induction motors)
Cheers in advance
Post by: david.wang28 on April 04, 2019, 02:12:03 pm
Hey guys,I think AC motors would be ideal for a vacuum cleaner, because it can handle varying voltages; you can vary the suction strength (correlating to the strength of the motor).
Just after a basic run-down on the pros and cons of an AC and DC motor, and what would be ideal for an appliance like a vacuum cleaner. (i.e. Brushless or induction motors)
Cheers in advance
Post by: david.wang28 on April 04, 2019, 02:18:42 pm
I am stuck on Q 1.3, Q 2.3 a) and Q 2.3 b) in the attachments below. Can anyone please help me out? Thanks :)
Post by: DrDusk on April 04, 2019, 08:55:35 pm
Hello,
I am stuck on Q 1.3, Q 2.3 a) and Q 2.3 b) in the attachments below. Can anyone please help me out? Thanks :)
Post by: david.wang28 on April 04, 2019, 11:03:54 pm
Ahhh, thanks man :)
Post by: Joseph.Ryan on April 13, 2019, 11:38:19 am
A double slit is illuminated by light of two wavelengths, 600nm and the other unknown. The two interference patterns overlap with the third dark band of the 600nm wavelength coinciding with the fourth bright band from the central band of the pattern for the light of unknown wavelength. What is the value for the light of unknown wavelength?
Answer: 450nm
Post by: DrDusk on April 13, 2019, 02:29:03 pm
I have a question which I think is fairly simple but I am not able to get the correct answer:Its just making use of the formula dsin(theta) = m*lambda
A double slit is illuminated by light of two wavelengths, 600nm and the other unknown. The two interference patterns overlap with the third dark band of the 600nm wavelength coinciding with the fourth bright band from the central band of the pattern for the light of unknown wavelength. What is the value for the light of unknown wavelength?
Answer: 450nm
For the first wave if we sub in the wavelength and take m as 3 as it is the 3rd dark band we get dsin(theta) = 1800
Then just do the same for the second wave. SInce we know what dsin(theta) is now(this will be the same for both waves) we can do m*lambda = 1800.
Now we take m and 4 as it is the 4th bright band giving us lambda = 1800/4 which equals 450nm
Post by: Joseph.Ryan on April 29, 2019, 02:32:23 pm
Post by: jamonwindeyer on April 29, 2019, 06:18:29 pm
Yes, but for the third dark spot it is destructive interference, which means m should be 2.5, not 3.
Cool! So for the overlap to occur, \(\theta\) is the same in both equations. Since \(d\) is a constant, that means \(d\sin{\theta}\) is the same for both uses of the equation. Those two uses are:
(Notice I have used \(m+0.5=2.5\) in the second one because the first dark band is arguably at \(m=0\), meaning the third dark band is at \(m=2\).
Solving:
And that should be it ;D ps - welcome to the forums!
Post by: Joseph.Ryan on April 30, 2019, 07:44:42 pm
The only problem is that my textbook said the answer was 450nm - should I just assume it is wrong?
Post by: jamonwindeyer on April 30, 2019, 10:48:26 pm
Thanks for the help!
The only problem is that my textbook said the answer was 450nm - should I just assume it is wrong?
You're welcome! And yep, I think so - Seems they've mistakenly done both calculations with the bright band as was done above. Great question regardless, but it looks like their working didn't match the question in this case :)
Post by: Jefferson on May 06, 2019, 05:15:15 pm
The moon is a satellite of Earth, which is a satellite of the Sun.
Then, is the moon a satellite of the Sun?
Post by: DrDusk on May 06, 2019, 05:26:33 pm
Hi everyone, Just a quick question.I would say this is more so an English question than a Physics question, but I don't think the Moon is a satellite of the Sun.
The moon is a satellite of Earth, which is a satellite of the Sun.
Then, is the moon a satellite of the Sun?
A satellite in a perfect world is defined as an object that has a defined circular trajectory around something else, and the moon doesn't have a defined circular trajectory around the sun.
And actually in reality the orbits are elliptical
Post by: david.wang28 on May 12, 2019, 09:59:35 pm
I have trouble with a question in the link below (I have no idea how to answer the question). Can anyone please help me out? Thanks :)
Post by: jamonwindeyer on May 12, 2019, 10:59:28 pm
Hello,
I have trouble with a question in the link below (I have no idea how to answer the question). Can anyone please help me out? Thanks :)
Hey! This is a loaded question, particularly Part A - Have you learned Proton-Proton Chain reactions and are you comfortable with them? That's where you need to start, want to make sure we're on the same page :)
Post by: david.wang28 on May 13, 2019, 10:54:38 am
Post by: jamonwindeyer on May 13, 2019, 12:10:59 pm
I understand proton-proton chain reactions and I know about them, but I don't understand how to do this question.
Cool! So in that case, you know that the P-P chain reactions turn four protons (Hydrogen nuclei) into a helium nucleus (two protons, two neutrons). To do this question properly you need to know the mass of a helium nucleus, which isn't given in the question, you might have it in a supplementary data sheet?
In any case, you calculate the difference in the masses between the Helium and the four protons - That gives you what we call the mass defect. We'll call this \(\Delta m\). Then calculate the energy released via:
Now, we're given the luminosity of the star in Watts. Remember that this is joules per second - Let's figure out how many reactions we need:
So we have this many reactions per second, which means we convert \(4m_p\) kilograms of protons into helium per second, where \(m_p\) is the mass of the proton. Mass per year is obtained by progressively multiplying to get mass per minute, hour, days and year!
Final question is just figuring out, with this rate of mass being converted per year, how long will it take to burn certain amounts of hydrogen. Remember the total start hydrogen mass is \(0.75\times1.99\times10^{30}=1.4925\times10^{30}\) kilograms, 75% of the mass of the sun! So figure out 12% of this and then determine how long that takes based on your answer to (b)! :)
Hopefully this outline is helpful! Let me know if you get stuck anywhere as you follow ;D
Post by: david.wang28 on May 13, 2019, 08:14:24 pm
Cool! So in that case, you know that the P-P chain reactions turn four protons (Hydrogen nuclei) into a helium nucleus (two protons, two neutrons). To do this question properly you need to know the mass of a helium nucleus, which isn't given in the question, you might have it in a supplementary data sheet?Thank you Jamon! :)
In any case, you calculate the difference in the masses between the Helium and the four protons - That gives you what we call the mass defect. We'll call this \(\Delta m\). Then calculate the energy released via:
Now, we're given the luminosity of the star in Watts. Remember that this is joules per second - Let's figure out how many reactions we need:
So we have this many reactions per second, which means we convert \(4m_p\) kilograms of protons into helium per second, where \(m_p\) is the mass of the proton. Mass per year is obtained by progressively multiplying to get mass per minute, hour, days and year!
Final question is just figuring out, with this rate of mass being converted per year, how long will it take to burn certain amounts of hydrogen. Remember the total start hydrogen mass is \(0.75\times1.99\times10^{30}=1.4925\times10^{30}\) kilograms, 75% of the mass of the sun! So figure out 12% of this and then determine how long that takes based on your answer to (b)! :)
Hopefully this outline is helpful! Let me know if you get stuck anywhere as you follow ;D
Post by: Joseph.Ryan on May 17, 2019, 01:35:54 pm
Post by: DrDusk on May 17, 2019, 06:53:26 pm
Hi, do we need to know how to derive the Lorenz transformations equations for the hsc exam?I love your enthusiasm but no :)
It's not that hard imo so don't let that stop you from trying :P
If your teacher actually knows his/hers Physics, they might even do it for you
Post by: _Himani_ on May 19, 2019, 09:32:39 pm
Post by: DrDusk on May 19, 2019, 10:05:28 pm
Can someone please explain this to me: "The interference occurs because the waves have travelled a different distance from each slit to the surface, depending on which point you are looking at on the surface." (It's from the "Young's double slit experiment" section in the Course Notes books).
Well if you draw a line from each slit to any arbitrary point on the surface, the lines will be of different length. Thus each wave has traveled a different distance to reach that point. As a result of this when the waves collide they will be slightly offset from each other, causing them to interfere. Depending on how much they are offset by they will either constructively or destructively interfere.
Post by: Coolmate on May 26, 2019, 10:18:09 pm
I'm in year 11 and I have come across these questions (questions attached below) and I am confused about how to go about answering it, would someone please be able to help me here!
Cheers :)
Post by: fun_jirachi on May 26, 2019, 11:03:05 pm
Gonna try guiding you instead of giving the answers to you straight up, since you need to going through these thought processes in an exam :)
6. a) Try using the equation W=Fs using the information given!
b) Remember that the work done is equal to the change in energy (this isn't technically correct in the real world, but it's a simplification we're using).
c) Equate the value from a) using E=0.5mv^2
d) Self explanatory, it's given in the question :)
e) Given acceleration, displacement, and initial velocity, try using one of the kinematic equations (SUVAT) to relate these three quantities to time!
7. a) Given that KE=0.5mv^2, calculate the initial and final kinetic energy and find the difference :)
b) Equate this value to work done to find the force acting over a distance that slows it down (W=Fs)
c) It's basically just a frictional force acting against the tires in the opposite direction to motion :)
d) similar to d) and e) from above :)
If you need any further help/full working, glad to provide it :) Hope this helps!
Post by: Coolmate on May 28, 2019, 10:09:52 am
Hey there!
Gonna try guiding you instead of giving the answers to you straight up, since you need to going through these thought processes in an exam :)
6. a) Try using the equation W=Fs using the information given!
b) Remember that the work done is equal to the change in energy (this isn't technically correct in the real world, but it's a simplification we're using).
c) Equate the value from a) using E=0.5mv^2
d) Self explanatory, it's given in the question :)
e) Given acceleration, displacement, and initial velocity, try using one of the kinematic equations (SUVAT) to relate these three quantities to time!
7. a) Given that KE=0.5mv^2, calculate the initial and final kinetic energy and find the difference :)
b) Equate this value to work done to find the force acting over a distance that slows it down (W=Fs)
c) It's basically just a frictional force acting against the tires in the opposite direction to motion :)
d) similar to d) and e) from above :)
If you need any further help/full working, glad to provide it :) Hope this helps!
Thanks fun_jirachi
I understand it now! ;D
Post by: david.wang28 on May 29, 2019, 09:58:45 pm
I am stuck on a projectile motion question. I've posted my working out, and I don't quite know how to do all of the question. Can anyone please help me out? Thanks :)
Post by: david.wang28 on May 29, 2019, 10:02:41 pm
I have another question on projectile motion. This time, I don't know how to figure out the angle in the last part of the question. Can anyone please help me out? Thanks :)
Post by: DrDusk on May 29, 2019, 11:15:36 pm
Hello,a)
I am stuck on a projectile motion question. I've posted my working out, and I don't quite know how to do all of the question. Can anyone please help me out? Thanks :)
Remember the horizontal velocity of the projectile does not change, and at the peak the y component of velocity will be zero. So the net velocity at the peak is just the initial horizontal velocity..
b) Your working seems fine..
c) We want the time of flight to be 3 seconds. To make things easier we should instead think of it as it should take 1.5 seconds to reach its peak.
Post by: DrDusk on May 29, 2019, 11:35:33 pm
Hello,For this part you must remember that the horizontal and vertical motions are separate from each other.
I have another question on projectile motion. This time, I don't know how to figure out the angle in the last part of the question. Can anyone please help me out? Thanks :)
Now lets look at the energies. The horizontal velocity is always constant which means the horizontal component of kinetic energy is also constant. This means we can say that all the initial vertical kinetic energy is converted to potential energy at its peak, because remember at the peak the vertical component of velocity is zero.
Just a piece of advice...It is a very important skill in physics that you are always thinking creatively and methodically. In your mind you should always constantly be going through any laws such as conservation of energy, momentum etc. Brainstorm all the rules that can apply.
Also because this is HSC Physics almost all of the time the so called 'hard' calculation problems boil down to just equating two formulas...
So basically if you absolutely have no idea what to do, just think of two formulas that apply really well and equate them as a last resort.
Post by: david.wang28 on May 30, 2019, 09:06:32 am
Post by: Coolmate on June 03, 2019, 06:53:47 pm
Would someone please be able to explain to me the common misconception, gravity is a force of attraction between 2 objects with mass. Including the General Theory of Relativity? and any advice about depth studies and how to get good marks?
Thanks :)
Post by: DrDusk on June 03, 2019, 07:09:25 pm
Hello!
Would someone please be able to explain to me the common misconception, gravity is a force of attraction between 2 objects with mass. Including the General Theory of Relativity? and any advice about depth studies and how to get good marks?
Thanks :)
Your in high school right...? General relativity is 4th year uni physics and it requires a far far advanced understanding.....
General relativity states that gravity is actually the bending of space-time. It states that every object with mass bends and warps the space-time around it which causes other objects to sort of 'fall' towards it, so in a sense gravity is not a force.
I'm not 100% sure with how this depth study stuff works but if your gonna do a depth study, general relativity is something you want to stay faaar away from. Try special relativity, its 10x easier.
Post by: david.wang28 on June 07, 2019, 04:45:12 pm
I have two questions that I am confused on in the attachments below. Can anyone please help me out? Thanks :)
Post by: Coolmate on June 13, 2019, 02:11:34 pm
I'm not 100% sure with how this depth study stuff works but if your gonna do a depth study, general relativity is something you want to stay faaar away from. Try special relativity, its 10x easier.Thanks DrDusk :)
Post by: mani.s_ on June 13, 2019, 05:18:35 pm
Hi, I have a physics depth study starting next week. It seems we have to design an experiment on how variations in pitch and loudness of sound change the characteristics of the sound wave. I have a deep understanding of the theory aspect of this part but I have no idea how to make practical, testing this question. Our practical has to give us quantitative data that we can analyse and evaluate. Any help is appreciated, Thanks :)
Post by: saloni.aphale on June 13, 2019, 07:05:20 pm
Post by: DrDusk on June 13, 2019, 08:29:52 pm
Can someone please help me with the question attached below.Let's make sense of what is fundamentally happening with the photoelectric effect.
Now doubling the intensity will increase the number of Photons of light which means more photons can undergo the photoelectric effect, meaning more photoelectrons should be ejected. So instantly rules out A and C. Now the reason D is incorrect is because from our calculations above we can clearly see that the light does not have enough energy to eject photoelectrons from metal Y, because the energy of the light is less than the minimum energy required, whereas for X it is greater. So the number of photoelectrons emitted cannot physically increase for Y, as it wont ever emit electrons at that frequency and wavelength of light in the first place!. So its B
Post by: saloni.aphale on June 16, 2019, 10:23:37 am
Let's make sense of what is fundamentally happening with the photoelectric effect.
Now doubling the intensity will increase the number of Photons of light which means more photons can undergo the photoelectric effect, meaning more photoelectrons should be ejected. So instantly rules out A and C. Now the reason D is incorrect is because from our calculations above we can clearly see that the light does not have enough energy to eject photoelectrons from metal Y, because the energy of the light is less than the minimum energy required, whereas for X it is greater. So the number of photoelectrons emitted cannot physically increase for Y, as it wont ever emit electrons at that frequency and wavelength of light in the first place!. So its B
Thank you so much, that helped a lot !!!
Post by: mani.s_ on June 18, 2019, 02:52:10 pm
Post by: DrDusk on June 18, 2019, 04:39:27 pm
Hi, I have a physics depth study starting next week. It seems we have to design an experiment on how variations in pitch and loudness of sound change the characteristics of the sound wave. I have a deep understanding of the theory aspect of this part but I have no idea how to make practical, testing this question. Our practical has to give us quantitative data that we can analyze and evaluate. Any help is appreciated, I have also attached my physics depth study notification below. Thanks!!!I don't suppose your school has logger software that allows you to map the sound wave on a computer?? This is what we used in uni and it's a great way to get accurate data.
