Post by: FlorianK on March 30, 2013, 08:10:52 pm
VCE PHYSICS Q&A THREAD
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If you have general questions about the VCE Physics course or how to improve in certain areas, this is the place to ask!
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Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.
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* EEEEEEP's practice questions
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• How to do well in Physics from SAMAD (Phys 50, ATAR 99.95)
• Paulsterio's Year 12 Physics Guide
• [Review] VCE Physics 3/4
• Extra formulae donated by VN users (good for cheat sheets)
• Harder physics questions
• Recreational Problems (Physics)
• TT's question thread - interesting read!
• Physics 3/4 Advice
* EEEEEEP's practice questions
Original post.
VCE Physics Question Thread!
Post by: ~T on April 01, 2013, 12:08:12 am
Seriously though, seconding a question that someone started as a thread, how much do we need to know about diodes? I'm confident with their applications and circuit questions, but what about the mechanics? No idea whether to spend time learning the workings of a diode or not.
Similarly, the Heinemann textbook is all like "Transistors aren't covered in this course" and now it's showing me a 'physics in action' segment that is telling me (very poorly) how a transistor works. I think I already understand them, but do I need to for the course?
And lastly, question 4 of chapter 4.1. I swear it can't actually happen. Unless I'm missing something... but it seems that when there is a grounded point on the circuit, there are two different currents at different points on the circuit. The answers say so too. Now as I'm typing I realise I'm not even sure about the point/function of a ground point. I know it's a reference point for the measurement of voltage, but does it affect the flow of current, or the circuit, in any way?
Many questions, many thanks in advance :)
Post by: b^3 on April 01, 2013, 01:10:03 am
As to the two different currents part, the second current is greater than the first current, that is because there will be some current leaking from the ground, as again, the current will flow towards the lower voltage. So from 0 V to -3 V, i.e. that is where your extra current is coming from.
For b, if we wanted to find the voltage dropped over each resistor, we would use a voltage divider, but we want the current through each, which is the same through each. So 8 V is dropped over 7 K worth of resistance, that is I=8/7K =1.14 mA.
As for the direction, the will all be flowing in the direction P, from higher potential to lower potential.
Hope that helps.
Post by: ~T on April 01, 2013, 11:40:49 am
As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
Post by: EspoirTron on April 01, 2013, 11:56:25 am
Post by: b^3 on April 01, 2013, 03:07:42 pm
Thank you both! I think I'm fine with diodes then, if that's all we need.Normally the ground is the lowest reference, which is why we can do what we normally do with other circuits. But since there is a potential that is lower than this reference in this case, we have to look at the situation differently, as we could find the two currents in the two resistors. But as the second was greater than the first, the current had to come from somewhere (again current flows from higher potential to lower potential, so that's why some current was drawn from there). In the normal case, where the lowest potential is the ground, the currents are flowing towards this direction, and so we can look at the problem normally.
As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
It's a little confusing, VCE physics doesn't explain it well (there was a fair bit of further electronics that I didn't fully understand (but was able to do the problems) until I got to uni and did an electrical engineering unit). Hope that makes sense.
Post by: Homer on April 08, 2013, 08:53:48 pm
Post by: FlorianK on April 08, 2013, 09:34:24 pm
could someone help me with the first twolet IB1 be the voltage drawn from the battery 1 and IB2 be the voltage drawn from battery 2
IR1 = IB1
IR2 = IR3 = 0.5 * IB1
IR4 = 2/3 * IB2
IR5 = IR6 = 1/3 * IB2
Ok, since both circuits will have different resistances IB1=/= IB2
The total resistance of circuit 1 is: [Rx is the resistance of any resistor in the circuit]
Rtotal = Rx + 1/2 * Rx [When 2 resistors of the same value are in parallel the resitance of this parallel circuit is half the resitance of each of them]
--> Rtotal = 1.5 * Rx
Ok the other one is a parallel circuit. The equation you should use for a parallel circuit, with 2 components is:
So here R1 is R4 so Rx and R2 is R5+R6 so 2*Rx
--> R1=Rx R2=2*Rx
we get:
So now to the current
IB1=V/(3/2*Rx) = 2/3 V/Rx
IB2=V/(2/3*Rx) = 3/2 V/Rx
So now putting 'values' into the equations from the top
IR1 = IB1 = 2/3 * V/Rx
IR2 = IR3 = 0.5 * IB1 = 1/3 V/Rx
IR4 = 2/3 * IB2 = V/Rx
IR5 = IR6 = 1/3 * IB2 = 1/2 V/Rx
So we see we have the highest Current in R4 and the lowest current in R5 and R6.
Hope I didn't make some stupid mistake :p
Post by: Homer on April 08, 2013, 09:48:51 pm
Post by: LazyZombie on April 10, 2013, 09:42:47 pm
Post by: Homer on April 10, 2013, 09:55:04 pm
Post by: LazyZombie on April 11, 2013, 09:22:10 pm
About rollercoasters again.
Explain what happens to your body on corners using at least one of Newtons Laws.
What points would I have to cover?
I wrote: There is no force acting on the body initially so it conforms to newtons first law, continuing to move in the original forward motion so there is a feeling of being 'pushed' outwards from the corner.
Plus a little birds eye diagram of a roller coaster rounding a corner and an arrow pointing tangent to the curve labelled velocity. (right thing to label?)
Post by: Homer on April 16, 2013, 07:28:29 pm
Post by: LazyZombie on April 16, 2013, 08:46:28 pm
in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distancekinetic energy is greater at a smaller distance.
by finding the area of the force distance graph, you can find the work done and so find the kinetic energy
Post by: Robert123 on April 17, 2013, 04:05:36 pm
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
Post by: LazyZombie on April 17, 2013, 08:46:40 pm
General questions about voltage;I'm not sure about the voltage thing but since voltage is relative, I think it is possible. But I'm prettyy sure its not necessary in the course. (not in the study design) I might get back to this later :)
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
As for the crumple zone and seatbelt
Inertia causes the body to continue moving in the original motion of the car in a collision and cause injury or fatality, and seatbelts help to prevent this from happening.
Seatbelts don't increase the time of collision. I think that's what your teacher means. (?)
Post by: availn on April 17, 2013, 08:53:02 pm
And yeah, seat belts aren't crumple zones, they don't increase the time of collision, they just make sure you don't shoot out of your car.
Post by: Lasercookie on April 17, 2013, 10:45:43 pm
Ok, thanks for that. I will ask my teacher what her opinion of what the answer "should" be (even though I find people on this forum more knowledgable and reliable with answer).Split this post and moved it to the Technical Score Discussion boards
Also, another question (this is more a general education question).
Does the only thing that matters with SACs is your ranking? Please only answer if you are 100% sure due to reading a report from VCAA or talking to an examiner?
If so, then does that means there is no different between averaging 90% and 100% on SACs if you go to a low scoring school?
Post by: [paradox]+ on April 25, 2013, 04:41:12 pm
^help anyone? thanks in advance
Post by: ~T on April 26, 2013, 05:16:06 pm
Jasper has a transistor radio with flat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.I'm not entirely sure on this one but I'll give you my thoughts...
^help anyone? thanks in advance
Basically, any amplification system works of a small varying AC signal (the radio signal here) and a larger DC input (the battery here) that amplifies the input signal to create a much larger output (the sound here).
So, if the batteries are running flat, then they will not supply enough power to the transistor(s) in order to amplify the larger voltages, and the linear amplification region may be smaller. The input voltages will still be the same, but the extreme values will no longer be amplified linearly, distorting the signal.
This only half makes sense in my head. But pretty much, there is no reason that the input signal would be larger (resulting in clipping outside of the region) so it must be that the linear region itself becomes smaller.
Post by: Alwin on April 26, 2013, 09:11:33 pm
Jasper has a transistor radio with flat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.
^help anyone? thanks in advance
For this one, I decided to add an image! just because.. I can haha
(http://www.electronics-tutorials.ws/amplifier/amp13.gif)
Quote
The input signal may be too large, causing the amplifier to be limited by the supply voltage.
This is true in the case of Jasper and his poor flat batteries. The DC power supply is the two parallel lines, where the voltage supplied is +Vcc and grounded at 0V. I won't go into the specifics of the transistor, since it is no longer on the study design, but simply the "in" signal is the current that enters the middle of the transistor and allows the larger current to flow through (top to bottom) from the DC power source.
Hence, if the DC source is not large enough then it cannont properly amplify the "in" signal because it cannot amplify a strong signal, ie it cannot amplify the strong +ve and strong -ve signals, ie smaller clipping range
Hope that makes sense!
Post by: Robert123 on April 27, 2013, 03:04:38 pm
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks :)
Post by: Alwin on April 27, 2013, 03:19:04 pm
What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks :)
There actually is no "purpose" for wanting negative gain. It is just a side effect of using a simple single transistor amplifier, the npn transistor inverts the input signal. If you want a non-inverting amplifier, you need a two-stage amplifier (two transistors). That's the only reason why a lot of amplifiers in physics questions have negative gain since we deal with simple amplifiers.
As for the difference, well there is none. Remember when a microphone or speaker works, it is a diaphragm that vibrates back and forth creating a sound wave. If the initial current is positive, then the diaphragm moves forward, then back, then forward etc. But, if the signal had been inverted, initially the current flows in the other direction and so the diaphragm moves back first then forwards then back again if you get what I mean. So, the frequency of sound produced / recorded it the exact same.
Hope that clears things up for you!
Post by: Robert123 on April 27, 2013, 03:42:51 pm
Post by: Homer on May 02, 2013, 07:08:52 pm
Thanks ;D
ANS: 309.1ohms
Post by: sin0001 on May 02, 2013, 07:48:45 pm
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:
Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
Post by: Alwin on May 02, 2013, 08:04:22 pm
Hey guys how would you calculate the value of R?
Thanks ;D
ANS: 309.1ohms
Well, first you look at the diodes:
Diode 1:
Since the current is 20 mA, V = 1. Hence, R = 1/0.020 = 50 ohm. Thus, you have a 50ohm 'resistor' and a 500 ohm resistor in parallel
Diode 2:
This diode is in reverse bias, hence no current flows. There are only the 2 100ohm resistors in parallel
Total resistance:
Total current:
From Ohm's law, R = V/I
Hope this makes sense!
Post by: Alwin on May 02, 2013, 08:08:37 pm
For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:
Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.
Hopefully you can see how the solutions got the answer from here!
Post by: sin0001 on May 02, 2013, 08:16:53 pm
I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.Ahhh yeah I had a mental blank lol
Hopefully you can see how the solutions got the answer from here!
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
Post by: Alwin on May 02, 2013, 08:53:31 pm
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
It's okay! just try not to have a mental blank in the exam ;)
Yeah, just R = V/I. I have an example in a previous post:
Re: Physics [3/4] Question Thread!
Post by: Homer on May 02, 2013, 09:36:09 pm
Post by: jono88 on May 03, 2013, 05:47:40 pm
Post by: Phy124 on May 03, 2013, 07:31:31 pm
Yield strength is the maximum stress which a material can withstand before it begins to deform plastically, rendering it unable to return to its original state.
Post by: Homer on May 13, 2013, 06:56:42 pm
EDIT: I actually figured out a way to do this now, but not sure if right or not
Post by: jono88 on May 13, 2013, 08:06:24 pm
Post by: Professor Polonsky on May 15, 2013, 08:17:33 pm
Post by: [paradox]+ on May 15, 2013, 10:50:24 pm
circuit question
Post by: Robert123 on May 22, 2013, 04:45:08 pm
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks
Post by: EspoirTron on May 22, 2013, 09:50:53 pm
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks
Remember that the formula for gain is constituted by 'change in' vout and vin. The reason you may get a 'negative' gain is that the amplifier is inverting and since the gradient of the graph is negative, you will get a negative answer. To my knowledge gain is only concerned with a magnitude; therefore, while using it to calculate vout it you simply multiply by the gain value.
To address your second question. The reason why in series that a resistor with a higher resistance will output more power is because that it will have a higher potential across it, if you recall in a series circuit each component has the same amount of current flowing through it; hence, to maintain the ratio of v2/v1 = r2/r1 the resistor with the high resistance must have a higher potential across it and consequently a higher power output. The reason why this isn't the case for resistors in parallel is that in parallel the resistors will have the same potential but varying current based on their resistance. Therefore, I believe the lower the resistance the higher the current; thus, the resistor with the lowest resistance will have the highest power output (in the parallel component).
I hope that helped you out! :)
Post by: sin0001 on May 22, 2013, 11:25:07 pm
E.g. Suppose there's a circuit involving 2 resistors (2 and 6 ohms) in parallel, the battery supplies, let's say, 8 V. Current would be 1.5 A, the 2 ohm resistor would receive 3/4 of the total current, so 9/8 A. Whereas the 6 ohm resistor would get 1/4 of the total current, so 3/8 A.
P(2 ohm)= I^2 x R = (9/8)^2 x 2 = 2.53 W
P(6 ohm) = (3/8)^2 x 6 = 0.84 W
So it can be seen that the resistor with the lowest resistance, in a parallel circuit, would have the highest power output.
Hope this helps!
Post by: ~T on May 27, 2013, 04:51:31 pm
therefore
Current is thus given
The point still stands of course, but none of that calculation is even required.
The smaller resistor will receive
The larger resistor will receive
As
Post by: Homer on May 27, 2013, 07:54:53 pm
Post by: FlorianK on May 27, 2013, 08:38:44 pm
Post by: tote.moore on May 28, 2013, 06:24:46 pm
If a certain amount of Power is transmitted through power lines into a house, why does the voltage of an appliance go up when another appliance is turned off?
Post by: Robert123 on May 29, 2013, 03:05:22 pm
What is RMS? How is it calculated? How much do we have to know about it for electronics & photonics?
Why do photodiodes have to be in reverse bias?
If a photodiode is in series with a resisitor, what would happen if the light intensity is doubled?
What are the 'reasons' for limits in an amplifier?
Any help will be greatly appreciated :)
Post by: lzxnl on May 29, 2013, 04:17:40 pm
2. If a photodiode is forward-biased, it is a regular diode. Only when reverse-biased does its behaviour depend on light like in the course. Then, the photocurrent is directly proportional to the intensity of light striking the diode. If the light intensity doubles, the current doubles.
3. I'm not great with transistors, sorry. My electronics knowledge is strictly limited to what's in the course.
Post by: Robert123 on May 29, 2013, 07:13:22 pm
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
Post by: availn on May 29, 2013, 08:02:17 pm
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
You cannot do this easily with kinematics, because a spring does not accelerate a mass at a constant rate. You should do this with energy, spring potential energy before equals kinetic energy after.
0.5kx2 = 0.5mv2
k · 0.22 = 0.15 · 4.22
∴ k = 66.15Nm-1
Post by: random_person on May 29, 2013, 08:04:22 pm
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
We have m=0.15kg, v=4.2ms-1, s=20cm=0.2m
EK=1/2 mv2
=1/2 0.15 x 4.22
=1.323 J
Us=1/2 ks2
therefore
1.323=1/2 k x 0.22
k=66.15Nm-1
Hope this helps
Post by: Sentar on June 03, 2013, 04:01:23 pm
Explain how it is possible to transmit the information contained in an electrical signal which has both a positive and negative component, when it is impossible to produce a beam of light with negative intensity?
Cheers
Post by: lzxnl on June 03, 2013, 04:29:53 pm
Post by: ~T on June 10, 2013, 11:19:35 pm
The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."
Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?
Post by: availn on June 11, 2013, 02:18:50 pm
Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."
The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."
Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?
Nope. My friend got hit by this too.
Magnetic North and North Magnetic Pole is at the earth's north.
Magnetic North Pole is at the earth's south.
Post by: ~T on June 12, 2013, 06:19:45 pm
Post by: jono88 on June 24, 2013, 07:43:46 pm
Post by: Homer on June 24, 2013, 07:55:32 pm
Post by: Phy124 on June 24, 2013, 08:18:47 pm
how would i find the tension in each cable?As the structure is in rotational equilibrium you can sum the moments (I think you refer to it as torque in yr 12 physics?) about the point of rotation and equate them to zero.
Hint: The weight force of the bridge and the two vertical components of the tension forces will be the forces that cause rotation. (the horizontal component of tension does not cause rotation as it acts "through" the point of rotation)
Spoiler
Post by: Homer on June 29, 2013, 11:54:24 am
Thanks
ANS: a) fh=390N, fv=680N
b) 330N
Post by: SocialRhubarb on June 29, 2013, 01:08:34 pm
The horizontal component is equal to
The vertical component is equal to
Umm, unless I've done something stupid I think your answers are slightly off.
Part B is a bit beyond the scope of VCE.
Essentially what you're doing is equating torques since the object isn't spinning off in one direction.
The net torque on the fence post is 0, so the torque from one source must be equal to the torque from the other.
Alternatively, using the answer of 390N,
Post by: Homer on June 29, 2013, 02:14:18 pm
anyways
A 6m ladder of mass 7.5 kg leans against a wall at 60 degrees to the horizontal. What is the torque exerted on the ladder by the weight of the ladder itself?
I know it should be 7.5g x 3 x sin(30), but i dont understand why they take theta as 30 instead of 60?
Post by: lzxnl on June 29, 2013, 03:13:09 pm
This is why I like thinking that torque = moment arm * force where the moment arm is the component of the radius that is perpendicular to the force. You'll never get it wrong this way.
Post by: Homer on July 02, 2013, 11:41:55 am
edit: accidentally clicked modify instead of quote, my bad (2/cos(c))
Post by: sin0001 on July 02, 2013, 11:57:53 am
It depends on how the angle is defined. The formulaIs the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?
Post by: Phy124 on July 02, 2013, 04:59:53 pm
The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2
edit: accidentally clicked modify instead of quote, my bad (2/cos(c))
Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?
I would think that torque is only relevant to the "Materials and their use in structures" detailed study.
Although your question wasn't directed at me, I would advise you to do some mid-year practice exams during these holidays to both reinforce key concepts and get a feel for VCAA physics exams.
edit: added answer to sin0001's question.
Post by: Homer on July 03, 2013, 10:11:24 am
hey just a question why is it 0.003 wouldnt it be
Post by: Homer on July 03, 2013, 11:45:34 am
Post by: Alwin on July 03, 2013, 12:58:15 pm
Also, What would be the tension in the cables of the cranes? Thankyou
Let the tensions of the cables be T1 and T2, corresponding the the crane numbers Crane 1 and Crane 2.
Note I have included units in my working, just a small personal habit.
Using Crane 2 as the pivot,
You can now either say,
or use Crane 1 as a pivot,
Either way, it works out that
So, then tensions on cables of the cranes 1 and 2 are 2,400,000N an 600,000N in that order
hey just a question why is it 0.003 wouldnt it beso
?
30cm2 is not the same as 30cm x 30cm = (0.3m)^2 = 0.09 m2
30cm2 is the same as 30cm x 1cm = (0.3m) x (0.01m) = 0.003 m2
The general rule is that 1cm2 = 0.0001m2
EDIT: answered both of homer's questions, rather than double post
Post by: Homer on July 05, 2013, 02:05:26 pm
Post by: SocialRhubarb on July 05, 2013, 03:00:17 pm
Post by: Nato on July 06, 2013, 11:16:12 pm
i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).
for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??
thank you guys
Post by: sin0001 on July 07, 2013, 12:09:18 am
Hey guys,Okay so the forces are identical, but what is the difference here? Something has to be different between you and the ball, because clearly one is moving while the other is stable. The difference is mass; you weigh a lot more than the ball you've kicked and according to the formula: a = F/m, you are going to experience negligible acceleration while the ball is going to experience much greater acceleration, because if 'm' increases, then acceleration of an object will decrease. Therefore, you are going to 'absorb' a reaction force of 100 N by only experiencing negligible acceleration, but this force will cause the ball to go 'flying'.
i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).
for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??
thank you guys
Hope it made sense!
Post by: SocialRhubarb on July 07, 2013, 05:30:34 pm
Post by: Homer on July 08, 2013, 02:45:18 pm
Post by: jssantucci on July 08, 2013, 03:58:21 pm
Thanks
Post by: SocialRhubarb on July 08, 2013, 05:07:45 pm
Let the primary current be
We know that
We also know that
Therefore,
Thus, since
Like I said before, I don't actually understand the physics behind this, but somehow the secondary circuit must somehow influence the primary circuit? Maybe someone else can help with that.
Post by: jssantucci on July 08, 2013, 05:21:27 pm
When there is no load connected to the secondary coil no current flows in the secondary and so all the flux is being generated by the primary coil. However, this flux will induce a voltage in the primary coil that will oppose the change of flux. That is, as the current increases, this ‘back EMF’ will tend to reduce the current. In a good transformer, this process is very effective and very little current will flow in the primary if none is flowing in the secondary.
Not really sure how this works but it seems to be the answer to the question. Could someone clarify the above idea about 'back EMF' for me?
Thanks
Post by: SocialRhubarb on July 08, 2013, 05:34:44 pm
Basically, in a simple DC motor with one loop, when you run a current through the loop, it runs through the magnetic field and generates a force, and your motor starts turning. But when your motor is turning, its flux is constantly changing. We know that a change in flux through a coil induces a voltage in that coil, so the changing flux actually induces an EMF which counteracts the voltage we applied to make the motor spin. This EMF we call a 'back EMF'.
Post by: Guest on July 08, 2013, 08:17:52 pm
Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.
Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?
The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?
Thanks, in advance
Post by: Robert123 on July 08, 2013, 09:21:34 pm
What would be the direction of the magnetic force and how do we work it out?
You missed an important piece of information there, the direction of the current (read 4 lines down from that picture in 4a & 4b).
From there, you use the right hand slap rule, fingers pointing to the South Pole and thumb down for 4a and up for 4b.
Clear?
Post by: lolipopper on July 08, 2013, 09:46:41 pm
the question is attached.
Lots of love. ahha.
sorry *i didnt notice the questions thread*
edit: damn sorry guys i just understood it. tell me if you want me to explain it.
Moderator edit [2/cos(c)]: Merged double post
Post by: Nato on July 09, 2013, 12:06:54 pm
as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
thanks guys
Post by: lolipopper on July 09, 2013, 01:07:49 pm
impulse/momentum question here:
as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.
thanks guys
you have it wrong my friend. By increasing time in the car case you increase your impulse:
the equation is ΣF= ∆P/∆T. So if we increase the time the Net force decreases.
say ∆P=4 and ∆T=1 then the net force is 4N. but if we increase the change in time to, lets say, 2, then the Net force is 2N.
similarly, say we have a force of 5N and ∆T=1 then the impulse will be 5. however if we increase the change in time to, lets say, 2, then the impulse is 10.
You understand broda?
Post by: lzxnl on July 09, 2013, 02:07:38 pm
you have it wrong my friend. By increasing time in the car case you increase your impulse:
the equation is ΣF= ∆P/∆T. So if we increase the time the Net force decreases.
say ∆P=4 and ∆T=1 then the net force is 4N. but if we increase the change in time to, lets say, 2, then the Net force is 2N.
similarly, say we have a force of 5N and ∆T=1 then the impulse will be 5. however if we increase the change in time to, lets say, 2, then the impulse is 10.
You understand broda?
Actually your interpretation is slightly off as well. Impulse is just change in momentum. For a given collision, their initial and final velocities are going to be the same, so regardless of the time taken for the collision, the change in momentum is the same. The time, however, affects the force and acceleration that will be felt by the object as mentioned.
impulse/momentum question here:
as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
thanks guys
More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.
Post by: Alwin on July 09, 2013, 05:15:48 pm
More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.
Since you're always soo nit-picky :P, ill add to this too: same impulse over the longer period of time means smaller average force (usually meaning smaller maximum force too). sorry, just had to go there :P
From a few posts ago:
Could someone please explain why when the secondary coil in a transformer is connected to a circuit with an open switch (i.e. one with no load) the energy used by the primary coil is zero? Wouldn't the power being dissipated by the primary coil always just be IxV regardless of what's going on in the secondary circuit? This is in relation to question 9 of chapter 10.6 in Heinemann.
Thanks
Since it's rather a lot to explain, I found a link that may interest you:
http://sound.westhost.com/xfmr.htm
Also, just something interesting from the Heinamann textbook:
Quote
To create an induced current we do not always need loops and magnets. Whenever a changing magnetic flux encounters a conducting material an induced current will occur. These currents are often called eddy currents and may result in lost energy in electrical machinery. The eddy currents produced in a moving conductor will themselves be subject to the IlB force.Eddy currents are present in the core of transformers too.
p369 of the Heinemann textbook
I quite like how it includes topics outside the 3/4 course (I first came across this stuff in Singapore coursework so not sure what level it is in australia) but you might find it interesting too! :D
Post by: jssantucci on July 09, 2013, 11:18:50 pm
for anyone interested, this link explains the answer to my question quite nicely.
Post by: Homer on July 12, 2013, 10:20:55 am
Post by: Phy124 on July 12, 2013, 06:34:45 pm
i know the top of the beam experiences compression and the bottom tension, but what about the centre? is it a little of both?In theory, the beam experiences no bending stresses at its centre. Although I believe in most cases there will be a very small amount, which is negligible.
These diagrams indicate the bending stress (RHS) and strain (LHS) patterns a beam which is symmetrical about its neutral axis experiences.
(http://www.fgg.uni-lj.si/kmk/esdep/media/wg07/f0820004.jpg)
As you can see when the beam is in its elastic state the bending stresses have a linear relationship with distance from the neutral axis. However, when the beam is plastically deforming the bending stress is constant throughout.
*Note: As far as I know for VCE Physics purposes you only need to know about the elastic state for bending stresses.
Post by: sin0001 on July 24, 2013, 06:16:28 pm
Also, is 'Forces on moving charges' part of the course?
Post by: lzxnl on July 24, 2013, 06:24:17 pm
Jks.
The force is actually given by a cross product: Force = iL x B
where L is the current length vector; it has magnitude equal to the length of the wire and has the same direction as the wire
and B is the magnetic field vector which has a magnitude and a direction.
The magnitude of this force is given by i*|L||B| |sin theta| where theta is the angle between L and B. As you've seen from VCE physics, if the current wire is parallel to the field, no force results.
Likewise, if it's perpendicular, the sine function returns the maximum value of 1. Hence the force is maximum when the external magnetic field is perpendicular to the conductor.
Read up on wiki if you're not sure about cross products in general.
Post by: sin0001 on July 24, 2013, 06:57:08 pm
Because VCE physics is inadequate and doesn't give you the full picture. Period.Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.
Jks.
The force is actually given by a cross product: Force = iL x B
where L is the current length vector; it has magnitude equal to the length of the wire and has the same direction as the wire
and B is the magnetic field vector which has a magnitude and a direction.
The magnitude of this force is given by i*|L||B| |sin theta| where theta is the angle between L and B. As you've seen from VCE physics, if the current wire is parallel to the field, no force results.
Likewise, if it's perpendicular, the sine function returns the maximum value of 1. Hence the force is maximum when the external magnetic field is perpendicular to the conductor.
Read up on wiki if you're not sure about cross products in general.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Post by: Conic on July 24, 2013, 07:00:33 pm
Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Post by: lzxnl on July 24, 2013, 07:56:27 pm
Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
Post by: SocialRhubarb on July 24, 2013, 08:25:12 pm
Post by: Conic on July 24, 2013, 08:41:58 pm
I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".They use them without knowing that they are using them :P
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
Post by: sin0001 on July 25, 2013, 11:08:14 pm
Post by: lzxnl on July 26, 2013, 04:24:27 pm
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V
Someone check my working.
Post by: sin0001 on July 26, 2013, 05:06:34 pm
They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.Thanks :)
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V
Someone check my working.
It mathematically makes sense but are we allowed to solve it using differentiation? Otherwise, is there some alternative way of solving this (i.e. how would one answer it graphically?)
Also, can you please explain the 'cos(wt)' part of the flux formula?
Post by: lzxnl on July 26, 2013, 05:50:31 pm
Graphically? It's like asking someone what the maximum rate of change of a sine curve is. Graphically? Good luck with that. I'd be stuck for an answer.
As for the cosine term...I'm not surprised you ask. VCE physics's definition of flux has a serious problem. It only considers the one special case where the magnetic field is perfectly perpendicular to the surface/parallel to the surface normal. Now for a general magnetic field, if you resolve it into components parallel and perpendicular to the surface, the component parallel to the surface will not pass through the surface as they are parallel, so no flux there. Only the perpendicular component contributes to the flux, so if theta is the angle between the perpendicular component of the magnetic field and the field itself, the perpendicular component has magnitude B cos theta. Therefore the flux is BA cos theta.
Now initially, theta = 0 in the diagram; the field lines are perfectly perpendicular to the surface. As it rotates with constant angular velocity w, the angle theta at any time would be given by wt + c where c is a constant. But when t = 0, theta = 0, so c=0. This is where the cos(wt) term comes from.
Post by: sin0001 on July 26, 2013, 06:47:57 pm
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!
Post by: lolipopper on July 27, 2013, 03:07:53 pm
I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!
Dont even start with the Hinemann book. It is the worst. Im using an old Jacaranda one to accompany my understanding and a few youtube videos.
Post by: Shyam995 on July 31, 2013, 01:06:36 pm
Just been wondering, as i love maths and physics, i thought this question can bring out some curiosity.
As i have been researching, i was actually fascinated about the whole concept of a "second".
As i have been told and have studied, the result:
"the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."
I have a possible rebuttal to this "too accurate" theory.
If this is a second, does that mean that, at the big bang, how the different phases of the explosion measure in seconds, is this deemed to be inaccurate?
- sorry for the defective expression, as all of you know already, physics requires a substantial amount of communication and reasoning, as it can be hard to express, especially in writing.
Post by: lolipopper on August 02, 2013, 08:37:10 pm
A double slit arrangement is illuminated by light that consists of two wavelengths, one of which is known to be 600nm. the interference pattern on a screen shows that the fourth dark fringe for the known wavelength coincides with the fifth bright fringe for the unknown wavelength. What is the unknown wavelength?
Post by: lzxnl on August 02, 2013, 09:21:32 pm
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm
Post by: lolipopper on August 02, 2013, 09:26:48 pm
Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm
are these sort of q's in vce?
Post by: lzxnl on August 02, 2013, 09:32:06 pm
Post by: Robert123 on August 07, 2013, 02:07:09 pm
So by the way I understand it is they change the current and voltage while (ideally) keeping the same power overall. A step up transformer increase voltage and decrease current while a step down transformer decrease voltage but increase current.
The difficulty I have is to relate this to Ohm's law (V=IR) which state that if you have an increase or decrease in voltage or current, you will get a proportional increase or decrease in the other. That is, you can't decrease one while increasing the other.
So, do transformers change the resistance? Could someone please explain to me how transformers work while taking into consideration ohms law.
Thank you
Post by: SocialRhubarb on August 07, 2013, 05:24:02 pm
Let's take a 5V RMS AC source, connected to a step-up transformer with a 1:10 ratio. A 100 Ohm resistor is connected to the secondary coil. Let's try to work out the current being drawn.