Post by: Aryan.S on July 03, 2019, 02:44:08 pm
Any help would be greatly appreciated!
Post by: DrDusk on July 03, 2019, 06:21:11 pm
Could some please provide me with an explanation for which type of radiation is the least suitable for medical procedures?Gamma radiation would obviously be the worst because it would go right through your body! Alpha and Beta radiation are much more highly ionizing but they are easily stopped by your skin and body, so its penetrating power is very low, meaning the damage done would be much less.
Any help would be greatly appreciated!
Basically safer radiation would be that which has lower penetration power
Post by: blyatman on July 05, 2019, 01:16:06 pm
Your in high school right...? General relativity is PHD level uni physics and it requires a far far advanced understanding.....
General relativity states that gravity is actually the bending of space-time. It states that every object with mass bends and warps the space-time around it which causes other objects to sort of 'fall' towards it, so in a sense gravity is not a force.
I'm not 100% sure with how this depth study stuff works but if your gonna do a depth study, general relativity is something you want to stay faaar away from. Try special relativity, its 10x easier.
GR is actually an honours-level physics course, and it doesn't require doing a PhD to understand it.
Might be worth talking about the very basic tenets of GR from a qualitative aspect, since SR has nothing to with gravity.
Post by: DrDusk on July 05, 2019, 04:26:05 pm
GR is actually an honours-level physics course, and it doesn't require doing a PhD to understand it.Ah I see. This is what I thought initially but I was unable to find the course in my degree outline which is weird..
Might be worth talking about the very basic tenets of GR from a qualitative aspect, since SR has nothing to with gravity.
Post by: blyatman on July 06, 2019, 12:11:01 am
Ah I see. This is what I thought initially but I was unable to find the course in my degree outline which is weird..
Yeh pretty sure it should be offered. I know it was offered at Usyd. Either way, can just learn it yourself by grabbing a textbook (recommend Gravity by James Hartle). Did 3yrs of research (undergrad and postgrad thesis) and published papers in GR but never took any courses in it lol, just learnt it by reading the textbook along with guidance from my supervisor.
Defs a fun topic to learn, and would've stayed in the field if there were jobs in it.
Post by: Coolmate on July 10, 2019, 07:09:55 pm
I am in Year 11 and am going to do my Physics prelim exams soon. I was wondering whether anyone has any good tips and what the most important things I should study that I can use for the exam, to achieve a high mark?
Also, does anyone have any past prelim papers for Physics?... (The new syllabus is making it hard to find them).
Thanks in advance,
Coolmate ;)
Post by: DrDusk on July 10, 2019, 08:17:44 pm
Hey Everyone,The most important thing is to make sure you have an understanding of the concepts. Do NOT rote learn the Math type questions for Physics, rather train your mind to think critically. Other than that practice your essay type questions, ask your teacher to mark them.
I am in Year 11 and am going to do my Physics prelim exams soon. I was wondering whether anyone has any good tips and what the most important things I should study that I can use for the exam, to achieve a high mark?
Also, does anyone have any past prelim papers for Physics?... (The new syllabus is making it hard to find them).
Thanks in advance,
Coolmate ;)
Good luck
Post by: Coolmate on July 10, 2019, 08:40:21 pm
The most important thing is to make sure you have an understanding of the concepts. Do NOT rote learn the Math type questions for Physics, rather train your mind to think critically. Other than that practice your essay type questions, ask your teacher to mark them.
Good luck!
Thankyou so much for your response and the help, I will do this! :D
Post by: classof2019 on July 24, 2019, 07:01:25 pm
Light with a frequency above the threshold frequency is incident on a metal surface. If light of the same intensity but double the frequency was used, how would the photocurrent (number of electrons emitted per second) and the kinetic energy of photoelectrons be affected?
Answer: The photoelectrons would have a higher kinetic energy and photocurrent would decrease.
I'm a bit confused because doesn't intensity, by definition, refer to the number of photons incident on a given area per second? And since intensity is the same, by the All or Nothing principle, shouldn't this mean the same number of electrons are liberated per second?
Any help would be greatly appreciated. Cheers!
Post by: DrDusk on July 24, 2019, 07:32:07 pm
Because it should increase and not decrease..
You are right in saying that the same amount of electrons will be ejected but the Photo current will still increase. The reason for this is we have I=Q/t. Suppose in a little thought experiment we take a stopwatch and measure a time T, and measure that a total of Q' electrons flow in time T. Let the frequency of the light be f and the frequency of the metal be f' Now I'll perform some derivations:
Now the remaining energy Ef becomes the Kinetic energy of the electron so we have:
We can assume 'd' is constant. Now if you look at the equation, INCREASING f increases the velocity factor, which in turn increases the measured current I
Now for a time interval T we defined above, we can use the formula T = distance/speed which gives us:
Clearly you can see increasing v increases the Photo current. In your mind you can picture it as an olympic running race. Suppose you take a time interval of 10 seconds for a 100 meter race. The slower the runners are, the less of them will pass through the finish line in 10 seconds. However the faster they are, more and more will start to pass through the finish line.
Just like that running race, more electrons will pass a given point in a time 't'. This increase the factor Q/t, which increases the current I = Q/t
Also just an interesting thing you can note from that equation. If the incoming frequency of the light " f " equals the frequency of the metal " f' ", then we have Current = 0. This is expected, Why?, well because when you shine a light with a frequency equal to the min frequency, the electrons exit with ZERO VELOCITY. Current is only formed when Electrons are MOVING.
Post by: classof2019 on July 24, 2019, 09:05:41 pm
Sorry, did you mean the photoelectric current would increase?
Because it should increase and not decrease..
You are right in saying that the same amount of electrons will be ejected but the Photo current will still increase. The reason for this is we have I=Q/t. Suppose in a little thought experiment we take a stopwatch and measure a time T, and measure that a total of Q' electrons flow in time T. Let the frequency of the light be f and the frequency of the metal be f' Now I'll perform some derivations:
Now the remaining energy Ef becomes the Kinetic energy of the electron so we have:
We can assume 'd' is constant. Now if you look at the equation, INCREASING f increases the velocity factor, which in turn increases the measured current I
Now for a time interval T we defined above, we can use the formula T = distance/speed which gives us:
Clearly you can see increasing v increases the Photo current. In your mind you can picture it as an olympic running race. Suppose you take a time interval of 10 seconds for a 100 meter race. The slower the runners are, the less of them will pass through the finish line in 10 seconds. However the faster they are, more and more will start to pass through the finish line.
Just like that running race, more electrons will pass a given point in a time 't'. This increase the factor Q/t, which increases the current I = Q/t
Also just an interesting thing you can note from that equation. If the incoming frequency of the light " f " equals the frequency of the metal " f' ", then we have Current = 0. This is expected, Why?, well because when you shine a light with a frequency equal to the min frequency, the electrons exit with ZERO VELOCITY. Current is only formed when Electrons are MOVING.
This makes complete sense to me, however the back-of-the-book answer does say that photocurrent decreases. Here is their justification:
"If the frequency was increased, each photon would have more energy. As the intensity remains constant, less photons will be incident on the surface per second and hence photocurrent will be reduced. Because energy is conserved and because it takes a specific amount of energy to remove an electron, the increased photon energy will cause the kinetic energy of the ejected electrons to increase".
What do you think?
Post by: DrDusk on July 24, 2019, 09:40:22 pm
" As the intensity remains constant, less photons will be incident on the surface per second and hence photocurrent will be reduced"See this is not wrong, it IS a valid statement, and I did think of this as well.
I did a bit of research and half the sources say that current should increase and the other half say it should decrease.
It appears there is no 'straight' answer to this question which is really weird. See I believe both arguments are technically correct, it just depends on which one outweighs the other in terms of its affect on the photo-current.
Still this is indeed really interesting. I'll consult my Professor for Quantum Physics and get back to you on this...
Post by: classof2019 on July 25, 2019, 09:35:37 pm
See this is not wrong, it IS a valid statement, and I did think of this as well.
I did a bit of research and half the sources say that current should increase and the other half say it should decrease.
It appears there is no 'straight' answer to this question which is really weird. See I believe both arguments are technically correct, it just depends on which one outweighs the other in terms of its affect on the photo-current.Maybe its just because I'm currently on 2 hours sleep, so my brain is just fatigued
Still this is indeed really interesting. I'll consult my Professor for Quantum Physics and get back to you on this...
Thank you for looking into it.
Just an idea ... could it perhaps have something to do with differing definitions of intensity? I.e. they seem to be referring to intensity as the amount of 'energy' passing through a given area per second (power per metre squared - consider the units of intensity, Wm-2), whereas I, and other textbooks, have interpreted intensity to be the number of photons per second?
Post by: DrDusk on July 25, 2019, 11:04:20 pm
Thank you for looking into it.
Just an idea ... could it perhaps have something to do with differing definitions of intensity? I.e. they seem to be referring to intensity as the amount of 'energy' passing through a given area per second (power per metre squared - consider the units of intensity, Wm-2), whereas I, and other textbooks, have interpreted intensity to be the number of photons per second?
My mind is actually functioning properly now that I'm fully awake and not sleep deprived.
Firstly that answer is most definitely wrong. Your are right, its due to their definition of intensity. I'll explain to you what their answer means and why its wrong. Even though its wrong, developing an understanding is crucial.
Okay so this is the justification of their answer:
We have E = hf for each photon. If the light contains 'n' photons, it must carry energy E = nhf. Now this is where they make their crucial mistake. There are two types of Particle physics, Classical and Quantum. In classical Physics the intensity of a wave as you said is defined by W/m^2. Using this definition we can clearly see Intensity is proportional to the Energy of a wave which is E = nhf. So if Intensity stays constant, E must stay constant. However h is a constant, so the only way to keep E constant with an increasing f is by decreasing n.
So now we have that the number of photons decrease. I hope you can see why this spells trouble, because in Quantum Physics decreasing the number of Photons decreases the intensity of the wave, but lets carry on. So photons decrease, meaning the rate of arrival for the photons will decrease. Which means the photo-current will decrease.
However as I mentioned above this is problematic.
The real definition of intensity in this case is proportional to the amount of photons, so 'n' must stay constant in the equation E = nhf. This means the energy of the light wave and subsequently each photon increases. This means each photon carries greater kinetic energy as I mentioned. However now that I'm not half asleep there is something more important to consider.
Greater kinetic energy per photon means each ejected electron will travel a greater distance before the next one is ejected. This effect should more or less cancel out the fact that they are travelling faster. This will lead to basically a constant current, or one that increases just slightly... It wont increase by a lot as I said in my first post.
Post by: not a mystery mark on July 29, 2019, 08:51:04 pm
Pls help. My trial is tomorrow and the end is neigh.
Post by: DrDusk on July 29, 2019, 09:20:45 pm
Hey squad! I have no idea what's happening - I can only figure this out to be anticlockwise, but the answer is clockwise??The loop must generate a current that opposes the change in flux. This is a very confusing statement so I'll make it easier for you.
Pls help. My trial is tomorrow and the end is neigh.
Basically it means the loop must create a current that fights to return the system back to its original setting. In the original setting the Magnet is stationary at some distance below the loop. The current in this loop will aim to get the magnet back into this position. The only way of doing this is to generate a magnetic field that doesn't attract the one of the magnet, i.e. it must repel the magnet. For this to happen a north pole must be generated by the loop at the bottom.
Now using the right hand coil rule, your thumb points downwards and your fingers will curl in the direction of the current, which will be clockwise.
Good luck for Physics tomorrow. Remember to think critically and outside the box, and if your stuck on a question stay calm and brainstorm any Laws that you can apply. The most common ones will be Conservation of Energy & Momentum.
Post by: not a mystery mark on July 29, 2019, 09:25:05 pm
The loop must generate a current that opposes the change in flux. This is a very confusing statement so I'll make it easier for you.
Basically it means the loop must create a current that fights to return the system back to its original setting. In the original setting the Magnet is stationary at some distance below the loop. The current in this loop will aim to get the magnet back into this position. The only way of doing this is to generate a magnetic field that doesn't attract the one of the magnet, i.e. it must repel the magnet. For this to happen a north pole must be generated by the loop at the bottom.
Now using the right hand coil rule, your thumb points downwards and your fingers will curl in the direction of the current, which will be clockwise.
Good luck for Physics tomorrow. Remember to think critically and outside the box, and if your stuck on a question stay calm and brainstorm any Laws that you can apply. The most common ones will be Conservation of Energy & Momentum.
OH MY GOD SORRY, THE ANSWER SAYS ANTICLOCKWISE. THIS WAS MY SAME TRAIN OF THOUGHT.
I'm guessing the answer on the book is wrong.
Also thank you heaps!!! I will try my best!
Post by: DrDusk on July 29, 2019, 09:28:12 pm
OH MY GOD SORRY, THE ANSWER SAYS ANTICLOCKWISE. THIS WAS MY SAME TRAIN OF THOUGHT.Books are always infamous for getting wrong answers its not even funny.
I'm guessing the answer on the book is wrong.
Also thank you heaps!!! I will try my best!
Remember though, as it begins to leave the loop, the current will flow Anti-clockwise
Post by: not a mystery mark on July 29, 2019, 09:33:22 pm
Books are always infamous for getting wrong answers its not even funny.I'm becoming sadly familiar with that by the day ahah.
Remember though, as it begins to leave the loop, the current will flow Anti-clockwise
Also yess, because of the system will induce a downwards South Pole to oppose it's motion away from the loop. #induction4dayzzz
Post by: DrDusk on July 30, 2019, 10:27:31 pm
Easy, hard, regardless how did you go? Any questions that required you to think outside the box or were most of them essay based?
I have still yet to see a real paper from the new syllabus.
Post by: not a mystery mark on July 30, 2019, 10:36:54 pm
How was the trial @Mystery mark?
Easy, hard, regardless how did you go? Any questions that required you to think outside the box or were most of them essay based?
I have still yet to see a real paper from the new syllabus.
Pretty well actually - felt a lot of it was somewhat formulaic though. I was a bit panicky during the beginning (as one does during the first few minutes) but all the information started surfacing when it needed too which was good. Overall I'm hoping for an 85 maybe?
The questions didn't look too essay focused compared to what I saw in past-old-syllabus HSC exams, although there is an added emphasis on reliability, validity, and accuracy of data. One question asked to refer to my depth study and analyse its reliability - wasn't too bad. Also a question like "How did the discovery of the Motor Effect impact society and the environment" which felt a bit strange haha
Currently working for the Extension 2 Math exam tomorrow and I am screaming ahaha. It's the final exam and after that I'm going to have a week's sleep. I'll be happy to send the paper whenever we get it back if you're interested. It was a nice test.
Post by: DrDusk on July 30, 2019, 10:53:34 pm
That's good to hear that you went well. Well I hope you achieve what you want!.
This is what I expected. NESA added 50 more formulas but still everything is just subbing into formulas, which is like what a 4u student says when they see a 2u paper. Sadly this wont ever change unless they add calculus..
Anyway best of luck for the rest of your trials. Now go on get back to studying.
Oh and IF you can I would love to see the paper. I'm quite interested in how it would look for the peeps this year.
Post by: not a mystery mark on July 30, 2019, 10:58:42 pm
Maths Extension 2 always had me sweating every exam, can't give any advice on that when I would freak out myself haha.
That's good to hear that you went well. Well I hope you achieve what you want!.
This is what I expected. NESA added 50 more formulas but still everything is just subbing into formulas, which is like what a 4u student says when they see a 2u paper. Sadly this wont ever change unless they add calculus..
Anyway best of luck for the rest of your trials. Now go on get back to studying.
[/quote
It hurts me that we can't use calculus in physics ahaha. Bro, I'm too scared to even use the quadratic formula in physics. Anyways, I shall get some rest for this exam :'') Goodnight man - will keep you updated with any cool physics questions or something else related.
Cheers again!!
Post by: Coolmate on August 01, 2019, 12:15:38 pm
Would someone please be able to step me through this question based on Snell's Law (file attached), I am a bit confused about Snell's Law and how to go about this question, thanks.
Cheers Coolmate 8)
Post by: DrDusk on August 01, 2019, 03:58:29 pm
a)
For this recall the formula that for light:
b)
The angle of incidence is taken as the angle the light ray makes with the normal. Now if its parallel to the normal, what would this angle be?
c)
d)
In terms of interpretation. Remember before I said that if it's parallel to the normal the angle is zero? Apply the same thing here. It basically means the light ray does NOT refract, because before it was travelling at 0 degrees, and it's still at 0 degrees with respect to the normal.
Post by: Coolmate on August 01, 2019, 10:24:31 pm
4)
a)
For this recall the formula that for light:
b)
The angle of incidence is taken as the angle the light ray makes with the normal. Now if its parallel to the normal, what would this angle be?
c)
d)
In terms of interpretation. Remember before I said that if it's parallel to the normal the angle is zero? Apply the same thing here. It basically means the light ray does NOT refract, because before it was travelling at 0 degrees, and it's still at 0 degrees with respect to the normal.
Hey DrDusk!
Thanks so much for your help! I understand it now and can do it. I was just wondering though whether the answer for 4a) would be a long number?; because I got an answer of, 2x108, is this correct?
Thanks again
Coolmate :D
Post by: redpanda83 on August 01, 2019, 10:28:50 pm
Hey DrDusk!looks correct
Thanks so much for your help! I understand it now and can do it. I was just wondering though whether the answer for 4a) would be a long number?; because I got an answer of, 2x108, is this correct?
Thanks again
Coolmate :D
Post by: Coolmate on August 01, 2019, 10:30:31 pm
looks correct
Cheers :)
Post by: mani.s_ on August 02, 2019, 11:06:48 pm
Post by: blyatman on August 03, 2019, 12:34:52 pm
Hi, can someone explain the concept of Entropy to me??? It's very abstract and confusing.
Google/wikipedia is your friend, and should be your first go to if you haven't already.
Entropy is typically interpreted as a measure of disorder (I'm not familiar with the new physics syllabus, but I would imagine that's probably all you'd need to know). There's a formula for it on wiki, but I doubt you'd ever need to use it unless you're taking a course on molecular gas dynamics or something similar. Basically, the second law of thermo states that entropy can never decrease, which limits the reactions and procceses that can occur.
Post by: classof2019 on August 09, 2019, 05:13:02 pm
I understand that kinetic energy increases and potential energy decreases, but after doing a practice question the solution suggested that total energy also decreases - why is this the case?
Thank you.
Post by: DrDusk on August 09, 2019, 05:39:48 pm
In calculation questions though you don't need to consider the effect of the retrograde rockets.
Post by: classof2019 on August 09, 2019, 05:44:41 pm
In theory yeah KE increases and Potential energy decreases, leading to energy conservation. However in reality the satellite must fire retrograde rockets(rockets in the opposite direction) which decrease the KE at the higher level orbit. So technically some KE is lost.
In calculation questions though you don't need to consider the effect of the retrograde rockets.
Thanks!
Just a q, are the mechanics behind orbital transfer a part of the new syllabus? Do we need to know how rockets transition between orbits or must we only perform calculations? Cause I've found no mention of it in numerous textbooks, yet the Excel book goes into it in quite some detail.
Post by: DrDusk on August 09, 2019, 05:53:46 pm
Post by: DrDusk on August 09, 2019, 06:11:06 pm
Can someone please explain what happens to the total energy of a satellite when it moves from a higher to lower orbit?Also another thing to note is that the value of 'g' is not constant as the satellite will usually undergo a large change in orbital radius. So you technically can't say that the Energy of the satellite is conserved.
I understand that kinetic energy increases and potential energy decreases, but after doing a practice question the solution suggested that total energy also decreases - why is this the case?
Thank you.
Post by: classof2019 on August 11, 2019, 04:42:56 pm
"When the Apollo astronauts switched off their rocket engine their initial kinetic energy carried them to the Moon. How did the apparent weight of the astronauts change as they travelled to the Moon?
(A) Their apparent weight remained zero until they switched on the rocket engines near the Moon.
(B) Their apparent weight decreased as they moved further from the Earth and then increased as they approached the Moon
(C) Their apparent weight increased as they moved further from the Earth and then decreased as they approached the Moon
(D) Their apparent weight decreased throughout the flight."
I'm a bit confused with the concept of apparent weight altogether.
Any help is greatly appreciated
Post by: blyatman on August 11, 2019, 05:11:08 pm
Can someone please help with this q?
"When the Apollo astronauts switched off their rocket engine their initial kinetic energy carried them to the Moon. How did the apparent weight of the astronauts change as they travelled to the Moon?
(A) Their apparent weight remained zero until they switched on the rocket engines near the Moon.
(B) Their apparent weight decreased as they moved further from the Earth and then increased as they approached the Moon
(C) Their apparent weight increased as they moved further from the Earth and then decreased as they approached the Moon
(D) Their apparent weight decreased throughout the flight."
I'm a bit confused with the concept of apparent weight altogether.
Any help is greatly appreciated
If you read the short wiki article on apparent weight, you should be able to answer this q.
The apparent weight is equal to the normal reaction force, and is more or less the force that one would "feel".
Post by: DrDusk on August 11, 2019, 06:17:09 pm
If you read the short wiki article on apparent weight, you should be able to answer this q.This.
The apparent weight is equal to the normal reaction force, and is more or less the force that one would "feel".
Apparent weight as the name suggests is the weight you 'feel'. The reason you feel your weight on the Earth is because relative to you the Earth is stationary and you are being accelerated downward by gravity onto the surface of the Earth.
However if you are in a rocket ship when no engines are being fired, the floor of the rocket will have an acceleration downward and so will you! So relative to the rocket ship you have zero acceleration, leading to you not feeling your weight, as compared to on Earth where your being accelerated against the floor your standing on by gravity.
Now try and answer that question.
Post by: not a mystery mark on August 13, 2019, 11:55:42 am
Note:
Column One is the experiment, Column Two has the Observation, Column Three is the conclusions drawn from each observation.
Post by: DrDusk on August 13, 2019, 04:53:22 pm
Hey, would somebody be able to explain my teachers answer to this question? Majorly confused. Why wouldn't the cathode rays be deflected by an electric field if they're electrons? Thanks!
Note:
Column One is the experiment, Column Two has the Observation, Column Three is the conclusions drawn from each observation.
Lol of course they would be deflected by an Electric field if they were charged particles.
The thing is Scientists back in the day were really bad at detecting deflections and so they were unable to see that it deflected due to the electric plates. Which led to the conclusion that the Cathode ray is a wave and not a particle for that experiment. This conclusion was later proven to be incorrect and deflections were observed.
Post by: not a mystery mark on August 13, 2019, 07:29:07 pm
Lol of course they would be deflected by an Electric field if they were charged particles.
The thing is Scientists back in the day were really bad at detecting deflections and so they were unable to see that it deflected due to the electric plates. Which led to the conclusion that the Cathode ray is a wave and not a particle for that experiment. This conclusion was later proven to be incorrect and deflections were observed.
Ah okay cool. This literally had my head hurting all day. Every time I would google it kept saying that JJ Thompson saw deflection. Hertz had to come fool everyone with his errors ahaha.
Thanks heaps, and also cool profile pic 8)
Post by: DrDusk on August 14, 2019, 03:51:33 am
Thanks heaps, and also cool profile pic 8)
Haha thanks, I'm making them for a Poster/Youtube channel where I can make videos for HSC/Prelim Physics, so I figured why not have it as a pic lol
Post by: Coolmate on August 15, 2019, 07:14:03 pm
I was just wondering if anyone would happen to have any example questions that I could complete for the Module 4, syllabus topic on Electricity and Magnetism dot point (I have attached the dot point) for practice? Also, if anyone has any good short answer/ extended written response questions that I could practice also? :)
Thanks in advance,
Coolmate 8)
Post by: DrDusk on August 15, 2019, 08:12:45 pm
Hi Everyone!
I was just wondering if anyone would happen to have any example questions that I could complete for the Module 4, syllabus topic on Electricity and Magnetism dot point (I have attached the dot point) for practice? Also, if anyone has any good short answer/ extended written response questions that I could practice also? :)
Thanks in advance,
Coolmate 8)
I don't see any attachment?
Post by: Coolmate on August 15, 2019, 08:21:05 pm
Post by: DrDusk on August 15, 2019, 10:10:15 pm
Oh! Sorry...... it's attached to this post ::)
Have you tried the sample paper posted by Blasonduo. If you scroll down you should find it. IIRC it has questions relating to this...
Post by: Coolmate on August 16, 2019, 03:37:13 pm
Thankyou, I just found it and the questions were great and very useful!
Cheers,
Coolmate 8)
Post by: louisaaa01 on September 03, 2019, 09:29:39 am
"Which fundamental quantity required that its unit of measurement be redefined following acceptance of the theory of special relativity?
(A) Luminous intensity
(B) Length
(C) Mass
(D) Time "
Any help would be very much appreciated!
Post by: blyatman on September 03, 2019, 09:53:38 am
Just came across this question and am very confused.
"Which fundamental quantity required that its unit of measurement be redefined following acceptance of the theory of special relativity?
(A) Luminous intensity
(B) Length
(C) Mass
(D) Time "
Any help would be very much appreciated!
I believe the answer is B, though I don't think any of them are right tbh. Luminosity intensity is clearly not related. The other variables (mass, length, and time) do not change within an inertial frame of reference: a meter is still a meter, a second is still a second, and a kilogram is still a kilogram.
Context: the previous definitions of a meter and kilogram were based off arbitrary objects stored in an arbitrary vacuum chamber in France. Every year, scientific institutes around the world would send their copies of a meter and kilogram to France to calibrate them against the official one in France. Cleaning the objects etc would often cause small layers to shed off, and thus the definition of a kilogram and meter changed every year. As a result, there needed to be a more robust way of defining the meter and the kilogram. I don't know the original definition of a second but you could look that up (it was also probably arbitrary).
1 second was redefined as the period it took for electrons to bounce a certain number of times between energy levels of the Caesium atom.
1 meter was redefined as a fraction of the distance that light travels in one second.
1 kg was only redefined up until very recently in the last year or two from what I recall (I'm not clear on the specifics).
The reason I believe the answer to your question is B is because the redefined unit of length is the only one whose definition directly depends on the speed of light. However, I don't know if it was necessarily REQUIRED its unit of measurement to be redefined. I guess one could argue that the meter ruler will contract according to observers in other inertial frames. However, in your frame it doesn't change and you could always use the meter ruler in your reference frame to measure lengths, so the old definition of the meter would still work in your inertial frame. Scientists just wanted a new definition of a meter that wasn't depended on some arbitrary object stored in a vacuum chamber in France. The constancy of the speed of light provided this solution, as it meant that the meter could now have its definition redefined in terms of universal constants.
Post by: Coolmate on September 03, 2019, 10:02:56 am
DISCLAIMER: I am in year 11 so I may not be correct! ;D
Yes, this is a confusing question, but I would guess that it is option "B"? ???
I chose this option because I think that length can be redefined depending on where you are standing and observing the event from. This is due to the Theory of Special Relativity, being said to have the length of an object moving at relativistic speeds can and will contract along the direction of motion. Whereas an observer not moving and just observing (relative to the moving object) will see the object to be shorter in length. :)
Therefore, length can be redefined due to the Theory of Special Relativity ;) P.S. I also feel as though this is the only unit which can actually change in regards to the Theory of Special Relativity!
The Physics Classroom Website Below explains further:
https://www.physicsclassroom.com/mmedia/specrel/lc.cfm
I hope this helps and please take my response with a grain of salt
Coolmate 8)
Post by: louisaaa01 on September 03, 2019, 05:03:47 pm
I believe the answer is B, though I don't think any of them are right tbh. Luminosity intensity is clearly not related. The other variables (mass, length, and time) do not change within an inertial frame of reference: a meter is still a meter, a second is still a second, and a kilogram is still a kilogram.
Context: the previous definitions of a meter and kilogram were based off arbitrary objects stored in an arbitrary vacuum chamber in France. Every year, scientific institutes around the world would send their copies of a meter and kilogram to France to calibrate them against the official one in France. Cleaning the objects etc would often cause small layers to shed off, and thus the definition of a kilogram and meter changed every year. As a result, there needed to be a more robust way of defining the meter and the kilogram. I don't know the original definition of a second but you could look that up (it was also probably arbitrary).
1 second was redefined as the period it took for electrons to bounce a certain number of times between energy levels of the Caesium atom.
1 meter was redefined as a fraction of the distance that light travels in one second.
1 kg was only redefined up until very recently in the last year or two from what I recall (I'm not clear on the specifics).
The reason I believe the answer to your question is B is because the redefined unit of length is the only one whose definition directly depends on the speed of light. However, I don't know if it was necessarily REQUIRED its unit of measurement to be redefined. I guess one could argue that the meter ruler will contract according to observers in other inertial frames. However, in your frame it doesn't change and you could always use the meter ruler in your reference frame to measure lengths, so the old definition of the meter would still work in your inertial frame. Scientists just wanted a new definition of a meter that wasn't depended on some arbitrary object stored in a vacuum chamber in France. The constancy of the speed of light provided this solution, as it meant that the meter could now have its definition redefined in terms of universal constants.
Hey lousiaaa01! :D
DISCLAIMER: I am in year 11 so I may not be correct! ;D
Yes, this is a confusing question, but I would guess that it is option "B"? ???
I chose this option because I think that length can be redefined depending on where you are standing and observing the event from. This is due to the Theory of Special Relativity, being said to have the length of an object moving at relativistic speeds can and will contract along the direction of motion. Whereas an observer not moving and just observing (relative to the moving object) will see the object to be shorter in length. :)
Therefore, length can be redefined due to the Theory of Special Relativity ;) P.S. I also feel as though this is the only unit which can actually change in regards to the Theory of Special Relativity!
The Physics Classroom Website Below explains further:
https://www.physicsclassroom.com/mmedia/specrel/lc.cfm
I hope this helps and please take my response with a grain of salt
Coolmate 8)
Hi blyatman and Coolmate!
Thank you so much for your help.
I completely agree with your logic, however I have just checked the 'official' answer and they suggest that it is C - mass.
I understand that in accordance with Einstein's first and second postulates, length, time and mass are all subject to relativistic effects - however do you have any idea as to why the answer could possibly be mass instead of length?
Post by: DrDusk on September 03, 2019, 05:41:31 pm
Hi blyatman and Coolmate!
Thank you so much for your help.
I completely agree with your logic, however I have just checked the 'official' answer and they suggest that it is C - mass.
I understand that in accordance with the second postulate, length, time and mass are all subject to relativistic effects - however do you have any idea as to why the answer could possibly be mass instead of length?
Nope, I also agree with it being B. Official answers are wrong. Even schools teach it as that Length is what was redefined initially, as said by Blyatman.