The voltage across the secondary coil is pretty straight forward to work out:
Applying Ohm's law:
Then we can find the current across the primary:
Notice that if we were to attach a 50 Ohm instead of a 100 Ohm resistor, we would double the current through the secondary coil and also double the current through the primary coil. This is probably one of the concepts students find hardest to grasp - changing the resistance across the secondary coil changes the current in the primary coil as well as the secondary coil. Why? It's a bit difficult to explain, but it's to do with the primary coil inducing a back EMF on itself. For this reason, you also can't treat the primary coil as a conventional 'circuit' - the coil induces a back EMF which almost totally opposes the voltage drawn, even if there are no physical resistors in its path. Also, the current produced by a power source is not fixed - it can vary and will vary depending on the external circuit, such as what kind of transformers and resistors it is attached to.
So Ohm's law still applies in circuits with transformers, but we also have to take into account the effect of electromagnetic induction, and remember that generally speaking the current generated by a power source is not constant, but varies depending on the circuit it's attached to.
Post by: Robert123 on August 07, 2013, 05:56:35 pm
I= 0.5/100= 5*10^-3A
P=VI
P= 0.5 * 5*10^-3= 2.5*10^-3 W
For the step up transformer in the example you provided, the power is 25W which is greatly differently to the result above (transformers are meant to have the same power). This is due to not taking into account the increase in current but that means ohms law is wrong. Correct?
So then, how can ohms law and transformers agree?
Post by: SocialRhubarb on August 07, 2013, 06:10:29 pm
Post by: ~T on August 11, 2013, 02:18:28 pm
Sometimes it talks about the extent of diffraction being most significant when the wavelength approaches the size of the gap, and other times it suggests/states the proportionality: extent of diffraction is proportional to wavelength/gap.
I presume the latter is most correct and - as such - wavelengths that are larger than the gap size will produce even more diffraction than wavelengths equal to the gap size?
Thanks
Post by: xenon2013 on August 11, 2013, 03:00:09 pm
A sprinter reaches his maximum speed vmax in 2.5 seconds from rest with constant
acceleration. He then maintains that speed and finishes the 100 meters in the overall
time of 10.40 seconds. Determine his maximum speed vmax.
Thanks
Post by: lzxnl on August 11, 2013, 04:21:37 pm
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s
Post by: Homer on August 11, 2013, 04:50:19 pm
Post by: xenon2013 on August 11, 2013, 04:57:17 pm
So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s
im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s
sorry if this is confusing but im having trouble understanding it all
Post by: lzxnl on August 11, 2013, 05:15:05 pm
im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s
sorry if this is confusing but im having trouble understanding it all
OK. I'll clarify my working.
a is the acceleration in the period where he is accelerating. The acceleration is only equal to a in that time. When he is travelling at max speed, the acceleration is zero.
He accelerated for 2.5 seconds. He ran for 10.4 seconds in total. Therefore the time in which he maintained the max speed is 7.9 seconds. What I subbed in was 1/2*at^2, the distance travelled during the acceleration, and 7.9*vmax, or 7.9*2.5a, which is the distance travelled at constant speed.
Post by: lolipopper on August 15, 2013, 06:47:40 pm
Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?
Post by: lzxnl on August 15, 2013, 07:11:02 pm
i gotta question:
Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?
Assuming that the coil is rotating at a constant rate:
The magnetic flux is actually given by BA*sin(theta) where theta is the angle between the magnetic field lines and the coil. As the rotation is at a constant rate, theta = wt where w is the angular velocity and t is time. I am assuming that at t = 0 there is no flux for simplicity.
Then, flux = BA*sin(wt)
By Faraday's law, emf = -rate of change of flux = -d(flux)/dt = -wBA*cos(wt)
So, for the magnitude to be a maximum, we want wt = some integer multiple of pi. This corresponds to the field being parallel to the coil.
When the field and coil are perpendicular, cos(wt) = 0, i.e. minimum magnitude of the emf.
Post by: sydneyboy on August 18, 2013, 07:32:20 pm
"Given Fnet = mv^2/r
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "
Not sure how to star, any pointers would be appreciated.
Post by: lolipopper on August 18, 2013, 08:54:00 pm
also when radius increases, increase the velocity so the acceleration remains same. and vice versa when radius decreases.
hope it helps.
Post by: lzxnl on August 18, 2013, 11:10:19 pm
We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:
"Given Fnet = mv^2/r
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "
Not sure how to star, any pointers would be appreciated.
v = radius * angular velocity
= rw
v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2
The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.
Post by: P0ppinfr3sh on August 21, 2013, 06:30:28 pm
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
Post by: lolipopper on August 21, 2013, 06:51:58 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
can you tell me which question it is? is it from hinemann?
Post by: SocialRhubarb on August 21, 2013, 06:55:35 pm
Post by: P0ppinfr3sh on August 21, 2013, 06:58:31 pm
can you tell me which question it is? is it from hinemann?Well question 5 and questions 8 of 10.4 involve peak induced voltage and magnetic flux; and yes, it's from Heinemann.
Find the average RMS voltage and multiply byto get the peak voltage.
How would you go about finding the average RMS voltage of a generator? I can find the average induced voltage but I don't think this is equivalent to the average RMS voltage.
Also, I might as well post the related question(s) from the textbook:
5. 100 turns, area=20cm^2, B=5.0mT, coil rotates at a rate of 15degrees per millisecond. What is the peak value of the induced EMF for this coil?.
ANS: 0.263V.
8. Peak voltage=8.0kV, N=1000 turns, each coil has a radius of 10cm, magnetic field strength=B, frequency=50Hz. Calculate the strength of the magnetic field required to produce a peak voltage of 8.0kV.
ANS: B=0.81T.
Thanks.
Post by: 09Ti08 on August 21, 2013, 07:07:12 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
Faraday's law: emf=-d(phi)/dt
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
Post by: P0ppinfr3sh on August 21, 2013, 07:37:59 pm
Faraday's law: emf=-d(phi)/dt
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
Yeah, that seems to make sense, but I don't really know how to apply it to the questions I've mentioned above. :-\
Post by: lzxnl on August 21, 2013, 08:08:04 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
Find the average RMS voltage and multiply by
Erm, RMS voltage and multiply by square root 2 only lets you find the RMS voltage of a sinusoidally voltage. Something whacky like a square wave in which the voltage is equal to V for half a period and -V for the other half would have an RMS value of V. Just an example.
Besides RMS has nothing to do with this question.
Faraday's law: emf=-d(phi)/dt
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
Indeed. Another instance of where VCE physics just fails. You have flux = BA*cos(theta)
but as theta is changing at a constant rate (constant rotation), theta = wt
so flux = BA*cos(wt)
differentiate: emf = BAw*sin(wt) with the sign depending on how the voltage is measured
So peak emf is BAw
Post by: Alwin on August 21, 2013, 08:26:56 pm
Yeah, that seems to make sense, but I don't really know how to apply it to the questions I've mentioned above. :-\
How would you go about finding the average RMS voltage of a generator? I can find the average induced voltage but I don't think this is equivalent to the average RMS voltage.
Also, I might as well post the related question(s) from the textbook:
5. 100 turns, area=20cm^2, B=5.0mT, coil rotates at a rate of 15degrees per millisecond. What is the peak value of the induced EMF for this coil?.
ANS: 0.263V.
8. Peak voltage=8.0kV, N=1000 turns, each coil has a radius of 10cm, magnetic field strength=B, frequency=50Hz. Calculate the strength of the magnetic field required to produce a peak voltage of 8.0kV.
ANS: B=0.81T.
5. This is what happens when you TRY to calculate peak voltage with RMS...
It rotates 15 degrees per millisecond. So, in 6 milliseconds it turns 90 degrees. Say if it start parallel to the field, then after 6 milliseconds it had turned to perpendicular to the field.
^ close but no cigar. You should note that from the textbook, a rate of 15° per millisecond (i.e. a frequency of 42 Hz) is not actually correct since not accurate enough.. it is actually 15.12 degrees per millisecond.. but that's still a bit off the correct answer
What you should actually do is use the formula:
8. Pretty much same thing,
The maximum EMF is equal to NBA2πf, so:
EDIT: didn't see nliu's post since was latexing. If you're wondering w is angular velocity, ω=2 pi f
Hope it helps / makes sense :)
Post by: P0ppinfr3sh on August 21, 2013, 08:38:27 pm
You're all awesome!
(I can now sleep peacefully)
Post by: SocialRhubarb on August 21, 2013, 09:57:19 pm
Post by: lzxnl on August 21, 2013, 10:13:58 pm
I guess sometimes you just have to use calculus, huh.
But that's too difficult for the delicate minds of VCE physics students.
Post by: BasicAcid on August 21, 2013, 10:47:52 pm
But that's too difficult for the delicate minds of VCE physics students.
Well you seem to be one of those guys who has no respect for the VCE design (for all subjects) even though you'll be one of the people who simply ace VCE.
May I ask why?
Also is it wrong that I never fully understood motion in physics (which we was the first thing we did in term 1) until I began kinematics in spesh a few weeks ago?
And do you know if we're allowed to use I and J systems in physics?
Post by: lzxnl on August 21, 2013, 11:03:59 pm
Well you seem to be one of those guys who has no respect for the VCE design (for all subjects) even though you'll be one of the people who simply ace VCE.
May I ask why?
Also is it wrong that I never fully understood motion in physics (which we was the first thing we did in term 1) until I began kinematics in spesh a few weeks ago?
And do you know if we're allowed to use I and J systems in physics?
Why? VCE physics is an absolute joke. As you've seen yourself, maths teaches physics better than physics does. I'm not surprised; I actually learnt forces, energy and kinematics firstly from spesh in year ten before doing physics, and it worked perfectly fine.
Why is physics a joke?
1. They cut down the maths. In a subject which is built on maths. It's like teaching a guy to build a house but you only give him sand. He needs to find the bricks somewhere, but if you're giving them to him...
2. You have frigging notes for the exam. Come on. You can literally summarise EVERYTHING you need for the exam in four sides of paper as long as you write small. I mean come on. That's really lazy. I can understand giving the formulas, but letting us bringing in a cheat sheet? It's a wonder 90% is still an A+ on the exams.
3. Why the frick are we repeating motion after doing it in year 11? Why the frick are we going over basic circuits when that should have been covered in year 8, at most year 10?
4. Why the bloody heck are we cutting a course down when we're increasing the study time for the course by removing the mid-year exam study period and making it available for class time? Hint. English and Maths all have three hours worth of exams in total, but they didn't cut anything down.
5. Actually test us on physical knowledge instead of boring subbing into equations. I don't know how many I used E = hc/wavelength for revision for my last SAC.
That rant enough for you?
If you use vectors in physics, it'll work, but there's no need to use something so advanced for VCE physics. I'm not kidding.
Post by: lzxnl on August 21, 2013, 11:28:21 pm
nliu1995, I am very intrigued by you at the moment.
Never seen anyone who seems to actually be capable of virtually destroying VCE with minimal effort.
But what? Spesh in year 10? In year 10 I was still learning what a parabola was lol...
I'm just a more outspoken member of the "dissatisfied with VCE" group. There are heaps of people like me. They just don't want to be so vocal for fear of offending people.
You should read pi's VCE Physics 3/4 review that's stickied. It's even better.
lol well, I spent too much on maths in earlier life. Parabolas was year five for me. I still remember fumbling around to understand what this f(-b/2a) meant in primary school.
Post by: lzxnl on August 21, 2013, 11:42:23 pm
I am simply speechless...
Man what the hell, although deep down know you're telling the truth, I simply cannot believe what you are saying.
Year 5... Pokemon Emerald/FireRed dominated my life throughout that year?
I only learnt what that f(-b/2a) thing meant last year lol.
You see, I am where I am now because I sacrificed gaming time for study. I looked like a dork then. I was even bashed one day for having a year eleven maths book in year six. But now, I'm laughing at those who laughed at me. It's well worth it :D
I think my decision to pre-learn (at least read upon) all the uni courses that I'll be doing next year in the long break makes sense now.
Post by: sydneyboy on August 25, 2013, 05:48:55 pm
v = radius * angular velocity
= rw
v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2
The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.
Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
Post by: lzxnl on August 25, 2013, 06:04:17 pm
Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
t = 2pi*r/v. You can only keep one of the three variables constant.
Post by: Alwin on August 25, 2013, 06:09:41 pm
I'm just a more outspoken member of the "dissatisfied with VCE" group. There are heaps of people like me. They just don't want to be so vocal for fear of offending people.
...
lol well, I spent too much on maths in earlier life. Parabolas was year five for me. I still remember fumbling around to understand what this f(-b/2a) meant in primary school.
with him on this^
I just cbbs ranting here + its a question thread not a bagging thread ;) (I just +1 him everytime ;) )
I'll just add for me parabolas in grade 3. as for the f(-b/2a) i 'proved' it by taking the mean of the two intercepts that can be found by the quadratic formula. Then proved it a couple years later with calculus, which is SO hard on physics... esp since on some exams they don't even bother to draw graphs correctly since we can count the squares XD
anyways:
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
you want it to increase your r.. say k=v^2/r, so when v increase you need r to increase (at a rate proportional to the square of the increase of v). Otherwise, v increases r stays same -> k increases (ie not constant = fail).
BUT you are testing the relationship between F and m. So, you want to keep everything else (r, T, v etc) constant.
you change the mass...
Post by: lzxnl on August 25, 2013, 06:36:13 pm
Post by: Alwin on August 25, 2013, 06:46:45 pm
To be honest, I had no idea what he wanted, as simply investigating the relationship between F and m is...trivial. F = ma arises from a definition of force in Newtonian mechanics and with a constant mass. So why would anyone measure the relationship between F and m? It's just measuring your experimental capabilities.
its not like vce physics would test maths skills :P
I've given it some thought, sydneyboy, and i think i have an idea of what your prac is testing: try different masses and plot a graph of F vs m?? It should be straight line since when v^2/r constant F ∝ m. If it's not straight, then you talk about errors like not keeping v^2/r constant...
Is this interpretation right?
Otherwise just post the prac sheet and we (esp nliu since he <3 vce physics :P ) will do our best to try help!! Or, ask your teacher for clarification.
EDIT (in spoiler):
SECRET TO SUCCEEDING IN ANY PRAC
Just copy what the smart people in your class are doing :P Or worse comes to worse make up/tweak your data results after the prac then write some cohesive BS about errors and stuff :D
Or just blame the teacher hahahaha. that always goes down well..
Or just blame the teacher hahahaha. that always goes down well..
Post by: lolipopper on August 25, 2013, 08:22:11 pm
Post by: sydneyboy on August 25, 2013, 08:49:06 pm
"Given Fnet = mv^2/r
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing." on a piece of paper.
@Alwin im sure your interpretation is right. I have 2 periods left to record data and plot the graph. I think it will be a tedious task to make everything constant.
Thanks for the help guys I'll have a crack tomorrow i have a much better idea of what to do now.
Post by: SocialRhubarb on August 25, 2013, 09:34:56 pm
It might help to know what equipment you've been given.
But honestly, I have no idea how to approach this. Obviously you want to keep v^2/r constant, but I suppose the question is how can you best do it, so as to minimise the error in your results. Especially since it's telling you to keep it constant "even when radius is changing".
My first thought was to swing it in a vertical arc, and then use gravity to kind of 'calibrate' your v^2/r value, but that really presents more problems than solutions, especially if you're swinging your mass by hand.
I suppose I'll have to go with what everyone else is saying and just say "keep the period and radius constant" even though it's not really answering the question of how to keep v^2/r constant with a changing radius.
Post by: sydneyboy on August 25, 2013, 09:43:13 pm
Post by: sydneyboy on August 25, 2013, 09:44:28 pm
Post by: lolipopper on August 27, 2013, 11:53:27 pm
Post by: Lasercookie on August 28, 2013, 04:10:34 pm
EDIT (in spoiler):I'm guessing you meant it tongue in cheek, but anyway :P it depends if your teacher is expecting you to get the correct results from the experiment though, a well designed practical sac would allow you to still do pretty well if you're able to explain how your data turned out to be messed up.SECRET TO SUCCEEDING IN ANY PRACJust copy what the smart people in your class are doing :P Or worse comes to worse make up/tweak your data results after the prac then write some cohesive BS about errors and stuff :D
Or just blame the teacher hahahaha. that always goes down well..
Post by: lolipopper on September 02, 2013, 08:00:09 pm
A proton and electron are accelerated from rest through the same potential difference. Given that their masses are in a ratio of 1836:1 , find the ratio of their de Broglie wavelengths.
Post by: SocialRhubarb on September 02, 2013, 08:14:50 pm
Using the formulas for momentum and energy of a particle,
Post by: saifh on September 02, 2013, 10:06:18 pm
1. Describe the photoelectric effect's implications for the wave model of light and how it forced the reintroduction of a particle model or photon model for light
Post by: Conic on September 02, 2013, 10:17:49 pm
Post by: saifh on September 02, 2013, 10:34:15 pm
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.
explains it, thanks!
Post by: lzxnl on September 02, 2013, 10:38:12 pm
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.
Erm...higher intensities DO produce more current. Another piece of evidence was the fact that intensities had no effect on the energies of the emitted electrons.
Post by: Conic on September 02, 2013, 10:39:27 pm
Erm...higher intensities DO produce more current. Another piece of evidence was the fact that intensities had no effect on the energies of the emitted electrons.True. I'll edit the post.
Post by: saifh on September 03, 2013, 12:18:12 pm
Explain the role of a reverse bias voltage and an ammeter in the photoelectric effect
Post by: Robert123 on September 03, 2013, 07:35:33 pm
QuestionA reverse bias voltage can be used to find the stopping voltage (ie. when current=0A). In doing so, the kinetic energy of the electrons (since voltage=energy/coulomb(1.609*10^19 electrons I believe). From this, we can determine the work function of the metal (amount of energy to liberate
Explain the role of a reverse bias voltage and an ammeter in the photoelectric effect
electrons=hf-KE)
Why is momentum measured in Ns for some questions? I know the units all add up but I thought that you just keep it as kgms^-1
Post by: ~T on September 09, 2013, 02:06:39 pm
Why is momentum measured in Ns for some questions? I know the units all add up but I thought that you just keep it as kgms^-1I presume it's from Newton's second Law, which in its true form is
Presuming a constant mass, and given
we get
...which should hopefully seem familiar!
But if we leave it in its original form and presume a constant force, then we can write
Rearranging yields
Thus, momentum is in Ns
Post by: Chazef on September 15, 2013, 03:04:10 pm
Post by: BasicAcid on September 15, 2013, 04:47:05 pm
Do we need to know about the thermal motion of electrons (I think that's what it's called)? A question came up about it in the second atarnotes exam from the study guide in regards to an incandescent globe
I'm pretty sure not but the guys at ATARnotes seem reliable.
Mind posting/PMing me the question? Maybe you're misinterpreting the question or something. :P
Post by: Robert123 on September 18, 2013, 04:43:07 pm
By having more modes in fibre optics, does that give any more advantages?
Post by: joey7 on September 23, 2013, 01:44:24 pm
If someone explain to me how this is wrong that would be great
Post by: availn on September 23, 2013, 01:57:32 pm
In the examiners report of last years second exam, in relation to the last question it stated "a common misconception was that the electrons moved around the orbit in a wave pattern"
If someone explain to me how this is wrong that would be great
The standing waves you draw around the nucleus are not the paths the electron takes, i.e. they do not continually move towards and away from the centre of orbit. The electrons are a matter wave, so the standing waves show the possible circumferences the electrons can use to orbit, because with these circumferences, constructive interference occurs.
Post by: SocialRhubarb on September 23, 2013, 02:33:05 pm
The electron exhibits both particle-like properties and wave-like properties. They have a rest mass, like a particle, but it's easiest to explain the orbitals of electrons using a wave model for electrons. The electrons can only form orbits when the 'wavelength' of the electron fits exactly into the circumference of their orbit, and they form a 'standing wave' structure. The electron cannot take any other orbital, or else it will 'destructively interfere' with itself, and so it can only exist at certain orbits and at certain energy levels. If you do chemistry, this is what causes the shells and sub-shells.
So the electron's orbit is not a wave-pattern, but instead we find the orbits by treating the electron as a wave.
Post by: lzxnl on September 23, 2013, 03:57:48 pm
The electrons don't orbit in a 'wave' pattern - electrons are the wave. This is probably a bit tricky for most people to understand, since it's difficult to comprehend something being a particle and a wave at the same time.
The electron exhibits both particle-like properties and wave-like properties. They have a rest mass, like a particle, but it's easiest to explain the orbitals of electrons using a wave model for electrons. The electrons can only form orbits when the 'wavelength' of the electron fits exactly into the circumference of their orbit, and they form a 'standing wave' structure. The electron cannot take any other orbital, or else it will 'destructively interfere' with itself, and so it can only exist at certain orbits and at certain energy levels. If you do chemistry, this is what causes the shells and sub-shells.
So the electron's orbit is not a wave-pattern, but instead we find the orbits by treating the electron as a wave.
At the same time...electrons in multi-electron atoms do not form orbitals solely when the wavelength of the electron fits the circumference of the orbit. There are electron-electron repulsions which cannot be neglected. This is why the Bohr model of the atom fails completely for even helium.
The VCE course isn't entirely right when they use the fitting standing waves argument to atoms in general. It is only valid for species with one electron, such as H, He+, Li 2+ etc...even Au 78+.
Just keep this in the back of your mind. VCE only requires the answer that electrons form standing waves around the nucleus, although they generally don't.
Post by: sin0001 on September 23, 2013, 04:27:28 pm
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
-For 6b, isn't the Impulse meant to calculated using the resultant force, which is 17 N up (i.e. 17 N x 1.5s)? In Itute's solutions, the force of 22 N was used o.O
-For question 16a, shouldn't the graph of the induced EMF begin under the x-axis from 0 to 1 seconds and be drawn above the x-axis for 2 to 4 seconds? Well Itute, their solutions, have drawn it starting from above the x-axis; who's wrong?
-And finally, for question 18, why is the total resistance of the two wires 5 ohms? I thought the two wires would represent a parallel circuit so that the current splits up equally amongst them, so that the total resistance becomes 1.25 ohms (using: 1/R= (1/r + 1/r))
Thanks, would appreciate any help!
Post by: SocialRhubarb on September 23, 2013, 04:35:30 pm
It asks for the impulse generated by the 22N force alone. A separate impulse is provided by gravity and together they make up the resultant change in momentum of the rocket.
16a.) Doesn't specify a direction for the voltmeter. Both solutions should be correct.
18.) It's not a parallel circuit, it's a series circuit.
Post by: Robert123 on September 27, 2013, 02:30:26 pm
How does the medium (air vs water) influence the Fringe Spacing?
Electrons of known energy are fired into Mercury vapour. The energy of the scattered electrons is then measured. When electrons of energy from 0eV to 4.8eV are fired into the mercury vapour, the energy of the scattered electrons equal the energy of the incident electrons. At 4.8eV the energy of some of the scattered electrons falls to zero. Which of the following statements best explains this observation?
The answer is...
Inelastic collisions within the atoms can occur for electrons of energy 4.8eV, but not at lower energies.
Could someone please explain this to me.
For electronics...
If a diode with a turn on voltage of 0.7V is in parallel with a 6.0V battery as well as a resistor, what would be the voltage drop across the resistor? Would it matter whether the diode is in forward bias or reverse bias ?
And for electric power...
Also, are we meant to know how to work the EMF generated from a moving object in a constant magnetic field using EMF=BLv as it was in the STAV 2013 practice exam?
Thanks
Post by: lzxnl on September 27, 2013, 02:40:08 pm
Some Questions related to light and matter...
How does the medium (air vs water) influence the Fringe Spacing?
Electrons of known energy are fired into Mercury vapour. The energy of the scattered electrons is then measured. When electrons of energy from 0eV to 4.8eV are fired into the mercury vapour, the energy of the scattered electrons equal the energy of the incident electrons. At 4.8eV the energy of some of the scattered electrons falls to zero. Which of the following statements best explains this observation?
The answer is...
Inelastic collisions within the atoms can occur for electrons of energy 4.8eV, but not at lower energies.
Could someone please explain this to me.
For electronics...
If a diode with a turn on voltage of 0.7V is in parallel with a 6.0V battery as well as a resistor, what would be the voltage drop across the resistor? Would it matter whether the diode is in forward bias or reverse bias ?
And for electric power...
Also, are we meant to know how to work the EMF generated from a moving object in a constant magnetic field using EMF=BLv as it was in the STAV 2013 practice exam?
Thanks
Question 1:
Medium affects the speed of the wave. Frequency cannot change => wavelength changes. Fringe spacings affected.
Question 2:
I don't quite like how it says "energy falls to zero". Kinetic plus potential? I think the point is that if electrons from 0 eV to 4.8 eV are fired, they do not have enough energy to knock out another electron, so they just rebound back with the same energy. At 4.8 eV, some of the mercury bound electrons are given enough energy to just escape, so those electrons will have zero energy.
Diode question:
If it's in parallel, diode in forward bias takes 0.7 volts. So does resistor, as the voltage drops are the same in parallel circuits.
If the diode is reverse biased, no current flows through the diode. All current flows through the resistor instead. The diode essentially has infinite resistance. Voltage across diode is 6.0V, and so is the drop across the resistor.
As for electric power...it's just a formula, another what, two lines on your cheat sheet? Can't be that bad.
Post by: BasicAcid on September 27, 2013, 05:10:16 pm
As for electric power...it's just a formula, another what, two lines on your cheat sheet? Can't be that bad.
That's literally 7 characters (EMF=Blv)
The word commutator takes up more space than that lol
Post by: lzxnl on September 27, 2013, 06:42:21 pm
That's literally 7 characters (EMF=Blv)
The word commutator takes up more space than that lol
You'd need an explanation of what the formula is and what it means.
Oh wait, wrong subject.
Post by: ~T on September 27, 2013, 07:53:27 pm
Question 3 asks "which of the lights had a greater intensity?" with options "blue," "ultraviolet" or "unable to determine." The answers say "unable to determine" because the intensity of the light isn't the only determining factor in the current - higher kinetic energy will also make a higher current, and this depends on the frequency of the incident light.
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
Post by: ~T on September 27, 2013, 08:00:16 pm
Post by: lzxnl on September 27, 2013, 10:26:00 pm
TSFX Exam 2 2010, Light and Matter question 3. Basically, photoelectric effect measured on a metal with blue light and with ultraviolet light. The graph of current vs. potential difference shows that blue light produces a higher current for the forward potentials.
Question 3 asks "which of the lights had a greater intensity?" with options "blue," "ultraviolet" or "unable to determine." The answers say "unable to determine" because the intensity of the light isn't the only determining factor in the current - higher kinetic energy will also make a higher current, and this depends on the frequency of the incident light.
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
I can see the issues here. What we're used to dealing with is the MAXIMUM kinetic energy of the electrons, and that is given by hf-W. However, not all of the electrons are at this max value. At a certain voltage, all of the electrons do get to this max value, which is why we eventually see the graph smoothen out. Before this, however, varying proportions of electrons are not at this max value. We can only work out what the maximum energy is; we don't know the energy distribution of the electrons though.
Post by: SocialRhubarb on September 27, 2013, 11:27:44 pm
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
If you have a laser shooting, say, 10000 photons of blue light per second, and a laser which is shooting 10000 photons of ultraviolet light per second, both on the same area, which laser has the higher intensity?
Well, the ultraviolet laser, since both the lasers emit the same number of photons, but the photons of ultraviolet light carry more energy per photon.
And that's really the issue here. Yes, from the graph we can tell that there are more photons of blue light being shone onto the metal. But intensity isn't only affected by the number of photons you shine, it's affected by the energy carried by each of those photons too. So while it is true that there are less photons of ultraviolet light, since each photon carries more energy than photons of blue light, it may be of a higher intensity, or lower intensity as well. It is unable to be determined.
Post by: lolipopper on October 03, 2013, 10:37:18 pm
Post by: lzxnl on October 03, 2013, 10:46:53 pm
So now resolve vertical components.
2T*sin 25 = mg
T=mg/(2 sin 25). Whatever that is.
Post by: lolipopper on October 03, 2013, 11:28:58 pm
Post by: lzxnl on October 03, 2013, 11:35:02 pm
If so...the tension will not be doubled. Even so. It's like one of those pulley questions. You could put a pulley there and it wouldn't change the force analysis.
Post by: lolipopper on October 04, 2013, 12:02:08 am
Post by: lzxnl on October 04, 2013, 01:30:18 am
Post by: lolipopper on October 04, 2013, 02:50:57 pm
Post by: lolipopper on October 04, 2013, 03:06:45 pm
Yep along with thousands of other physics students.
I've got a fair few friends who have physics as their last exam as well.
We're all planning on keeping our cool together after they say pens down in the physics exam (we don't want to be reported to VCAA or anything haha) then we're gonna go absolutely ballistic outside.
Planning on getting smashed on the footy oval, hopefully it's a nice day as well considering it'll be halfway through November!
for me its gonna be a full day of sleep and PS4 session.
Post by: BasicAcid on October 04, 2013, 03:25:58 pm
for me its gonna be a full day of sleep and PS4 session.
Plenty of time for that after the day in my signature, I'm buying GtaV on the 14th lol
Post by: lzxnl on October 04, 2013, 06:56:28 pm
Post by: barydos on October 04, 2013, 07:12:23 pm
Questions: http://imgur.com/a/UhAma
Answers: http://imgur.com/wab3ASk
1) How come the graph needed to be inverted?
2) How come it clips at 3.0 V?
Thanks
Edit: Also, it's fine for explanation questions to be answered in dot points, right? Or are we required to write full sentences?
Post by: BasicAcid on October 04, 2013, 09:05:44 pm
Just got some electricity questions from the VCAA 2007 exam
Questions: http://imgur.com/a/UhAma
Answers: http://imgur.com/wab3ASk
1) How come the graph needed to be inverted?
2) How come it clips at 3.0 V?
Thanks
Edit: Also, it's fine for explanation questions to be answered in dot points, right? Or are we required to write full sentences?
Yeah dot points are fine and wtf, I haven't done any VCAA physics exams yet (except for the 2013 sample one... That seemed incredibly easy) but I actually have no idea why it's inverted o_o
Oh well, another thing to add to my cheat sheet when nliu answers haha
Lucky. For me, after the physics exam, all I can say is three down, three to go.
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
Post by: Alwin on October 04, 2013, 10:37:09 pm
Oh well, another thing to add to my cheat sheet when nliu answers haha
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
Im sorry to disappoint for I'm not nliu (@ nliu you have no idea what autocorrect on my iPod suggested for your nick LOL)
1) Anonymiza youre actually doing a paper from an older study design when students where actually required to know how things worked D; D; D; IKR THE SHOCK
2) Most simply put, what you have us a single transistor amplifier. Because you have an NPN transistor, when a large input signal comes in through the base, a "smaller" output voltage signal I recorded. Note if it was a double transistor amplifier then it would be non inverting
3) I really don't want to go into capacitors specifically, only because the textbook brushes on n-p and p-n junctions (not in stuy design) but no capacitors or the saturation point or cutoff point of transistors.