What paper is this question taken from?
Post by: blyatman on September 03, 2019, 07:49:03 pm
Hi blyatman and Coolmate!No it's definitely not mass, the answers are wrong.
Thank you so much for your help.
I completely agree with your logic, however I have just checked the 'official' answer and they suggest that it is C - mass.
I understand that in accordance with the second postulate, length, time and mass are all subject to relativistic effects - however do you have any idea as to why the answer could possibly be mass instead of length?
Mass dilation isn't even a thing, despite what the HSC says (but that's a topic for another day).
Post by: DrDusk on September 03, 2019, 07:54:21 pm
Mass dilation isn't even a thing, despite what the HSC says (but that's a topic for another day).Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?
Post by: blyatman on September 03, 2019, 08:20:33 pm
Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?Nah, your energy goes to infinity, but that should not be interpreted as your mass going to infinity. The HSC teaches \(E=mc^2\) where \(m=\gamma m_0\) is the dilated mass (and \(\gamma\) being the Lorentz factor), but rather it should be \(E = \gamma m_0 c^2\). In other words, rather than thinking of mass being scaled by \(\gamma\), it's more correct to think of the energy as being scaled by \(\gamma\). Likewise, momentum should be treated as \(p=\gamma m_0v\) rather than using \(p=mv\) with \(m\) being the dilated mass.
University courses in relativity often accentuate this difference to rectify what students learnt in the HSC. Recently, the physics teachers at Matrix had a long debate whether they should be teaching relativistic mass since they knew it was wrong, but ultimately decided to do so since it's what's examined in the HSC.
https://en.wikipedia.org/wiki/Mass_in_special_relativity
"The term relativistic mass tends not to be used in particle and nuclear physics and is often avoided by writers on special relativity, in favor of using the body's total energy." There's a section named "Controversy" at the bottom which goes into a bit more detail.
I was also like wtf when I first learnt this at uni. But even this doesn't take the cake as the most "wrong" thing I've learnt in HS physics.
Post by: DrDusk on September 03, 2019, 08:59:17 pm
Nah, your energy goes to infinity, but that should not be interpreted as your mass going to infinity. The HSC teaches \(E=mc^2\) where \(m=\gamma m_0\) is the dilated mass (and \(\gamma\) being the Lorentz factor), but rather it should be \(E = \gamma m_0 c^2\). In other words, rather than thinking of mass being scaled by \(\gamma\), it's more correct to think of the energy as being scaled by \(\gamma\). Likewise, momentum should be treated as \(p=\gamma m_0v\) rather than using \(p=mv\) with \(m\) being the dilated mass.
University courses in relativity often accentuate this difference to rectify what students learnt in the HSC. Recently, the physics teachers at Matrix had a long debate whether they should be teaching relativistic mass since they knew it was wrong, but ultimately decided to do so since it's what's examined in the HSC.
https://en.wikipedia.org/wiki/Mass_in_special_relativity
"The term relativistic mass tends not to be used in particle and nuclear physics and is often avoided by writers on special relativity, in favor of using the body's total energy." There's a section named "Controversy" at the bottom which goes into a bit more detail.
I was also like wtf when I first learnt this at uni. But even this doesn't take the cake as the most "wrong" thing I've learnt in HS physics.
OH YES that controversy section actually makes so much sense!
So this means essentially the formula m = gamma m_0 is actually not correct, but rather the correct definition is through the formula for momentum.
That is interesting, I'm definitely going to read up on this
Post by: louisaaa01 on September 04, 2019, 11:41:18 am
I'm having a bit of trouble resolving forces on banked tracks. Basically, I can't seem to reconcile what I've learnt in Maths Ext 2 and Physics.
In Physics, we learn to resolve the forces parallel and perpendicular to the inclined plane. The plane is inclined at an angle θ to the horizontal. You wind up with:
Normal force (perpendicular to plane) = mgcosθ
Force parallel to plane = mgsinθ
However, in Maths, we consider the vertical and horizontal components of N (normal force) which is perpendicular to the banked track.
Neglecting friction,
Ncosθ = mg
Nsinθ = mv2/r
Yet, this essentially implies that N = mg/cosθ
But in Physics we've been taught that N = mgcosθ
Which one is correct? How do we reconcile these two equations?
Any help is very much appreciated!!
Post by: ^^^111^^^ on September 04, 2019, 12:28:57 pm
Wait what??? Doesn't mass dilate as your speed becomes higher due to E = mc^2 + (pc)^2 ?Sorry I may be wrong, but shouldn't it be:
E2= (mc2)2 + (pc)2
Post by: fun_jirachi on September 04, 2019, 12:52:43 pm
Hi,
I'm having a bit of trouble resolving forces on banked tracks. Basically, I can't seem to reconcile what I've learnt in Maths Ext 2 and Physics.
In Physics, we learn to resolve the forces parallel and perpendicular to the inclined plane. The plane is inclined at an angle θ to the horizontal. You wind up with:
Normal force (perpendicular to plane) = mgcosθ
Force parallel to plane = mgsinθ
However, in Maths, we consider the vertical and horizontal components of N (normal force) which is perpendicular to the banked track.
Neglecting friction,
Ncosθ = mg
Nsinθ = mv2/r
Yet, this essentially implies that N = mg/cosθ
But in Physics we've been taught that N = mgcosθ
Which one is correct? How do we reconcile these two equations?
Any help is very much appreciated!!
Correct me if I'm wrong, but I think you can't resolve these two equations, because they both describe different cases of motion on a slope. In your 'Physics' example, the motion that is experienced is down the slope (as a result of the parallel force), while in your 'Extension 2' example, the motion that is experienced is circular motion (as a result of the net force). If we were considering Circular motion in Physics on a banked track (disregarding friction), the only force acting on the object would be the horizontal net force acting towards the centre of motion ie. exact same as X2. Similarly, if we were considering an object moving up or down the slope (disregarding friction) the only component we 'see' is the parallel force up and down the slope ie. the net force, ie. same as Physics.
Hope this helps :)
Post by: louisaaa01 on September 04, 2019, 03:30:26 pm
Correct me if I'm wrong, but I think you can't resolve these two equations, because they both describe different cases of motion on a slope. In your 'Physics' example, the motion that is experienced is down the slope (as a result of the parallel force), while in your 'Extension 2' example, the motion that is experienced is circular motion (as a result of the net force). If we were considering Circular motion in Physics on a banked track (disregarding friction), the only force acting on the object would be the horizontal net force acting towards the centre of motion ie. exact same as X2. Similarly, if we were considering an object moving up or down the slope (disregarding friction) the only component we 'see' is the parallel force up and down the slope ie. the net force, ie. same as Physics.
Hope this helps :)
This is really helpful, thank you so much :) Was just misled because the official answer to a Banked Track trial question used these inclined plane equations! Anyway, really appreciate your assistance.
Post by: classof2019 on September 14, 2019, 12:33:46 pm
Post by: DrDusk on September 14, 2019, 06:15:31 pm
Can the NESA Sample Answers in the Marking Guidelines for each past HSC paper be relied upon as an indicator of what would have achieved full marks? Because I've heard that apparently not all answers included in the sample answers would have achieved full marks in a given question.
Usually the marking guidelines give the best possible answer, but you are right. Sometimes it cannot be relied upon.
Post by: classof2019 on September 27, 2019, 05:10:37 pm
A power plant generates 150 MW of electricity for a town 20 km away. The transmission wires have a resistance of 0.01 W per km. The voltage drop between the plant and the town is 50 V.
(a) Calculate the power loss between the plant and the town.
(b) Calculate the voltage transmitted by the plant.
Cheers.
Post by: louisaaa01 on September 27, 2019, 05:19:41 pm
Post by: fun_jirachi on September 27, 2019, 05:44:26 pm
Can someone please help with this?
A power plant generates 150 MW of electricity for a town 20 km away. The transmission wires have a resistance of 0.01 W per km. The voltage drop between the plant and the town is 50 V.
(a) Calculate the power loss between the plant and the town.
(b) Calculate the voltage transmitted by the plant.
Cheers.
Hey there!
For part a), consider the fact that P=VI, and that losses are lost through resistive heating ie. P=I2R. This leads us to deduce that the power of the town is going to be (V-50)I, while the power at the plant will be VI. We are given R as equal to 0.01 ohms/km x 20km ie. R = 0.2.
Then, we can solve for I in the equation (V-50)I=VI-I2R. :)
For b), to find the voltage transmitted by the plant to the town is simply going to be subbing in your new value for I back into the equation in a) and solving for V. Note that the voltage that actually gets transmitted is going to be V-50, since some power is lost.
Hope this helps :)
Hey! Just a question, when calculating the amount of magnetic flux through a coil, do you need to multiply by the number of turns in the coil?
Correct me if I'm wrong, but I think you do! :)
Post by: classof2019 on September 27, 2019, 05:51:43 pm
Hey there!
For part a), consider the fact that P=VI, and that losses are lost through resistive heating ie. P=I2R. This leads us to deduce that the power of the town is going to be (V-50)I, while the power at the plant will be VI. We are given R as equal to 0.01 ohms/km x 20km ie. R = 0.2.
Then, we can solve for I in the equation (V-50)I=VI-I2R. :)
For b), to find the voltage transmitted by the plant to the town is simply going to be subbing in your new value for I back into the equation in a) and solving for V. Note that the voltage that actually gets transmitted is going to be V-50, since some power is lost.
Hope this helps :)
Hey, thanks for looking into it - though, follow up q, does this rely on the assumption that current is constant? And if so, how can we assume that current doesn't decrease?
Post by: Jackson.Sprigg on September 27, 2019, 06:11:05 pm
Correct me if I'm wrong, but I think you do! :)
I'm not 100% sure if you're right about that or not but I can't find anywhere actually talking about it so I'm just replying to find out the truth! My understanding is that flux is BA or the product of area and magnetic field strength. So if I were to consider a 2 dimensional coil with multiple turns the area would not actually be increasing? EMF does depend on the number of turns but now I'm just confused about flux.
Post by: Aryan.S on September 28, 2019, 07:56:40 pm
Post by: akjen on October 03, 2019, 04:38:33 pm
Post by: DrDusk on October 03, 2019, 04:46:07 pm
This might be something obvious but I don't understand why the answer to this is C. Please help thanks
The keyword/line is "the hydrogen separated". This tells you that the hydrogen atom itself must contain some separate particles which are charged and hence experienced a force due to an electric field. The other answers are a bit ridiculous and really have nothing to do with it.
Either way questions like these are quite poorly designed because really it's testing your English skills imo.
Post by: akjen on October 03, 2019, 05:25:32 pm
The answer is A.
Post by: mani.s_ on October 17, 2019, 10:54:21 pm
Post by: stewartw20 on October 18, 2019, 10:41:43 am
Guys please show me how to answer this, I'm absolutely clueless :-\
Post by: AngelWings on October 18, 2019, 01:33:23 pm
I NEED URGENT HELP NOW!!! PLEASE ANSWER ME ASAP! YOU MUST HELP ME!I believe you’ve already been told that everyone on AN is a volunteer, so not everyone will be available at all times. Please avoid using red bold font or people may feel less inclined to answer. In addition, the red bold font is even less likely to get you a faster response. (The quote below applies to the whole of AN and not just one board.)
Guys please show me how to answer this, I'm absolutely clueless :-\
It should be noted that forum etiquette requires users to respect each other and realise that all question-answerers on ATAR Notes are volunteers, as was stated in a very recent post above. Exam time means extra stress but by no means does being stressed give a person the right to demand what they need from others, especially if those others are answering queries out of the goodness of their hearts. I wanted to make it clear that this attitude will not be tolerated here and that, regardless of individual circumstances, everyone deserves respect. Bold red does not equate to respect.
Post by: stewartw20 on October 18, 2019, 01:51:04 pm
Post by: tuna on October 23, 2019, 08:03:10 pm
Could someone please help me, I'm having difficulty understanding the attached projectile motion question.
The suggested answer uses s=ut+1/2at^2 , using s=150m, t=21s, a=1.6ms^2. The given answer is 9.7ms^1.
Shouldn't acceleration be negative as the force is applied downward?
Post by: fun_jirachi on October 23, 2019, 10:08:36 pm
You're correct in that the acceleration should be downward ie. a negative value. This causes the change in position in the vertical direction ie. the vertical velocity to increase towards a downwards direction. What we surmise from this is that given the astronaut projects the stone from the origin, is that while the change in vertical position is 150m, it actually finishes 150m below the origin ie. s=-150. Substituting in the rest of the values given in the question will give you that initial vertical velocity was 9.657m/s upwards. :)
Hope this helps!
Post by: stewartw20 on October 24, 2019, 01:20:19 pm
Can you please answer my question?
Post by: louisaaa01 on October 28, 2019, 03:29:01 pm
Thank you.
Post by: mani.s_ on October 28, 2019, 09:46:18 pm
Thanks :)
Post by: fun_jirachi on October 28, 2019, 10:03:48 pm
-snip-
I believe you’ve already been told that everyone on AN is a volunteer, so not everyone will be available at all times. Please avoid using red bold font or people may feel less inclined to answer. In addition, the red bold font is even less likely to get you a faster response. (The quote below applies to the whole of AN and not just one board.)
This was a fair comment from AngelWings; I was really not inclined to answer because it felt like more of a pushy demand as opposed to a legitimate question. However, I will go ahead and answer this anyway; but be wary that I or other people may not answer your other questions in the future if this continues.
Basically we have a 'paradox' between the observations of A and B. As B is an outside observer watching the barn doors close simultaneously and watching the ladder move at a relativistic speed relative to the barn, B will see the ladder contract to a length of about 8.72m (using the length contraction formula - the question asks for calcs), thus seeing the ladder fit completely inside the barn with the doors shut the instant the button is pressed. The paradox arises when we then consider the observations of A; instead, the barn moves towards the ladder A is carrying at 0.9c, and shrinks to a size of about 6.54m (again using the length contraction formula) as observed by A. Clearly, the barn is far too small to even fit the ladder, and thus a paradox arises at the instant the doors close; how can we possibly have the ladder simultaneously fitting in the barn and becoming a part of the doors in the same instant of time?
However, the answer to the question is in fact that the ladder will never touch the doors; because events observed to be simultaneous by A will not be simultaneous for B. B will observe the ladder entering the barn, then being enclosed for an instant, then exiting, while A will observe the ladder entering, the front door shutting and opening, then during that instant, A will have already moved a fraction and will then observe the back door opening and shutting, and then the ladder will exit.
Hopefully this makes sense :)
Can someone please help out with this multiple choice question?
Thank you.
Basically we have the formulae \(F=\frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}\) (for the force between two point charges) and similarly \(E=\frac{1}{4\pi \epsilon_0} \frac{q_1}{r^2}\) (for the magnitude of the electric field produced by a point charge). The first one is on the reference sheet, while the second one can be derived reasonably easily.
I'm honestly not sure why the answers are like that; from what I've calculated (just subbing in values into the first formula), none of them match. Also, the units are wrong, as they're asking for a force, not a field strength, which is odd. I'd think this question is a dud actually :)
Hope this helps :)
Post by: harry.braithwaite on October 28, 2019, 10:08:31 pm
I would suggest changing radius, which is probably the easiest way to go about doing the experiment. For a specific radius (can then change radius to use as independent variable) I would recommend hollowing a pen and passing a string through, which gives you a nice handle to hold. Start spinning above your head (making the plane of revolution horizontal) at such a speed that you keep that marked length at the tip of the pen and get someone to film you as you keep that orbital radius constant (If you fail to keep it constant for a run you can just discard that value), the experiment can be made very accurate if you keep it at the one radius for 10-20 revolutions, so that when looking back through the footage you can find the average orbital period for each revolution (which should be the same anyway if your radius is kept constant).