It's suffices to say you WON'T get a question like this. it's nice to know how an amplifiers work so you can answer questions like this, but 2013 VCE physics just doesnt require this standard.
EDIT: Sorry if I sound a bit too cynical of the physics subject. I put it down to my old age and that I did quite a few older exams (2009 was when they made the cuts/change ) just to spice up my revision last year :P
Post by: barydos on October 04, 2013, 11:08:38 pm
In sorry to disappoint for I'm not nliu (@ nliu you have no idea what autocorrect on my iPod suggested for your nick LOL)
1) Anonymiza youre actually doing a paper from an older study design when students where actually required to know how things worked D; D; D; IKR THE SHOCK
2) Most simply put, what you have us a single transistor amplifier. Because you have an NPN transistor, when a large input signal comes in through the base, a "smaller" output voltage signal I recorded. Note if it was a double transistor amplifier then it would be non inverting
3) I really don't want to go into capacitors specifically, only because the textbook brushes on n-p and p-n junctions (not in stuy design) but no capacitors or the saturation point or cutoff point of transistors.
it's suffices to say you WON'T get a question like this. it's nice to know how an amplifiers work so you can answer questions like this, but VCE physics just doesnt require this standard.
1) LOL
2,3) I can see why you got 49 last year! Thanks a lot, much appreciated :)
Post by: lzxnl on October 04, 2013, 11:28:23 pm
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
Yep, English the day after. Which is going to be one hell of an exam.
And please don't give me that. At the rate I'm going at, I'll be surprised if I even get a 50 in anything.
Post by: barydos on October 05, 2013, 11:59:49 am
Post by: ~T on October 05, 2013, 12:05:42 pm
Post by: barydos on October 05, 2013, 12:15:23 pm
I believe both are acceptable, I usually just writeor whatever the gain is if it is inverting, otherwise just the number.
Thanks :)
Another question: for the magnetic field graph, is it better to have the corresponding induced EMF graph to be joined (with vertical lines) or not?
(see this image: http://imgur.com/a/Gt5E2)
Post by: ~T on October 05, 2013, 12:17:43 pm
Post by: barydos on October 05, 2013, 12:29:55 pm
Again, I'm not strictly certain on this, but I usually do dotted vertical lines. I doubt they are pedantic enough to pick up on either of your two questions though :)
That's good to hear haha thanks!
Post by: lzxnl on October 05, 2013, 12:39:41 pm
Post by: ~T on October 05, 2013, 09:40:44 pm
Post by: barydos on October 06, 2013, 12:57:31 pm
Spoiler
At the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."
Spoiler
Young’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
Post by: Aurelian on October 06, 2013, 01:12:50 pm
In the VCAA 2009 paper there was a 3-mark question:Assessment Report Comment:SpoilerAt the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."Is a response like that enough for the whole 3 marks??SpoilerYoung’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Personally I would add what the particle model would predict for the experiment, and perhaps mention that nature of the interference pattern actually observed as well (i.e. alternating dark and light fringes). I do think that some examiners would only award you two marks for the response given there by the examiner's report...
Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
Why not just write down the working to be safe? Usually you'll only need one or two very short lines =) You'll also minimize the chance of making a silly mistake, as well as enabling the examiner to give you one of the two marks if you get the answer wrong but have valid working.
Post by: barydos on October 06, 2013, 01:27:32 pm
Personally I would add what the particle model would predict for the experiment, and perhaps mention that nature of the interference pattern actually observed as well (i.e. alternating dark and light fringes). I do think that some examiners would only award you two marks for the response given there by the examiner's report...
Why not just write down the working to be safe? Usually you'll only need one or two very short lines =) You'll also minimize the chance of making a silly mistake, as well as enabling the examiner to give you one of the two marks if you get the answer wrong but have valid working.
Alright, I'll keep that in mind, thanks.
And yeah aha I know I know, I was just curious about it more than anything else :P
Post by: ~T on October 06, 2013, 01:29:55 pm
I guess my query is:
No working but correct answer = full marks regardless of the number of marks?
Post by: Alwin on October 06, 2013, 01:35:09 pm
In the VCAA 2009 paper there was a 3-mark question:Assessment Report Comment:SpoilerAt the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."Is a response like that enough for the whole 3 marks??SpoilerYoung’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
If I remember correctly the marking scheme for such questions went like this:
Your 1st Question
Your 1st Question
| Explain the wave model of light and what it predicts (interference pattern of light and dark) | 1 Mark |
| Explain the particle model of light and what it predicts (no such pattern, just one light band) | 1 Mark |
| Young's DS Experiment supported the wave model of light because a fringe interference pattern was observed on the screen | 1 Mark |
You have to remember that most assessment reports are brief because they don't want students to memorise answers from the reports. So, they just touch on important points
Your 2nd Question
Your 2nd Question
| Correct formula with wrong values or wrong conversion substituted in | 0 Marks |
| Correct formula with right values substituted in but wrong answer | 1 Mark |
| Correct formula with right values substituted in and correct answer | 2 Mark |
| No/wrong formula and wrong answer | 0 Marks (no duh) |
| Wrong formula/working but somehow correct answer | 1 Mark (you showed bad physics in your working) |
| No formula and correct answer | 1 Mark I'm pretty sure for 2013 because for short answer questions it says: |
"In questions worth more than 1 mark appropriate working should be shown." |
The moral of the story? Just show some working :))
Even if it's a MC question where anything that gets you the right answer goes (even closing your eyes and guessing without righting a single formula) you should write some working. Why? It's helps when you go back and double check your answer so you can make sure you used the right method and substituted correctly!
:D
Post by: BasicAcid on October 06, 2013, 02:42:46 pm
You have to remember that most assessment reports are brief because they don't want students to memorise answers from the reports. So, they just touch on important points
Or better yet copy down the assessor's report onto your A3 cheat sheet
Post by: lolipopper on October 06, 2013, 02:55:19 pm
"In questions worth more than 1 mark appropriate working should be shown."
where does it say that?
Post by: Alwin on October 06, 2013, 03:06:34 pm
where does it say that?
in the 2013 trial paper is says:
(http://i.imgur.com/9KVgf3d.png)
Post by: ~T on October 06, 2013, 03:48:07 pm
Post by: Alwin on October 06, 2013, 03:54:14 pm
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it."You should clean your room"
"You must clean you room"
Either way you get yelled at if you don't do it :P
But seriously, better safe than sorry. Plus, what if you did the question all on your calculator get the answer wrong. Don't even get any marks for method or working. It's just safer this way :))
I always showed full working last year, but if someone on AN wants to try it with no working at all this year be my guest and let us know how it went :) (no, this is not a personal challenge at you nliu :P)
Post by: lzxnl on October 06, 2013, 04:39:19 pm
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it.
If you're willing to take the risk, be my guest.
"You should clean your room"
"You must clean you room"
Either way you get yelled at if you don't do it :P
But seriously, better safe than sorry. Plus, what if you did the question all on your calculator get the answer wrong. Don't even get any marks for method or working. It's just safer this way :))
I always showed full working last year, but if someone on AN wants to try it with no working at all this year be my guest and let us know how it went :) (no, this is not a personal challenge at you nliu :P)
Alwin, I'm not going to potentially jeopardise my physics score just to test the system :P
Post by: lolipopper on October 06, 2013, 10:15:52 pm
in the 2013 trial paper is says:
(http://i.imgur.com/9KVgf3d.png)
Dont be smartass alright. It was a genuine question.
i just got the physics marking scheme from my teacher. it says "where questions dont specify that the working must be shown, if the correct answer is in the box the student receives full marks". Maybe you should get some suffice information yourself before absolutely classifying other peoples question invalid and, from the sound of your tone, stupid.
Post by: Phy124 on October 07, 2013, 12:15:03 am
Post by: barydos on October 07, 2013, 06:04:42 pm
Are we still required to know about incandescent (filament) light globes?
E.g. "explain, in terms of electron behaviour how light is produced in an incandescent (filament) light globe?"
Post by: lzxnl on October 07, 2013, 06:12:34 pm
In case you do, something on the lines of the thermal vibrations of the electrons creates an electromagnetic wave, light, would probably suffice.
Post by: barydos on October 07, 2013, 06:30:28 pm
I've seen those questions before in past exams. Given the cutting of the physics course down, it's possible that you're not asked to know this.
In case you do, something on the lines of the thermal vibrations of the electrons creates an electromagnetic wave, light, would probably suffice.
Thanks nliu :)
Post by: lolipopper on October 07, 2013, 09:00:34 pm
http://www.vcaa.vic.edu.au/Documents/exams/physics/2009physics1-w.pdf
Post by: lzxnl on October 07, 2013, 09:12:54 pm
It looks to me that Q, the wave from the RF source to the modulator is the carrier wave, high frequency, so A is at Q. Then R appears to be the modulated wave signal C. S is the original electric signal B.
Post by: SocialRhubarb on October 07, 2013, 09:16:50 pm
So in this case we're transferring the sound coming through the microphone on the back of a radio wave.
Into the question now:
at P, all that we have is the signal from the microphone. It's a sound wave which has been converted into an electrical signal, and the other waves which we're required to label are radio waves. In general, the frequencies of light, even in the form of radio waves, will be much higher than the frequencies of sound. So if we're trying to pick a waveform which will match the sound, we'll probably pick B, as its frequencies seems the lowest.
At Q, all the we're receiving is the 'carrier wave', the wave that will be altered to carry the sound wave. In this case it is a radio wave. The wave hasn't yet reached the modulator, so what we're looking for is a flat, uniform wave, which in this case will be A.
At R, the wave has been altered to carry the signal. The answer will be C, as you can see that the waveform has been altered to 'carry' the original sound sample on top of the radio waves being emitted. You can imagine that if you multiplied the waves in A and B together, you might get something that looks like C.
At S, the wave has been demodulated to form the original electrical signal from the microphone, so we're back to B again.
Post by: Jaswinder on October 07, 2013, 11:48:28 pm
Post by: Robert123 on October 10, 2013, 06:48:59 pm
Could someone please explain the what "numerical aperture" and "acceptance angle". I can usually get MC questions related to them right by applying the formula but I have no genuine idea what the numbers even mean.
Cheers
Post by: ~T on October 11, 2013, 07:11:57 pm
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
Post by: BasicAcid on October 11, 2013, 07:22:21 pm
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
Wow are you serious lol
Do you think you could link us/post the marking principles?
Post by: sin0001 on October 11, 2013, 07:33:05 pm
To nliu and Alwin and potentially others, I got my hands on the VCE Physics Exam marking principles today. No working, correct answer = full marks regardless of number of available marks.So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
Post by: ~T on October 11, 2013, 07:55:33 pm
http://www.2shared.com/document/0BRMUi2k/General_marking_principles.html
P.S. Enjoy <3
P.P.S. I believe this was from the chief examiner during a 2012 teachers conference
Post by: Phy124 on October 11, 2013, 08:18:11 pm
So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.
Why would anyone put just an answer when there are multiple marks allocated?
If you get the answer wrong and have no working out, then you get zero, it's as simple as that.
However, if you have the incorrect answer but have working out is certainly possible to obtain marks.
In a subject where you hear numerous stories of people mistyping things into their calculators and obtaining an incorrect final answer, I really don't understand why you would simply write an answer and nothing else, especially considering it takes, what, an extra 5 seconds to write? ???
Post by: ~T on October 11, 2013, 08:29:10 pm
Post by: sin0001 on October 11, 2013, 09:57:51 pm
Why would anyone put just an answer when there are multiple marks allocated?I'm the type of person that will show working even if the questions is worth 1 mark, there's no way I'm ever gonna write an answer without the working shown. I was just curious about whether the examiners have concrete rules for allocating marks
If you get the answer wrong and have no working out, then you get zero, it's as simple as that.
However, if you have the incorrect answer but have working out is certainly possible to obtain marks.
In a subject where you hear numerous stories of people mistyping things into their calculators and obtaining an incorrect final answer, I really don't understand why you would simply write an answer and nothing else, especially considering it takes, what, an extra 5 seconds to write? ???
Post by: Jaswinder on October 12, 2013, 03:59:53 pm
2) What would be the range of answers expected for 2005 VCAA Exam 2 question 2 from Light and Matter Section
3) Are we expected to know about the mechanism by which light is produced in an incandescent light?
4) In a transformer, Joan increases the load on the secondary side of the transformer. Suddenly, it
stops working. She suspects that the fuse in the primary circuit has blown and intends to replace it.
In order to replace the fuse as safely as possible, which of the following is the best precaution for Joan to
take?
A. stand on a rubber mat
B. switch off the mains supply
C. disconnect the transformer from the mains supply
D. remove the load from the transformer
Why C instead or B?
Thanks
Post by: Stevensmay on October 12, 2013, 05:59:22 pm
In order to replace the fuse as safely as possible, which of the following is the best precaution for Joan toThis is just my thought process.
take?
A. stand on a rubber mat
B. switch off the mains supply
C. disconnect the transformer from the mains supply
D. remove the load from the transformer
Why C instead or B?
Switch could fail and not actually disconnect power to the fuse.
Unplugging it guarantees there will be no power to the fuse, therefore no chance of electrocution.
I think this would be a case of, one answer is slightly better than the other.
Post by: Robert123 on October 12, 2013, 06:36:01 pm
Photonics (detailed study) question,Bumping this
Could someone please explain the what "numerical aperture" and "acceptance angle". I can usually get MC questions related to them right by applying the formula but I have no genuine idea what the numbers even mean.
Cheers
Energy level diagram question:
E(eV)
0 N= infinite
-1.38 N=5
X N=4
-1.94 N=3
Y N=2
-5.13 N=1
A photon of 3.61eV is absorb by the ion. Moments later, photons with energy 0.42eV and 1.5eV are ejected from the atom. Use the information provide to calculate the energy levels at n=2 and n=4.
For n=1 to n=4,
5.13-3.61= 1.52eV
This agree with the energy level for n=3 since
N=4 to n=3
1.94-1.52= 0.42eV
The issue I'm having is getting the right n=2 since the 1.5eV could come from n=3 (giving a value of -3.44eV) or from n=4 (giving a value of -3.02eV). The answer gives -3.02eV but how is this more right than -3.44eV since both could be ejected.
So which one is 'more right'? If vcaa ask a question like this would they give both answers as correct or just one?
Post by: SocialRhubarb on October 12, 2013, 11:23:38 pm
If the jump from N=4 to N=2 gives us a 1.5 eV photon, it can no longer produce a photon with an energy of 0.42 eV, hence it cannot jump from N=4 to N=2 to produce a 1.5 eV photon.
Post by: Jaswinder on October 13, 2013, 08:26:20 am
Post by: Stevensmay on October 13, 2013, 11:51:53 am
We can see that as we increase w,
As we decrease w, more diffraction occurs as
This also explains why we need to use very thin slits/apertures for this experiment, as if they are too big any diffraction is not evident.
Post by: lolipopper on October 13, 2013, 12:16:14 pm
Our equation for the extent of diffraction iswhere w is the width of the slit.
We can see that as we increase w,becomes smaller, thus less diffraction occurs.
As we decrease w, more diffraction occurs as
This also explains why we need to use very thin slits/apertures for this experiment, as if they are too big any diffraction is not evident.
what about the intensity of the pattern? and what effect does it have on the fringe spacing?
Post by: BasicAcid on October 13, 2013, 03:46:58 pm
what about the intensity of the pattern? and what effect does it have on the fringe spacing?
No effect on the intensity and as the diffraction is larger, the diffraction patterns are also larger, i.e. the fringe spacing is larger
Post by: barydos on October 15, 2013, 09:25:20 pm
This question came from exam 1 of the VCAA 2006 paper:
Spoiler
(http://i.imgur.com/MVC5OOU.png)
Spoiler
(http://i.imgur.com/iWEThyJ.png)
I'd just like to know what your responses might be, because I feel the provided comment is insufficient.
Post by: Alwin on October 15, 2013, 09:42:23 pm
offtopic
Dont be smartass alright. It was a genuine question.Sorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
i just got the physics marking scheme from my teacher. it says "where questions dont specify that the working must be shown, if the correct answer is in the box the student receives full marks". Maybe you should get some suffice information yourself before absolutely classifying other peoples question invalid and, from the sound of your tone, stupid.
I was just showing the source of my "information". I can only talk about the pre-2012 exam marking scheme, and as Tim...blahhh has confirmed from more accurate sources Re: Physics [3/4] Question Thread! it turns out I was wrong. Not going to blame the practise exam, it's just that it turns out I was wrong and I'm sorry if my response was inconsiderate.
Good luck to everyone with physics :)
Post by: sin0001 on October 15, 2013, 10:17:45 pm
Post by: lolipopper on October 15, 2013, 11:10:43 pm
offtopicSorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
I was just showing the source of my "information". I can only talk about the pre-2012 exam marking scheme, and as Tim...blahhh has confirmed from more accurate sources Re: Physics [3/4] Question Thread! it turns out I was wrong. Not going to blame the practise exam, it's just that it turns out I was wrong and I'm sorry if my response was inconsiderate.
Good luck to everyone with physics :)
im sorry too bro. all good. and good luck to you too.
With questions talking about the extent of diffraction: do we assume that max. diffraction occurs when wavelength/width ratio ~ 1 or when this ratio is a large value?
yes, the size of the aperture must be approximately the same as that of the wavelength for significant diffraction to occur.
Mod Edit - Merged double posts (Phy124)
Post by: joey7 on October 16, 2013, 08:46:01 pm
"The shape of racing cars is designed to minimise air resistance. To a reasonable approximation the force (F) or air resistance is proportional to the square of the speed (v). That is: F=kv^2 where k is the drag coefficient.:
Which if the following choice is a possible unit for K
A. kgm^-1
B. s^2m^2
C. N
D. Nm^2s^-1
E. S^2M^-2
Post by: SocialRhubarb on October 16, 2013, 08:51:46 pm
v2 has the units of m2 s-2.
k=F/v2=kg m s-2 m-2 s2=kg m-1
So the answer is A, kg m-1.
Post by: lolipopper on October 17, 2013, 08:35:25 pm
-what is modulation and demodulation?
Post by: Jaswinder on October 17, 2013, 09:53:53 pm
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics1w06.pdf
thanks
Post by: barydos on October 17, 2013, 09:54:22 pm
Post by: lzxnl on October 17, 2013, 10:04:56 pm
Post by: barydos on October 17, 2013, 10:08:56 pm
The corresponding constant DC voltage that provides the same average power output.
Thanks!
Post by: lzxnl on October 17, 2013, 10:14:00 pm
what is a good answer to the following:
-what is modulation and demodulation?
Modulation is the conversion of an electric signal into a light signal. In the VCE course, we only look at amplitude modulation. In this form of modulation, the shape of the electric field signal is embedded in the shape of the carrier wave light signal, which generally has a very high frequency. When the electric field is negative, the light signal's amplitude is just lower than its average amplitude, which corresponds to the zero of the electric field signal. The purpose of the modulation is to allow the electric signal to be literally carried by the carrier way across a distance to where it is demodulated, or converted back into an electric signal, for use.
As an example, let's take two circuits, one with an LED and one with a photodiode. In the circuit with the LED, the current through the LED determines brightness of the light emitted. This light contains information about the current of the LED circuit; it is modulated. When it hits the photodiode, the intensity of the light then determines the current of the photodiode circuit; the light becomes demodulated. That's the example I use to remember what modulation does.
Post by: Robert123 on October 18, 2013, 09:26:58 pm
Basically a moving trolley with a spring on its front collide into a stationary trolley. The spring compress then uncompressed in the collision which is elastic. How would the total momentum of the system varies with time. The answers says that momentum is always conserved in a collision so it will be a straight line with no variation when the springs come into play. But, my intuition tells me that some of the momentum will sort of be temporarily stored within the spring and decrease, could someone please explain to me why I'm wrong. Cheers :)
Post by: barydos on October 18, 2013, 10:36:28 pm
Spoiler
(http://i.imgur.com/TN8UFLV.png)
(http://i.imgur.com/m9c0257.png)
(http://i.imgur.com/m9c0257.png)
Spoiler
So for max induced voltage, we would use that -NBA/t formula. The difficult for me was finding 't'
If the square was moving at 4.0 cms^-1, I thought the max change in flux would have been when it reaches the middle of the face of the magnet (aka 6cm in?) so t would be 6/4 = 1.5 seconds. But the answer suggests that it just takes 0.5 seconds meaning the square only just goes completely in the field.
Could someone just clarify if I'm being silly or not, and whether or not the square requires to be in the middle or not?
Post by: lzxnl on October 18, 2013, 10:44:59 pm
Remember, the formula flux = BA refers to the area that is exposed to the magnetic field.
Now, the maximum voltage induced would then be due to the change in flux from when the square first enters the field to when it is fully immersed in the field. The side length is 2 cm, and the speed of motion is 4 cm/s, so the time we want is 0.5 s.
Just one loop only.
The change in flux is 3.7*10^-3 T * (2 cm)^2 = 14.8 * 10^-7 T m^2 (note change in units)
So dividing by the time, our max voltage is 29.6* 10^-7 V, so 2.96 uV which is the unrounded form of the answer given.
The trick is identifying what the time means. The time is the period of time over which the change in flux occurs. Note that when the square is in the middle of the magnetic field, the entire square will have remained in the field for a period of time, so the flux isn't changing; the voltage is zero in that case.
Honestly mate, I don't fully understand half the crap we're learning in physics and I got 96% on that sample VCAA 2013 exam.
All you need to do is sub in values.
We know speed = distance/time so therefore time = distance/speed = 0.02/0.04 = 0.5, all values directly from the question subbed into our formulas from our A3 cheat sheet.
Then sub time = 0.5 into the EMF equation and wallah, you get the answer.
Another example of how physics is becoming ridiculous; you don't even need any understanding of the mechanics of the formulas to get the marks.
Post by: SocialRhubarb on October 18, 2013, 10:49:44 pm
VCAA 2010 Q16:
Basically a moving trolley with a spring on its front collide into a stationary trolley.
Spoiler
I think this is quite a difficult question to grasp conceptually, and difficult to answer without simply parroting the line that momentum is always conserved. But essentially what is happening is that forces are acting on the spring in both directions, and the spring is exerting a force on both trolleys. Because momentum is a vector quantity, that is, it has direction, exerting equal forces in opposite directions results in no change in momentum. The problem is it is a little more complicated than that, as the spring would not exert equal forces on both the trolleys, but essentially what would happen is that it would exert a slightly greater force on the stationary trolley than the moving trolley behind it, and the 'unbalanced' force actually contributes to the change in momentum of the spring itself. I expect that VCAA might expect you to briefly touch on some of these points and simply mention that momentum is always conserved.
I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.
I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.
Post by: lzxnl on October 18, 2013, 11:01:33 pm
SpoilerI think this is quite a difficult question to grasp conceptually, and difficult to answer without simply parroting the line that momentum is always conserved. But essentially what is happening is that forces are acting on the spring in both directions, and the spring is exerting a force on both trolleys. Because momentum is a vector quantity, that is, it has direction, exerting equal forces in opposite directions results in no change in momentum. The problem is it is a little more complicated than that, as the spring would not exert equal forces on both the trolleys, but essentially what would happen is that it would exert a slightly greater force on the stationary trolley than the moving trolley behind it, and the 'unbalanced' force actually contributes to the change in momentum of the spring itself. I expect that VCAA might expect you to briefly touch on some of these points and simply mention that momentum is always conserved.
I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.
Why don't we look at a different reference frame, that of the "stationary trolley", so that the situation becomes a moving trolley collides with a trolley with a spring on it?
Let the now moving trolley be trolley A and the stationary trolley with a spring be trolley B.
Now, as trolley A runs into the spring, A exerts a force on the spring, which compresses it. Similarly, the spring exerts a retarding force on A which slows it down. The spring, meanwhile, also exerts a force on trolley B, which reduces the compression of the spring, effectively transferring momentum from A to B. As the spring compresses, it slows A down more and more. When it can be compressed no further, it pushes A backwards and this pushes B forwards. Even when the spring isn't being compressed and is 'stationary', the situation becomes similar to having two blocks; here, A effectively pushes directly onto B. When the spring unwinds itself, it pushes A backwards and B forwards. Momentum is conserved, but there are a number of forces at play here.
Note; conservation of momentum can be seen to be a consequence of Newton's third law in classical mechanics, which is partly the argument I'm making here.
Post by: SocialRhubarb on October 18, 2013, 11:12:05 pm
Also, how does the spring exerting a force on trolley B reduce the compression of the spring? Surely at that stage of your argument the spring is still being compressed.
Post by: lzxnl on October 18, 2013, 11:21:38 pm
The point about the force on trolley B is that the compression of the spring is occurring because trolley A is pushing onto the spring. Trolley B is in the other direction of the applied force, so pushing on that, and trolley B's resultant motion, relaxes the spring. It's probably most easily demonstrated with a wall. If you ran a trolley with a spring on it against a wall and the wall doesn't move, the spring is compressed a lot, because although it pushes on the wall, the wall doesn't budge. In my scenario, the spring pushes on trolley A and makes it move, and as the trolley is attached to the spring, the spring is stretched in the opposite direction it is being compressed.
Post by: barydos on October 18, 2013, 11:28:31 pm
The square doesn't need to be in the middle; it just needs to be in a position such that the entire square is in the magnetic field. This is because the magnetic field is uniform.
Remember, the formula flux = BA refers to the area that is exposed to the magnetic field.
Now, the maximum voltage induced would then be due to the change in flux from when the square first enters the field to when it is fully immersed in the field. The side length is 2 cm, and the speed of motion is 4 cm/s, so the time we want is 0.5 s.
Just one loop only.
The change in flux is 3.7*10^-3 T * (2 cm)^2 = 14.8 * 10^-7 T m^2 (note change in units)
So dividing by the time, our max voltage is 29.6* 10^-7 V, so 2.96 uV which is the unrounded form of the answer given.
The trick is identifying what the time means. The time is the period of time over which the change in flux occurs. Note that when the square is in the middle of the magnetic field, the entire square will have remained in the field for a period of time, so the flux isn't changing; the voltage is zero in that case.
Another example of how physics is becoming ridiculous; you don't even need any understanding of the mechanics of the formulas to get the marks.
Ahhh that makes so much sense. Thanks for the clarification!
Post by: SocialRhubarb on October 18, 2013, 11:29:01 pm
In your example of the wall, the spring is still exerting a force. In fact, it's probably exerting an even greater force. But the force doesn't 'decrease the compression'. The force increases the compression of the spring. You can imagine that if you had no force you'd have no compression. Which is why it's strange to say that the "force on trolley B ... reduces the compression of the spring".
But of course, the force accelerates the trolley, which increases its velocity, reducing its speed relative to the spring, reducing its compression, which I think is what you're trying to say?
Post by: lzxnl on October 18, 2013, 11:37:08 pm
But of course, the force accelerates the trolley, which increases its velocity, reducing its speed relative to the spring, reducing its compression, which I think is what you're trying to say?
Yes, that was my point that I failed to articulate clearly. Thank you for clarifying it :D
Post by: barydos on October 19, 2013, 11:07:01 am
Spoiler
(http://i.imgur.com/RXyjDio.png)
What happens between the two?
Post by: Jaswinder on October 19, 2013, 11:49:31 am
Post by: bonappler on October 19, 2013, 12:01:31 pm
The answer is A, but shouldnt it be the other way around? SO negative current at start then positive? (because of the EMF being (-) change in flux?) ThanksI don't think it matters, remember the motion has to be relative.
How do you do this question?Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.Spoiler(http://i.imgur.com/RXyjDio.png)
What happens between the two?
Post by: lolipopper on October 19, 2013, 12:26:38 pm
How do you do this question?Spoiler(http://i.imgur.com/RXyjDio.png)
What happens between the two?
its explained in the pic
Post by: barydos on October 19, 2013, 01:17:15 pm
I don't think it matters, remember the motion has to be relative.Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.
its explained in the pic
Thanks both of you :D
Another question haha
How do you draw the graph of question 11: http://www.vcaa.vic.edu.au/Documents/exams/physics/2011physics2-w.pdf
I thought it would've contained some "rectangular graphs" because constant change in flux = constant voltage? but apparently the graph is a curve.
This leaves me wondering why the answer to electric power q8: http://www.vcaa.vic.edu.au/Documents/exams/physics/2008physics2-w.pdf is option C.
I would have thought they'd produce the same graphs (looking like q8 option C)
Help please
Post by: SocialRhubarb on October 19, 2013, 03:22:08 pm
Post by: barydos on October 19, 2013, 09:37:02 pm
The difference in the questions is that the question in the 2008 VCAA exam states that the field can be considered 0 outside the poles, while it cannot be assumed so in the 2011 paper. In terms of realism, the curved graph is a more accurate representation, but sometimes for simplicity's sake, we make assumptions like no field outside the poles.
Didn't even notice that small detail, thanks :)
Post by: SocialRhubarb on October 19, 2013, 10:44:39 pm
If you think about it, if the net force was towards the centre of the chain, the person in the device would accelerate upwards, towards the centre of the chain, but they clearly instead follow a circular path.
We find the magnitude of the tension by realising that the net force must be the centripetal force, and so the vertical component of the tension must negate the person's weight force, while the horizontal component of the tension provides the centripetal force. With these two figures, we can calculate the magnitude of the tension in the chain with Pythagoras' theorem.
Post by: lzxnl on October 19, 2013, 11:10:36 pm
As for astronauts in outer space, there is no contradiction. The net force is directed inwards in a direction that is perpendicular to the imaginary axis they are revolving around.
You can see that the tension has a horizontal component, which is the net force mv^2/r, and a vertical component to cancel out the weight force. As these components are perpendicular, use Pythag.
Post by: barydos on October 20, 2013, 11:20:31 am
Post by: lolipopper on October 20, 2013, 02:19:32 pm
When drawing a circuit, how do we know when the switch has to be opposite a thermistor or LDR rather than a variable resistor (or vice versa)?
you have to see when the heater must be turned on. If it is to be turned on when the voltage increases then against the element whose resistance increases (Voltage increases across it as well) and vice versa.
Post by: barydos on October 20, 2013, 09:48:28 pm
you have to see when the heater must be turned on. If it is to be turned on when the voltage increases then against the element whose resistance increases (Voltage increases across it as well) and vice versa.
That makes sense, thank you!
Post by: Robert123 on October 21, 2013, 08:07:44 pm
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.
Post by: random_person on October 21, 2013, 08:57:37 pm
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.
You're right!!!!!!?????????
Post by: Thu Thu Train on October 21, 2013, 09:57:36 pm
Post by: Stevensmay on October 21, 2013, 10:03:53 pm
Post by: Thu Thu Train on October 21, 2013, 10:15:05 pm
The answer you have gotten is correct. One upside of physics is that there often aren't very many technicalities to a question, it's just a matter of plugging numbers in.
(http://i.imgur.com/nUqLD8D.gif)
Post by: Stevensmay on October 21, 2013, 10:23:15 pm
The only thing that has some degree of 'better watch out for that' I find is light.
Motion is essentially 'I've got these letters, let me look at my cheatsheet to see which formula has them.'
Electronics is 'can I read a graph, a formula and use V=IR.'
Light actually takes some learning, but only because VCAA tend to ask more theory orientated questions.
Post by: lolipopper on October 21, 2013, 11:55:56 pm
Lets be honest here, this is VCAA style physics.
The only thing that has some degree of 'better watch out for that' I find is light.
Motion is essentially 'I've got these letters, let me look at my cheatsheet to see which formula has them.'
Electronics is 'can I read a graph, a formula and use V=IR.'
Light actually takes some learning, but only because VCAA tend to ask more theory orientated questions.
i havent dropped a single mark for any unit 4 exam that i have done for physics. However i tend to drop 1 or 2 marks on the tension and relative motion questions for Unit 3. This is why only a quarter of my summary sheet is light and matter and electricity.The rest is motion.
Post by: Robert123 on October 22, 2013, 07:39:39 am
I knew that vicphysics solutions has some other errors in their answer but when I check itute and they both have the same answer, it threw me off completely
Post by: lolipopper on October 23, 2013, 09:35:05 pm
is relative motion still in Unit 3?
and also are the specific properties of materials to be known in "Detailed study : Materials and Structure"
Post by: Alwin on October 23, 2013, 09:52:43 pm
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.
Oh this question :) VCAA copy + pasted that q from the 2009 exam iirc, but changed the numbers. SO the solutions companies posted up were wrong too because they copied their solutions from 2009 too lol
here, from a set of 'model' solutions I made myself for the 2013 practice exam :) :
(http://i.imgur.com/LrNa17g.png)
@lollipopper: when I did physics last year it wasn't, I think it was in Special Relativity detail study for some reason (apparently too hard for normal physics students?). And I don't recall there being any specifics needed to be rote learned for Materials I think. If there are you don't even need to learn it, just chuck it on your cheat sheet if anyone knows what needs to be memorised this year :D (sorry, it's been over a year and a half for me since materials so forgotten heaps =\)
Post by: Jaswinder on October 24, 2013, 07:41:00 pm
Post by: sin0001 on October 25, 2013, 11:22:23 am
Does it matter where you place the diode in a series circuit?Current would first pass through the resistor, resulting in a drop, and THEN it'll be stopped by the diode in reverse-biased.
For example, you have the positive end of supply connected to reverse biased diode then diode connected to resistor then resistor connected back to supply.
The entire voltage supply drops across the diode as no current flows through it hence no drop across resistor.
But how about if the position of the diode and the resistor are swapped? Would the same thing happen?
What happens to the current that flows across the resistor first?
Post by: sin0001 on October 25, 2013, 11:49:12 am
Post by: Robert123 on October 25, 2013, 12:38:14 pm
But what I'm asking is what if the current flows through the resistor first, then the reverse biased diode stops the current flow?For the current to flow, you would need a complete circuit (ie doesn't flow in one half then stops)
Assuming the reverse bias diode has no leakage current, it would probably just act like an incomplete circuit.
The best way to find out is just to actually test it in real life.
Post by: SocialRhubarb on October 25, 2013, 12:54:40 pm
Robert123 is right - the diode in reverse bias would act just like a break in the circuit.
You can't have current flowing halfway through a circuit and then stopping.
Post by: SocialRhubarb on October 25, 2013, 01:38:45 pm
I think in this case it may actually be helpful to at least think of the force diagram. There are three forces involved here - the force from the rod, the force from the wire and the weight force of the lamp, and we want the lamp to have a net force of 0. Now, which ways are the forces acting?
Weight force is quite simple, obviously straight down. But the rod is being compressed, so will be exerting a force diagonally upwards on the lamp, while the wire is in tension, so will be exerting a force to the left.
If we resolve these forces, the vertical component of the rod's force must balance out the lamp's weight force, and the horizontal component of the rod's force must balance out the tension in the string.
Spoiler
Post by: Jaswinder on October 25, 2013, 04:03:33 pm
how to do 1b from 2012 Exam 2? About the orientation the magnet will take?
bumb :)
Post by: Thu Thu Train on October 25, 2013, 04:25:12 pm
bumb :)
Figure out the where the 'north' and 'south' end of the solenoid is. Your smaller magnet will act as a compass needle which(without any other magnetic fields) points at magnetic north. Placing the magnet near point Q and allowing it freely rotate means that the north end of the magnet will rotate to face the south end of the solenoid.
Spoiler
(the answer is C)
Post by: Jaswinder on October 25, 2013, 05:56:15 pm
Figure out the where the 'north' and 'south' end of the solenoid is. Your smaller magnet will act as a compass needle which(without any other magnetic fields) points at magnetic north. Placing the magnet near point Q and allowing it freely rotate means that the north end of the magnet will rotate to face the south end of the solenoid.Spoiler(the answer is C)
wouldn't the magnetic field of earth have any effect on it?
Post by: Jaswinder on October 25, 2013, 07:26:23 pm
Post by: BasicAcid on October 25, 2013, 08:05:13 pm
also just to be 100% sure, we can take in 2 double sided a4 sheets into the exam yeh? :)
They need to be stuck together, but yes.
Post by: Jaswinder on October 25, 2013, 08:31:35 pm
Post by: Alwin on October 25, 2013, 08:34:46 pm
stuck together? how do i do that? you can't bound 2 A4 sheets can you? would stapling them together work?
you can try photocopying double sided onto A3 if you want :)
Post by: Jaswinder on October 25, 2013, 08:38:40 pm
Post by: Robert123 on October 29, 2013, 08:18:43 pm
Their reasoning for their answer is...
"when the switch was closed a magnetic field built up to the left. To oppose this, the induced current must produce a magnetic field to the left."
Wouldn't the induced current produce a field to the right?
Post by: angus_grant on October 30, 2013, 05:35:15 pm
"when the switch was closed a magnetic field built up to the left. To oppose this, the induced current must produce a magnetic field to the left."
Yeah must have been a typo, without looking at the question id say the flux increased in the left direction, and so to oppose this, a current was induced that had a flux in the right direction, then do the right hand rule to find current flow (guessing its a solenoid question or something similar)
hope this helps :)
Post by: Robert123 on October 30, 2013, 07:23:23 pm
Yeah must have been a typo, without looking at the question id say the flux increased in the left direction, and so to oppose this, a current was induced that had a flux in the right direction, then do the right hand rule to find current flow (guessing its a solenoid question or something similar)
hope this helps :)
I'm not sure if it was a typo, since it was a MC choice question, the answer would be the complete opposite. IF it is not too much of a hassle could be please have a look at it at http://www.vcaa.vic.edu.au/Documents/exams/physics/2007physics2.pdf page 11.
Post by: SocialRhubarb on October 30, 2013, 07:30:32 pm
Okay, so when the switch is closed, the current is increasing, creating an increasing magnetic field. The field's direction points towards the left. At the second coil, the direction of the magnetic field is also pointing towards the left. The current induced in the coil will be such that it opposes the change, hence we want to induce a field towards the right. Creating a field pointing towards the right requires a current to flow from Y to X.
The current only flows momentarily, as once the current maintains a steady value there will no longer be any change in flux and hence no induced current.
So the answer is B.
Post by: Robert123 on October 30, 2013, 07:43:16 pm
Highly doubt there would be an error in VCAA solutions. If something's wrong they will revise it.
Okay, so when the switch is closed, the current is increasing, creating an increasing magnetic field. The field's direction points towards the left. At the second coil, the direction of the magnetic field is also pointing towards the left. The current induced in the coil will be such that it opposes the change, hence we want to induce a field towards the right. Creating a field pointing towards the right requires a current to flow from Y to X.
The current only flows momentarily, as once the current maintains a steady value there will no longer be any change in flux and hence no induced current.
So the answer is B.
But wouldn't a current flowing from Y to X produce a magnetic field to the left inside the solenoid? Using the RH grip rule on the leftest loop, your fingers would wrap around pointing to the left on the solenoid side
Post by: lzxnl on October 30, 2013, 07:54:50 pm
The answer B is perfectly fine.
Post by: Alwin on October 30, 2013, 08:14:06 pm
> Close switch
> Magnetic field increasing to the left
^ using right hand grip rule
> Same as moving towards a south pole
> So south pole induced at the end closest to first coil
^ using Lenz's law
> Current flows from Y to X
^ using right hand grip rule again
This alternate method works best when there is on magnet and coil moving towards each other / in circles / whatever vcaa wants lol
Post by: Jaswinder on October 30, 2013, 08:55:20 pm
Post by: Stevensmay on October 30, 2013, 09:40:07 pm
http://tele-information.com/wp-content/uploads/2012/10/Amplitude-modulation-AM.gif
Demodulation is the removal of the carrier wave to leave only the data wave. Reverse the above image.
Post by: Jaswinder on October 30, 2013, 10:24:31 pm
also this one, i get q to p but answers say p to q
Post by: BasicAcid on October 30, 2013, 10:36:52 pm
Post by: lzxnl on October 30, 2013, 10:37:53 pm
Post by: Jaswinder on October 30, 2013, 10:39:16 pm
North pole moves to left=>induced magnetic field points to the the right=>use right hand rule, current is from q to p in the EXTERNAL circuit, but p to q through the ammeter.
whats the external circuit and how is it different to this one :-\ :-[
Post by: lzxnl on October 30, 2013, 10:51:54 pm
Post by: Robert123 on October 31, 2013, 02:04:43 pm
A loop is sitting in a magnetic field that is going out of the page, it is quickly pulled to the left. Which direction will the current flow (clockwise or anticlockwise)?
Looking at the left hand side of the loop, you can use the right hand slap rule to determine the direction of the current( fingers going out of the page, palm to the right to oppose the force to the left, current will be clockwise). However, if you do the same analysis on the other side, the thumb would be pointing the same way so current will be anticlockwise. Which one is correct and why?
Post by: Stevensmay on October 31, 2013, 09:25:47 pm
Having some difficulties with lenzs law.
A loop is sitting in a magnetic field that is going out of the page, it is quickly pulled to the left. Which direction will the current flow (clockwise or anticlockwise)?
Looking at the left hand side of the loop, you can use the right hand slap rule to determine the direction of the current( fingers going out of the page, palm to the right to oppose the force to the left, current will be clockwise). However, if you do the same analysis on the other side, the thumb would be pointing the same way so current will be anticlockwise. Which one is correct and why?
As the loop is moved out of the field, the amount of flux it encloses decreases. Lenz law then tells us that there will be an induced current that will oppose this decrease in flux by increasing it.
To increase flux we want to generate a B-field that is in the same direction as the original B-field, out of the page. Using our right hand grip rule, we can see that to generate a B-field going out of the page within the loop, the current must be flowing anti clockwise.
Can you please check this with your answer? I'm not totally happy with my reasoning.
Post by: Robert123 on November 01, 2013, 07:43:58 am
Also, with that reasoning, does it matter if the coil is move to the LEFT or to the RIGHT?
Post by: lolipopper on November 02, 2013, 12:17:38 am
also what is the range of wavelength-gap ratio that we should assume that allows diffraction, because in the same exam Q5 i got the ratio to be 0.667 and thought this was much lower than 1, hence no diffraction.
http://www.vcaa.vic.edu.au/Documents/exams/physics/2007physics2.pdf
Post by: SocialRhubarb on November 02, 2013, 01:48:53 am
I think incandescent lightbulbs are off the course, but you should understand the principles behind a vapour lamp, and how they work.
Quote from the study design: explain the production of atomic absorption and emission spectra, including those from metal
vapour lamps.
We were told that anything above 0.1 will give you noticeable diffraction. A ratio of 0.667 would give you about 45 degrees of diffraction, definitely enough to be noticed.
Post by: ~T on November 02, 2013, 10:40:21 am
Does the diffraction become even more significant, or does it gradually decrease again, or does it stop entirely? (perhaps because the wave is "too large"???)
Post by: lzxnl on November 02, 2013, 10:46:40 am
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.
*begins rant about VCE physics
Post by: sin0001 on November 02, 2013, 10:57:39 am
Don't see many company solutions, and even VCAA, mentioning diffraction for these questions.
Post by: SocialRhubarb on November 02, 2013, 12:01:03 pm
The thing is that it's a double slit diffraction experiment that you're being asked about. They've given you two slits, why only talk about single slit diffraction? It's clear that the question is directing you to talk about interference patterns, and if it asks you specifically what Young's double-slit experiment shows, the most notable thing about it is the interference pattern it shows. If you wanted an experiment to show simple diffraction of light, you probably wouldn't use Young's double-slit experiment.
I'm not sure what point you're trying to make about electrons and de Broglie wavelengths. The fact that electrons can diffract, and also interfere, speaks about the wave nature of matter, not the nature of light.
Post by: sin0001 on November 02, 2013, 12:06:57 pm
I guess?Ah okay, what I was trying to ask was whether *only* mentioning interference would suffice, or is it better to also talk about diffraction.
The thing is that it's a double slit diffraction experiment that you're being asked about. They've given you two slits, why only talk about single slit diffraction? It's clear that the question is directing you to talk about interference patterns, and if it asks you specifically what Young's double-slit experiment shows, the most notable thing about it is the interference pattern it shows. If you wanted an experiment to show simple diffraction of light, you probably wouldn't use Young's double-slit experiment.
I'm not sure what point you're trying to make about electrons and de Broglie wavelengths. The fact that electrons can diffract, and also interfere, speaks about the wave nature of matter, not the nature of light.
Post by: ECheong on November 02, 2013, 12:26:59 pm
On the whole though, I'd personally focus on constructive and destructive interference as that's typically what they want from an answer (judging by examiners reports). As long as its good physics you can't be pinged :)
Post by: BasicAcid on November 02, 2013, 12:28:44 pm
For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.
*begins rant about VCE physics
So actually answering Tim...Blah's question, there will still be diffraction if wavelength > slit size but just to a lesser extent?
Post by: SocialRhubarb on November 02, 2013, 12:32:21 pm
Post by: lolipopper on November 02, 2013, 12:33:38 pm
Post by: BasicAcid on November 02, 2013, 12:34:54 pm
Umm, no, there will be more diffraction if wavelength > slit.
I thought maximum diffraction occurred when wavelength/slit = 1?
Post by: lolipopper on November 02, 2013, 12:43:42 pm
Could someone verify this?
I'm a bit worried now after reading this
And I don't understand any of nlius explanation
nliu you sexy boy, explain it to us in simple terms.
Post by: ECheong on November 02, 2013, 12:44:43 pm
If you imagine an aperture and a wave coming through it, it's not possible for the wave to diffract backwards at an angle >90degrees. This is because if you treat each single point on the edge of a wave as a point source (huygen's principle) there's no way for the point source to send a wave 'around the corner' of the aperture and start sending waves backwards.
Reading over this I've realised it's a pretty difficult concept to explain in words LOL, pictures would be better.
Tl;dr maximum diffraction at lamda/d =1 because you don't get any more diffraction if lamda/d>1
Post by: SocialRhubarb on November 02, 2013, 12:46:24 pm
Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.
I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.
Post by: lolipopper on November 02, 2013, 12:52:01 pm
I haven't read anything outside of nliu1995's points, but they seem to make physical sense.
Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.
I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.
so the whole screen will be just bright, correct?
Post by: lzxnl on November 02, 2013, 01:01:38 pm
The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.
The reason why we say in VCE physics that "incomplete diffraction" occurs is that the width of the middle intensity maximum is small. After the first intensity minimum, the intensity drops off significantly.
Read up on diffraction on wiki for diagrams, but be warned, this subject can get maths-heavy. The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.
There's a neat diagram on wiki too; it'll make my point clearer.
Post by: Alwin on November 02, 2013, 01:05:28 pm
For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.
*begins rant about VCE physics
nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.
Anyways, for those interested from page 496:
kinda tangential, only read it if you want to
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.
What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.
So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.
What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.
So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.
EDIT: beaten, but hope you get the gist
This is the page he is referring to: http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction
From that page:
| Numerical approximation of diffraction pattern from a slit of width equal to wavelength of an incident plane wave in 3D spectrum visualization: | Numerical approximation of diffraction pattern from a slit of width equal to five times the wavelength of an incident plane wave in 3D spectrum visualization | Numerical approximation of diffraction pattern from a slit of width four wavelengths with an incident plane wave. The main central beam, nulls, and phase reversals are apparent. |
| (http://upload.wikimedia.org/wikipedia/commons/thumb/4/46/Wavelength%3Dslitwidthspectrum.gif/220px-Wavelength%3Dslitwidthspectrum.gif) | (http://upload.wikimedia.org/wikipedia/commons/thumb/0/0a/5wavelength%3Dslitwidthsprectrum.gif/220px-5wavelength%3Dslitwidthsprectrum.gif) | (http://upload.wikimedia.org/wikipedia/commons/thumb/3/3c/Wave_Diffraction_4Lambda_Slit.png/220px-Wave_Diffraction_4Lambda_Slit.png) |
Post by: lzxnl on November 02, 2013, 01:13:26 pm
nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.
Anyways, for those interested from page 496:
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.
What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.
So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.
Ah Bragg's law. It's similar, although in that case you've got to remember where the two comes from; the path difference is 2 d sin theta, and that's because there are two layers in the crystal lattice and the 2 comes from a return journey from the first layer to the second layer. Look it up on wiki for a diagram.
Alwin, you'll confuse people xP
In double slit interference, the condition is d sin theta = m lambda for CONSTRUCTIVE interference because d sin theta is the path difference.
For single slit diffraction, the argument works differently. Again, look at some diagrams; they'll be better than what I can draw.
But essentially, if we split up the slit into half, to form two minislits of size d/2, then we have a double slit scenario with path difference d/2 * sin theta. This is equal to half a wavelength for minimum, so d/2*sin theta = w/2. d sin theta = w
If we split the slit into quarters, then quarter 1 can interfere destructively with quarter 2 and then 3 with 4. The slit width is now d/4, so the condition is d/4 * sin theta = w/2. d sin theta = 2w. Etc if we split the slit into 6, 8, 10, 12.
That's sort of where it comes from.
Define "max diffraction" though.
Post by: ECheong on November 02, 2013, 01:21:35 pm
So essentially for this exam (which is all I care about):
- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction
Correct?
At the core of it, pretty much haha :)
Post by: lzxnl on November 02, 2013, 01:23:47 pm
Post by: Alwin on November 02, 2013, 01:31:00 pm
So essentially for this exam (which is all I care about):
- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction
Correct?
pretty much.
*back to discussion with nliu* :P
For anyone who's interested in what we're talking about, this is the easiest explanation I can find:
scan of a textbook
(http://i.imgur.com/Lx5mhbq.jpg)
(http://i.imgur.com/hr0FPyH.jpg)
(http://i.imgur.com/hr0FPyH.jpg)
See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.You referring to the ones in my post?
Post by: lzxnl on November 02, 2013, 03:27:23 pm
Post by: joey7 on November 02, 2013, 04:06:23 pm
Post by: Stevensmay on November 02, 2013, 04:47:36 pm
What happens to magnetic field (B) when the current increases, and which equation tells us thisMisread question.
So we can see that as the current increases, the magnetic field needs to increase or the time needs to be reduced.
Post by: lzxnl on November 02, 2013, 04:52:00 pm
What equation tells you that?
Biot-Savart law:
Oh wait, not part of course. My bad.
Post by: Alwin on November 02, 2013, 05:16:20 pm
Magnetic field strength is directly proportional to the current.
What equation tells you that?
Biot-Savart law:
Oh wait, not part of course. My bad.
really? :OOOOOOOOOOOOO dang, they must have really cut down on the course. I swear Biot-Savart's Law was in it last year :o
JOKES, no one panic please, just nliu deciding to demonstrate his knowledge again =P
Post by: lzxnl on November 02, 2013, 07:36:29 pm
So we can see that as the current increases, the magnetic field needs to increase or the time needs to be reduced.
This is only for an induced current, not for currents ttat create magnetic fields
Post by: joey7 on November 02, 2013, 07:50:09 pm
Also, I came across a question involving two parallel current carrying wires, where the distance between them was reduced by a factor of 2, In the answers it said field varies inversely with distance squared. Again how am I meant to know this?
Post by: Stevensmay on November 02, 2013, 08:04:49 pm
This is only for an induced current, not for currents ttat create magnetic fields
Skipped over part of the question, thanks.
Post by: lzxnl on November 02, 2013, 11:18:00 pm
Wait, so if a question talks about doubling the current how am I meant to know the effect on the field?
Also, I came across a question involving two parallel current carrying wires, where the distance between them was reduced by a factor of 2, In the answers it said field varies inversely with distance squared. Again how am I meant to know this?
The awkward bit is...your answers are wrong. For parallel current carrying wires, the field strength is inversely proportional to the distance from the wire. I don't want to prove this here as 1. I would have to do some messy integration, or 2. I would have to introduce another physical law which would make no sense here.
You don't needed to know anything about magnetic field strengths in VCE, although maybe you need to know that increasing the distance generally decreases field strength (but not how).
That logic. We're taught that currents create a magnetic field, that increasing currents create larger magnetic fields and how magnetic forces operate, but we're not taught how to calculate magnetic field strengths. VCAA logic.
Post by: lolipopper on November 02, 2013, 11:48:54 pm
Post by: lzxnl on November 02, 2013, 11:56:44 pm
Post by: lolipopper on November 03, 2013, 12:49:56 pm
It's a parabola; the particle's horizontal velocity component is constant but the vertical component increases in magnitude. Just draw something like that and I think you'll be fine.
thanks but are these types in the course, because it similar to relative motion using frames of references and stuff?
Post by: Robert123 on November 03, 2013, 02:04:24 pm
thanks but are these types in the course, because it similar to relative motion using frames of references and stuff?I think it is sort of on the course since you have to be aware that for projectile motion, the horizontal component of the velocity is constant
Post by: sin0001 on November 03, 2013, 06:32:50 pm
Post by: Robert123 on November 03, 2013, 09:24:59 pm
Check out this link, it gives all the details regarding marking
http://www.vicphysics.org/documents/events/conf2009/Generalmarkingprinciples.doc
Post by: Henreezy on November 09, 2013, 03:31:00 pm
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.
Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)
I'm so confused.
Post by: Robert123 on November 09, 2013, 03:35:44 pm
F(a on b)= -F (b on a) that is, newton's 3rd law
From this, we can sort of sub in and solve so we get
F(earth's gravity on object)= -F(object's gravity on earth)
So what actually occur is the gravitation force from the ball actually pulls the earth up but since the earth is so massive, nothing is observed on the earth's behalf.
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
http://www.youtube.com/watch?v=Iuv6hY6zsd0
Post by: Stevensmay on November 09, 2013, 03:36:25 pm
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.
Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)
I'm so confused.
So we agree that the action force is the gravitational attraction exerted on the spheres by the earth.
To find our reactionary force, we simply swap the two objects around.
The reaction force is the gravitational attraction exerted on the earth by the spheres. This is effectively pulling the earth closer to the spheres, thus the direction is up.
Post by: Stevensmay on November 09, 2013, 03:39:50 pm
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
http://www.youtube.com/watch?v=Iuv6hY6zsd0
Someone had the exact same question as you.
http://physics.stackexchange.com/questions/76692/is-coherent-light-required-for-interference-in-youngs-double-slit-experiment
Post by: lzxnl on November 09, 2013, 04:56:52 pm
Henreezy, the reaction force rule is given by
F(a on b)= -F (b on a) that is, newton's 3rd law
From this, we can sort of sub in and solve so we get
F(earth's gravity on object)= -F(object's gravity on earth)
So what actually occur is the gravitation force from the ball actually pulls the earth up but since the earth is so massive, nothing is observed on the earth's behalf.
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
http://www.youtube.com/watch?v=Iuv6hY6zsd0
Torch light does allow interference to occur, as you have waves at each slit interfering with each other, but you don't get a particularly convenient interference pattern as you can see in the video.
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.
Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)
I'm so confused.
The tension is merely a force that exists to hold the object there. It doesn't have to exist if the object is falling.
However, as long as there is a gravitational force, there is a reaction to that force, and that force is always the same as the gravitational force in magnitude, but opposite in direction by Newton 3. That's the difference. In this case, as people before me have pointed out, the reaction is the attraction of the Earth by the object.
Post by: Jaswinder on November 10, 2013, 10:39:11 am
02 -For 14 I thought A, as the left of the battery equipped circuit would become north it would create flux going right to left in the other circuit and to oppose that you would need a current going from X to Y?
Thanks
Post by: eddybaha on November 10, 2013, 10:56:01 am
2. you are looking for the direction of current through the AMMETER, yes a bit tricky. so when the switch closes, it causes flux to thread the secondary coil to the left. Using lenz's law you need a magnetic field threading the coil to the right to oppose this change in flux. Then using grip rule you get your answer.
Post by: This-is-not-me on November 10, 2013, 01:03:34 pm
Post by: eddybaha on November 10, 2013, 01:41:33 pm
Post by: Alwin on November 10, 2013, 01:44:41 pm
Can someone help me with the concept in part of b of this question?
One (vce level reply) would be to say that electrons need to form stationary waves to be stable so the orbit circumference has to be a whole number multiple of the wavelength of the electron's de broglie wavelength.
If an electron orbited at radius such that the circumference was not a whole number multiple, destructive interference would occur and the electron would 'drop' to a lower stable orbit.
Thus, only certain energy levels, orbits, are possible for the electrons of the hydrogen atom... or something along those lines
inb4 nliu comments how they're not levels so much, but bands but that's beyond vce
Post by: lzxnl on November 10, 2013, 02:19:11 pm
One (vce level reply) would be to say that electrons need to form stationary waves to be stable so the orbit circumference has to be a whole number multiple of the wavelength of the electron's de broglie wavelength.
If an electron orbited at radius such that the circumference was not a whole number multiple, destructive interference would occur and the electron would 'drop' to a lower stable orbit.
Thus, only certain energy levels, orbits, are possible for the electrons of the hydrogen atom... or something along those lines
inb4 nliu comments how they're not levels so much, but bands but that's beyond vce
Actually, I'm happy with calling them energy levels; I'm not happy with the standing wave explanation in general :P
It only works for any atom with one electron. AKA H, He+, Li 2+ etc
Alwin, I think you might have say that as only specific wavelengths are permitted, only specific energies are allowed. This is VCE. We can't make any jumps in our thought processes apparently.
Post by: Robert123 on November 11, 2013, 09:56:30 pm
Post by: lzxnl on November 11, 2013, 10:00:04 pm
Post by: BasicAcid on November 11, 2013, 10:55:52 pm
Is diffraction stronger when wavelength equals slit width or when wavelength is greater than slit width?
For VCE purposes, they are the same strength.
Post by: Infinity Plus on November 12, 2013, 04:22:56 pm
Post by: eddybaha on November 12, 2013, 04:31:31 pm
Post by: saifh on November 12, 2013, 04:48:02 pm
Post by: Infinity Plus on November 12, 2013, 04:55:49 pm
definitely noticeable diffraction at 1 but vcaa will either give you something obviously small like 0.000001 if theres no diffraction
Thank you man!
Post by: Robert123 on November 12, 2013, 05:02:20 pm
Post by: Alwin on November 12, 2013, 05:37:02 pm
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA
It's not hard... but I'm pretty sure that that sounds like a detail study (synchrotron irrc) question not a core studies question...
Pretty sure it's not something you need to know.
Post by: lzxnl on November 12, 2013, 05:42:28 pm
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy. Use E=Vq if you ever come across it.
It's not hard... but I'm pretty sure that that sounds like a detail study (synchrotron irrc) question not a core studies question...
Pretty sure it's not something you need to know.
Technically, in the photoelectric effect, if you have electron energy=hf-W, then the stopping voltage would be 1/q*(hf-W) which explains the gradient of the curve. It's not entirely out of the course.
Post by: Alwin on November 12, 2013, 05:44:48 pm
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy. Use E=Vq if you ever come across it.
Technically, in the photoelectric effect, if you have electron energy=hf-W, then the stopping voltage would be 1/q*(hf-W) which explains the gradient of the curve. It's not entirely out of the course.
k
coz last year when I asked about this (forces on charged particles) my teacher said it wasn't strictly on the course. Who knows, maybe this could be your lucky year guys XD
Post by: jono88 on November 12, 2013, 06:26:48 pm
-explain the production of atomic absorption and emission spectra, including those from metal vapours?
I don't understand these 2 dot points, could someone elaborate further?
Post by: jono88 on November 12, 2013, 06:31:16 pm
The electromagnetic radiation that is emitted from a free atom only occurs at a number of discrete or separate wavelengths or frequencies. This is usually represented as a set of lines.
Visible spectrum of hydrogen spectrum
Interpretation of Spectra
After Einstein, frequency is now linked to energy by E = hf. Each line in the spectra now corresponds to a transition of an electron from one energy level to another level.
In fact, the discrete lines in the spectra must mean that electrons in atoms can only exist at specific energy levels, and the spectral lines represent transitions between these levels.
When an atom is excited electrons are sent to a higher energy level. From there, the electron drops back to the ground state by any of the possible paths, each drop emitting a photon of specific energy and frequency.
That right or completely wrong?
Post by: Robert123 on November 12, 2013, 06:31:49 pm
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy.Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?
I have a fairly good idea in deriving the formulas but if I do have to know it, it would be easier to just plug and play VCE style
Post by: ECheong on November 12, 2013, 06:48:35 pm
Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?
I have a fairly good idea in deriving the formulas but if I do have to know it, it would be easier to just plug and play VCE style
If you consider the electronvolt (eV), 1 eV is the amount of energy you get when you accelerate an electron through a potential difference of 1 volt. So, I guess what you can do is find the energy the electron has in eV, and then you can find the potential difference it was accelerated through.
Re: X-rays, you'd just need to find the velocity at which the electrons have a debroglie wavelength the same as xrays and then use E=0.5mv^2 to find energy, convert to eV and you'll know the potential that electron was accelerated through :)
Post by: Alwin on November 12, 2013, 07:26:02 pm
-Identify rms voltage as an AC voltage which produces the same power in a resistive component as a DC voltage of the same magnitude
-explain the production of atomic absorption and emission spectra, including those from metal vapours?
I don't understand these 2 dot points, could someone elaborate further?
RMS: Vp indicates the voltage peak of an AC voltage vs power graph. Basically:
It works on the prinicple the power output should be the same, but that's based on a perfect sine graph. Here's a generic proof if you were interested :P
http://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-sine-wave-with-a-dc-offset/
Absorption and emission in physics is pretty straight forward, specific/discrete/quantum (if you want to be fancy) amount of energy needed to excite electrons to a higher state, and they release this energy when they fall back to a lower state.
Most calculations in physics involve finding the energy in eV, wavelength of photon emitted and whether a photon of another different energy will have any impact on the atom in question.
Note that electron absorption is not on the course... but fun to read about :D
Post by: iitwinz on November 12, 2013, 08:22:12 pm
Post by: Alwin on November 12, 2013, 08:29:40 pm
hey guys, did we have to know the skin effect and optical fibres and how they work? I don't remember learning about it at all and haven't seen any on the exam.