If you are keeping your radius constant, then the mass on the string isn't moving up and down and isn't accelerating which enables you to equate centripetal (mv^2/r) and weight (mg) and them from their determine your relationship between force and orbital radius.
As for validity, the only thing that isn't valid about this method would be the friction of the tip of the pen hole and the string and if you are spinning ensuring that you keep it flat. Also just with depth studies, it isn't necessarily about making the most accurate or most valid experiment anyway, as long as you can identify the sources of error and their impact. Just another thing, never try and make your depth study too complex, much better to do something simple very well than fail to comment on everything for example I wrote 4500 words on launching a marble at different angles, and someone in the previous year got full marks for dropping masses at 1m, 2m and 3m heights which goes to show you can tease words out of anything so do something simple so that you can really get across the actual physics.
Post by: louisaaa01 on October 30, 2019, 09:36:12 am
I was doing this multiple choice question:
An object's gravitational potential energy is tripled. Assuming all other variables are constant, this gain in energy is accompanied by a change in radius equivalent to:
a. Tripling the radius
b. Reducing the radius by a third.
c. Doubling the radius.
d. Halving the radius.
I was just a bit confused, for, ordinarily I would have said a (which is given as the correct answer) since there is an increase in gravitational potential energy with radius. However, given the equation U = -GMm/r, this suggests an inversely proportional relationship between the magnitude of U and radius (tripling U would actually make it more negative, thus decreasing it) - which suggests b. I was just wondering what the correct approach is?
Post by: theolast_ on October 30, 2019, 10:20:50 am
I was doing this multiple choice question:
An object's gravitational potential energy is tripled. Assuming all other variables are constant, this gain in energy is accompanied by a change in radius equivalent to:
a. Tripling the radius
b. Reducing the radius by a third.
c. Doubling the radius.
d. Halving the radius.
I was just a bit confused, for, ordinarily I would have said a (which is given as the correct answer) since there is an increase in gravitational potential energy with radius. However, given the equation U = -GMm/r, this suggests an inversely proportional relationship between the magnitude of U and radius (tripling U would actually make it more negative, thus decreasing it) - which suggests b. I was just wondering what the correct approach is?
Hi. If you consider the equation U=-Gmm/r, reducing r by one third doesn't actually work either because that takes r down to 2r/3. With 2r/3, the GPE only increases (in magnitude) by 1.5. So you can rule out option b.
Because it says that the GPE is tripled and that this was a gain in GPE, you can assume that they are using the equation for GPE in a uniform field, U=mgh. This implies U is in linear proportion to the radius, which makes sense for option A.
Overall, the question is quite vague.
Post by: david.wang28 on October 30, 2019, 03:31:16 pm
I ran into this question on advanced mechanics; my working out was very contrasting compared to the answers. I wanted to clarify which was the right way of doing the question; can anyone please help me out? Thanks :)
Post by: DrDusk on October 31, 2019, 05:13:19 pm
Hello,I would encourage you to do it the way in the book. The reason is sometimes they might ask a specific sig fig or decimal place, and for your answer your already substituting values in before the final answer which means your decimals or sig figs have the chance of being a bit off. This CAN lead to marks being taken off depending on how lenient they are. It's always best to be safe than sorry imo! =)
I ran into this question on advanced mechanics; my working out was very contrasting compared to the answers. I wanted to clarify which was the right way of doing the question; can anyone please help me out? Thanks :)
Post by: mani.s_ on November 01, 2019, 01:35:27 pm
Post by: louisaaa01 on November 01, 2019, 05:49:43 pm
Post by: DrDusk on November 01, 2019, 09:15:04 pm
Hi, I was just wondering, how do you define torque direction? For instance, if an object is moving anticlockwise, is torque just anticlockwise? Because I just read that the Right Hand Grip Rule is used to determine torque direction, but I had never come across this before. Thank you!You've answered your question yourself. Use Right Hand Grip Rule. =)
Post by: classof2019 on November 01, 2019, 09:15:08 pm
Cheers
Post by: DrDusk on November 01, 2019, 09:27:56 pm
How do you do this question? I have come across conflicting sources - some say that the motor force is the centripetal force (which I thought it was since it acts perpendicular to current / the magnetic field) while others say it isn't as motor force is actually parallel to velocity (and rather centripetal force is sum of magnetic force, drag, etc).The latter is correct. Motor force is not what results in the circular motion, but rather the fact that the magnetic field of the wire interacts with the one of the Magnet known as Magnetic force. It doesn't matter whether the motor force is perpendicular to the current/magnetic field. The definition of Centripetal force is that it must be parallel to velocity.
Cheers
I remember getting this question in an exam and I proved my teachers answer wrong lol
Post by: rorygolledge on November 02, 2019, 05:41:35 pm
Everywhere I look seems to give vague answers on what the actual experimental evidences were for both theories and I would like something clear if anyone is able to explain.
Post by: louisaaa01 on November 03, 2019, 04:29:13 pm
Need help with the syllabus dotpoint "analyse the experimental evidence that supported the models of light that were proposed by Newton and Huygens"
Everywhere I look seems to give vague answers on what the actual experimental evidences were for both theories and I would like something clear if anyone is able to explain.
Hi!
Foucault's rotating mirror experiment is a good starting point. Essentially, Newton's corpuscular model relied on the idea that light travelled faster in denser media, while Huygens' wave model revolved around light slowing down in denser media. Foucault's experiment proved the latter, thus providing support for Huygens' model (and invalidating Newton's model - which predicted the opposite).
You could also look at Young's Double Slit experiment - phenomena such as diffraction and interference could only be explained sufficiently by the wave model. Polarisation experiments / Malus' law could also only be satisfactorily explained by a wave model (though Huygens' model inevitably required some modification). Newton attempted to explain both phenomena, though his model proved insufficient.
Post by: louisaaa01 on November 04, 2019, 03:05:13 pm
I just had a question in relation to the attached sample q from NESA.
Why is the initial induced emf negative? I've applied the Right Hand Grip Rule and I've concluded that since the top of the solenoid must induce a South pole, current flows from the positive terminal of the voltage probe, down through the solenoid, to the negative terminal - why is this regarded a negative rather than positive voltage?
Thanks.
Post by: DrDusk on November 06, 2019, 07:35:53 pm
Hi,I suggest you look up conventional current and Non-Conventional current. When we use right hand grip rule, our fingers curl in the direction of the current flow. This would be true IF current was the flow of positive charge, however current is actually the flow of NEGATIVE charge. Hence you really need to flip the direction because Non-Conventional current is the real direction a current flows i.e. it's the flow of electrons NOT positive charge.
I just had a question in relation to the attached sample q from NESA.
Why is the initial induced emf negative? I've applied the Right Hand Grip Rule and I've concluded that since the top of the solenoid must induce a South pole, current flows from the positive terminal of the voltage probe, down through the solenoid, to the negative terminal - why is this regarded a negative rather than positive voltage?
Thanks.
Post by: 006896 on November 07, 2019, 09:38:48 pm
Could I please receive help for this multiple choice question?
Which fundamental quantity required that its unit of measurement be redefined following acceptance of the theory of special relativity?
A: luminous intensity
B: length
C: mass
D: time
I thought B, C and D were all redefined after special relativity, but the answer is C. Why?
Thanks
Post by: DrDusk on November 07, 2019, 10:06:31 pm
Hi,Oh no there was a massive discussion on this question already. Out of curiosity where are you getting it from? because the answer is incorrect. The answer IS (B), not C.
Could I please receive help for this multiple choice question?
Which fundamental quantity required that its unit of measurement be redefined following acceptance of the theory of special relativity?
A: luminous intensity
B: length
C: mass
D: time
I thought B, C and D were all redefined after special relativity, but the answer is C. Why?
Thanks
Also it cannot be (D), because the question asks that the "Unit of measurement was redefined". Our unit of time was NOT redefined, just how we think of it. We still use seconds and what not to define the unit of time. However the unit of length WAS redefined because now we measure a meter on the distance light travels in a certain period of time.
Post by: not a mystery mark on November 10, 2019, 01:50:25 pm
Post by: louisaaa01 on November 10, 2019, 04:22:30 pm
Hey guys: Thoughts on this emf question???
Hey not a mystery mark,
I've done this question a few times in my studies and I keep getting the same answer - one that isn't even an option. I'll show you my working/approach - let me know if your answer is similar!
Magnetic flux transitions from a maximum to a zero every 90o, i.e. every quarter of a rotation.
The coil completes 46.8 rotations/second - so 1 rotation is 1/46.8 s and thus 1/4 rotation occurs every 1/(4 x 46.8 ) = 5/936 s
Maximum flux = BA = 0.05 x 0.162 = 0.00128 Wb
So, there is a change in flux of 0.00128 Wb every 5/936 s
Using Faraday's Law, to find the induced emf, we multiply the rate of change of flux with the number of turns in the coil.
So induced emf = 48 V (we ignore the negative as this just accounts for Lenz's Law) - which isn't listed as an option.
I've had a look at the official answer (which I believe is A from memory) and I think the issue lies in the fact that, to get 12V, this relies on magnetic flux changing from a maximum to zero over one complete rotation - this actually isn't the case as the flux will be maximum when the coil is perpendicular to the field lines and minimum when it is parallel. In one complete rotation, the net change in magnetic flux is actually zero as the coil returns to the same position!
Post by: not a mystery mark on November 10, 2019, 04:39:36 pm
Hey not a mystery mark,
I've done this question a few times in my studies and I keep getting the same answer - one that isn't even an option. I'll show you my working/approach - let me know if your answer is similar!
I've had a look at the official answer (which I believe is A from memory) and I think the issue lies in the fact that, to get 12V, this relies on magnetic flux changing from a maximum to zero over one complete rotation - this actually isn't the case as the flux will be maximum when the coil is perpendicular to the field lines and minimum when it is parallel. In one complete rotation, the net change in magnetic flux is actually zero as the coil returns to the same position!
YESS!! It's exactly the same, I couldnt find a way to get 12V. Ahaha - bro problems with the new syllabus. It's already so confusing - stuff like this is drilling my brain harder.
Thanks so much!!
Post by: not a mystery mark on November 10, 2019, 06:38:39 pm
Can somebody explain this solution??
Normally I just use
And I'm pretty sure their pytharoas's theorem isn't for real. Eughhhhhhh
Pls help
Post by: louisaaa01 on November 10, 2019, 06:49:47 pm
Okay, I'm writing one last question for today.
Can somebody explain this solution??
Normally I just usebut this has something completely difference. I then assumed that it would be because of friction, but the working out doesn't use it at all? So now I'm just mega confused.
And I'm pretty sure their pytharoas's theorem isn't for real. Eughhhhhhh
Pls help
Hey not a mystery mark,
Funnily enough I had exactly the same problem with this question and turns out their solution is completely wrong! They actually issued an amendment to this solution (which still had errors). The main reason that they're wrong is that they combine equations for motion up and down an inclined plane (from Year 11) with circular motion around a banked track - which you can't do.
Keep in mind that the equation you gave for v only works when no friction is acting. In this case, we have both friction (down the slope) and normal force (perpendicular to the track and thus at θ to the vertical). It may help to draw a force diagram. We begin by resolving forces vertically and horizontally:
Vertically: Ncosθ - Fsinθ = mg
Horizontally: Nsinθ + Fcosθ = mv2/r
We have the values for F, θ, m and g (given in parts (a) and (c)) so we can substitute these into the vertical equation to find N.
Knowing N, we rearrange the horizontal equation and make v the subject. We can substitute all our known values to find v. It should end up being 17.62ms-1.
The mathematics involved actually resembles what we do in Maths Ext 2 - due to the error in the question, it ends up being harder than intended. I hope this clarifies things though!
Post by: not a mystery mark on November 10, 2019, 06:58:30 pm
Hey not a mystery mark,
Funnily enough I had exactly the same problem with this question and turns out their solution is completely wrong! They actually issued an amendment to this solution (which still had errors). The main reason that they're wrong is that they combine equations for motion up and down an inclined plane (from Year 11) with circular motion around a banked track - which you can't do.
Keep in mind that the equation you gave for v only works when no friction is acting. In this case, we have both friction (down the slope) and normal force (perpendicular to the track and thus at θ to the vertical). It may help to draw a force diagram. We begin by resolving forces vertically and horizontally:
Vertically: Ncosθ - Fsinθ = mg
Horizontally: Nsinθ + Fcosθ = mv2/r
We have the values for F, θ, m and g (given in parts (a) and (c)) so we can substitute these into the vertical equation to find N.
Knowing N, we rearrange the horizontal equation and make v the subject. We can substitute all our known values to find v. It should end up being 17.62ms-1.
The mathematics involved actually resembles what we do in Maths Ext 2 - due to the error in the question, it ends up being harder than intended. I hope this clarifies things though!
You are always there to save the day!!! TYSMM, you clarified that so well dang. Also yea the Ext 2 squad - one of the best subjects.
Thank you again for saving my hsc ahaha
Post by: mani.s_ on November 14, 2019, 02:29:41 pm
Thanks :)
Post by: DrDusk on November 14, 2019, 02:49:27 pm
Hi, I have a question, say I have a conical pendulum and the length of the string is kept constant. If I increase the radius of the pendulum, will period increase or decrease???Depends if your just increasing radius and decreasing density to keep mass constant or increasing radius and mass at the same time.
Thanks :)
Post by: mani.s_ on November 14, 2019, 03:12:08 pm
Depends if your just increasing radius and decreasing density to keep mass constant or increasing radius and mass at the same time.The mass also stays constants, so only increasing the radius.
Post by: DrDusk on November 14, 2019, 04:07:07 pm
The mass also stays constants, so only increasing the radius.What are your thoughts on the matter?
Post by: mani.s_ on November 15, 2019, 05:08:20 pm
What are your thoughts on the matter?I did an experiment with a conical pendulum where the string length and the mass of the pendulum were kept constant. I increased the radius in intervals of 10 cm up to 50cm. At each radius, the period seemed to decrease. It wasn't by much like around 0.2-0.3 seconds. But not sure if that's supposed to happen. Using the period I'm going to find the centripetal force and relate it to the radius.
Post by: DrDusk on November 15, 2019, 06:56:33 pm
I did an experiment with a conical pendulum where the string length and the mass of the pendulum were kept constant. I increased the radius in intervals of 10 cm up to 50cm. At each radius, the period seemed to decrease. It wasn't by much like around 0.2-0.3 seconds. But not sure if that's supposed to happen. Using the period I'm going to find the centripetal force and relate it to the radius.Sorry I meant have you done any calculations. Have you got any ideas on what kind of approach you should take to solving the problem mathematically?
Perhaps you could balance the forces on the mass or something and go from there. Have you tried that? =)
Post by: Coolmate on December 19, 2019, 12:42:46 pm
I have been given a physics depth study recently, and I am wondering what exactly differentiates a Band 5 depth study from a Band 6 depth study? ???
Is it more in depth research? or including formula's within your explanations?
Any advice is much appreciated! :D
Coolmate 8)
Post by: louisaaa01 on December 21, 2019, 07:43:20 pm
Hi Everyone, ;D
I have been given a physics depth study recently, and I am wondering what exactly differentiates a Band 5 depth study from a Band 6 depth study? ???
Is it more in depth research? or including formula's within your explanations?