It's been put into one of the detailed studies :) not in the core section of the exam
Post by: iitwinz on November 12, 2013, 09:30:34 pm
It's been put into one of the detailed studies :) not in the core section of the exam
cool thanks man :P
Post by: Lachjames on November 12, 2013, 09:45:31 pm
Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)
Any advice would be great, thanks :)
Post by: eddybaha on November 12, 2013, 09:59:20 pm
Post by: ~T on November 12, 2013, 09:59:34 pm
Hi guys, on the 2011 Exam 2 Electric Power Q.11, I've been talking with my friend from school and we can't figure the graph out. We looked at the VCAA solution and we get what they're saying but we were thinking something else - I put our train of thought in the attachment below.I'm a little confused but I think I see where you're coming from. The only thing is that the flux doesn't reduce when the magnet is inside the loop. Flux still "runs through" the inside of a magnet, and the field lines are still there when this magnet is in the centre of the loop.
Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)
Any advice would be great, thanks :)
At every point with the magnet approaching, going through, and leaving the loop, there is a leftwards flux, yes? It never changes direction. Now, given this constant direction, all that matters is whether this flux is increasing or decreasing. The field lines are closest together when near the magnet, so the flux will be greatest when the magnet is inside the loop. It will increase up until this point, and then decrease away as the magnet leaves.
I hope I've clarified, but I'm not sure it was super clear sorry :-\
Post by: Lachjames on November 12, 2013, 10:05:32 pm
Post by: papertowns on November 13, 2013, 01:29:12 am
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.
What would be the power of the beam?
Post by: silverpixeli on November 13, 2013, 06:50:21 am
If anyone is here :)
I'm not sure, sorry, hopefully somebody catches this before the exam :(
I tried rotating it both ways in my head and both times got current from X to Y initially, not sure if it's a bad question or if I just stuffed up the logic somewhere.
PLeaseee:
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.
What would be the power of the beam?
P=E/t
E = number of photons x single photon energy
= 7.54 x 10^19 x hf
= 7.54 x 10^19 x 6.63 x 10^-34 x 4 x 10^14
so P=(that)/1=that
EDIT: I used the Js version of h because we want energy in Joules so that we can have Power in Watts (Joules/second), if we had have used the eVs value of h (the 4.14x10^-15 one) we would have got energy in eV meaning power in eV/s which isn't a standard unit for power, afaik.
Hopefully that makes sense, the key idea is that the energy of a beam is the number of photons times the energy of one photon :)
Good luck today everyone!
Post by: Robert123 on November 13, 2013, 07:31:23 am
If anyone is here :)
Ok, to solve this we use lenz's law and the right hand slap rule. Let start off by analysining what would happen if we rotate it clockwise.
Firstly side XY would go up and due to lenzs law we know it would create a force in the opposite direction (that is down).
So using our right hand slap rule, palm down and fingers point to the right, we can see the current will flow X to Y. Therefore, we can deduce that it must rotate the opposite way (anticlockwise).
You can do the same analysis for anticlockwise to confirm it.
Post by: Rod on December 17, 2013, 09:43:51 pm
Post by: Stevensmay on December 17, 2013, 09:44:48 pm
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?
Feel free to post whatever.
Post by: Einstein on March 01, 2014, 11:44:44 am
Post by: katie101 on March 01, 2014, 05:31:53 pm
Post by: Thorium on March 05, 2014, 09:23:45 pm
What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
m=0.2 kg
But the ans says to use kx=mg, and the ans is 0.4 kg.
I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?
Post by: RKTR on March 05, 2014, 09:31:18 pm
Post by: lzxnl on March 05, 2014, 09:43:41 pm
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."
What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
m=0.2 kg
But the ans says to use kx=mg, and the ans is 0.4 kg.
I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?
Let's think about this reasonably. The only thing you know about the spring when it is still is that the net force is zero, so you use forces.
Now, let's see what happens when you attach a mass to the relaxed spring and let go. Obviously, the spring is going to start moving. Initially, the spring only has gravitational potential energy. By equating mgx with 1/2 kx^2, all you are doing is finding when all of the gravitational potential energy goes completely into spring energy. By conservation of energy, this means the kinetic energy at this point (where all of the potential energy is spring) is still the same at the st art, where the kinetic energy is zero. This is where the object is momentarily stationary. HOWEVER, the forces are clearly not balanced as the spring is still headed upwards.
The point is, if you just let a spring go, it will oscillate forever assuming no resistive drag forces. Note how when you attach a spring, the forces are unbalanced. In order to balance the spring, the person holding the mass must physically move it with the spring to the point of stable equilibrium. This action drains energy from the spring and explains why you cannot just equate the energies.
Post by: Rishi97 on March 09, 2014, 01:31:37 pm
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
a) How far from the centre of the moon is this new location in terms of the radius of the moon?
b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude
Cheers ;)
Post by: lzxnl on March 09, 2014, 01:48:53 pm
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
a) How far from the centre of the moon is this new location in terms of the radius of the moon?
b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude
Cheers ;)
So...F = k/r^2 where k is some constant.
If the force quarters, the distance r must double, so this new location is now two moon radii from the centre of the moon.
Now, three moon radii above the surface = four moon radii from centre of moon (r is ALWAYS measured from the centre of the moon), so we've quadrupled the distance r with regards to the initial distance, so the force drops to 1/16, or 10 N.
Post by: Capristo on March 11, 2014, 10:27:50 pm
Just a mass attached to a string moving in a horizontal circular motion. (Diagram attached). At the bottom of the vertical line of string there is some weights of mass M. At the end of the diagonal section of string that moves in the circular path, there is a rubber stopper of mass m.
The tension, F, is constant. Show the relationship between L and T (period) by completing the following.
Write a proportionality expression for L
Show an equation with constant, K. (L=K.....)
Write an expression for K in terms of pi, m, M and g (gravity).
Thanks!
Post by: clıppy on March 22, 2014, 07:53:24 pm
a) Find the magnitude of the tension in the towing cable
b) After traveling a distance of 100m the cable breaks. Determine the distance required for the car to come to a complete stop from the moment the cable breaks
I've found part a as 1043N which is correct but I have no idea how to find the answer to part b (15m)
Post by: RKTR on March 29, 2014, 02:57:45 pm
speed of weight is 4.0ms^-1 at the top
to find speed at the bottom ,i let change in ke = change in gpe . is this right?
Post by: rhinwarr on March 29, 2014, 03:17:16 pm
Post by: RKTR on March 29, 2014, 03:46:57 pm
Yes, KE final = KE initial + change in GPE, then solve for speed.
erm if the question says the person does not add any further energy into the system as he twirls it, is the method still correct? tension at top is found . will tension at bottom in this case be the same ?
Post by: rhinwarr on March 29, 2014, 04:03:15 pm
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight
Post by: RKTR on March 29, 2014, 04:14:02 pm
That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight
thanks . i was confused because my friend says tension will be the same since the person doesnt add any energy.
Post by: RKTR on March 31, 2014, 08:24:07 pm
had this as one of my sac questions today. 1. which ship will get hit 1st? 2.explain by referring to relevant physics principle. can anyone explain? i keep thinking there's not enough info.
Post by: rhinwarr on March 31, 2014, 09:11:53 pm
Vertical acceleration is the same for both so if the horizontal distances were the same (which they are not), ship B would get hit first as it reaches half the height of ship A's projectile.
However, since we need to take into account horizontal velocity (which remains constant for both), then the projectile for ship B actually has to travel twice as far so should land at the same time as ship A's projectile.
I'm not really sure so better check with someone else.
Post by: RKTR on March 31, 2014, 10:34:55 pm
using x=ut+1/2 at^2
0= vsin(theta)t -5t^2
t=vsin(theta)/5
v for both are the same . when theta is larger, sin theta is also larger , time is also increased. so B will get hit first. ohno already lost 3 marks out of 40
Post by: lzxnl on March 31, 2014, 10:54:03 pm
Post by: RKTR on March 31, 2014, 10:56:43 pm
There's not enough information. You haven't been told about the speeds and angles of each missile.
really? that's what i thought. but today the question on the sac was which ship will get hit 1st? not multiple choice. and how about the explanation i heard from other people and found online in my previous post?
Post by: lzxnl on March 31, 2014, 11:51:22 pm
Let the speed of the first projectile be v1 and the angle be s1
Then, the vertical displacement is 0 = v1*sin s1 t - 1/2 gt^2 => t = 2v1 sin s1/g
Hmm. The only way the two times are the same is if v1 sin s1 = v2 sin s2 and that's not always true. They haven't given you any numbers, so you really can't make a judgement.
Post by: 09Ti08 on April 01, 2014, 11:57:43 am
*for the angle: it is obvious that ship A is closer to the battleship, so the angle must be bigger for the shell to hit it.
Post by: lzxnl on April 02, 2014, 08:41:41 pm
Post by: 09Ti08 on April 02, 2014, 09:15:39 pm
I think the force exerted by a canon is quite large, but we don't know the distance of the two shells from the mouth of the cannon and their masses and sizes... These numbers do matter, as I need them for calculations. I can't really do anything if I assume too much...
So I think they were fired from two identical canon balls from the same position. That would make sense.
Post by: lzxnl on April 02, 2014, 10:54:05 pm
In any case, we can't conclusively say that the two balls will hit each boat at the same time.
Post by: 09Ti08 on April 02, 2014, 11:04:04 pm
Anyway, the problem is getting unsolvable. However, because it's a question on a sac, we have to come up with an answer by making simplifications and/or assumptions.
Post by: lzxnl on April 02, 2014, 11:08:28 pm
Post by: rhinwarr on April 06, 2014, 02:47:15 pm
Post by: alchemy on April 07, 2014, 11:37:43 am
Can someone help with this question please? I'm not sure how to rearrange the circuit. The answer is supposed to be 83.3 ohms but from the circuit I drew I get 116.7 ohms.
I don't do Physics, so I hope this is correct. I think you found the effective resistance for the entire circuit, whereas the question asks to find it between P and R. Check the attached image for my drawing of the circuit, which is pretty much similar to yours. The path marked in red is the area the question concerns. The effective resistance of the first resistor (in parallel) is 1/[(1/100)+(1/100)]=50. The effective resistance of the other resistors from Q to R and the one below them is 1/[(1/100)+(1/100)+(1/100)]=100/3=33.3. Thus, the total effective resistance is 50+33.3=83.3.
Post by: Yacoubb on April 22, 2014, 09:38:26 pm
1. Describe the origin of the centripetal force that causes an object to follow a circular path.
2. I want to know whether I've approached this correctly.
Saturn has at least eighteen natural satellites, two of which are Titan and Tethys.
mass of Titan = 1.35 x 10^23 kg
radius of Titan = 2.6 x 10^6 m
period of Titan's orbit = 1.38 x 10^6 s
radius of Titan's orbit = 1.22 x 10^9 m
mass of Tethys = 7.4 x 10^20 kg
radius of Tethys orbit = 2.9 x 10^8 m
Calculate the gravitational field strength of the surface of titan.
g = GM/r^2.
G = 6.67 x 10^-11
M = 1.35 x 10^23 kg
r = 2.6 x 10^6 m
Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?
Help would be much appreciated. Thank you :)
Post by: Bronzebottom64 on April 23, 2014, 04:52:37 pm
Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?
Help would be much appreciated. Thank you :)
Yes that is the correct value for the radius, the gravitational field strength is calculated purely on the object in question. Any gravitational field on the surface of a planet in the solar system (Earth) will also have external gravitational fields acting upon them as is the nature of matter in the known universe. The question is a bit ambiguous, however I believe what it's asking is what you have done. :)
Post by: EspoirTron on April 23, 2014, 05:34:17 pm
I have a few questions:
1. Describe the origin of the centripetal force that causes an object to follow a circular path.
2. I want to know whether I've approached this correctly.
Saturn has at least eighteen natural satellites, two of which are Titan and Tethys.
mass of Titan = 1.35 x 10^23 kg
radius of Titan = 2.6 x 10^6 m
period of Titan's orbit = 1.38 x 10^6 s
radius of Titan's orbit = 1.22 x 10^9 m
mass of Tethys = 7.4 x 10^20 kg
radius of Tethys orbit = 2.9 x 10^8 m
Calculate the gravitational field strength of the surface of titan.
g = GM/r^2.
G = 6.67 x 10^-11
M = 1.35 x 10^23 kg
r = 2.6 x 10^6 m
Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?
Help would be much appreciated. Thank you :)
For the first part to your question: the centripetal force is always supplied by a real force. This could include the tension in a string or the frictional force between the tyres of a car and the road.
Let's first take the example of the tension in the string scenario. Say if I have a bucket of water attached to a string, as I begin to swing it in the circular path the tension will act towards the center of the circle, which is by consequence the centripetal force.
The second example is one that has popped up a few times on exams. So consider a 'normal' road; that is, one that hasn't been rained on or has much due so it's pretty dry. If a car wanted to go in a circular path around this dry road it would be quite possible. This is because there is a frictional force between the surface of the road and the car tyres which acts towards the center the circular path, and this too is the centripetal force. Often in calculations on horizontal surfaces you will denote this as the net force.
Caution: One of the reasons why you can't maintain such motion on an icy track is because there is little friction between the car tyres and the surface of the road. So as you speed up it becomes difficult to control the car and the car will leave the circular path at a tangent to the path.
Post by: Yacoubb on April 23, 2014, 06:45:54 pm
For the first part to your question: the centripetal force is always supplied by a real force. This could include the tension in a string or the frictional force between the tyres of a car and the road.
Let's first take the example of the tension in the string scenario. Say if I have a bucket of water attached to a string, as I begin to swing it in the circular path the tension will act towards the center of the circle, which is by consequence the centripetal force.
The second example is one that has popped up a few times on exams. So consider a 'normal' road; that is, one that hasn't been rained on or has much due so it's pretty dry. If a car wanted to go in a circular path around this dry road it would be quite possible. This is because there is a frictional force between the surface of the road and the car tyres which acts towards the center the circular path, and this too is the centripetal force. Often in calculations on horizontal surfaces you will denote this as the net force.
Caution: One of the reasons why you can't maintain such motion on an icy track is because there is little friction between the car tyres and the surface of the road. So as you speed up it becomes difficult to control the car and the car will leave the circular path at a tangent to the path.
Yes that is the correct value for the radius, the gravitational field strength is calculated purely on the object in question. Any gravitational field on the surface of a planet in the solar system (Earth) will also have external gravitational fields acting upon them as is the nature of matter in the known universe. The question is a bit ambiguous, however I believe what it's asking is what you have done. :)
Thanks to you both!
Post by: Rishi97 on April 27, 2014, 12:41:39 pm
Post by: Rod on April 27, 2014, 12:56:32 pm
Does anyone have any good ideas for EPI on electronics? I want to do somthing awesome/interesting :)Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.
If we were to do another EPI I would base it on that, looks so fun.
Good luck Rishi :)
Post by: Rishi97 on April 27, 2014, 01:10:12 pm
Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.
If we were to do another EPI I would base it on that, looks so fun.
Good luck Rishi :)
Yeah that does sound fun Rod. But we haven't learnt anything about modulated signals at school. So would I still be able to do it?
Post by: Rod on April 27, 2014, 01:23:33 pm
Yeah that does sound fun Rod. But we haven't learnt anything about modulated signals at school. So would I still be able to do it?Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.
All my teacher when through was 'a low frequency input signal in increased by a carrier signal, and then we get our modulated signal.....'. And showed us one diagram. So yeah lol he didn't go through it with us a single bit but we still understand it.
It's just a suggestion, I would probably do it.
Best of luck!
Post by: Rishi97 on April 27, 2014, 01:41:15 pm
Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.
All my teacher when through was 'a low frequency input signal in increased by a carrier signal, and then we get our modulated signal.....'. And showed us one diagram. So yeah lol he didn't go through it with us a single bit but we still understand it.
It's just a suggestion, I would probably do it.
Best of luck!
I'll see some other options and ask my teacher as well. Thanks heaps Rod :) :) :)
Post by: knightrider on April 27, 2014, 09:01:10 pm
Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy
Post by: Alwin on April 30, 2014, 02:26:04 pm
how would you do these questions
Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy
Hint: 1 eV = 1.6×10−19 J
Eg for (a) we have: 8.5*10^-12 / (1.6×10−19) = 53 052 829.4 eV (or equivalent scientific notation)
You can try the rest yourself :)
Post by: Rishi97 on April 30, 2014, 04:37:57 pm
I have no idea what to investigate. Any ideas would be greatly greatly appreciated :)
A massive thanks :D Pls help me
Post by: Stevensmay on April 30, 2014, 06:06:05 pm
ok so I got my physics EPI project info today and I found out that I have to design a circuit using either resistors, led or diodes. We had to buy this short circuit kit from jaycar and our EPI can only be made using the components which are spring connectors, battery holders, and all the basic electronic components.What does 'all the basic electronic components' actually entail?
I have no idea what to investigate. Any ideas would be greatly greatly appreciated :)
A massive thanks :D Pls help me
Post by: Rishi97 on April 30, 2014, 06:11:05 pm
Post by: Stevensmay on April 30, 2014, 06:25:56 pm
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)
I asked for clarification just because it didn't seem like much sorry :).
Maybe a row of LED's with different brightness? Honestly I can't think of much you can do with that (at this level).
Post by: Rishi97 on April 30, 2014, 06:51:39 pm
I asked for clarification just because it didn't seem like much sorry :).
Maybe a row of LED's with different brightness? Honestly I can't think of much you can do with that (at this level).
Yeah..,that's why it's so hard to choose an investigation. So when you said have a row of LED's, what exactly would I be measuring?
Post by: Stevensmay on April 30, 2014, 07:11:51 pm
Post by: Rishi97 on April 30, 2014, 07:16:49 pm
The effect of doubling resistance on brightness or something? Or different configurations of resistors eg series/parellel
sounds great but how would I prove that the brightness has improved? Don't I need like some sort of measuring device?
Post by: Orb on April 30, 2014, 08:08:38 pm
bit stuck on this question:
In a laboratory class at school, Lee is given a spring with a stiffness of 15.4 N m–1 and unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new length of the spring is 1.10 m when the mass is stationary. Assuming the spring has no mass, what was the value of the mass he attached?
Thanks!
Post by: Rishi97 on April 30, 2014, 08:22:26 pm
Hey guys,
bit stuck on this question:
In a laboratory class at school, Lee is given a spring with a stiffness of 15.4 N m–1 and unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new length of the spring is 1.10 m when the mass is stationary. Assuming the spring has no mass, what was the value of the mass he attached?
Thanks!
I got 0.18kg. Do you have the answers?
Post by: Orb on April 30, 2014, 08:25:18 pm
I got 0.18kg. Do you have the answers?
it's 1.08.. :/
Post by: Rishi97 on April 30, 2014, 08:40:15 pm
it's 1.08.. :/
ohhh. I'm waay off then. This is what I did....can someone please tell me where I'm going wrong
Us= 1/2kx2
=1/2 x 15.4 x 0.72
= 3.78 J
Since Us is 3.78 J, gpe must be 3.78
3.78 = mgh
And then I found mass by using the height as 1.1.
Post by: Stevensmay on April 30, 2014, 08:58:51 pm
The mass will pull the spring down with a force equal to its mass * gravity.
Therefore the force exerted on the spring is
We also know that
Therefore
x will be equal to the change in the springs length, so 1.1-0.4 = 0.7
k = 15.4
Substituting in and rearranging gives 1.08.
Post by: Zealous on April 30, 2014, 09:03:25 pm
I'm getting a different answer of 1.72...Looks alright?
The mass will pull the spring down with a force equal to its mass * gravity.
Therefore the force exerted on the spring is
We also know that
Therefore
Substituting in and rearranging gives 1.72 which is wrong.
(I've left out all the negative signs so this isn't strictly true.)
If we define downwards as being positive, our change in x (of the spring) will be a positive value as well as our acceleration of gravity so I don't think we'll encounter any negative answers.
Post by: Stevensmay on April 30, 2014, 09:09:15 pm
Post by: Orb on April 30, 2014, 09:09:57 pm
Looks alright?
If we define downwards as being positive, our change in x (of the spring) will be a positive value as well as our acceleration of gravity so I don't think we'll encounter any negative answers.
Thanks guys!
Post by: Alwin on April 30, 2014, 10:22:59 pm
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)
Are you allowed to use a multimeter (or voltmeter/ammeter)??
This would considerably help with designing an experiment because you could measure the voltage and current (and hence calculate power) across components rather than buying some fancy device to measure light intensity of an LED.
If you were allowed these measuring devices, a simple (but hard depending if you're allowed a variable resistor or not) is you could vary the current in the circuit by adding resistors and measure the voltage across the diode (with an LED to dissipate excess power). Then plot voltage vs current and find the switch on voltage
(note that this could be VERY hard if you only have large-sized resistors, so google the diode before hand to see what the Vs is)
Otherwise, as others have suggested, you can create a variety of circuits and analyse. It all depends on what your teacher wants :)
(Perhaps even find someone who did Physics last year from your school and ask what they did, but don't do the exact some thing :P)
GLHF :D
Post by: Rishi97 on May 01, 2014, 03:58:41 pm
Great news!!! I found what I want to do my EPI on.
I'm going to measure the effects of varying reistance has on the power outage of a light bulb. I'm also measuring current as the voltage will remain the same
It's simple but I didn't want to do anything over the required. Like I said, I'm not that good at physics :-\
Post by: speedy on May 16, 2014, 10:10:32 pm
a) V1 decreases due to the negative gradient of the graph and I thought that V2 would stay the same due to it being in photocurrent mode but the answer says it increases.
b) This is where I really had no idea. The units for the graph is mW/m2 but we're given 3 W/m2 so I don't understand how we can use that graph. We can determine the current to be 15 microAmps from the other graph. But yeah that's all I get.
Question:
Spoiler
(http://i.imgur.com/hihb0Ye.png)
Graphs:
V1
Spoiler
(http://i.imgur.com/ML0BTk3.png)
Spoiler
(http://i.imgur.com/oVvopFg.png)
Post by: RKTR on May 16, 2014, 11:49:07 pm
Post by: speedy on May 17, 2014, 12:21:52 am
3Wm^-2 =3000mWm^-2
Oh my god I am so stupid. For some reason I was thinking the units for the graph were mega watt / m^2... Thanks for the help!
Post by: Einstein on May 17, 2014, 12:34:15 pm
Post by: coolrezaee on May 19, 2014, 10:21:35 pm
Does anyone know any good youtube channels to watch vce physics 1/2 videos?
This might help http://www.youtube.com/user/lowdownonphysics
Post by: Rishi97 on May 20, 2014, 06:53:35 pm
Post by: Rod on May 20, 2014, 08:27:35 pm
Is anyone going to the TSFX Physics lecture this weekend???nah, are you?
Post by: Rishi97 on May 20, 2014, 09:38:16 pm
nah, are you?
Yeppp :)
Post by: speedy on May 21, 2014, 09:05:52 pm
Spoiler
(http://i.imgur.com/usMS3pS.png)
Generally with these graphs, to find gain must you find the gradient (rise/run) or can you just go Vout/Vin for any value? Because I was taught the latter but for this question that is wrong (must use gradient).
Post by: rhinwarr on May 22, 2014, 04:30:31 pm
Post by: Nato on May 22, 2014, 04:50:09 pm
Quick question:Spoiler(http://i.imgur.com/usMS3pS.png)
Generally with these graphs, to find gain must you find the gradient (rise/run) or can you just go Vout/Vin for any value? Because I was taught the latter but for this question that is wrong (must use gradient).
In this situation, you can't just do Vout/Vin.
That works for when the line passes through the origin.
but in this case, the graph has been shifted - so when you simply do Vout/Vin, you are not taking the shifting of the graph into consideration.
Post by: speedy on May 23, 2014, 10:33:59 am
Another question:
I've seen the formula V1/V2 = R1/R2 on some notes and stuff but I don't understand how to use it, and when it would be used as opposed to the regular voltage divider formula.
Post by: speedy on May 31, 2014, 09:11:09 am
This is the answer, the top being the modulated wave... So how come there are "two" signal waves? Would it be wrong to just draw a regular sinusoidal wave?
(http://i.imgur.com/9mdRc8d.png)
Post by: PB on May 31, 2014, 06:11:10 pm
Ok first things first, just do me a favour and completely ignore the answers because I think it is wrong. Either that, or I just don't understand it.
Now, this is an example of amplitude modulation, not frequency modulation, I think we gathered that much.
Secondly, I think the carrier wave is a monoamplitude sinusoidal wave with the same frequency as the mondulated wave.
Thus, the signal wave would have to be the other component of the modulated wave. And the only other wave that can summate with the carrier wave I just described to give you the modulated wave shown would be the positive sin looking graph in the answers (not the negative sin graph).
Why? because only a maximum on the signal wave would boost the positive and negative amplitudes of the carrier wave to give you that first section of that modulated wave. and only a negative minimum on the signal wave would squeeze the positive and negative amplitudes of the carrier wave to give you that section at 2ms of the modulated wave (that part which looks like a noose had been wrung tightly over the wave)
Post by: PB on May 31, 2014, 06:15:40 pm
It would be illogical to have the two signal waves looking like what the answer has shown you because they would simply cancel each other out upon wave summation. Thus, the 'modulated wave' and the carrier wave would look exactly the same. Which is why I don't understand what they are trying to say there...
So how come there are "two" signal waves?
Post by: Yacoubb on June 01, 2014, 09:33:46 am
Anyone else doing Materials and Structures for their detailed study? It actually looks like a decent detailed study. We then have to do data analysis on materials and structures for this SAC.
Is unit 4 Phys more difficult or easier than U3?
Post by: Rishi97 on June 01, 2014, 10:21:39 am
Silly mistakes will be the death of me in this subject. Argh.
Anyone else doing Materials and Structures for their detailed study? It actually looks like a decent detailed study. We then have to do data analysis on materials and structures for this SAC.
Is unit 4 Phys more difficult or easier than U3?
We're doing materials and structures for our detailed study as well.. although, I would much prefer to do sound. It 'sounds' more interesting. hehe lol
Post by: Thorium on June 01, 2014, 10:40:23 am
We're doing materials and structures for our detailed study as well.. although, I would much prefer to do sound. It 'sounds' more interesting. hehe lol
I love to do materials and structures as it hss to do with my dream job, civil engineering. But in vain, the teacher told i have to do sound coz it is related to the wave model of light.
Post by: speedy on June 01, 2014, 12:32:59 pm
Ok first things first, just do me a favour and completely ignore the answers because I think it is wrong. Either that, or I just don't understand it.
Now, this is an example of amplitude modulation, not frequency modulation, I think we gathered that much.
Secondly, I think the carrier wave is a monoamplitude sinusoidal wave with the same frequency as the mondulated wave.
Thus, the signal wave would have to be the other component of the modulated wave. And the only other wave that can summate with the carrier wave I just described to give you the modulated wave shown would be the positive sin looking graph in the answers (not the negative sin graph).
Why? because only a maximum on the signal wave would boost the positive and negative amplitudes of the carrier wave to give you that first section of that modulated wave. and only a negative minimum on the signal wave would squeeze the positive and negative amplitudes of the carrier wave to give you that section at 2ms of the modulated wave (that part which looks like a noose had been wrung tightly over the wave)
Alright thanks heaps. My answer was just the positive sin wave so yeah that's good. However, generally I still don't understand why the modulated wave has both a crest and a trough at the same time.
Post by: PB on June 01, 2014, 03:17:04 pm
Post by: hyunah on June 02, 2014, 08:59:56 pm
thanks in advance
Post by: Alwin on June 02, 2014, 09:23:46 pm
can someone help me with this question 886?
thanks in advance
hi :) (assuming you are doing sound as your detail study and know standing waves etc)
So the pitch is most important in this question. Now, we know that the pitch is essentially the frequency and that v = λ f
First of all consider the piano:
The piano creates music by the means of a hammer hitting a string, which would create a standing wave along the string of constant λ. Since we know that the speed of sound is greater, this implies that f is also 10% larger. Hence, the piano will sound off-pitch (ie need to be retuned)
Now consider the flute:
The flute creates music by means of a standing wave in the air column, kinda like those questions about standing waves in open pipes. Now, since the pipe has a constant length, this means that λ is also constant for say the fundamental frequency. But since the speed of sound is 10% greater, the flute will also sound too high pitched since the frequency of sound produced is 10% greater.
So, I would choose [A] ...although it has been a while since I touched sound :P
Post by: hyunah on June 02, 2014, 09:40:43 pm
i am also wondering why at short wavelengths the sound is more likely to travel in straight line?
thanks in advance :)
Post by: lzxnl on June 02, 2014, 10:46:39 pm
Smaller wavelengths => smaller wavelength/gap size ratio => less 'diffraction' occurring
Really, a smaller wavelength/gap size ratio just means the wave spreads out less
Post by: Einstein on June 04, 2014, 03:37:23 pm
Post by: Thorium on June 04, 2014, 04:05:21 pm
for units 1/2 this year, i scored 92% on my radioactivity sac, 100% on Flight (dual highest) and just recently electricity 98% (highest). Despite this, i have chosen to drop physics, im not enjoying it but i am getting really good marks, if you were in my position what would you do? do you think ive made the right decision to drop out of it for Accounting.
If you get similar or better marks in accounting, then go for it
Post by: Einstein on June 04, 2014, 06:13:30 pm
Thanks
Post by: b^3 on June 04, 2014, 06:20:43 pm
how and what is the answer to this question, from electricity sac.They need to be a multiple of the elementary charge, which is approximately
Thanks
So if you divide each of the charges by the elementarycharge, you should get an integer.
Doing this gives
A. 20
B. 5
C. 37.5
D. 90
So option C is the only one that isn't an integer multiple of the elementary charge, and so that charge could not exist. i.e. Option C.
Post by: Einstein on June 04, 2014, 07:23:57 pm
Post by: RKTR on June 04, 2014, 11:11:06 pm
how is this derived?already forgot about this. had to refer to my year 11 textbook. Sv is the unit for dose equivalent while Gy is the unit for absorbed dose.
the symbol before the Sv and Gy is micro which is 10^-6 . To determine the most damaging, you have to convert all them to dose equivalent. dose equivalent = absorbed dose x quality factor.
A 20 x 10^-6
B 30 x 10^-6 x (1) =30 x 10^-6
C 30 x 10^-6
D 20 x 10^-6 x( quality factor of 10-20) =(200 to 400) x 10^-6
dose equivalent of D is highest so it is the most damaging.
Post by: Stew_822 on June 14, 2014, 12:33:28 pm
I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.
It's question 32 and 33 of A+ Notes for physics.
I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V
So here's the answers from the book and my answers afterwards:
Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.
Thanks,
Stewart
Post by: speedy on June 14, 2014, 05:30:43 pm
Hi guys,
I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.
It's question 32 and 33 of A+ Notes for physics.
I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V
So here's the answers from the book and my answers afterwards:
Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.
Thanks,
Stewart
Gain is an absolute value and is usually given as positive regardless of it inverting or not. However the assessors accept either positive or negative values.
What you must realise is that this amplifier is biased - its transfer characteristics graph doesn't pass through the origin. When you simply multiply Vin by the gain, you are treating the amplifier as if it is unbiased. To take the biasing into account, the graph must be used.
Post by: Stew_822 on June 14, 2014, 09:55:46 pm
Is that the only way you can find V_out, by reading it off the graph? Or is there a mathematical method also that takes into account the biased nature of the amplifier. Final question, if there is a mathematical method do I need to know it or will the graph be given in the exam?
Many thanks for your help,
Stewart
Post by: Rishi97 on June 21, 2014, 12:01:24 pm
Many thanks :)
Post by: lzxnl on June 21, 2014, 06:52:22 pm
Post by: Orb on June 23, 2014, 01:35:28 pm
where m is the co-efficient of friction,
What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?
Post by: Jacyan on June 23, 2014, 02:01:23 pm
When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,
What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?
Fr is in newtons, N is in newtons, and the co-efficient of friction has no units.
Post by: Orb on June 23, 2014, 02:07:12 pm
I get a result that's 10x more than the actual answer :/
Post by: Ancora_Imparo on June 23, 2014, 07:17:31 pm
So how should I approach this question...
I get a result that's 10x more than the actual answer :/
As we are told the system is moving left, let left be positive. Let the tension in the left rope be
Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up
Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:
For M2:
You also know that
Thus:
For M3:
Adding Eq. 1 and Eq 3. gives:
Adding Eq. 2 and Eq 4. gives:
[EDIT: Whoops, got the fraction the wrong way around the first time]
Hope that helps.
Post by: Orb on June 24, 2014, 10:32:47 am
As we are told the system is moving left, let left be positive. Let the tension in the left rope beand the tension in the right rope be
.
Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up
Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:g - T1 = (M1)a)
--- Eq. 1
For M2:
You also know that
Thus:g = (M2)a)
(5.4)g = (5.4)a)
--- Eq. 2
For M3:g = (M3)a)
--- Eq. 3
Adding Eq. 1 and Eq 3. gives:--- Eq. 4
Adding Eq. 2 and Eq 4. gives:(negative sign shows that it's decelerating)
Hope that helps.
I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...
Solution for the question was 5.6 ms ^-2
Post by: Zealous on June 24, 2014, 11:01:53 am
I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...I'll give this a shot. Hopefully my explanation makes sense.
Solution for the question was 5.6 ms ^-2
First let's look at the forces due to gravity. We have a force due to gravity of M3 which is
Now let's consider the frictional force of M2:
Friction acts against a motion, so since M2 was initially moving left, the force of friction will be acting in the right/clockwise direction.
If we consider moving right/clockwise to be positive, the sum of the forces will be:
M3 and M2 are causing the system to move right so they are positive, and M1 is working against both of them so is negative.
Hence:
Post by: Ancora_Imparo on June 24, 2014, 12:03:31 pm
Zealous, in the post above, also got the same, but just used
Post by: knightrider on June 28, 2014, 09:37:29 pm
Post by: Homer on June 28, 2014, 09:49:58 pm
b)50-30=20km north
Post by: Stew_822 on June 28, 2014, 09:55:06 pm
So 50 + (-30) = +20km and since we have said positive is north, the full answer is 20km north.
Post by: knightrider on June 28, 2014, 10:40:20 pm
but in my book it says displacement=final position-initial position
in this case it would be 30-50=-20 which would be south
Can anyone explain how this works
Post by: Yacoubb on June 29, 2014, 01:19:36 am
Thankyou for your help guys
but in my book it says displacement=final position-initial position
in this case it would be 30-50=-20 which would be south
Can anyone explain how this works
So total distance would be 50km + 30km = 80km (no direction needed as this is a scalar quantity).
Now, let's make south positive, and thus north negative.
-50km + 30km = -20km
So, we know that the displacement is 20km NORTH.
Let's try making south negative, and thus north positive.
-30km + 50km = 20km
So, displacement = 20km north (given that this value is positive, denoting north).
Hope this helps (:
Post by: Stew_822 on June 29, 2014, 10:02:01 am
Code: [Select]
- - - - - >
<- - -
So he moves 50km right (north) then 30km south. His initial position is 0, then +50, then he moves back 30km so his final position is +20km.Using that formula, displacement = final position - initial = +20 - 0 = +20km and since positive then its north.
An easier way to do these questions where they list all of the distance is to say that displacement = vector sum of all movements, ie. how its been calculated in the last few posts.
Post by: knightrider on June 30, 2014, 11:08:45 pm
Post by: knightrider on July 09, 2014, 08:16:41 pm
Post by: PB on July 09, 2014, 10:56:41 pm
Can anyone help with this questionWell the instantaneous velocity would require you to find the gradient at point t=35s. So that requires differentiation - which is basically impossible because you have no formula to differentiate!
In this case, I reckon you should try draw a tangent from the point t=35s and find the gradient of that line :P
A bit of a shoddy technique but its VCE Physics. Besides, VCAA is unlikely to give this kind of question because there would be too much variation in the answers obtained. Hope that helps
Post by: knightrider on July 09, 2014, 11:58:11 pm
how would you do this question
Post by: Rishi97 on July 10, 2014, 11:29:35 am
What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me
Thanks :)
Post by: PB on July 10, 2014, 12:02:30 pm
Thanksseeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement). You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.
how would you do this question
Post by: PB on July 10, 2014, 12:14:40 pm
A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohmsJust wondering what the answer was?
What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me
Thanks :)
Post by: Rishi97 on July 10, 2014, 12:28:25 pm
Just wondering what the answer was?
289 W
Post by: PB on July 10, 2014, 01:08:27 pm
289 WYep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip. Ip= Vp/R =170/100 =1.7
Pp=1.7*170 = 289W
Post by: Rishi97 on July 10, 2014, 01:37:08 pm
Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip. Ip= Vp/R =170/100 =1.7
Pp=1.7*170 = 289W
Thanks heaps for that PB :)
Post by: knightrider on July 10, 2014, 02:21:09 pm
seeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement). You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.
Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them out
Post by: PB on July 10, 2014, 02:56:14 pm
Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them outOk, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s x 1 s = 1m (seconds cancel out to give you metres - the position!). t=2 ----> 1m/s x 2s = 2m. t=3 ----> 3 and a half squares so 3.5m So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots. Do you understand how that works?
Post by: knightrider on July 10, 2014, 03:40:25 pm
Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s x 1 s = 1m (seconds cancel out to give you metres - the position!). t=2 ----> 1m/s x 2s = 2m. t=3 ----> 3 and a half squares so 3.5m So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots. Do you understand how that works?
yes i do thanks so much :)
Post by: PB on July 10, 2014, 04:34:38 pm
yes i do thanks so much :)
Cool beans :) BTW, I have a physics lecture on next Tuesday which teaches you how to deal with exactly these kinds of stuff and includes shortcuts for some questions and much much more. Would you be interested in coming?
I have held this lecture before and it seems to have been very beneficial for those who attended! Alll the details can be found in the link in my signature below. Please do check it out! It will be the last time I will be holding it, so don't miss out!
Post by: Rishi97 on July 10, 2014, 07:17:18 pm
Which is better and why?
Post by: knightrider on July 10, 2014, 08:04:31 pm
What is the acceleration of the train 10 s after starting?
What is the acceleration of the train 40 s after starting?
Post by: Stew_822 on July 11, 2014, 11:06:00 am
eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.
This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs
Does that make sense? Hope it helps,
Stewart
Post by: lzxnl on July 11, 2014, 11:56:33 am
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?
They do completely different things. Slip rings don't change current direction => AC
Split rings do reverse current direction => DC
Hopefully you learn how split rings reverse the current direction
Post by: knightrider on July 11, 2014, 01:01:02 pm
The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.
eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.
This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs
Does that make sense? Hope it helps,Stewart
i still dont get how you got 20/20 = 1m/s^2. because it is a curve so you cant do rise on run and if you did that it would be 20/10=2m/s^2 the answers say something else to both the answers can anyone help
Post by: Stew_822 on July 11, 2014, 01:45:00 pm
What is the answer?
Check these links for finding the gradient of a non-linear graph, you have to draw a line and find the gradient of that line as the other link explains. This is the acceleration.
http://cstl.syr.edu/fipse/graphb/unit8/unit8a.html
http://www.columbia.edu/itc/sipa/math/slope_nonlinear.html
Post by: PB on July 11, 2014, 02:18:17 pm
Post by: Rishi97 on July 11, 2014, 04:13:45 pm
WHat is the acceleration?
Using the a=gsin(angle) formula, I am getting the answer of 5m/s/s but the answer says 3.3m/s/s
What am I doing wrong? Why doesn't the formula work?
Post by: rhinwarr on July 11, 2014, 04:49:57 pm
a = F/m = 20/12 = 1.7m/s/s
So the 'net' acceleration would be 5-1.7=3.3m/s/s
Post by: PB on July 11, 2014, 05:16:44 pm
Post by: PB on July 11, 2014, 06:28:12 pm
SLIP rings are used to "collect" the AC current generated from generators while SPLIT rings facilitate the constant rotation of motors.
You should try to understand how these rings achieve those feats though, and not just memorise the answers :)
Post by: Yacoubb on July 11, 2014, 07:49:13 pm
how would you do these questions
What is the acceleration of the train 10 s after starting?
What is the acceleration of the train 40 s after starting?
With velocity-time graphs, acceleration can be calculated from the gradient.
So:
(a) 20m/s / 10s = 2m/s/s
(b) 40m/s / 40s = 1m/s/s
We can validate this because we can see that the graph eventually plateus; an indication of the fact that the object is travelling at constant speed (at which gradient = 0).
Post by: Rishi97 on July 11, 2014, 10:25:02 pm
Oh and Rishi, for your split/slip ring question. The full definitions can be found in my cheat sheet under the section Electric Power. And neither is "better" per se, rather they both have different functions.
SLIP rings are used to "collect" the AC current generated from generators while SPLIT rings facilitate the constant rotation of motors.
You should try to understand how these rings achieve those feats though, and not just memorise the answers :)
Thanks PB
Yeah I'll check your cheat sheet on this info. :)
Post by: knightrider on July 11, 2014, 11:24:05 pm
Post by: knightrider on July 11, 2014, 11:25:09 pm
how would you draw an acceleration–time graph for the bus
Post by: Stew_822 on July 12, 2014, 11:25:38 am
With velocity-time graphs, acceleration can be calculated from the gradient.Hello there,
So:
(a) 20m/s / 10s = 2m/s/s
(b) 40m/s / 40s = 1m/s/s
We can validate this because we can see that the graph eventually plateus; an indication of the fact that the object is travelling at constant speed (at which gradient = 0).
The acceleration you've calculated in (a) is the average acceleration for the first 10 seconds. The question is asking the accerelation at ten seconds after starting, meaning that you need to draw a line at a tangent to the gradient at that point and calculate the gradient from that line, as the acceleration isn't linear since the graph isn't linear. You can see that there is a lot of initial acceleration, but it dies off later. So although the average acceleration for the first 10s is 2m/s/s, the acceleration at say 2s is greater than this and the acceleration at say 8s is less than this (and obviously then the acceleration at 10s is less).
I hope this makes sense and helps. When I calculated the gradient, I didn't bother drawing a line - I basically guestimated. I can draw you guys a pic if you like.
Cheers,
Stewart
Post by: Stew_822 on July 12, 2014, 11:28:02 am
how would you draw an acceleration–time graph for the busHave you read the chapter?
You find the accelation of each linear portion of the graph (you can find it using rise over run as this v-t graph is linear) and graph the acceleration against time.
Post by: knightrider on July 12, 2014, 01:09:30 pm
Have you read the chapter?
You find the accelation of each linear portion of the graph (you can find it using rise over run as this v-t graph is linear) and graph the acceleration against time.
Yea thanks but how would you find the acceleration for 6 seconds or 8 seconds for the bus
Post by: Stew_822 on July 12, 2014, 03:40:42 pm
Post by: knightrider on July 14, 2014, 04:18:28 pm
Post by: knightrider on July 14, 2014, 10:47:23 pm
Q1 Hint: What does the gradient of a velocity-time graph tell you?SpoilerAcceleration
Q2,3,4 Hint: What does the area under a velocity-time graph tell you?SpoilerDisplacement
If you still need help just post again
Thanks for tips but i still need help
Post by: knightrider on July 14, 2014, 11:43:03 pm
Post by: Stew_822 on July 15, 2014, 10:25:56 am
When asking questions and for help, it is better if you let us know what you are struggling with specifically, rather than just saying you need help. For example, you could tell us how you attempted the problem and that way we would know where a mistake may have been made and point you in the right direction.
I remember that question with the bus and Anna from last year, I had to ask my teacher.
You can usually find the worked solutions by googling "heinemann physics 11 worked solutions chapter X" where X is obviously the chaper you're up to.
I hope this helps.
Cheers,
Stewart
Post by: hyunah on July 15, 2014, 07:04:21 pm
An electron moving north enters a magnetic field
that is directed vertically upwards.
If the electron’s motion was inclined upwards
at an angle, as well as travelling north, what
would be the path of the electron?
Post by: rlenora on July 15, 2014, 09:52:30 pm
can someone please help me?
An electron moving north enters a magnetic field
that is directed vertically upwards.
If the electron’s motion was inclined upwards
at an angle, as well as travelling north, what
would be the path of the electron?
Use right hand palm rule (slap rule :P)
Direction of magnetic field: up (use fingers of right hand to point up)
Direction of current, i.e. direction of positive charge: south (opposite to direction of electron) (point thumb to point south on right hand)
Direction of force on electron: West (given by the direction of the palm of right hand)
Hence electron will initially move West, and since force applied on the electron will always be perpendicular to its motion (like it is in circular motion), the electron will travel in a circular path.
We do not have to worry about the component of the electron's motion upwards (i.e. parallel to the magnetic field), since components of current (velocity of the electron in this case) parallel to the magnetic field doesn't result in a force being applied on the electron.
Hope this helps :)
Let me know if you'd like clarification on part!
Post by: rlenora on July 15, 2014, 10:27:41 pm
Hello knightrider,
When asking questions and for help, it is better if you let us know what you are struggling with specifically, rather than just saying you need help. For example, you could tell us how you attempted the problem and that way we would know where a mistake may have been made and point you in the right direction.
I remember that question with the bus and Anna from last year, I had to ask my teacher.
You can usually find the worked solutions by googling "heinemann physics 11 worked solutions chapter X" where X is obviously the chaper you're up to.
I hope this helps.
Cheers,
Stewart
Exactly!
This way, we can help you out better than just giving you the answers knightrider :)
Post by: hyunah on July 15, 2014, 11:18:02 pm
the part which I didn't get is the inclines angle bit and yet as well as travelling north, does that when my hand is like slanted north - west or north east?
Post by: rlenora on July 15, 2014, 11:51:44 pm
So all you have to worry about is that the electron is moving north.
Which is the same as if a positive charge was moving south.
and since the thumb points in the direction of the conventional current (i.e. direction of movement of positive charge), thumb will need to point south.
So, fingers up. thumb south. palm faces west. Force applied on electron is to the west :)
Post by: knightrider on July 18, 2014, 11:27:47 pm
For these 2 equations how do you know when to use the one with the plus 1/2 and the one with -1/2
What does the plus and minus mean and how would you know to use which formula
Post by: Thorium on July 19, 2014, 02:37:32 am
hi guys,
For these 2 equations how do you know when to use the one with the plus 1/2 and the one with -1/2
What does the plus and minus mean and how would you know to use which formula
Apparently the second one is not officially recognised (maybe only by my school), but it is quite useful in some situations.
They are both used under one main condition, that is, when acceleration is constant. You can notice that one involves u (initial velocity) and the other involves v (final velocity). So if you have one quantity missing out of x, u, t, a, then you use the equation x=ut+1/2at^2. If you have one quantity missing out of x, v, t, a, then you use the equation x=vt-1/2at^2.
Hope that helps :)
Post by: knightrider on July 19, 2014, 09:07:57 am
Apparently the second one is not officially recognised (maybe only by my school), but it is quite useful in some situations.
They are both used under one main condition, that is, when acceleration is constant. You can notice that one involves u (initial velocity) and the other involves v (final velocity). So if you have one quantity missing out of x, u, t, a, then you use the equation x=ut+1/2at^2. If you have one quantity missing out of x, v, t, a, then you use the equation x=vt-1/2at^2.
Hope that helps :)
Didn't realize the equations were different i thought they bith had u's that were negative and positive thankyou for your help :)
Post by: knightrider on July 19, 2014, 05:29:13 pm
I have worked out the initial speed to be 20m/s and the maximum height to be 20m
how would you find the speed of the cork as it returned to its
starting point?
Post by: Butterscotch on July 20, 2014, 04:09:17 pm
Hi how would you do the following question
I have worked out the initial speed to be 20m/s and the maximum height to be 20m
how would you find the speed of the cork as it returned to its
starting point?
An easy way to go about solving such types of questions is to ignore the rise of the cap. Let's just look at the fall of the cork from the maximum point.
You already have:
u = 0 ms^-1
a = 9.8 ms^-2
t = 2 s (remember we're only looking at the return journey of the cork)
v = ?
You'd plug these into the formula: v = u + at
Your answer rounded off would be 20 ms^-1.
EDIT: Orrr an alternative (also one which saves time) is one that PB has said below :)
Post by: PB on July 21, 2014, 11:58:40 am
edit: speed, not velocity
Post by: lzxnl on July 21, 2014, 02:46:12 pm
Well, technically you don't even have to do any working out... i mean, the cork decelerates on the way up and accelerates by the same amount on the way down. It is bound to have the same velocity at the end of the flight as it did at the beginning.
You've invoked the conservative nature of the gravitational force, and a rigorous proof of that isn't going to be given before second year university :P
Technically, it should reach the starting height slower than its initial speed because of the non-conservative nature of air resistance. In addition, PB I think you've made a typo there that I'll highlight. I trust you can see what is wrong with it :P
Post by: yang_dong on July 21, 2014, 09:11:35 pm
thank you
Post by: PB on July 22, 2014, 01:04:09 am
You've invoked the conservative nature of the gravitational force, and a rigorous proof of that isn't going to be given before second year university :P
Technically, it should reach the starting height slower than its initial speed because of the non-conservative nature of air resistance. In addition, PB I think you've made a typo there that I'll highlight. I trust you can see what is wrong with it :P
Well yeah of course, but I didn't want to confuse anyone else given that this is VCE Physics we are talking about and not real life physics.
But your second point is valid. My apologies people, it should really say speed instead of velocity because 'same velocity' would imply that the cork has the same direction of travel too..which is of course false - it is the opposite on the way down.
Post by: PB on July 22, 2014, 01:19:14 am
why no force is produced when the wire is parallel to the magnetic field?
thank you
Interesting phenomenon and sadly way outside the course content of VCE Physics :P
If it were to be asked though, just prove it using the formula for the force acting on current carrying wire "F=nILBsin(theta)". When M-field and current is parallel, theta is 0 which makes the whole left side of the equation equal 0. Thus, force is zilch when wire is parallel to M-field.
Always good to link it back to a formula to prove a point.
Post by: yang_dong on July 22, 2014, 11:28:00 am
can someone please help me with question 26a)
thank you
Post by: Saikyo on July 27, 2014, 05:30:11 pm
In an elastic collision between two objects of mass m1 and m2, show that the speed of approach (u2-u1) is equal to the speed of seperation (v2+v1). The symbols u1, u2, v1 and v2 each represent speeds, not velocities.
there is NO WAY that this is a vcaa type question right guys??
[sorry for the bad drawing i made from the photo for the diagram in the question!]
Post by: lzxnl on July 27, 2014, 08:30:08 pm
This is just proving the conservation of momentum (if initial total masses are equal to final total masses, then total initial speeds are equal to total final speeds)
VCAA won't give anything this vague, it will always have actual figures for you to plug into equations
It's not even proving the conservation of momentum; that is a principle you'd have to use to prove it
I wish VCAA gave things without figures.
Post by: Brunette15 on July 28, 2014, 07:25:23 pm
Can someone please help me out with this question for light:
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.
Post by: PB on July 29, 2014, 12:01:08 am
Hi everyone 8) ,A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Can someone please help me out with this question for light:
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.
Post by: Brunette15 on July 29, 2014, 12:51:47 pm
A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.
Thanks so much! :D
Post by: hyunah on July 30, 2014, 10:27:03 pm
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground
thanks you and please
Post by: RKTR on July 31, 2014, 01:00:40 am
a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:(a) v^2=u^2+2ax
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground
thanks you and please
at max height,v=0
0=24.5^2+2(-10)x
x=30m
but this is the displacement from the platform therefore max height=30+16=46m
(b) v=u+at
0=24.5-10t
t=2.45s
(c) x=ut+1/2 a t^2
from max height to ground x=-46
-46=1/2(-10)(t^2)
t=3.03s
(d) total time = ans of (b) + ans of (c)
=2.45+3.03
=5.48s
(e)v^2=u^2+2ax
v^2=2(-10)(-30)
v^2=600
speed=24.5 m/s
(f)v^2=u^2+2ax
v^2=2(10)(46)
v^2=920
speed=30.3 m/s
Post by: hyunah on July 31, 2014, 11:23:43 am
someone please help me here too!
A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:
i) how much further it will travel before ocming to rest
ii) how many more seconds it takes before coming to rest
thanks and please
Post by: speedy on August 03, 2014, 01:05:42 pm
Take this image -
Spoiler
(http://www.geocities.ws/motorac2002/field_dir_poles_e.gif)
Post by: lzxnl on August 03, 2014, 02:16:24 pm
Post by: speedy on August 03, 2014, 06:04:49 pm
Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.
What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.
Post by: lzxnl on August 03, 2014, 07:17:20 pm
What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.
The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.
As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.
Post by: speedy on August 03, 2014, 07:33:04 pm
The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.
As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.
Yeah I completely agree. This is really annoying because he's a very good teacher and does know his stuff. I'll bring it up with him tomorrow, tell him what you said, show him the textbook and a few questions. Thank you :)
Also with that, for a magnet approaching a loop from the left, would you say that the field is increasing to the left, thus the induced current must produce a field increasing to the right, which would be coming out of the loop in the same direction the magnet was moving?
Also, just another question I came across, how do you find the average voltage produced by a coil that has been turned one complete cycle? You use faraday's law with the time for one quarter of the cycle, right?
Post by: Brunette15 on August 04, 2014, 08:02:50 pm
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?
mass of alpha is 6.67x10^-27kg
I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.
Post by: hyunah on August 04, 2014, 08:17:21 pm
what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?
thank you
Post by: Bestie on August 05, 2014, 10:29:50 pm
I know it has something to do with momentum but I don't quite get it....
Post by: knightrider on August 16, 2014, 11:38:14 pm
A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?
Post by: lzxnl on August 17, 2014, 10:56:07 am
How would you do this question
A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?
Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.
Post by: knightrider on August 17, 2014, 12:00:57 pm
Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.
Thanks would this be right :)
using m1u1+m2u2=m1v1+m2v2
(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer
Post by: Zealous on August 17, 2014, 04:54:10 pm
Thanks would this be right :)The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
using m1u1+m2u2=m1v1+m2v2
(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer
Post by: Alwin on August 17, 2014, 05:42:21 pm
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?
mass of alpha is 6.67x10^-27kg
I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.
Think about the charge on an alpha particle ;) It's a bit of a trick question because you have to had remembered what the charge is from year 11, or do it the way you did it with year 12 methods
what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?
Hmm interesting question but usually not examinable. A quick run down of things that can make transformers less ideal:
Eddy currents
Non ideal joins (side note is that Wilson Transformers actually makes all their transforms by hand and its an amazing process to get maximum efficiency)
Heat in the coil (usually transformers are oil-convection cooled)
Whether or not the core is laminated (this goes back to the eddy currents issue)
And the last one I can think of is flux leakage, but tbh I don't really get this either
But basically in essence the most important one as you've pointed how is the frequency as this will have an effect on eddy currents (imagine it as a sort of like a residue current in the coil)
question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....
I assume you mean something like this :)
(http://www.tpub.com/neets/book5/32NE0420.GIF)
I'll start with B first, because it's the most obvious: Force on side is going up and the force on the other is going down, so the coil spins, seems simple enough :)
But then when it's in points A and C, the force on the top is going upwards and the force on the bottom is acting downwards so I get there can be some confusion as how the coil keeps on turning. In fact, it's because of the commutator. At that instant the coil briefly loses connection with the batter (the 'split' part of the split ring commutator) so there is no current flowing and thus no force acting on the coil. Now, the coil was originally spinning (see point B) so it will continue to spin as there is no braking or frictional force acting on the coil. The terminals of the coil then touch the other side of the split ring commutator (reversing the current) and now the forces are reversed as well meaning the coil keeps on spinning.
So in essence, the coil's spinning momentum just before the perpendicular position is what carries to through the vertical position and allows the direction of the current to be changed via the commutator
Post by: Stick on August 18, 2014, 12:08:26 pm
Post by: Zealous on August 18, 2014, 05:08:56 pm
Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)
Hey Stick,
From what I've calculated, Rachel would be 6m from the centre pivot point, which is not actually on the see-saw. Yeah - doesn't make sense to me also.
Post by: Stick on August 18, 2014, 05:11:51 pm
Post by: knightrider on August 18, 2014, 09:05:34 pm
The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
Thanks Zealous :)
Can you help with these questions?
A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.
a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?
Post by: Alwin on August 19, 2014, 12:46:36 pm
Thanks Zealous :)
Can you help with these questions?
A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.
a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?
Sorry, I'm not Zealous but I hope you'll accept my help too :P
To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)
Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,
Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive
Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time
Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again
Post by: knightrider on August 19, 2014, 06:13:39 pm
Sorry, I'm not Zealous but I hope you'll accept my help too :P
To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)
Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,
Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive
Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time
Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again
Thank you so much Alwin i really appreciate your help, you are very nice :)
Post by: allstar on August 20, 2014, 10:44:45 pm
thanks
Post by: Conic on August 21, 2014, 11:38:07 pm
Post by: chemdeath on August 22, 2014, 02:28:18 am
explain this observation in terms of energy and heat flow? (
Post by: Bestie on August 22, 2014, 12:00:17 pm
help please :)
Post by: allstar on August 22, 2014, 12:04:40 pm
but what happens if it was initially constant velocity and then accelerate in the opposite direction, is the gradient of acceleration positive or negative in this case?
Post by: Conic on August 22, 2014, 03:42:19 pm
(http://i.imgur.com/SZGEgH8.png)
Post by: knightrider on August 25, 2014, 08:59:57 pm
A removalist is loading five boxes onto a truck.
Each has a mass of 10 kg and a height of 30 cm. The
tray of the truck is 1.5 m above the ground and the
removalist is placing each box on top of the previous
one.
What is the total work done on the boxes in lifting
all the boxes onto the truck as described?
Post by: JHardwickVCE on August 25, 2014, 09:11:07 pm
As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.
For the first box:
x=1.5m
W=(100)(1.5) = 150N
You will have to lift each box an extra 0.3m
Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Post by: lzxnl on August 25, 2014, 11:20:09 pm
Work (W) = Force (F) * Displacement (x)
As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.
For the first box:
x=1.5m
W=(100)(1.5) = 150N
You will have to lift each box an extra 0.3m
Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Your answer is correct, but be careful of the working.
For instance, if you were to actually try to exert a force equal to the weight force on the box, the net force on the box would be zero. Problem. In reality, we pull up with a greater force than the weight force to get the box moving, and then we reduce the force when we want to slow it down, so the force isn't constant. W = Fx only works for constant forces.
What you should actually do is note that the kinetic energy at the beginning and the end is zero, so the work done is the change in gravitational potential energy. The working out is the same, the reasoning isn't.
Post by: knightrider on August 25, 2014, 11:40:45 pm
Work (W) = Force (F) * Displacement (x)
As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.
For the first box:
x=1.5m
W=(100)(1.5) = 150N
You will have to lift each box an extra 0.3m
Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Your answer is correct, but be careful of the working.
For instance, if you were to actually try to exert a force equal to the weight force on the box, the net force on the box would be zero. Problem. In reality, we pull up with a greater force than the weight force to get the box moving, and then we reduce the force when we want to slow it down, so the force isn't constant. W = Fx only works for constant forces.
What you should actually do is note that the kinetic energy at the beginning and the end is zero, so the work done is the change in gravitational potential energy. The working out is the same, the reasoning isn't.
Thx lzxnl and jhardwickvce :)
But remember jhardwickvce that work is measured in joules or newton metres
in your working out you wrote newtons at the end :)
By the way jhardwickvce and lzxnl how different is physics 1/2 to 3/4 physics and how are you guys finding it/found it
Post by: Phenomenol on August 26, 2014, 12:13:30 am
Thx lzxnl and jhardwickvce :)
But remember jhardwickvce that work is measured in joules or newton metres
in your working out you wrote newtons at the end :)
By the way jhardwickvce and lzxnl how different is physics 1/2 to 3/4 physics and how are you guys finding it/found it
I know I'm not jhardwickvce nor lzxnl but I hope you don't mind me answering this question.
Motion in 3/4 builds upon what was learnt in 1/2 very well. Some new concepts involve circular motion, more intricate block/ramp systems and newtons's law of gravity.
Electricity involved familiar stuff too. Some additions included diodes, thermistors, photodiodes, light-dependent resistors in circuits. A common question is to figure out where to place a cooling/heating element in a circuit.
Light is a different story entirely (at least from what I learned in 1/2). In 1/2 we did a lot of ray tracing and calculations based on focal lengths of lenses/mirrors, but in 3/4 the content shifted towards wave/particle duality of light and the experiments that proved such light properties (photoelectric effect, interference patterns). I found this topic the driest and most repetitive.
Electromagnetism was not something we were taught in 1/2 so concepts like flux and field took a bit of time to get comfortable with, but otherwise it's fine.
Post by: knightrider on August 26, 2014, 06:29:57 pm
I know I'm not jhardwickvce nor lzxnl but I hope you don't mind me answering this question.
Motion in 3/4 builds upon what was learnt in 1/2 very well. Some new concepts involve circular motion, more intricate block/ramp systems and newtons's law of gravity.
Electricity involved familiar stuff too. Some additions included diodes, thermistors, photodiodes, light-dependent resistors in circuits. A common question is to figure out where to place a cooling/heating element in a circuit.
Light is a different story entirely (at least from what I learned in 1/2). In 1/2 we did a lot of ray tracing and calculations based on focal lengths of lenses/mirrors, but in 3/4 the content shifted towards wave/particle duality of light and the experiments that proved such light properties (photoelectric effect, interference patterns). I found this topic the driest and most repetitive.
Electromagnetism was not something we were taught in 1/2 so concepts like flux and field took a bit of time to get comfortable with, but otherwise it's fine.
Thankyou Phenomenol :)
Post by: knightrider on August 28, 2014, 04:06:30 pm
A rope that is at 35° to the horizontal is used to pull a
10.0 kg crate across a rough floor. The crate is initially
at rest and is dragged for a distance of 4.00 m. The
tension in the rope is 60.0 N and the frictional force
opposing the motion is 10.0 N.