Any advice is much appreciated! :D
Coolmate 8)
Hey Coolmate,
It really depends on the parameters of your depth study and the associated marking criteria but everything you've mentioned certainly differentiates a Band 5 from a Band 6. Here are a couple more pointers:
- If you need to assess the validity + accuracy + reliability of the primary/secondary investigation, you really need to know the difference between these terms, and not just acknowledge them under one 'umbrella'. Critically evaluate all aspects of your depth study.
- Clearly established inquiry question and a valid investigation (primary OR secondary) that addresses this question. Conclusion made that clearly reflects your findings.
- As you mentioned, include all appropriate Physics formulae. Perhaps derive such formulae if it is relevant to your depth study (e.g. if it is investigating circular motion)
- Your depth study should also be well-structured and clear.
- Research should indeed be in-depth, as you've pointed out, but try not to digress from the question at hand.
This list isn't exhaustive, but should point you in the right direction :) I'd highly recommend reading your marking criteria if you have access to it and pay attention to the modality used in each mark bracket (e.g. a Band 5 might be "thorough" but a Band 6 might be "extensive")
Hope I could be of some assistance!
Post by: shekhar.patel on January 04, 2020, 04:06:04 pm
I am getting answer as 0.9m. But it is wrong, I am not quite sure why.
Post by: DrDusk on January 05, 2020, 02:39:50 pm
A catapult launches a 2.0 kg rock at 3.2 m/s, 30° elevation, from a 7.0 m high rooftop determine the total horizontal displacement.Do you mind showing your working out? Usually especially for projectile motion it's just a small error such as a negative sign being somewhere it shouldn't.
I am getting answer as 0.9m. But it is wrong, I am not quite sure why.
Post by: mani.s_ on January 06, 2020, 06:21:29 pm
A catapult launches a 2.0 kg rock at 3.2 m/s, 30° elevation, from a 7.0 m high rooftop determine the total horizontal displacement.I'm getting 10.13m, is that correct?
I am getting answer as 0.9m. But it is wrong, I am not quite sure why.
Post by: blasonduo on January 06, 2020, 07:50:26 pm
I'm getting 10.13m, is that correct?
Hey! By doing this question, I get a value of 3.80 meters. So something has gone wrong with the answers and your working!
An easy, intuitive way to see if you're on the right track is to visualise it. 0.9m seems way too small for the height of the cliff (this would correlate with a time of flight of about 0.32 seconds!), but 10.13m seems too big for the speed the rock (time of flight for this would be around 3.66 seconds, dropping the rock from the height of 7 meters would be around 1.2 seconds, so adding another 2.4 seconds seems a bit too much for the speed of the rock)
If you'd want to post your working, I might be able to see where you went wrong!
Post by: Coolmate on January 07, 2020, 01:10:12 pm
Quote from: louisaaa01 link=topic=164552.msg1150132#msg1150132
Hey Coolmate,
It really depends on the parameters of your depth study and the associated marking criteria but everything you've mentioned certainly differentiates a Band 5 from a Band 6. Here are a couple more pointers:
- If you need to assess the validity + accuracy + reliability of the primary/secondary investigation, you really need to know the difference between these terms, and not just acknowledge them under one 'umbrella'. Critically evaluate all aspects of your depth study.
- Clearly established inquiry question and a valid investigation (primary OR secondary) that addresses this question. Conclusion made that clearly reflects your findings.
- As you mentioned, include all appropriate Physics formulae. Perhaps derive such formulae if it is relevant to your depth study (e.g. if it is investigating circular motion)
- Your depth study should also be well-structured and clear.
- Research should indeed be in-depth, as you've pointed out, but try not to digress from the question at hand.
This list isn't exhaustive, but should point you in the right direction :) I'd highly recommend reading your marking criteria if you have access to it and pay attention to the modality used in each mark bracket (e.g. a Band 5 might be "thorough" but a Band 6 might be "extensive")
Hope I could be of some assistance!
Thanks louisaaa01 for your in-depth response! It is really helping me as I am doing my depth study! ;D
Thanks again, Coolmate 8)
Post by: Coolmate on January 07, 2020, 07:14:59 pm
I am a bit confused about the questions below, that are in my depth study and what kind of specific information would be required for each of these questions. If anyone could please provide any helpful tips, that will be much appreciated:
1) Keeping Low Earth Satellites in Orbit
Use Physics principles to explain how Low Earth Satellites are able to travel in circular motion to orbit the Earth at a uniform speed and how and why these orbits are maintained.
2) Explaining weightlessness
For manned space satellites, discuss why astronauts live in conditions of ‘apparent’ weightlessness.
Your discussion should include common misconceptions about weightlessness.
Thanks in advance, ;D
Coolmate 8)
Post by: mani.s_ on January 07, 2020, 08:16:55 pm
Hey! By doing this question, I get a value of 3.80 meters. So something has gone wrong with the answers and your working!hmmm I just did the question again and I also got 3.7955m which rounds to 3.8m. I don't know how I did it incorrectly the first time. I think for the speed, I took 32m/s instead of 3.2 when calculating the time using S=ut + 1/2at^2. Thank you so much for spotting my error and correcting it :)
An easy, intuitive way to see if you're on the right track is to visualise it. 0.9m seems way too small for the height of the cliff (this would correlate with a time of flight of about 0.32 seconds!), but 10.13m seems too big for the speed the rock (time of flight for this would be around 3.66 seconds, dropping the rock from the height of 7 meters would be around 1.2 seconds, so adding another 2.4 seconds seems a bit too much for the speed of the rock)
If you'd want to post your working, I might be able to see where you went wrong!
Post by: blasonduo on January 07, 2020, 08:30:47 pm
Hi everyone! ;D
I am a bit confused about the questions below, that are in my depth study and what kind of specific information would be required for each of these questions. If anyone could please provide any helpful tips, that will be much appreciated:
1) Keeping Low Earth Satellites in Orbit
Use Physics principles to explain how Low Earth Satellites are able to travel in circular motion to orbit the Earth at a uniform speed and how and why these orbits are maintained.
2) Explaining weightlessness
For manned space satellites, discuss why astronauts live in conditions of ‘apparent’ weightlessness.
Your discussion should include common misconceptions about weightlessness.
Thanks in advance, ;D
Coolmate 8)
Hey! I'd hate to be that person, but I genuinely believe for questions like these, it's best for your learning if you give it a shot and try your best at explaining it. From there, I'll happily see if you need anything added.
Again, if I just told you what needs to be included, the chances of it sticking (the things you omitted of course) are much lower. I look forward to seeing what you come up with! :))
Post by: mani.s_ on January 08, 2020, 03:50:03 pm
Physics In Focus says to define it parallel to the field as shown in the first image that I have attached, but I'm not sure why.
Post by: DrDusk on January 08, 2020, 04:22:22 pm
Hi, for projectile motion in uniform electric fields, do we define theta perpendicular to the field or parallel to the field??? Which one should we use for theta when we want to break down the velocity vector into its components (parallel and perpendicular)???They're the exact same thing, like actually the exact same thing. Do it how you always do it.
Physics In Focus says to define it parallel to the field as shown in the first image that I have attached, but I'm not sure why.
Post by: shekhar.patel on January 14, 2020, 09:32:15 pm
Post by: sweetiepi on January 14, 2020, 09:47:29 pm
Hi, for the following question I am confused why the answers are as such. Any help would be great. Thanks.With current, we follow the right-hand grip rule (thumb in direction of current, fingers going out are the dots, and fingernails coming towards your thumb is the crosses going in).
With electron flow, this is the reverse. There are two ways to do this:
- left hand rule (exactly like the right-hand grip rule)
- or, the opposite directions as to the right hand rule suggests!
Hope this helps! :)
Post by: shekhar.patel on January 14, 2020, 09:56:32 pm
With current, we follow the right-hand grip rule (thumb in direction of current, fingers going out are the dots, and fingernails coming towards your thumb is the crosses going in).
With electron flow, this is the reverse. There are two ways to do this:
- left hand rule (exactly like the right-hand grip rule)
- or, the opposite directions as to the right hand rule suggests!
Hope this helps! :)
Thanks a lot. I understand it now.
Post by: shekhar.patel on January 15, 2020, 10:14:37 pm
Post by: blasonduo on January 16, 2020, 12:20:47 am
Hi, In this questions is I am confused about the angle that theta should be.
Hey!
The first thing you should note is that if we rotate this wire such that it is vertical, nothing here changed, the magnetic field still passes directly through the wire, meaning on this axis, the angle in which it is held does not matter, so the value for sine is 90. (because the angle that the rod and the magnetic field makes will always be 90, no matter which way we orientate our wire on this axis we spoke about earlier)
Hope this helps!
Post by: Coolmate on January 16, 2020, 09:42:12 am
Hey!
The first thing you should note is that if we rotate this wire such that it is vertical, nothing here changed, the magnetic field still passes directly through the wire, meaning on this axis, the angle in which it is held does not matter, so the value for sine is 90. (because the angle that the rod and the magnetic field makes will always be 90, no matter which way we orientate our wire on this axis we spoke about earlier)
Hope this helps!
Hi blasonduo, just wondering, is it 90 degrees for the value of theta for every question of this type?
Post by: blasonduo on January 16, 2020, 12:19:57 pm
Hi blasonduo, just wondering, is it 90 degrees for the value of theta for every question of this type?
Hey! Not necessarily. For example, if the magnetic field was pointing up, the axis of rotation would matter. When I went through these type of questions, they tended to ask a mixture of both questions where it either did or didn't matter. This is why it is extremely important to do the "rotation test". You are always looking for the angle between the current and magnetic field. If both of these are pointing up, the angle is 0, and so it would be sin(0) = 0.
Hope this helps! Keep asking away if it doesn't. :)
Post by: DrDusk on January 16, 2020, 03:39:44 pm
Hi, In this questions is I am confused about the angle that theta should be.
Hey! Not necessarily. For example, if the magnetic field was pointing up, the axis of rotation would matter. When I went through these type of questions, they tended to ask a mixture of both questions where it either did or didn't matter. This is why it is extremely important to do the "rotation test". You are always looking for the angle between the current and magnetic field. If both of these are pointing up, the angle is 0, and so it would be sin(0) = 0.Just to add onto this. In an exam if you ever feel confused as to what the angle should be get your pen and make it the magnetic field and make your ruler the wire. Arrange it as it shows in the diagram and observe the angle that you need.
Hope this helps! Keep asking away if it doesn't. :)
Post by: Coolmate on January 16, 2020, 03:53:39 pm
Hey! Not necessarily. For example, if the magnetic field was pointing up, the axis of rotation would matter. When I went through these type of questions, they tended to ask a mixture of both questions where it either did or didn't matter. This is why it is extremely important to do the "rotation test". You are always looking for the angle between the current and magnetic field. If both of these are pointing up, the angle is 0, and so it would be sin(0) = 0.
Hope this helps! Keep asking away if it doesn't. :)
Just to add onto this. In an exam if you ever feel confused as to what the angle should be get your pen and make it the magnetic field and make your ruler the wire. Arrange it as it shows in the diagram and observe the angle that you need.
Thanks blasonduo and DrDusk 8)
Post by: Coolmate on January 17, 2020, 06:45:03 pm
I am having trouble with trying to calculate the orbital velocity of a satellite in a Low Earth Orbit. I know I need to use the formula:
$$v=\sqrt{\frac{GM}{r}}$$
But, I know I need to use the constants 'G' and 'M', but am unsure about the value of 'r'.
Could someone please point me in the right direction? :)
Thanks in advance,
Coolmate 8)
Post by: Erutepa on January 17, 2020, 08:52:28 pm
Hi everyone, ;Dhere 'r' is the distance between the center of mass of the orbiting body and the body it is orbiting around.
I am having trouble with trying to calculate the orbital velocity of a satellite in a Low Earth Orbit. I know I need to use the formula:
$$v=\sqrt{\frac{GM}{r}}$$
But, I know I need to use the constants 'G' and 'M', but am unsure about the value of 'r'.
Could someone please point me in the right direction? :)
Thanks in advance,
Coolmate 8)
Post by: Coolmate on January 18, 2020, 04:22:14 pm
here 'r' is the distance between the center of mass of the orbiting body and the body it is orbiting around.
Hey Erutepa, thankyou for the help! ;D I have figured out the radius to be: Radius of Earth + Low Earth Orbit = the desired value
Thanks again!
Coolmate 8)
Post by: shekhar.patel on January 20, 2020, 08:23:18 pm
Post by: Coolmate on January 20, 2020, 08:38:54 pm
If you follow this link, there are some videos on some concepts. I think these are for the new syllabus? (Could someone please verify):
https://www.youtube.com/channel/UCZRlS4Zld_pZ_O1J_ctZfqA/search?query=physics
Also:
Jamon has divided his video series into ten main areas:
- Video 1: Studying for Physics
- Video 2: Long Response Questions
- Video 3: Mathematical Questions
- Video 4: Orbits
- Video 5: Motors
- Video 6: Transformers
- Video 7: Models of Light
- Video 8: The Atom
- Video 9: Exam Preparation
- Video 10: Exam Technique
I hope this helps!
Coolmate 8)
Post by: shekhar.patel on January 20, 2020, 08:44:23 pm
Hey shekhar.patel! ;D
If you follow this link, there are some videos on some concepts. I think these are for the new syllabus? (Could someone please verify):
https://www.youtube.com/channel/UCZRlS4Zld_pZ_O1J_ctZfqA/search?query=physics
Also:
Jamon has divided his video series into ten main areas:
- Video 1: Studying for Physics
- Video 2: Long Response Questions
- Video 3: Mathematical Questions
- Video 4: Orbits
- Video 5: Motors
- Video 6: Transformers
- Video 7: Models of Light
- Video 8: The Atom
- Video 9: Exam Preparation
- Video 10: Exam Technique
I hope this helps!
Coolmate 8)
Hi , Coolmate, thanks for the help, much appreciated
Post by: r1ckworthy on January 20, 2020, 09:08:31 pm
Hey shekhar.patel! ;DCan confirm that they are updated for the new syllabus ;D
If you follow this link, there are some videos on some concepts. I think these are for the new syllabus? (Could someone please verify):
https://www.youtube.com/channel/UCZRlS4Zld_pZ_O1J_ctZfqA/search?query=physics
Also:
Jamon has divided his video series into ten main areas:
- Video 1: Studying for Physics
- Video 2: Long Response Questions
- Video 3: Mathematical Questions
- Video 4: Orbits
- Video 5: Motors
- Video 6: Transformers
- Video 7: Models of Light
- Video 8: The Atom
- Video 9: Exam Preparation
- Video 10: Exam Technique
I hope this helps!
Coolmate 8)
Post by: Coolmate on January 20, 2020, 09:47:00 pm
Can confirm that they are updated for the new syllabus ;D
Awesome, thanks r1ckworthy :D
Post by: Coolmate on January 24, 2020, 03:28:07 pm
I have a question about the Van Allen Belts. Is a Low Earth Orbit within the Inner Van Allen Belt and if so how would the radiation affect the satellites within a Low Earth Orbit?💫
Thanks in advance,
Coolmate 8)
Post by: milie10 on February 06, 2020, 10:33:32 pm
Unsure about how to do this question- could someone please help?
(https://i.imgur.com/nIVwVkA.jpg)
Thanks so much!