Determine the energy lost from the system as heat
and sound due to the frictional force.
Post by: allstar on August 30, 2014, 01:33:30 pm
maximum speed of 60 km/h. Find the shortest time the bus can take to travel between
two bus stops 1 km apart on a straight stretch of road.
please?
Post by: Brunette15 on August 30, 2014, 02:29:36 pm
A winch uses a steel cable to lift a large piece of machinery
from a loading dock. While supporting this load the length
of the steel cable increases to 1.001 times its original
value. What is the tensile strain on the cable when
supporting this load? Express your answer as a
percentage 8)
Post by: Rishi97 on August 30, 2014, 05:14:55 pm
Can someone please help me calculate this question for structures and materials:
A winch uses a steel cable to lift a large piece of machinery
from a loading dock. While supporting this load the length
of the steel cable increases to 1.001 times its original
value. What is the tensile strain on the cable when
supporting this load? Express your answer as a
percentage 8)
Hey
Ok when I did this question, I just made up random numbers to get an idea of what the question is trying to say.
Lets say that the original length is 5m
When the cable increases to 1.001 times its original length, then 5x1.001 = 5.005
That means that the change in length is 0.005
Sub into the strain=change in length/ original length and you should get an answer of 0.001
Since the answer says to give as a percentage, 0.001 x 100 = 0.1%
That should be right :)
Post by: qwerty04 on August 30, 2014, 09:48:53 pm
Why is a vacuum needed in a photocell?
Why do photoelectric cells need to be stored in a dark space?
Post by: Brunette15 on September 02, 2014, 06:33:19 pm
Post by: Phenomenol on September 02, 2014, 09:34:51 pm
Is the yield strength the same as the elastic limit of a material? If not what is the difference? 8) :-\
I believe they mean the same thing.
Post by: speedy on September 04, 2014, 04:58:59 pm
Post by: Rishi97 on September 08, 2014, 08:53:02 pm
In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/
Thanks
Post by: Phenomenol on September 08, 2014, 09:00:43 pm
Hey everyone
In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/
Thanks
Don't use obscure ones - use ones that the examiner can clearly see are rearrangements of a formula on the formula sheet, or when very simple substitutions are made. For example, e = hc/lambda is completely acceptable. (substituting f = c/lambda into the photon energy equation)
Post by: lzxnl on September 08, 2014, 09:09:28 pm
Hey everyone
In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/
Thanks
Which formulas would you even want to use? I never felt the need to use extra formulas explicitly.
Post by: Rishi97 on September 08, 2014, 09:32:57 pm
Which formulas would you even want to use? I never felt the need to use extra formulas explicitly.
just ones for projectile motion such as quicker formulas to find range and time
Post by: lzxnl on September 08, 2014, 10:13:22 pm
In any case, deriving it wouldn't take you more than half a minute.
Post by: speedy on September 08, 2014, 10:17:33 pm
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.
It's not - that's all the motion ones:
Spoiler
(http://i.imgur.com/1yiTWX9.png)
I always thought that if you used a different formula, and got the answer wrong, you wouldn't get any marks for working. May be wrong though, but I would say always derive it first.
Post by: Rishi97 on September 08, 2014, 10:32:17 pm
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.
Nope not on there. Could you by any chance give me tips on how to derive? And from what formula?
Post by: lzxnl on September 08, 2014, 10:38:10 pm
Anyway, I'm assuming you mean the formula for no change in height projectile motion.
In the vertical direction, y=0. So ut + 1/2 at^2 = 0
We don't want t=0 = 2u + at = 0
Letting up be positive, a = -g
t = 2u/g. But u = v sin theta => t = 2v sin theta / g
Distance travelled = v cos theta * t = 2v sin theta cos theta v /g = v^2 sin(2 theta)/g by trig identity
Post by: allstar on September 17, 2014, 10:39:05 pm
an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
Post by: Yacoubb on September 17, 2014, 10:50:52 pm
any help is greatly appreciated :)
an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
V= u -10t
t = -(v-u)/10
t = u-v/10 *2
t= u-v/5
Not sure if this is your answer but that's my jab! :-)
Post by: lzxnl on September 17, 2014, 10:59:12 pm
any help is greatly appreciated :)
an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
What's the proportionality constant?
You'd have to break this question up into two parts as the air resistance changes direction, and even then it's a question involving calculus which wouldn't be asked for a VCE physics student.
V= u -10t
t = -(v-u)/10
t = u-v/10 *2
t= u-v/5
Not sure if this is your answer but that's my jab! :-)
Not constant acceleration; there's air resistance here
Post by: Yacoubb on September 17, 2014, 11:04:50 pm
Not constant acceleration; there's air resistance here
Would this be right if we were told to ignore air resistance?
Post by: lzxnl on September 17, 2014, 11:47:11 pm
Post by: Yacoubb on September 17, 2014, 11:55:58 pm
Your working missed a nuance in the question. The question says 'speed v'. But it's returning downwards, so its final velocity is actually -v, not v.
Ah yes, I see. Thanks lzxnl!
Post by: maurlock on September 21, 2014, 01:01:16 pm
Thanks!
Post by: Conic on September 21, 2014, 02:01:40 pm
which means that E is correct. Because
Post by: maurlock on September 22, 2014, 05:50:39 pm
The angle used in the formula is the angle between the force and the lever. In this case, as the angle with the ground is 65, the angle with the lever is 25, so the torque is given byOooh I see, thank you so much!
which means that E is correct. Becauseit appears that the cosine of the angle is used, but you need to make sure you use the right angle. The cosine of the angle with the ground happens to be equal to the sine of the angle between the weight force and the lever, which is what is actually used in the calculation.
Post by: speedy on September 23, 2014, 05:01:06 pm
Spoiler
(http://i.imgur.com/syISyRb.png)
Post by: Yacoubb on September 23, 2014, 06:15:52 pm
Why is Fbox on trailer > Ftrailer on box?Spoiler(http://i.imgur.com/syISyRb.png)
I could be wrong, but I think you're not looking at the diagram correctly. The arrow from the trailer upwards is actually from the bottom of the box (where the trailer makes contact with the box), and the arrow downwards is from the bottom of the trailer downwards, so in reality the proportion of upwards to downward arrow size is the same. It would have to be according to Newton's 3rd Law of Motion...
Post by: speedy on September 23, 2014, 06:42:25 pm
I could be wrong, but I think you're not looking at the diagram correctly. The arrow from the trailer upwards is actually from the bottom of the box (where the trailer makes contact with the box), and the arrow downwards is from the bottom of the trailer downwards, so in reality the proportion of upwards to downward arrow size is the same. It would have to be according to Newton's 3rd Law of Motion...
Well I just drew that on so idk. I guess it comes down to interpretation of the image in the question:
Spoiler
(http://i.imgur.com/JcbEd2w.png)
If we say that the box is within the trailer, then I agree with you. If we say that it just sits on top then clearly the arrow downwards is larger. I don't know why the answer doesn't show both the trailer and box - makes the origin very ambiguous.
When I did the question I took it to be sitting on top of the trailer (see diagram), but I upon looking at it again, it does appear to go deeper.
Post by: allstar on September 23, 2014, 06:55:02 pm
Post by: Zealous on September 23, 2014, 07:02:12 pm
for the gain of an amplifier, (even if it is an inverting amplifier), should it always be a positive value?I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.
Post by: Yacoubb on September 23, 2014, 07:28:24 pm
I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.
Yeah I always just find the absolute value. Negative gain...it does sound awkward lol.
Post by: allstar on September 23, 2014, 07:31:00 pm
just another one:
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
Post by: Yacoubb on September 23, 2014, 07:31:57 pm
thank you guys!
just another one:
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
Yep! :-)
Post by: allstar on September 23, 2014, 07:38:55 pm
Post by: Yacoubb on September 23, 2014, 07:43:03 pm
but the ans says i can't? why?
Does it say photodiode?
Post by: allstar on September 23, 2014, 07:49:17 pm
i dunno probably just another dodgy question
can i ask another one?
what is the purpose of slip rings in AC generator?
Post by: Zealous on September 24, 2014, 05:15:51 pm
can i ask another one?Slip Rings maintain contact with the rotating coil in an AC generator as a way to transfer the current out from a coil into a circuit for use.
what is the purpose of slip rings in AC generator?
I've got a quick Electric Power question.
How come the magnetic field lines diverge in this image? Can't they all be parallel? (VCAA 2008)
(http://i1282.photobucket.com/albums/a531/Ovazealous/screenshot305_zps56e7fd60.png)
Current flows in an anti-clockwise direction, by the way.
Post by: Makutar on September 24, 2014, 07:43:39 pm
Post by: Zealous on September 24, 2014, 08:20:26 pm
The lines diverge because the magnetic field (B) gets smaller the further away from the loop you go.Thanks for the response.
Just on that, wouldn't the magnetic field strength have no effect on the direction?
Post by: silverpixeli on September 24, 2014, 09:34:30 pm
Just on that, wouldn't the magnetic field strength have no effect on the direction?
field strength is represented by field line density (closer lines equals stronger field) so if they stayed parallel, you'd be saying that the strength of field is the same no matter how far away from the loop you go
this isn't the case, and the only way for the strength to dissipate in a field-line drawing is if the lines spread out. the direction IS consistent with these lines and does spread out
something else that might help you is this; i assume you used a right hand rule to determine that an anticlockwise current (as viewed from the side the arrows point in this case) causes the field to point the way it does. this actually comes from the 'grip rule' for a single current-carrying wire, and happens to also work for a loop.
if you imagine gripping the wire anywhere around the loop, with your right thumb pointing in the direction of the conventional current (anticlockwise) you'll see that inside the loop, the current around the wire points in that direction, (this is the same everywhere in the loop).
the circular field around a wire wrapped in a loop gives the net result of having a field which is stronger inside the loop because all points around the loop contribute to that direction. but their contributions are only parallel in the plane of the loop, and everywhere else (including outside) they are weaker and not parallel.
As per this image, just showing the lines in one plane;
Spoiler
(https://online.science.psu.edu/sites/default/files/phys010/W5electron/500px-VFPt_dipole_magnetic3.svg_.png)
it's not very important to draw the lines that do loop back around and it's pretty hard to draw clearly on an exam, but they are there
Post by: silverpixeli on September 24, 2014, 09:47:36 pm
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
LDR's have too slow of a response rate to be at all effective in this application. you need a component that is fast enough to respond to the rapid changes in signal, but an LDR's response rate is never much faster than milliseconds, and is usually order 0.1s
For this reason, they're more useful as light sensors as in those circuits the response time is acceptable
In actual fact, sometimes they're used in audio compression for softening audio signals (utilising their delayed response)
Post by: Brunette15 on September 26, 2014, 05:32:45 pm
Post by: kinslayer on September 26, 2014, 05:38:18 pm
The equation of motion to use here is s = ut + 1/2gt^2 where u = 0 and t = 0.45s.
s = 1/2*(9.8 m/s^2)(0.45 s)^2 = 0.99 m
The tail-light was 0.99 m above the ground before it fell off.
edit: that is with g accurate to two significant figures... if you use g = 10 m/s^2 you get 1.01 m. I never did 3/4 physics, is it common to use g = 10 m/s^2? ???
Post by: silverpixeli on September 27, 2014, 12:30:44 am
edit: that is with g accurate to two significant figures... if you use g = 10 m/s^2 you get 1.01 m. I never did 3/4 physics, is it common to use g = 10 m/s^2? ???
VCE Physics exams have g as 10m/s^2 on the formula sheet and in the study design, probably to save on calculations and make it more about the physics. The mechanics section of Specialist Maths uses 9.8m/s^2, and VCE physics questions aren't marked wrong if you use 9.8 instead of 10 (nor in fact any error in sigfigs unless you give like 7 decimal places or something silly) afaik.
Post by: Brunette15 on September 27, 2014, 01:45:34 pm
The ans is B...
Post by: Zealous on September 27, 2014, 02:37:56 pm
Can someone please help me with question 12 for structures...I have looked at the worked solutions on itute and i don't understand how you can neglect the torque of one of the beams when answering this question :/Torque is calculated as
The ans is B...
When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to
Post by: Brunette15 on September 27, 2014, 04:23:08 pm
Torque is calculated as- the force multiplied by the distance from a "pivot point".
When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to. So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.
Ah yes...thankyou! :D
Post by: lzxnl on October 03, 2014, 10:14:34 pm
Torque is calculated as
When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to
Not quite. The torque is a product of a force and a distance, but this distance isn't just the distance from the pivot point to the force. That only holds if the force acts perpendicular to the vector from the pivot point to the point of application of the force.
Think about a see-saw. Logically, it would move the most if you pushed straight down on the seesaw, not if you pushed diagonally down on it. The angle matters too.
Post by: silverpixeli on October 03, 2014, 10:47:47 pm
Post by: yang_dong on October 07, 2014, 11:00:44 am
The forward driving force is provided by the friction exerted by the ground on the rear wheel. It is the action – reaction pair to the friction force of the wheel pushing backwards on the ground
Retarding forces acting on the bicycle include air resistance (drag) and friction on the wheels (especially when the brakes are applied).
So what does friction do? How is part of the driving force and the retarding force? So if the bicycle is travelling where there is no air resistance, it won’t ever stop? Because the retarding force is trying to stop it, but instead it actually helps it to move, but as it moves the friction on the wheels try to stop it, and then the cycle repeats itself?
Post by: Bestie on October 07, 2014, 03:23:06 pm
thanks in advance :)
Post by: Rishi97 on October 07, 2014, 05:28:57 pm
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?
thanks in advance :)
I'm not too sure of the "real" reason but I'm a more mathematical person so I think of it using the formula
W=Fxcos(angle)
If the angle is 90 which as you said is perpendicular, the work will equal 0
Post by: myanacondadont on October 07, 2014, 06:01:20 pm
Post by: silverpixeli on October 07, 2014, 06:49:26 pm
Can someone please clarify this concept?As I understand it depends on what the wheels are trying to do;
The forward driving force is provided by the friction exerted by the ground on the rear wheel. It is the action – reaction pair to the friction force of the wheel pushing backwards on the ground
Retarding forces acting on the bicycle include air resistance (drag) and friction on the wheels (especially when the brakes are applied).
So what does friction do? How is part of the driving force and the retarding force? So if the bicycle is travelling where there is no air resistance, it won’t ever stop? Because the retarding force is trying to stop it, but instead it actually helps it to move, but as it moves the friction on the wheels try to stop it, and then the cycle repeats itself?
If you're pedalling and trying to turn the wheels, consider the contact point of each wheel with the ground. It's trying to push the ground back because that's the way the wheel is turning when you try to pedal, and the reaction force pushes the entire bike/rider system forward. In addition to this there's a little bit of 'rolling resistance' which is a force opposing the propulsion due to the surface but for most surfaces and certainly most VCE physics questions this can be ignored.
If you're braking it's a slightly different story. Brakes apply a torque to the wheels to stop them from rotating rather than a torque to make them rotate which is the case above. As the wheels are slowing down, the situation is reversed. The wheel pushes against the ground the other way at the point of contact with the surface, and the surface pushes back and you slow down.
At all times, there's going to be air drag which is proportional to your velocity.
Hope that helps!
Post by: PB on October 07, 2014, 09:26:18 pm
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?Work is what I like to call - "useful energy". The consumed energy that actually contributed towards an object's propagation in a certain direction.
thanks in advance :)
Lets say that there are a billion forces acting on an object which all add up to a net force accelerating in a northerly direction. Now lets pick one of these forces which happens to be acting in a NW direction. We can say that this force has done some work because it has a north bound force component which has contributed some energy to the object's propagation in the northerly direction.
However, I would call a westerly-bound force a lazy, if not counter productive, force. All it is doing is expending all its energy trying to drag the object to the west instead of to the desired north. Simply because it doesn't have a north bound force component to contribute some work in the northerly direction.
Hence, no work is done by a force acting in a direction perpendicular to the object's movement.
Post by: qwerty04 on October 07, 2014, 11:07:21 pm
I have a some questions about a dodgy momentum prac report.
The aim of this experiment was to investigate whether momentum is conserved in elastic and/or inelastic two dimensional collisions.
An air table provides a surface with minimal friction for the pucks to move across.
Multi-image photography of the motion of the pucks on the air table can be obtained by illuminating the air table with a stroboscopic lamp.
(For the elastic collision)
Puck A was launched at Puck B (initially stationary)
A program was used to obtain the motion of the pucks at different time intervals in coordinates scaled to metres.
Strobe was set a 6 flashes per second, so the time interval between each image (coordinate points) was 1/6 of a second.
Now the annoying part......
the analysis wants a graph of momentum against time of each puck on the same axes
Puck A has a nice slope, momentum is increasing and then hits Puck B and starts decreasing cause momentum must be conserved.
it looks weird cause the puck B is initially stationary, so for the first 3 time intervals it is 0 and then goes up.
So now we state a relationship between momenta before and momenta after judging by the graph shapes.
Anyways, the next step is to adjust quantities of the graph to somehow show that momentum is conserved (of course some error will be present) so somehow get a relationship or two things for before and after, which when compared will be very close and hence momentum conserved.
Oh and so far i have tried a bunch of things but this one seemed the least wrong m*delta(s) vs time
Post by: silverpixeli on October 08, 2014, 12:17:55 am
total momentum will be the sum of the momenta of the two pucks, i.e. m1 v1 + m2 v2 which should remain the same across the collision
sounds like a pretty epic prac though, stroboscopes!
Post by: Rod on October 08, 2014, 07:49:24 am
Just did 2006 VCAA, and I found motion section so hard, much harder than any neap, insight etc commercial exam motion section. How is everyone else feeling about motion 06? Will the motion section be of that difficulty in 14? Does the difficulty stay the same from 06-13?
thanks
Post by: silverpixeli on October 08, 2014, 09:23:35 am
Hi everyone
Just did 2006 VCAA, and I found motion section so hard, much harder than any neap, insight etc commercial exam motion section. How is everyone else feeling about motion 06? Will the motion section be of that difficulty in 14? Does the difficulty stay the same from 06-13?
thanks
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks
all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)
should be fine!
Post by: Rod on October 08, 2014, 04:09:19 pm
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marksThank god.
all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)
should be fine!
I did so many commercial exams, and thinking I would now be able to ace VCAA, I did 2006 and got so freaking depressed haha. Would you reccomend me going over all those questions? Can they come up?
Thanks
Post by: knightrider on October 08, 2014, 09:09:34 pm
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks
all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)
should be fine!
How can you tell how many people got full marks for particular exams
Post by: Rod on October 08, 2014, 09:28:45 pm
How can you tell how many people got full marks for particular examsUsually it says in the assessors report. For the 2013 one it didnt, but for years like 2009 where 100s and 100s people got 100% they did
Post by: silverpixeli on October 08, 2014, 10:57:51 pm
How can you tell how many people got full marks for particular exams
Yes some older physics/specialist reports seemed to have this information, but I'm going based off the spreadsheet that my physics teacher gave me towards the end of last year which was his info on grade distributions and everything so idk he may have gotten the information from some other source related to being a VCE physics teacher. he had all sorts of info like that.
Post by: allstar on October 08, 2014, 10:59:54 pm
I guess im stuck between those three big words: weight = mg
weightlessness? - whats that?
aparent weightlessness - when normal = 0?
Post by: silverpixeli on October 08, 2014, 11:43:29 pm
can someone please help me with this question?
I guess im stuck between those three big words: weight = mg correct, force due to gravity
weightlessness? - whats that? actually having weight = 0, i.e. m=0 or g=0, doesnt truly happen at any times here
aparent weightlessness - when normal = 0 yep
so 1.8g refers to the gravitational force the plane 'feels' even though the actual force on it is always based on g. it's to do with apparent weight. near the start and end of the maneuver the plane is in the bottom half of a vertical circular path and the force on the wings towards the centre of this circle is kinda the normal force from the air, providing this greater force.
same story with zero g, apparent weightlessness because there is no normal force as the plane is at the top of another vertical circle, its weight force is all that is needed to cause circular motion.
at all times the weight of the plane is unchanged and it is not truly weightless because weight is not zero.
dont think you'd need to talk about circles because that's not what it asked for, i just mentioned it to help explain!
Post by: knightrider on October 08, 2014, 11:46:58 pm
Post by: Rod on October 08, 2014, 11:50:50 pm
Yes some older physics/specialist reports seemed to have this information, but I'm going based off the spreadsheet that my physics teacher gave me towards the end of last year which was his info on grade distributions and everything so idk he may have gotten the information from some other source related to being a VCE physics teacher. he had all sorts of info like that.Hey silver sorry for the questions!
Would you reccomend me going through the qs for motion for 06? Or leave it?
Post by: silverpixeli on October 09, 2014, 08:30:40 am
Where would you find the A+ cutoffs and things like that on the vcaa website for physics
go to VCAA website and search "grade distributions 2013" or whatever year you want, you should find a page with links to the distributions for each subject that year
Hey silver sorry for the questions!
Would you reccomend me going through the qs for motion for 06? Or leave it?
all the questions except 9* are still relevant, but it's an overall long section with some difficult ones. i think going through it with that mindset (that it's a source of challenging questions) could be beneficial! only if practice at harder questions is what you're after though.
it's all valid and in the course, it's just that motion exams usually don't contain so many difficult questions, there are normally only a few spread between a lot of more basic ones.
*question 9 is about reference frames, interesting physics but no longer in the course
Post by: allstar on October 09, 2014, 11:18:41 am
but at the top of the flight, wouldn't it still experience centripetal acceleration/therefore centripetal force downwards in that situation? in the same direction as its weight force? but its weight force is zero? unknow? im confused....
Post by: silverpixeli on October 09, 2014, 11:43:24 am
thank you! silverpixeli
but at the top of the flight, wouldn't it still experience centripetal acceleration/therefore centripetal force downwards in that situation? in the same direction as its weight force? but its weight force is zero? unknow? im confused....
its weight force is always mg and so never zero (like unless you have no gravity or no mass, so realistically a plane would always have weight)
at the top of the flight, it does experience centripetal motion and therefore there must be a centripetal force, which is the weight force in this situation (a centripetal force has to be a real force, it's not so much a type of force like friction or gravitational attraction as a label for a real force that causes circular motion. an actual force (gravity, friction, tension) is called centripetal if it causes circular motion)
because at the top of the curve the weight force is all that is needed to cause circular motion (weight force is acting as centripetal) there is no need for a normal reaction force on the wings, meaning Normal force = 0 (this is called apparent weightlessness)
Post by: allstar on October 09, 2014, 10:01:50 pm
the plane has a weight force because it has mass and experiences 9.8N (g) on the Earth.
weightlessness can never happen, cause for this to happen it requires g=0, which is like never? so it does not apply to any situation?
apparent weightlessness occurs bcause there is no normal reaction force acting on the plane because only the weight force is needed? but how do we know this?
Post by: silverpixeli on October 09, 2014, 11:10:23 pm
so for that question I would ans it like this?
the plane has a weight force because it has mass and experiences 9.8N (g) on the Earth.
weightlessness can never happen, cause for this to happen it requires g=0, which is like never? so it does not apply to any situation?
apparent weightlessness occurs bcause there is no normal reaction force acting on the plane because only the weight force is needed? but how do we know this?
yeah that's pretty much right, i just looked back at the question and noticed it wants answers directly about the passengers not the aircraft itself but it's all the same
- the astronauts have a weight at all times, because of the gravitational field of the earth
- weightlessness doesn't apply, the astronauts have weight, though they do 'feel' weightless at the top of the arc
- this is apparent weightlessness, as there is no normal reaction force acting at this point (the only force acting is weight, which accounts fully for the plane's circular motion)
as for how we know only the weight force is needed, the diagram says 'zero-g' for the top of the arc. when we talk about something's 'g-force' we mean how much apparent weight they're feeling compared to actual weight. one-g is normal conditions, 1.8-g is feeling heavier than normal, and zero-g is feeling weightless which means Normal = 0
Post by: myanacondadont on October 11, 2014, 09:53:36 am
Post by: silverpixeli on October 11, 2014, 10:15:03 am
Hi all, I was doing the 2008 VCAA exam 1. In the electronics and photonics section it shows a npn transistor and stuff like that. That's no longer in our course right? Just want to confirm. I don't even know how to go about the question.
yeah transistors were taken out of the study design and circuits involving them are no longer examinable
Post by: Brunette15 on October 11, 2014, 03:32:11 pm
I am unsure how they found the change in time. I presume they converted the 4m/s using a distance but I can't interpret what the distance would be. :-[
Post by: Rishi97 on October 11, 2014, 03:54:41 pm
Can someone please help me with this question from the 2006 vcaa exam 2?
I am unsure how they found the change in time. I presume they converted the 4m/s using a distance but I can't interpret what the distance would be. :-[
Hey Brunette
Speed=Distance/time
the speed they used is obviously 4cm/s but the distance is 2cm. They used the length of the loop.
so 2/4 = 0.5s
let me know if this makes sense or not
Post by: qwerty04 on October 13, 2014, 06:09:37 pm
when the slope of a momentum vs time graph is equal to 0, this shows that momentum in conserved right.
So for explaining this:
Do i say, because the slope of p vs t graph is force, if it is equal to 0, there is no net external force acting on the system and hence it is isolated and hence momentum is conserved??
Btw what i have is a graph of p vs t, with two points before collision and two after, the teacher said to put in two lines of best fit separately for before and after collsion and then compare them....
The slope values of both lines of best fit are very close to 0 but i am just having trouble explaining/comparing them to come to a conclusion that momentum is conserved.
Post by: myanacondadont on October 13, 2014, 07:21:08 pm
Anyway; does anyone have any 2013/2014 practice exams they could spare? My teacher gave us a total of 6 commercial exams, and I've gone and completed all the VCAA ones (some of which I'm now onto doing twice) soooooo yeah. If anyone has any I'd greatly appreciate it
Post by: maurlock on October 20, 2014, 08:10:20 pm
How do you know whether to use the formula net force=Fn+Fg, or net force=Fn-Fg?
Thanks!
Post by: myanacondadont on October 20, 2014, 08:33:58 pm
edit: Should just add that we know it is Fn-Fg since net centripetal force is always directed to the center.
Post by: lzxnl on October 20, 2014, 10:10:42 pm
Hey guys, quick question
How do you know whether to use the formula net force=Fn+Fg, or net force=Fn-Fg?
Thanks!
Don't remember formulas. Think about the direction of the net force and where the normal, weight forces are pointed.
For a car on top of the hill, the net force must be directed down towards the centre of the hill. It therefore makes sense to define down as positive (you can choose whatever direction you want to be positive). Therefore, the normal force, pushing up on the car, is going to be negative. You'll have net force = weight force - normal force as the normal force is opposing the net force
Post by: maurlock on October 20, 2014, 10:31:43 pm
Post by: Brunette15 on October 22, 2014, 07:20:48 pm
I understand the method to work it out but i always get confused when the resistance illustrated is the total resistance or you have to add it? How do i know what the total resistance is for these types of questions?
Post by: Zealous on October 22, 2014, 08:30:42 pm
Can someone please help me with part ii to this question from the 2007 exam 2?
I understand the method to work it out but i always get confused when the resistance illustrated is the total resistance or you have to add it? How do i know what the total resistance is for these types of questions?
Maybe it will help to imagine the resistance of the wires as constant resistors in a circuit. In this case, we can imagine the resistance of the cables to be 2x2 ohm resistors for a total of 4 ohms resistance in the circuit. As V=IR, the voltage drop across the resistance of the wire will be V=0.5x4=2. The remaining 10V (12-2) will be across the voltmeter V1.
Post by: knightrider on October 22, 2014, 09:15:39 pm
Post by: myanacondadont on October 23, 2014, 09:22:08 am
Post by: faredcarsking123 on October 23, 2014, 02:46:11 pm
hey I was wondering whether pn junctions and those type of things are in our study design. I remember my teaching doing something brief on them mid-year but every practice exam I've done I've not witnessed anything to do with them. Theyre also in my textbook - n and p type semiconductors and all that stuff.
No they aren't.
Post by: bts on October 24, 2014, 08:49:02 am
Thanks in advance :)
Post by: allstar on October 24, 2014, 11:35:50 am
Whats the difference between elastic and isolated systems?
Post by: lzxnl on October 24, 2014, 08:19:09 pm
Can someone please help me with this question.
Whats the difference between elastic and isolated systems?
Isolated = no external forces
Elastic systems? I haven't heard that term used. If you mean elastic as in collisions, you can have an isolated system in which collisions are not elastic, namely that of light and matter, or a collision that involves heat transfer. Friction can still act in an isolated system.
Post by: myanacondadont on October 24, 2014, 08:23:18 pm
Isolated = no external forcesYeah I think he means elastic collisions. Elastic collisions (in terms of physics 3&4) are collisions in which no kinetic energy is lost. (Ek(initial) = Ek(final)). And then inelastic is when kinetic energy is lost.
Elastic systems? I haven't heard that term used. If you mean elastic as in collisions, you can have an isolated system in which collisions are not elastic, namely that of light and matter, or a collision that involves heat transfer. Friction can still act in an isolated system.
Post by: speedy on October 29, 2014, 07:49:13 pm
Post by: Thorium on October 29, 2014, 07:56:39 pm
This is a stupidly simple question, but I've got myself confused, when you find the 'direction' of a projectile, which angle to you take?
Usually when the object is moving up or down, we take the acute angle that the direction has with the horizontal.
Hope that makes sense, and best of luck with the exams.
Post by: speedy on October 29, 2014, 11:10:11 pm
Usually when the object is moving up or down, we take the acute angle that the direction has with the horizontal.
Hope that makes sense, and best of luck with the exams.
Alright thanks :)
Resistance here is ~19kOhm right?
Spoiler
(http://i.imgur.com/K6OR6yw.png)
Answers says it is 1kOhm
Post by: silverpixeli on October 30, 2014, 01:04:30 pm
Resistance here is ~19kOhm right?
Answers says it is 1kOhm
careful of the scale of the illumination axis, the question gives you
but yeah if it was the point you thought, you are reading the graph correctly and ~19 would have been correct
Post by: speedy on October 30, 2014, 01:59:33 pm
careful of the scale of the illumination axis, the question gives youbut the graph is in
but yeah if it was the point you thought, you are reading the graph correctly and ~19 would have been correct
Oh ffs lol, thanks a lot :)
Post by: Bestie on October 31, 2014, 11:16:00 pm
And also...
Q6c) shouldn't there be kinetic energy at points q and p? If not why?
Q15b)was I supposed to give it to two sig fig cause I used 18v which is two sig fig? The assessors gave it to 3 sig fig?
Q15c) how do I know whether to use the RMS voltage or the voltage peak?
Post by: speedy on October 31, 2014, 11:41:12 pm
Spoiler
(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
(http://i.imgur.com/rNvoTSd.png)
Post by: lzxnl on October 31, 2014, 11:44:11 pm
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
It seems REALLY brief, but it actually does summarise essentially why Young's model supported the wave theory of light. You should, however, explain why it suggests interference.