Post by: jamonwindeyer on February 09, 2020, 12:53:22 am
Hi Everyone, ;D
I have a question about the Van Allen Belts. Is a Low Earth Orbit within the Inner Van Allen Belt and if so how would the radiation affect the satellites within a Low Earth Orbit?💫
Thanks in advance,
Coolmate 8)
Hey! Sorry that this answer is a little late, but the inner Van Allen belt appears to be at a 1000km altitude or above in most areas, which would traditionally be considered higher than a LEO. Wikipedia (lol) tells me that some places it can have concentrations at lower altitudes though so in some places maybe? This is outside syllabus scope though :)
Hi!
Unsure about how to do this question- could someone please help?
(https://i.imgur.com/nIVwVkA.jpg)
Thanks so much!
Your working is on the right track! We get the EMF by relating it to the change in flux, which you have spotted. Note the initial flux is zero (no magnetic field). So the change in flux is basically just whatever flux is from the current, divided by the time taken to switch that on (0.1s). The one missing piece for you I think is the formula which gives you a magnetic flux produced by a current in a solenoid. First, remember the difference between flux and flux density, \(\phi=BA\). That in mind, the formula you need is a Prelim formula:
So in summary the steps are:
- Figure out the strength of the magnetic field produced by the larger solenoid using the formula above (reference your Prelim notes!)
- Figure out how much flux this introduces in the smaller solenoid using \(\phi=BA\). Remember the \(A\) should be the area for the smaller solenoid because that is where the current is induced!
- Use this flux (also the change in flux, remember the initial flux is zero!) in \(\epsilon=\frac{\Delta \phi}{\Delta t}\) to determine the resultant induced emf.
Hopefully this is enough to let you have a go - It's a tough question (B6 level) so awesome job for wanting to explore it and understand it properly :)
Post by: Coolmate on February 09, 2020, 12:30:32 pm
Hey! Sorry that this answer is a little late, but the inner Van Allen belt appears to be at a 1000km altitude or above in most areas, which would traditionally be considered higher than a LEO. Wikipedia (lol) tells me that some places it can have concentrations at lower altitudes though so in some places maybe? This is outside syllabus scope though :)
Thanks Jamon for the reply! ;D
Also, I was just wondering if someone could please help me on how I should structure a response, in a Physics way, that asks me to "Discuss two things".
Thanks again!
Coolmate 8)
Post by: milie10 on February 10, 2020, 12:07:14 am
Your working is on the right track! We get the EMF by relating it to the change in flux, which you have spotted. Note the initial flux is zero (no magnetic field). So the change in flux is basically just whatever flux is from the current, divided by the time taken to switch that on (0.1s). The one missing piece for you I think is the formula which gives you a magnetic flux produced by a current in a solenoid. First, remember the difference between flux and flux density, \(\phi=BA\). That in mind, the formula you need is a Prelim formula:
So in summary the steps are:
- Figure out the strength of the magnetic field produced by the larger solenoid using the formula above (reference your Prelim notes!)
- Figure out how much flux this introduces in the smaller solenoid using \(\phi=BA\). Remember the \(A\) should be the area for the smaller solenoid because that is where the current is induced!
- Use this flux (also the change in flux, remember the initial flux is zero!) in \(\epsilon=\frac{\Delta \phi}{\Delta t}\) to determine the resultant induced emf.
Hopefully this is enough to let you have a go - It's a tough question (B6 level) so awesome job for wanting to explore it and understand it properly :)
thanks so much!! this helped a lot :)
Post by: mani.s_ on February 14, 2020, 08:00:39 pm
Post by: fun_jirachi on February 14, 2020, 11:34:37 pm
Hi, could someone confirm my answer and working out for this question? Thanks
You're very very close!
It's just a trick in the wording - it asks how much more work is required to move the object (note that 'more work' is very important!). This just means you've got to subtract 1MJ from 1.75MJ as calculated.
Hope this helps :)
Post by: mani.s_ on February 15, 2020, 12:14:14 am
Post by: mani.s_ on February 17, 2020, 04:31:10 pm
Thanks
Post by: Coolmate on February 17, 2020, 08:36:59 pm
Could someone please help me break down this part of my marking criteria for my depth study and what it means:
Questioning and predicting
--> Identifying scientific inaccuracies
Identifies and proposes valid scientific mistakes and develops appropriate improvements, (asks questions and makes evidence based predictions) (5 MARKS)
Thanks in advance 8)
Post by: jamonwindeyer on February 17, 2020, 09:11:16 pm
Hi, I was wondering if I needed to know the slingshot effect for the HSC??? Would I be able to use to explain stuff for long responses?
Thanks
Hey! Nah, the slingshot effect isn't necessary knowledge for the new syllabus (you could use it as an example of a collision for Year 11 momentum stuff, but you won't be assessed on that in Year 12) :)
Post by: jamonwindeyer on February 17, 2020, 09:13:03 pm
Hi Everyone, ;D
Could someone please help me break down this part of my marking criteria for my depth study and what it means:
Questioning and predicting
--> Identifying scientific inaccuracies
Identifies and proposes valid scientific mistakes and develops appropriate improvements, (asks questions and makes evidence based predictions) (5 MARKS)
Thanks in advance 8)
This wants you to be critical! What is missing from experiments you are referencing? What errors were made, if any? How could this error have been caught? Basically, for every 'key experiment/finding' you talk about, or key source you reference, don't just take it at face value. Consider it critically and decide whether anything is 'wrong.' If it is, that's okay, but make it a discussion point! :)
Post by: Coolmate on February 17, 2020, 09:59:15 pm
This wants you to be critical! What is missing from experiments you are referencing? What errors were made, if any? How could this error have been caught? Basically, for every 'key experiment/finding' you talk about, or key source you reference, don't just take it at face value. Consider it critically and decide whether anything is 'wrong.' If it is, that's okay, but make it a discussion point! :)
Hey Jamon! :)
Thankyou so much for the clarification! ;D
Also, when it says "Makes evidence based predictions", I am analysing a movie so does this mean I should research and provide evidence to support my answer?
Thanks again,
Coolmate 8)
Post by: mani.s_ on February 17, 2020, 10:43:12 pm
Hey! Nah, the slingshot effect isn't necessary knowledge for the new syllabus (you could use it as an example of a collision for Year 11 momentum stuff, but you won't be assessed on that in Year 12) :)thank you so much Jamon, you the man!!! :)
Post by: jamonwindeyer on February 18, 2020, 07:44:07 pm
Hey Jamon! :)
Thankyou so much for the clarification! ;D
Also, when it says "Makes evidence based predictions", I am analysing a movie so does this mean I should research and provide evidence to support my answer?
Thanks again,
Coolmate 8)
Correct! "This isn't right, because Reputable Source X tells us that..." or "This is a flawed viewpoint, because Law of Physics that makes it flawed. Or something like that :) you need the BECAUSE in there, the justification, based on evidence :)
Post by: Coolmate on February 19, 2020, 05:21:49 pm
Correct! "This isn't right, because Reputable Source X tells us that..." or "This is a flawed viewpoint, because Law of Physics that makes it flawed. Or something like that :) you need the BECAUSE in there, the justification, based on evidence :)
Thanks Jamon! This makes a lot more sense now ;D
Post by: Coolmate on February 24, 2020, 08:47:32 pm
Just a quick question, I was just wondering if anyone knew of any High Earth Orbiting satellite examples
Thanks ;D
Post by: milie10 on February 24, 2020, 10:58:53 pm
Hi Everyone,
Just a quick question, I was just wondering if anyone knew of any High Earth Orbiting satellite examples
Thanks ;D
Hi Coolmate!!
Geostationary (equatorial) satellites or Geosynchronous satellites (not equatorial, but have the same period and altitude as a geostationary satellite) are the general names of high earth orbiting satellites. Characteristics include:
- Altitude of 36000km
- Orbital period of 24 hours/day
- Used for communication
I'm not sure if you wanted specific real life examples of these satellites- maybe someone else can pop in for that! :) However, I believe these low and high earth satellites only need to be related to their properties (altitude, orbital period and orbital plane) and uses based on the syllabus.
Hope that helps! :D
Post by: Coolmate on March 29, 2020, 05:48:05 pm
Currently, I am structuring my notes around the syllabus dot points and I am just wondering how other people structure/ structured their HSC Physics Notes, for example:
For Module 5 and Module 6, did you just put the concepts around the specific dot point on one page, then put example calculations next to/ on the next page?
Thanks :)
Coolmate 8)
Post by: BlackFrost on May 14, 2020, 10:53:17 am
World-class hurdlers raise their centre of mass as little as possible when they jump over the hurdles.
Why?
My response: Each time a hurdler jumps over the hurdles, their kinetic energy decrease until it totally becomes the gravitational potential energy at the highest point. Thus, by raising their centre of mass as little as possible, their kinetic energy will be higher, thus their velocity increases, allowing them to cross the hurdles in less time.
Please comment on my response; if I have used the wrong terminology or something doesn't make sense.
Thx :)
Post by: Coolmate on May 29, 2020, 11:27:01 pm
So i've a got question regarding Module 2, and I'm wondering if my response makes sense.
World-class hurdlers raise their centre of mass as little as possible when they jump over the hurdles.
Why?
My response: Each time a hurdler jumps over the hurdles, their kinetic energy decrease until it totally becomes the gravitational potential energy at the highest point. Thus, by raising their centre of mass as little as possible, their kinetic energy will be higher, thus their velocity increases, allowing them to cross the hurdles in less time.
Please comment on my response; if I have used the wrong terminology or something doesn't make sense.
Thx :)
Hey BlackFrost! :D
Sorry if this is a late response.
Great answer, I would suggest a few things though:
🔆 Define Key Terminology; such as Gravitational Potential Energy
🔆 Include the GPE formula, you learn in Yr 11; U = mgh --> This supports your response (and including a relevant formula into your response may be part of the marking guidelines)
🔆 Try to avoid saying, "Thus", twice in a sentence, as you can replace this with other synonyms like, "Thereby"
I hope this helps!
Coolmate 8)
Post by: r1ckworthy on May 30, 2020, 11:19:33 am
Hey Everyone,
Currently, I am structuring my notes around the syllabus dot points and I am just wondering how other people structure/ structured their HSC Physics Notes, for example:
For Module 5 and Module 6, did you just put the concepts around the specific dot point on one page, then put example calculations next to/ on the next page?
Thanks :)
Coolmate 8)
Hey Coolmate,
Sorry for the incredibly late response ;D. I would say that the way you structure notes completely depends upon you. For myself, I didn't bother with writing down example calculations as they made my notes excessively long, and I could always figure out the maths behind a question. But if you are someone who struggles with maths, then, by all means, go for it! It all depends on your strengths and weaknesses. My weakness was memorisation of the larger concepts and facts, so I would make my notes as concise and explanatory as possible so that I could get a document to revise from and get the bigger picture. The key with notes is that they provide you with the big picture so that you know how those tiny little details fit into the overarching concept. This helps a ton with memorisation, at least for me.
Hope that helps! Again, sorry for the late response!
R1ckworthy
Post by: Coolmate on May 30, 2020, 02:14:03 pm
Hey Coolmate,
Sorry for the incredibly late response ;D. I would say that the way you structure notes completely depends upon you. For myself, I didn't bother with writing down example calculations as they made my notes excessively long, and I could always figure out the maths behind a question. But if you are someone who struggles with maths, then, by all means, go for it! It all depends on your strengths and weaknesses. My weakness was memorisation of the larger concepts and facts, so I would make my notes as concise and explanatory as possible so that I could get a document to revise from and get the bigger picture. The key with notes is that they provide you with the big picture so that you know how those tiny little details fit into the overarching concept. This helps a ton with memorisation, at least for me.
Hope that helps! Again, sorry for the late response!
R1ckworthy
Hey r1ckworthy,
Thankyou for your response! I think I might do a combination of both a few example questions and write about the big concepts as well. :)
Thanks again,
Coolmate 8)
Post by: Coolmate on May 31, 2020, 07:09:21 pm
Could someone please help me with this question (attached), I know I have to use Wein's Law, but am a bit confused...
Thanks!
Coolmate 8)
Post by: Einstein_Reborn_97 on May 31, 2020, 08:08:32 pm
Hey everyone :)Hi Coolmate,
Could someone please help me with this question (attached), I know I have to use Wein's Law, but am a bit confused...
Thanks!
Coolmate 8)
Yes, you'll have to use Wien's Law.
A wavelength of 263 nanometres is in the ultraviolet part of the electromagnetic spectrum (between about 350nm - 10nm). Therefore, we won't be able to see it as our eyes cannot see wavelengths smaller than about 350 nanometres.
Hope that helps, let me know if you have any further questions! ;)
Post by: Coolmate on May 31, 2020, 08:18:59 pm
Hi Coolmate,
Yes, you'll have to use Wien's Law.
A wavelength of 263 nanometres is in the ultraviolet part of the electromagnetic spectrum (between about 350nm - 10nm). Therefore, we won't be able to see it as our eyes cannot see wavelengths smaller than about 350 nanometres.
Hope that helps, let me know if you have any further questions! ;)
Ohhh! Ok, thankyou so much Einstein_Reborn_97! This makes so much sense now, I think I just got confused with the wording of the question. ;D
Thanks again!
Coolmate 8)
Post by: Coolmate on June 01, 2020, 08:42:53 pm
When converting from nanometres into metres is it x10^-9 or x10^-7?
Thanks,
Coolmate 8)
Post by: Einstein_Reborn_97 on June 01, 2020, 08:46:29 pm
Hey everyone,Hey Coolmate,
When converting from nanometres into metres is it x10^-9 or x10^-7?
Thanks,
Coolmate 8)
It's \(\times10^{-9}\)
Nine and nano both start with n - that should help you remember ;)
Post by: Coolmate on June 01, 2020, 08:51:43 pm
Hey Coolmate,
It's /(\times10^{-9}\)
Nine and nano both start with n - that should help you remember ;)
Thanks
Post by: Coolmate on June 16, 2020, 03:00:39 pm
I have a practical assessment task for Physics soon (finding slit separation distance) and there is a part that wants us to look at "Experiment Analysis", does this just essentially mean answering questions on, Reliability, Validity and Accuracy of the experiment or something else?
Thanks in advance!
Coolmate 8)
Post by: Chocolatepistachio on June 18, 2020, 09:11:35 pm
I don’t do physics but I think it means analysing your results and coming to a conclusion based on that and interpretation of the results and explain what the results mean, reliability and how reliable it was, improvements that can be made to the experiment, whether the results were consistent with expectations and did or did not support the hypothesis
Post by: Coolmate on June 21, 2020, 08:10:25 pm
Hi,
I don’t do physics but I think it means analysing your results and coming to a conclusion based on that and interpretation of the results and explain what the results mean, reliability and how reliable it was, improvements that can be made to the experiment, whether the results were consistent with expectations and did or did not support the hypothesis
Thanks Chocolatepistachio! This was very helpful.
Coolmate 8)
Post by: JaimeFreeze on June 23, 2020, 07:29:12 pm
Thanks!
Post by: twelftholmes on June 29, 2020, 10:04:09 pm
I have a question from Jamon Windeyer's awesome physics topic tests if that's okay
Here is the Q: https://imgur.com/a/EfhvX8o
Here is my attempted answer: https://imgur.com/a/kXLoz0e
So basically, in the back of the answers Jamon has put that we can solve this either using conservation of energy and work, or calculate the acceleration and then approach it as a projectile motion question. I don't feel confident with latter and my class hasn't done module 5 yet (we're doing the modules in the order 6, 8, 7, 5 oop) so I chose the former which I feel like I understand much at this point. However I got it wrong and I'm wondering if it's because of an error in my working out or understanding using the method I did, or if it's because I'm better off approaching it like a projectile motion question (like Jamon did in the back of the book, however I didn't understand that which is why I'm here haha). If someone wouldn't mind helping me figure out how to get the correct answer that'd be awesome. (using the method I did if that's possible).
Thanks in advance!
Post by: blasonduo on June 29, 2020, 11:21:27 pm
Hey!!
I have a question from Jamon Windeyer's awesome physics topic tests if that's okay
Here is the Q: https://imgur.com/a/EfhvX8o
Here is my attempted answer: https://imgur.com/a/kXLoz0e
So basically, in the back of the answers Jamon has put that we can solve this either using conservation of energy and work, or calculate the acceleration and then approach it as a projectile motion question. I don't feel confident with latter and my class hasn't done module 5 yet (we're doing the modules in the order 6, 8, 7, 5 oop) so I chose the former which I feel like I understand much at this point. However I got it wrong and I'm wondering if it's because of an error in my working out or understanding using the method I did, or if it's because I'm better off approaching it like a projectile motion question (like Jamon did in the back of the book, however I didn't understand that which is why I'm here haha). If someone wouldn't mind helping me figure out how to get the correct answer that'd be awesome. (using the method I did if that's possible).
Thanks in advance!
Hey!
Just looking at the question given, it originally says half a meter apart, but the diagram shows it as a meter apart. (So there's a typo in the question!)
What distance did Jamon use? This may be the error, because at a quick glance I don't see anything wrong with your working!
EDIT: So I've checked out how Jamon has done his answer. In the first part he uses d = 0.5, but in the second part of the working he uses d = 1.
In the line that reads:
\[\ v^2 = 0^2 + 2 \times 3.52\times 10^{15}\]
There should be an additional factor of 0.5.
Doing this does indeed give the same answer that you have worked out :)
Post by: twelftholmes on June 30, 2020, 06:17:03 pm
Post by: shekhar.patel on August 04, 2020, 09:58:55 pm
How did we determine the direction of torque on an object like in this question. Is there a rule?
Thank you.
Post by: Bri MT on August 05, 2020, 09:20:04 am
Hi,
How did we determine the direction of torque on an object like in this question. Is there a rule?
Thank you.
Hey, if you look at the diagram you should be able to see that the component of the force perpendicular to the radius is pointed down. Downwards force to the right of the pivot point = clockwise rotation
Post by: mrsc on August 08, 2020, 05:43:24 pm
Post by: fun_jirachi on August 08, 2020, 09:12:19 pm
The galvanometer detects changes in current - so when we look at the transformer equation \(\frac{V_P}{V_S} = \frac{I_S}{I_P} = \frac{N_P}{N_S}\), we can deduce that to increase the deflection in the galvanometer ie. increasing \(I_S\), we can do a few things: increase the number of primary coils, decrease the number of secondary coils, increase the primary voltage, etc. We also can rule out C and D because the primary function of a resistor is to reduce current (not what we want here!) and copper cores perform worse as compared to iron cores. Hence, the answer is B.
Hope this helps :)
Post by: Coolmate on August 11, 2020, 07:44:31 pm
Could someone please explain to me why the answer is 'C'? Does it involve deriving another formula, or the like out of this formula: \[dsin\theta = m\lambda\]
Thanks in advance!
Coolmate 8)
Post by: 1729 on August 11, 2020, 08:01:50 pm
Hey Physics Gang! ;)The answer is C because it's a dark spot, it means they're out of phase, so p2 would have to be longer than p1 by a factor of (2n(lambda)+1)/2 to create perfectly destructive interference. As you can see by the center bright spot, that's what happens when p1 and p2 are the same length. This means that the 2 dark spots closest to the center will be half a wavelength out of phase which is the closest difference that creates a dark spot. This means that it would be lambda/2, and since the dark spot we want is one dark spot further, the difference would have to be 3(lambda)/2
Could someone please explain to me why the answer is 'C'? Does it involve deriving another formula, or the like out of this formula: \[dsin\theta = m\lambda\]
Thanks in advance!
Coolmate 8)
Post by: Coolmate on August 12, 2020, 09:21:10 am
The answer is C because it's a dark spot, it means they're out of phase, so p2 would have to be longer than p1 by a factor of (2n(lambda)+1)/2 to create perfectly destructive interference. As you can see by the center bright spot, that's what happens when p1 and p2 are the same length. This means that the 2 dark spots closest to the center will be half a wavelength out of phase which is the closest difference that creates a dark spot. This means that it would be lambda/2, and since the dark spot we want is one dark spot further, the difference would have to be 3(lambda)/2
Thanks 1729!
This makes much more sense now :)
Coolmate 8)
Post by: Coolmate on August 24, 2020, 02:44:17 pm
Could someone please step me through how to solve this question (attached) from the 2012 HSC paper (Q27, 2012)
Thanks in advance,
Coolmate 8)
Post by: fun_jirachi on August 24, 2020, 05:44:54 pm
One of the first things that pops out to me is that you have that the horizontal displacement is 45m - indicating that from \(s_x = u_xt\), we have that \(45 = u\cos 60 \times t\). Secondly, we have that from \(s_y = u_yt - \frac{1}{2}gt^2\) - thus we also have that \(34 = u\sin 60 \times t - \frac{gt^2}{2}\). Rearranging and substituting the first equation into the second yields an equation in \(u\), specifically
The thought process I followed here was to start with the easiest equations that you can substitute info from the question into - it's important to ask yourself what do you know? It's also important to try and eliminate one variable ASAP in equations of multiple variables (you can always sub back in later to find the other once you've found a definitive value for one of them). If all else fails, literally write down everything they give you ie. \(s_x = 45, s_y = 34\) and see what equations 'match up' - much like questions towards the back end of maths exams and stuff (just try something!) Definitely look to practice more of these types of questions (or change the numbers a bit) if this is the type of thing you struggle with :)
Post by: Coolmate on August 25, 2020, 01:12:21 pm
Hey :)
One of the first things that pops out to me is that you have that the horizontal displacement is 45m - indicating that from \(s_x = u_xt\), we have that \(45 = u\cos 60 \times t\). Secondly, we have that from \(s_y = u_yt - \frac{1}{2}gt^2\) - thus we also have that \(34 = u\sin 60 \times t - \frac{gt^2}{2}\). Rearranging and substituting the first equation into the second yields an equation in \(u\), specificallyRearrange and solve from here to get the answer of 30ms-1
The thought process I followed here was to start with the easiest equations that you can substitute info from the question into - it's important to ask yourself what do you know? It's also important to try and eliminate one variable ASAP in equations of multiple variables (you can always sub back in later to find the other once you've found a definitive value for one of them). If all else fails, literally write down everything they give you ie. \(s_x = 45, s_y = 34\) and see what equations 'match up' - much like questions towards the back end of maths exams and stuff (just try something!) Definitely look to practice more of these types of questions (or change the numbers a bit) if this is the type of thing you struggle with :)
Hey fun_jirachi, :)
Thankyou for the assistance, it really helped me today! (I just came out of my Physics Trial exam and your help was very helpful)
Thanks again!
Coolmate 8)
Post by: jessica2166 on September 29, 2020, 11:19:56 am
The total energy of a satellite is equal to the sum of its gravitational potential energy and its orbital kinetic energy. However, the work done to put that satellite into orbit is less than this.
Explain how this does not contravene the law of conservation of energy.
Post by: mrsc on October 06, 2020, 07:28:19 pm
Post by: nicholasbikolas on October 31, 2020, 12:05:24 pm
Can someone help me with this?
The total energy of a satellite is equal to the sum of its gravitational potential energy and its orbital kinetic energy. However, the work done to put that satellite into orbit is less than this.
Explain how this does not contravene the law of conservation of energy.
ok so basically our GPE is negative and approaching zero as we leave Earth. Our kinetic energy is positive and decreasing as we leave Earth. We have TWO components of energy when we are orbiting the Earth:
1) GPE of orbit
2) KE of orbiting
if we consider getting up into orbit, our kinetic energy (work done) converts into potential energy (our kinetic energy makes our GPE less negative). there is only the GPE component being established here.
so. to get up to orbit + ESTABLISH an orbit will cost much more energy than just to get up there??
hope this helps
Post by: BakerDad12 on October 31, 2020, 12:16:25 pm
Can someone help me with this?
The total energy of a satellite is equal to the sum of its gravitational potential energy and its orbital kinetic energy. However, the work done to put that satellite into orbit is less than this.
Explain how this does not contravene the law of conservation of energy.
This was in our trials as well. I still don't properly understand it.
Post by: Coolmate on November 01, 2020, 11:12:10 pm
Could I please get some help with the following Module 8 questions:
🚀Question 1
(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-0/p280x280/123140426_407470780657586_1492828447652801282_n.png?_nc_cat=101&ccb=2&_nc_sid=ae9488&_nc_ohc=DvZDddkuRcEAX_71Dpx&_nc_ht=scontent-syd2-1.xx&oh=88a9b8ab1501823a3802a545b159a1e5&oe=5FC42556)
🛰Question 2
(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-0/p280x280/123058689_356355752141382_24845712250901749_n.png?_nc_cat=101&ccb=2&_nc_sid=ae9488&_nc_ohc=al1MXdX7E20AX8ITbmH&_nc_ht=scontent-syd2-1.xx&oh=e977178af55cf1761c35376e36b9ef56&oe=5FC38FD3)
☄Question 3
(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-0/p280x280/123140835_2806863866250934_7235110435417664429_n.png?_nc_cat=105&ccb=2&_nc_sid=ae9488&_nc_ohc=gMSo4897qrwAX9LVue3&_nc_ht=scontent-syd2-1.xx&oh=547ea854a5aea31d1f4da85c706b7936&oe=5FC2B8CA)
Thanks in advance!
Coolmate 8)
Post by: fun_jirachi on November 02, 2020, 02:39:18 am
We can then use the formula \(E= \frac{hc}{\lambda}\) to find the value of the energy change, which just so happens to be equivalent to the difference between levels 1 and 3 - hence, the answer is A.
2. Try to target these sorts of questions in the exam! The decay constant is equal to \(\frac{\ln 2}{t_\frac{1}{2}}\). observe from the graph that the initial sample is 5g, so the half life is the time it takes to go down to 2.5g, which is about 29 years. Putting it into the calculator with that formula will get you an answer close to D.
3. Recall that when beta decay occurs, a neutron 'splits' into a proton and an electron (simplistically, kinda true). The electron is released (beta particle!) and the proton latches on to the nucleus. This means that there is one more proton and one less neutron. In heavier elements the ratio of protons to neutrons is about 1:1.5, so this will cause the ratio to go closer to 1:1 ie. towards the n=p line. Hence, the answer is A.
Post by: Coolmate on November 05, 2020, 03:08:00 pm
1. Recall that when a photon is absorbed, electrons 'jump' up energy levels. We can thus rule out B and D.
We can then use the formula \(E= \frac{hc}{\lambda}\) to find the value of the energy change, which just so happens to be equivalent to the difference between levels 1 and 3 - hence, the answer is A.
2. Try to target these sorts of questions in the exam! The decay constant is equal to \(\frac{\ln 2}{t_\frac{1}{2}}\). observe from the graph that the initial sample is 5g, so the half life is the time it takes to go down to 2.5g, which is about 29 years. Putting it into the calculator with that formula will get you an answer close to D.
3. Recall that when beta decay occurs, a neutron 'splits' into a proton and an electron (simplistically, kinda true). The electron is released (beta particle!) and the proton latches on to the nucleus. This means that there is one more proton and one less neutron. In heavier elements the ratio of protons to neutrons is about 1:1.5, so this will cause the ratio to go closer to 1:1 ie. towards the n=p line. Hence, the answer is A.
Hey fun_jirachi!
Thankyou for your help!
I just had a question regarding Q3 --> Have I correctly solved the problem by doing this:
\(\frac{\ln 2}{(29)\frac{1}{2}}\)
= 0.04780325383
= 0.048
= 0.04780325383
= 0.048
The final number wasn't any of the answers, but was closest to 'D'
Thanks again!
Coolmate 8)
Post by: fun_jirachi on November 05, 2020, 04:37:46 pm
Remember that the subscripted half actually means half-life (the half denotes half-life as opposed to regular t) - it does not mean multiply the denominator by half (which is what I assume you did, considering that your answer is roughly twice the size of D). Just be careful with these sorts of things - if an answer doesn't look right, it probably isn't! In general, look for an error percentage that seems reasonable if interpreting from a graph or some other data set, or an exact result if you're given actual numbers. Otherwise, the method seems okay :)
Post by: Coolmate on November 05, 2020, 05:57:42 pm
Hey there Coolmate :)
Remember that the subscripted half actually means half-life (the half denotes half-life as opposed to regular t) - it does not mean multiply the denominator by half (which is what I assume you did, consider that your answer is roughly twice the size of D). Just be careful with these sorts of things - if an answer doesn't look right, it probably isn't! In general, look for an error percentage that seems reasonable if interpreting from a graph or some other data set, or an exact result if you're given actual numbers. Otherwise, the method seems okay :)
Ohhh, ok I see that now ;D
When doing just ln(2)/(29), I got an answer of 0.024, which is the closest one. This makes sense now, thanks again for your help :)
Coolmate 8)
Post by: Maroon and Gold Never Fold on August 07, 2021, 08:29:21 pm
the universe underwent as described by the Big Bang theory. (4 marks)
I wasn't sure about this question as there are no answers. My answer included: 1. initial bang (release of energy) 2. inflation (universe cool with pair production) and 3. production of elements (lithium, hydrogen and helium). Not sure if this is right or I have missed anything.
Post by: crippledbyenglish on November 18, 2021, 04:54:08 pm
Hi! I have been doing a few practice papers for physics and have a bunch of questions I am confused about! Does anyone know:
1. What is the best definition for an uncontrolled/controlled chain reaction. I would have said the number of fissions subsequently initiated in comparison to the number of original fissions but have also see it defined in terms of rate.
2. Why can't a particle in free fall move straight past Earth? What if gravity was pulling in two opposing directions and these cancelled out?
3. How does precision work for experiments and stuff I can find a standard answer literally nowhere
4. In back emf graphs why is back emf proportional to motor speed if emf changes periodically as the rotor rotates
If you could explain it would be greatly appreciated!
Post by: NightOwl123 on April 23, 2022, 01:08:01 pm
Post by: K.Smithy on April 23, 2022, 05:21:40 pm
Can someone please explain to me why the answer is c. Using Lenz's law the coil would have a north pole face the north pole of the magnet. This would be an anticlockwise current and as the magnet passes through the current it would change in the opposite thereby resulting in clockwise current. Using this logic I've gotten answer d?
Hey NightOwl123!
The way I approached this is by simply considering how the magnetic field of the magnet would interact with the coil. Firstly, we know that the magnetic field lines produced by a magnet go out of north and into south. So we can see that as the north face of the magnet passes through the coil, the magnetic field will wrap over the top of the coil.
Now, let's ignore the magnet for a second. Suppose that it was the wire itself that was generating this magnetic field. That is, a magnetic field which is directed out of the page at the centre of the coil, and directed into the page above the coil. For this magnetic field to exist, a current running through the coil needs to run in the anticlockwise direction. We can see this by using the right-hand rule. Now, if we bring the magnet back into the picture we know that it isn't really that has generated this field, rather it is the magnet. According to Lenz's law, the coil will want to oppose the magnetic field of the magnet. To do this, the opposing current will be produced. That is, a clockwise current through the coil.
The opposite is true when the South Pole passes through the coil - resulting in an induced current that is directed anticlockwise.
I hope this is helpful!
Katelyn
Post by: Daron Kutch on May 28, 2022, 03:21:56 pm