Post by: Zealous on October 31, 2014, 11:48:57 pm
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
I guess you could also talk about diffraction as a wave property?
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?
Just need some clarification thanks. :D
Post by: speedy on October 31, 2014, 11:54:52 pm
I guess you could also talk about diffraction as a wave property?
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy?
Eh, I wouldn't, as VCAA count Young's slit experiment as producing an 'ideal' interference pattern - not a true one. Ie. it doesn't take into account diffraction. (This is from my teacher, who is an assesor). You wouldn't lose marks for it though because it is 'good physics'.
It just seemed crazily simple for 3 marks. I also said what particle model would predict and that bright/dark bands were constructive/destructive interference.
Yeah funny you say that, because the exam I just did (2009 Exam 2 - Q11) assumes an electron is in the 1st excited state for the answer, which then is promoted higher.
Truthfully, it can happen, but it is very unlikely, at this level I have been told to assume it is impossible. But as I just mentioned, this question kinda throws it out.
Post by: speedy on October 31, 2014, 11:57:03 pm
It seems REALLY brief, but it actually does summarise essentially why Young's model supported the wave theory of light. You should, however, explain why it suggests interference.
Yeah that's what I thought, I definitely wouldn't be this blunt on an exam lol.
Post by: lzxnl on November 01, 2014, 01:41:13 am
I guess you could also talk about diffraction as a wave property?
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?
Just need some clarification thanks. :D
Sigh. This gets complicated :P you have all these issues about which energy is it more favourable for it to relax to, or will it preferentially absorb energy...don't worry about that xP
Fluorescence is essentially when excited electrons go via intermediate energy levels back down to the ground state. So to be honest, I don't exactly know how particles determine which energy level to go to. Need to ask a quantum physicist there :P
Post by: Yacoubb on November 01, 2014, 08:04:16 am
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
For these questions, I mentioned that interference is a typical property of waves. Young's double-slit experiment demonstrated that light interferes constructively and destructively to produce light and dark bands respectively. Thus, Young's experiment supports the wave-model.
Post by: speedy on November 01, 2014, 08:59:12 am
For these questions, I mentioned that interference is a typical property of waves. Young's double-slit experiment demonstrated that light interferes constructively and destructively to produce light and dark bands respectively. Thus, Young's experiment supports the wave-model.
You should also explain why it doesn't support the particle model - for any of these questions, you have to explain why it 'disproves' the other too. (You might have, but just going off what you wrote ^^ :) )
Post by: Yacoubb on November 01, 2014, 09:01:25 am
You should also explain why it doesn't support the particle model - for any of these questions, you have to explain why it 'disproves' the other too. (You might have, but just going off what you wrote ^^ :) )
Yeah no you're right. On my cheat sheet I've got how X supports the wave-model but not particle, and vice versa. :) Definitely important in fully answering the question!
Post by: Yacoubb on November 01, 2014, 10:48:17 am
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
So for a 3 mark question, would we be expected to say for instance:
1 mark - whether it supports the wave or particle model
1 mark - why it supports the particle model (or vice versa)
1 mark - why it doesn't support the wave model (or vice versa)
??
Post by: lzxnl on November 01, 2014, 10:55:20 am
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
Post by: silverpixeli on November 01, 2014, 10:59:52 am
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?
Just need some clarification thanks. :D
Sigh. This gets complicated :P you have all these issues about which energy is it more favourable for it to relax to, or will it preferentially absorb energy...don't worry about that xP
Fluorescence is essentially when excited electrons go via intermediate energy levels back down to the ground state. So to be honest, I don't exactly know how particles determine which energy level to go to. Need to ask a quantum physicist there :P
This is a question I had all of last year! My physics teacher explained it by saying that the electrons are not excited for long enough for there to be any significant chance for a collision with another photon (unless you have a loooot more photons coming in) and then my first year lecturer mentioned that they're only out of ground state for a few nanoseconds.
So my understanding is that an already-excited electron COULD accept another photon with the right energy to promote it higher (or ionise) but the chance of that collision occurring is just really really slim.
As for which levels they fall back down to, as lzxnl said, that's beyond the scope of the course and you just need to know that it can fall down to anything below it on its way back to ground state!
EDIT: I actually JUST learned how to determine whether a particular path downwards towards n=1 ground state is allowed! Including a simple-english explanation for anyone interested.
Spoiler
It's to do with electron orbitals, which, at a particular energy level, basically 'point' different ways, and you can only move to one which points in a similar way to you. At ground state, there's only a single orbital and it points in direction '0' so if you're at (n=6) and pointing in the maximum direction, '5', you can't reach ground straight away you'd need to go to (n=5) where there's a '4' direction, etc. But if you're in (n=6) and you're only pointing in direction '1', you can jump straight down into (n=1)'s direction '0'.
If you want to learn more, look up orbitals after exams are done and once you understand those, look up the selection rule for emission and absorption!
If you want to learn more, look up orbitals after exams are done and once you understand those, look up the selection rule for emission and absorption!
Post by: silverpixeli on November 01, 2014, 11:10:21 am
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
While probably more fair, I think this would have a negative effect (certainly in the year the change occurred) because people would spend time memorising formulas and answers by rote instead and that would take away from solving physics problems. I do agree that the provided formula sheet should be all you need in a physics exam, and if the change was made then they could improve the current one. But it's not really an issue, I mean you're going to do better if you have some intuition behind a concept than if you 'just have it on your cheat sheet' for obvious reasons.
So for a 3 mark question, would we be expected to say for instance:
1 mark - whether it supports the wave or particle model
1 mark - why it supports the particle model (or vice versa)
1 mark - why it doesn't support the wave model (or vice versa)
??
Looks good to me, I'd say;
The experiment supports the wave model for light, as the pattern that is observed on the screen can be explained by diffraction and interference, both wave phenomena. It contradicts the particle model, which does not account for diffraction or interference, instead predicting two bright bands of light directly behind the slits.
Since the question seems to be geared towards a comparison, I'd chuck in a quick diagram of the two expectations if I had time! That's probably more than needed for 3/3, but it can't hurt to give more than required (as long as it's all correct)
Post by: Yacoubb on November 01, 2014, 12:18:13 pm
Perfect. :)
Thank you :)
Quote
Looks good to me, I'd say;
The experiment supports the wave model for light, as the pattern that is observed on the screen can be explained by diffraction and interference, both wave phenomena. It contradicts the particle model, which does not account for diffraction or interference, instead predicting two bright bands of light directly behind the slits.
Since the question seems to be geared towards a comparison, I'd chuck in a quick diagram of the two expectations if I had time! That's probably more than needed for 3/3, but it can't hurt to give more than required (as long as it's all correct)
Yeah so would that be sufficient to mention how the wave-model is supported? Mentioning that observations of diffraction/interference are consistent with waves, supporting the wave-model, and then describing how the particle-model would predict only two bright bands behind the slits?
I just want to be sure how much to put in, not omitting anything important :)
Also, are incandescent globes on the course? I've been doing pre-2006 exams and in the sections for Light and Matter, it's got questions on incandescent lights and I'm not sure whether or not they're still on the course. :/ thanks!
Post by: speedy on November 01, 2014, 04:40:43 pm
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
Lol I never knew that aha... pretty funny...
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
I agree with you, but from what I've seen (in my class), if people don't actually understand the content, they do badly, regardless of the fact that they have a full cheat sheet with size 6 font lol.
Also, are incandescent globes on the course? I've been doing pre-2006 exams and in the sections for Light and Matter, it's got questions on incandescent lights and I'm not sure whether or not they're still on the course. :/ thanks!
Production of incoherent light was moved to the photonics detailed study.
Post by: speedy on November 01, 2014, 11:52:38 pm
eg. "thus induced EMF will produce a current that flows in a direction which produces a field that opposes this change in flux"
or
"thus a field will be produced that opposes this change in flux"
(This might seem stupid but I usually write the former, I want to know what others write :) )
Post by: PB on November 02, 2014, 01:15:59 am
Also, the former would probably impress the examiner more too as it shows that you know WHY an opposing field is created, not just that it happens because of lenz's law.
Post by: yang_dong on November 02, 2014, 05:27:50 pm
Post by: magneto on November 02, 2014, 05:50:49 pm
N = mg - ma?
Post by: bts on November 02, 2014, 06:06:29 pm
please help :)
Post by: Zealous on November 02, 2014, 06:34:35 pm
what mass should i use 33, 44 or 77??Depends on how you want to approach the question.
N = mg - ma?
Simplest way is to use ma=mg-N with the 33kg mass where N is the force exerted by B on A (working against the gravitational force of A).
So
Post by: lzxnl on November 02, 2014, 07:19:34 pm
Explain why adding a soft iron core increases the strength of an electromagnet.
please help :)
To put it simply, the iron itself is magnetic, so you're adding another magnet on top of an existing magnet.
Post by: Yacoubb on November 02, 2014, 07:31:32 pm
Explain why adding a soft iron core increases the strength of an electromagnet.
please help :)
I believe that soft iron has high magnetic permeability, and so by adding another magnet, you're increasing the strength of this electromagnetic.
Post by: Bestie on November 02, 2014, 08:46:38 pm
can someone please help me with question 3,4 and 14 in the attachment? all multiple choice questions
thank you
Post by: speedy on November 02, 2014, 09:13:16 pm
STAV 2013
can someone please help me with question 3,4 and 14 in the attachment? all multiple choice questions
thank you
These questions are weird as lol...
3) Don't have the image?
4) Bombardment of electrons is off the study design now... This doesn't seem to be a question aligned with the current course, or maybe my knowledge is lacking lol. My guess is A though.
14) Is it D? Although it wouldn't exactly half the current, it is the best way.
Post by: Bestie on November 03, 2014, 08:36:12 pm
question 3 is above question 4 in the same attachment.
ans to question 4 is B?
Post by: speedy on November 03, 2014, 10:05:28 pm
hey speedy! :)
question 3 is above question 4 in the same attachment.
ans to question 4 is B?
As in there is no diagram.
Yeah, maybe someone else could help, but bombardment with electrons has been removed from the study design so it's not something you're meant to know.
Post by: rui97 on November 05, 2014, 11:05:02 pm
for significant diffraction to occur, does lambda/slit width have to be greater than or equal to one or does it need to have the same order of magnitude? I havent been able to get a definite answer, what do you guys think?
Post by: lzxnl on November 06, 2014, 09:17:59 am
hey guys
for significant diffraction to occur, does lambda/slit width have to be greater than or equal to one or does it need to have the same order of magnitude? I havent been able to get a definite answer, what do you guys think?
For noticeable diffraction, they need to be of the same order of magnitude. For complete diffraction, you need your slit to be smaller than the wavelength.
Post by: rui97 on November 06, 2014, 08:22:44 pm
For complete diffraction, you need your slit to be smaller than the wavelength.
Is complete diffraction the same as significant diffraction?
Post by: lzxnl on November 06, 2014, 11:28:44 pm
Post by: Brunette15 on November 07, 2014, 02:31:15 pm
Post by: RKTR on November 07, 2014, 02:35:51 pm
Are there any main difference's between a split ring commutator and slit ring commutator that we should note? So far on my cheat sheet I have only put that split rings are for DC and produce a DC output, and slip rings are for AC and produce an AC output... :-\Split ring reverse direction of current every half turn but slip ring maintains it.
Can anyone confirm?
Post by: Thorium on November 07, 2014, 02:46:04 pm
Split ring reverse direction of current every half turn but slip ring maintains it.
Can anyone confirm?
Yep that is right.
In addition, in generators, slip rings are used to produce AC EMF. This type of generator is also called alternator. And split ring is used to reverse current every half cycle, so that the produced EMF remains in the same direction. (So DC)
The Voltage vs time graph of an alternator looks like a sinusoidal graph.
If a split ring is used, the negative values of the sinusoidal graph are flipped, i.e. they become positive.
Hope that helps :)
Post by: speedy on November 07, 2014, 03:11:32 pm
Post by: myanacondadont on November 07, 2014, 03:14:10 pm
For the cheat sheet, do we have to stick the two A4 pieces together?
Afaik they have to be bound together by tape. Otherwise you can use a single piece of A3 doublesided.
Post by: myanacondadont on November 07, 2014, 04:27:49 pm
What should I chuck on my cheat sheet for modulation? I don't think we touched on it during the year but the exams seem to look to ask questions on it. I was thinking of just putting what an information signal/unmodulated signal/modulated signal/carrier wave looks like.
If anyone is doing structures & materials; What's the difference between strain energy and toughness? Are they both the same thing or is one multiplied by volume? Also, when reading a f-x graph do we calculate the area under the graph (similar to a stress-strain graph) for strain energy?
Post by: silverpixeli on November 07, 2014, 05:59:37 pm
Got a few questions:
What should I chuck on my cheat sheet for modulation? I don't think we touched on it during the year but the exams seem to look to ask questions on it. I was thinking of just putting what an information signal/unmodulated signal/modulated signal/carrier wave looks like.
If anyone is doing structures & materials; What's the difference between strain energy and toughness? Are they both the same thing or is one multiplied by volume? Also, when reading a f-x graph do we calculate the area under the graph (similar to a stress-strain graph) for strain energy?
modulation: yeah just a diagram is fine, make sure you know the few key points they look for in worded questions like
- the reason for modulating signals for transmission
- the process, from modulating to sending to demodulating and what waves are involved (signal wave, carrier wave, modulated carrier wave, etc)
strain energy: the energy of an object under load due to its strain which is given by the area under stress-strain graph, then multiplied by volume. the area in question might not be the whole graph, all the way to failure.
toughness: the maximum strain energy density of a material, a property of the material that describes how much energy a unit of volume or that material can take to failure. because of this, it's always calculated as the full area under the stress strain graph, and because it's not object-specific, it's material specific, you don't multiply by volume.
hope that helps!
Post by: speedy on November 07, 2014, 06:26:29 pm
Spoiler
(http://i.imgur.com/3A5PKeJ.png)
I usually just used the first and last points that are given to estimate Planck's constant -> would you lose marks for this (let's say they weren't exactly on the line of best fit)?
Edit: Yup, not marks was confirmed later in the report.
Spoiler
(http://i.imgur.com/kD4UL22.png)
I'm still kinda unsure though, because my points were on the line of best fit, but I'm sure my answer was the "accepted value"
Post by: jumcakes on November 07, 2014, 11:30:18 pm
furthermore would the purpose of modulation be to transmit information more effectively since higher frequency waves diffract less (hence why we impose the signal wave onto the carrier wave)
Post by: lzxnl on November 08, 2014, 09:13:47 am
does anybody have a suitable definition for single slit interference? I have it such that when light passes through a slit it diffracts and the waves from each side interfere with each other creating bright/dark bands but am still a bit unclear.
furthermore would the purpose of modulation be to transmit information more effectively since higher frequency waves diffract less (hence why we impose the signal wave onto the carrier wave)
It's not like you really need a definition for this and I haven't tried to define it myself before so I'll see what I come up with here.
Huygen's wave model of light assumes that in a light wave, every point on the edge of the light wave acts as a new site of propagation of the light wave. In other words,you can think of light as originated from every single point on a light wave. Therefore, in single slit diffraction, you can think of light as radiating from every single point in the slit. Considering light being emitted from different parts of the slit and considering their path differences allows for analysis of the single slit diffraction phenomenon.
Yes, modulation transmits information more effectively, but I would have thought simply that it would have been because the shape of the original signal is more readily maintained in a high frequency wave than in a low frequency wave. Just what I always thought last year.
Post by: speedy on November 08, 2014, 09:35:31 am
It's not like you really need a definition for this and I haven't tried to define it myself before so I'll see what I come up with here.
Huygen's wave model of light assumes that in a light wave, every point on the edge of the light wave acts as a new site of propagation of the light wave. In other words,you can think of light as originated from every single point on a light wave. Therefore, in single slit diffraction, you can think of light as radiating from every single point in the slit. Considering light being emitted from different parts of the slit and considering their path differences allows for analysis of the single slit diffraction phenomenon.
Do we need to know about Huygen's wavelet model? It hasn't been assessed before has it?
Post by: lzxnl on November 08, 2014, 09:39:37 am
Post by: speedy on November 08, 2014, 10:15:43 am
Heck no. Hence why I said you don't need a definition of this.
Ahk, yeah I just recall my teacher mentioning it and wanted to clarify :)
Post by: Rod on November 08, 2014, 12:51:38 pm
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf
Here is my working out:
Let / = wavelength (because I don't know how to do it on this LOL)
So since it is the second brightest band, path different must be 1/ because central maxima (the first brightest band) is 0/.
Therefore 1/ = PD
1/ = 1.4X10^3X10^-9
/ = 1.4X10^3 nm
Therefore, since we have the wavelength, we can find the path difference of the first dark band
(n-1/2)/ = PD
1-1/2/ = PD
1/2 x 1.4x10^3x10^-9 = PD
PD = 7x10-7 m
-----------------------------------------
Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks
Post by: Rod on November 08, 2014, 12:58:28 pm
23.B
How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks
Post by: Zealous on November 08, 2014, 01:10:21 pm
Hey guys, need help with question 22.c
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf
Here is my working out:
....
-----------------------------------------
Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks
The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula
And just another quick q sorry;
23.B
How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks
The formula for de Broglie matter wavelength is
You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
Post by: Rod on November 08, 2014, 01:58:38 pm
The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formulaOh I see! So the central maxima does not count as a bright band? Thanks!, we are looking at the 2nd bright band so n=2, and we get:
.
The formula for de Broglie matter wavelength is. So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.
You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
And I get 23.b now. Thanks! Awesome explanations ....
Post by: Zealous on November 08, 2014, 02:13:54 pm
Oh I see! So the central maxima does not count as a bright band? Thanks!Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.
It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance (
Post by: Rod on November 08, 2014, 02:26:21 pm
Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.Thanks again
It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance (), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
Sorry for bothering but would it be okay if you please explaind two more
2b and 6c
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf
Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
Post by: Zealous on November 08, 2014, 02:45:42 pm
2b and 6c2b:
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf
Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
Look at the two components of the system separately.
So for m1, we've got the gravitational force working in one direction and tension in the string working in the other:
For m2, the only force acting on m2 is the tension in the string which is pulling it to the left:
The tension in a string is always the same, so we can sub in m2a=T into the first equation to get:
So this is the simultaneous equation method, where you look at the forces acting on each block - alternatively you can look at the whole entire system as the whole, imagine both blocks are attached together and there's only a force of gravity acting on it, find the acceleration then look at individual blocks (the method in the examiner report).
6c:
So they've assumed that when the mass is at position Q, that this will be a point of 0 spring potential which is incorrect. Because the spring has already been stretched by 0.5m, it is incorrect to say there's no energy there. So to prove that it's not 0, do a calculation and show that
Post by: Yacoubb on November 08, 2014, 02:49:17 pm
Thanks again
Sorry for bothering but would it be okay if you please explaind two more
2b and 6c
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf
Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
2b. To calculate tension, we firstly need to calculate the acceleration of m2. So:
W = mg = 2*10 = 20N.
Fnet = ma.
20 = (6+2)a
a = 2.5 m/s^2
T = ma = 2.5 * 6 = 15N.
______________________________________________________________________________________________
6c. The spring potential is not 0 because the spring is being stretched 1m if you can see in the second diagram (so it has been stretched by 0.5m). Thus the strain potential energy is not equal to 0, and this is the mistake made in the collated data.
Edit: just realised Zealous beat me to the punch :)
Post by: Rishi97 on November 08, 2014, 03:59:10 pm
Thanks :)
Post by: myanacondadont on November 08, 2014, 05:48:52 pm
What is an oscilloscope?
Thanks :)
When used in VCE physics it refers to a voltmeter/ammeter thingo in the sense it reads the voltage/current. But it produces a graph of the voltage/current aswell on a screen. (That's all I know). Someone can probably offer a better explanation.
Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.
Post by: Zealous on November 08, 2014, 05:52:55 pm
Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
Post by: myanacondadont on November 08, 2014, 05:58:16 pm
I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
Thankyou! Another question too :D
When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.
And when it's connected with slip rings to a DC power source, does it just align itself with the external field?
Post by: Zealous on November 08, 2014, 06:04:19 pm
When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.
I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.
And when it's connected with slip rings to a DC power source, does it just align itself with the external field?Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
Post by: myanacondadont on November 09, 2014, 11:53:23 am
I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
Thankyou - I think I have some understanding wrong. What's the scenario when a coil will oscillate 90 degrees back and forth?
Post by: Rod on November 09, 2014, 02:11:57 pm
http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/physics_assessrep_12.pdf
Having trouble understanding 2d.
Agree:
Old -> 2\
New -> 1\
Don't understand:
How can we conclude that the path differences are the same?
Thanks
Post by: Zealous on November 09, 2014, 03:44:08 pm
How can we conclude that the path differences are the same?You can make that conclusion because the position Y is in the exact same spot with both patterns. The light from both slits has still travelled the exact same distance to reach point Y, so the difference in the path they take will still be the same.
(http://i1282.photobucket.com/albums/a531/Ovazealous/ROF_zps0a8084a5.png)
If you see the diagram, there has been no change to the path of the light to reach point Y, the only thing that has changed is the spacing of the pattern because of the new wavelength of light.
Post by: Plitzer on November 09, 2014, 05:04:15 pm
Post by: Rishi97 on November 09, 2014, 05:19:08 pm
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks :)
Basically, slip rings allow the AC current induced in the loops to be transferred to the LOAD. Generators that are fitted with slip rings produce AC
Post by: Mujteba on November 09, 2014, 05:43:11 pm
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks :)
Also compared to the commutator, slip-rings remain stationary and do not rotate with the coil. The only place they are seen as far as the physics course is concerned is on the AC generator. One problem with slip rings though is because of the metal to metal contact between the coil and the rings they are very prone to wear and also cannot handle vibration too well as the coil can temporarily lose contact with the slip rings and cause sparks. This is why in many motor vehicles an alternator is used which rotates the magnet rather than the coil and hence eliminates the need for slip rings.
Went a bit off track but hope that covered all bases ;D
Post by: Plitzer on November 09, 2014, 06:06:37 pm
Also compared to the commutator, slip-rings remain stationary and do not rotate with the coil. The only place they are seen as far as the physics course is concerned is on the AC generator. One problem with slip rings though is because of the metal to metal contact between the coil and the rings they are very prone to wear and also cannot handle vibration too well as the coil can temporarily lose contact with the slip rings and cause sparks. This is why in many motor vehicles an alternator is used which rotates the magnet rather than the coil and hence eliminates the need for slip rings.How is the current directed in the coil, though? Like a split-ring commutator reverses the current every 180degrees but slip rings maintain an AC current within the coil? How does this keep it rotating? Maybe I should have been more specific as this is really what I want to know. That gave me heaps of useful information though - so thank you!
Went a bit off track but hope that covered all bases ;D
Post by: lzxnl on November 09, 2014, 06:20:33 pm
As for the current in the coil, if you have an alternator, the current is generated by the motion of the coil through principles you're hopefully aware of by now. Its direction is determined by whether the flux is increasing or decreasing.
Post by: speedy on November 09, 2014, 06:36:24 pm
Post by: Yacoubb on November 10, 2014, 08:56:28 am
For the exam, are you not allowed to write anywhere in the border? Or is it only on the side closest to the centre?
You can't write outside the border. So only within the vicinity of that box. :)
Post by: davomac on November 10, 2014, 09:11:59 am
Post by: silverpixeli on November 10, 2014, 10:46:25 am
Would somebody be able to explain modulation and demodulation? Thanks :)
You have some information you want to transmit, and it's in the form of a low-frequency wave. We'll call this the signal.
You know that a low frequency signal wont make it very far if you try to transmit it through a wire or the air, or something, so you want to encode your information onto a high frequency wave that we'll call the carrier. This might be a light wave, for example. The modulated wave wont suffer as much of a loss when transmitted over a distance, so your information goes further.
Modulation is the process of encoding the information into the carrier wave, and in VCE physics we talk about intensity/amplitude modulation which means we encode the changes in signal into changes in amplitude of the high-frequency carrier. (there's also frequency modulation where changes are mapped onto changes in frequency).
Demodulation is the reverse process, where the information is extracted from the modulated wave (after transmission), so basically you read the changes in amplitude of the incoming wave and that's your signal.
Post by: Yacoubb on November 10, 2014, 10:53:44 am
Modulation is the process of encoding the information into the carrier wave, and in VCE physics we talk about intensity/amplitude modulation which means we encode the changes in signal into changes in amplitude of the high-frequency carrier. (there's also frequency modulation where changes are mapped onto changes in frequency).
So if we were using a light beam, would modulation be that we encode the changes in the signal into changes in the light brightness?
Post by: myanacondadont on November 10, 2014, 11:40:27 am
Post by: silverpixeli on November 10, 2014, 11:44:07 am
So if we were using a light beam, would modulation be that we encode the changes in the signal into changes in the light brightness?
yeah, intensity (brightness) is proportional to amplitude squared, so essentially you're making it brighter/dimmer in accordance with the information signal you're trying to send.
Is it necessary to put a negative in front of voltage gain if the amplifier is inverting? Do you lose marks if you don't?
Do gain as the absolute value of the gradient (no minus sign either way), and then if necessary mention that it's an inverting amplifier when they ask. Putting the negative there is fine too, afaik.
Post by: myanacondadont on November 10, 2014, 11:48:35 am
It says that there is zero emf when there is no change in flux (i.e gradient of the graph is 0) and list the possible times as 0.5,1,1.5. Wouldnt 0, 2 and 2.5 be included?
Nevermind I'm an idiot. It specifies a time restriction, gotta watch out for those aye.
Post by: jumcakes on November 10, 2014, 04:03:51 pm
Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?
Post by: silverpixeli on November 10, 2014, 04:09:50 pm
How should we deal with the negative sign in Faraday's law when computing the average emf induced?
Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?
If you have a negative change in flux, the negatives will cancel. Otherwise you get a negative number of volts, but it doesn't matter because in VCE you figure out the direction of current analytically with right hand rules and you wont be tested on the specifics of which way is 'positive voltage' because that's not the focus of these questions.
If you do find yourself going through the working and get a negative answer at the end, don't sweat, just leave it or say the magnitude of the voltage is |your answer| = your answer without the negative.
Post by: myanacondadont on November 10, 2014, 06:38:45 pm
Or am I missing something?
Post by: Zealous on November 10, 2014, 06:41:12 pm
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
Or am I missing something?
VCAA didn't put out any solutions - interestingly the sample exam is just a compilation of old VCAA questions.
iTute has some solutions here: http://www.itute.com/wp-content/uploads/2013-2016-vcaa-physics-sample-exam-solutions.pdf
Post by: theshunpo on November 10, 2014, 08:15:35 pm
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
Or am I missing something?
In addition to Zealous's link i found this while looking through VicPhysics once, hope it helps.
Post by: rui97 on November 10, 2014, 10:01:09 pm
The answer to Q13 is 19.6 btw
Post by: silverpixeli on November 10, 2014, 10:16:43 pm
so in short, it's a valid formula but we aren't given what x is in that situation
this means you need to use conservation of energy here!
Post by: lzxnl on November 10, 2014, 10:33:54 pm
If you're told energies or speeds, use conservation of energy. If you're told something isn't moving, using forces (net force = 0)
Post by: Bestie on November 11, 2014, 01:03:48 pm
whats the difference between dynamic microphones and dynamic loudspeaker loudspeakers? VCAA separated the two in the study design???
Post by: chickenfries on November 11, 2014, 02:40:36 pm
VCAA 2011 Exam 2 q11 electric power
http://www.vcaa.vic.edu.au/Pages/vce/studies/physics/exams.aspx
thank you
Post by: myanacondadont on November 11, 2014, 03:48:09 pm
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?
However I always get 'em wrong :(
Post by: Plitzer on November 11, 2014, 04:16:04 pm
It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?
However I always get 'em wrong :(
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?
Post by: myanacondadont on November 11, 2014, 04:24:03 pm
The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?
No :( Sorry. The answer for part c is 26.7N and the answer for part d is 73.4N
Post by: myanacondadont on November 11, 2014, 04:37:29 pm
Why is that?
Post by: Camo15 on November 11, 2014, 04:41:58 pm
EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?
And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?
Why is that?
The overall torque clockwise due to the pole is 1.5*100 because that's its horizontal length from the pivot, I believe. It's the same reason why the torque due to the 20kg mass is 3*200, even though the mass is at the end of the 6m rod
Post by: Zealous on November 11, 2014, 05:21:20 pm
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?
However I always get 'em wrong :(
Look at all three blocks as a whole to find the acceleration of the system: m=21kg, F(applied)=140N, friction=42N, acceleration=14/3
Force of B on C, set up an equation for block C individually:
Force of B on A, look at Box A by itself, you've got the push force, force of B on A and the frictional forces:
And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?
Why is that?
Torque is calculated as distance multiplied by perpendicular component of force, the 100N of weight is not perpendicular to the pole PQ, so we multiply the 100N of weight by the component of the pole which is perpendicular to the weight force which is 1.5m (using a bit of trigonometry).
What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?Modulators: you're mainly looking for transducers (electro-optic) which convert electrical signals to some physical signal, where their output can be adjusted based on an input signal. So things like an LED (can't think of any others right now).
Demodulators: take the opposite, so opto-electric devices such as LDR's and photodiodes. They will basically take the variations in the signal and convert it back into an electrical signal.
Carrier Waves: VCAA usually uses light waves in their scenarios, or even the brightness of an LED.
EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.
(http://questions.transtutors.com/Transtutors001/Images/Transtutors001_e5ae4d3b-0bf7-4ba8-9555-a895c99d2e5b.PNG)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.
Post by: Camo15 on November 11, 2014, 05:44:12 pm
Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.
(http://questions.transtutors.com/Transtutors001/Images/Transtutors001_e5ae4d3b-0bf7-4ba8-9555-a895c99d2e5b.PNG)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.
Ok, so the rebound force is working against the horizontal component of the tension?
Say that there's a cable with a tension of 5N that's got an upward component of 4N and a horizontal component of 3N, what would I use to figure out the rebound force?
If anything I've said is completely wrong let me know haha
Post by: Marrogi12 on November 11, 2014, 07:58:42 pm
Post by: S61778 on November 11, 2014, 08:16:14 pm
Could someone please explain how standing waves support the existence of discrete energy levels in the atom?
Post by: Camo15 on November 11, 2014, 09:15:44 pm
Hi,
Could someone please explain how standing waves support the existence of discrete energy levels in the atom?
Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.
Post by: allstar on November 11, 2014, 09:36:49 pm
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Post by: Camo15 on November 11, 2014, 09:46:14 pm
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
What's the answer to that?
I think it has to do with the fact that you can't take ugh from a certain point.
Post by: allstar on November 11, 2014, 09:47:47 pm
how would i use their formula f=kx?
Post by: theshunpo on November 11, 2014, 09:55:29 pm
ans says k =33?
how would i use their formula f=kx?
When it is stationary, the change in x is 30cm (0.3m). At this point The netforce =0, which means the force kx-mg=0. using x=0.3 you should be able to work out k=33
Post by: speedy on November 11, 2014, 09:55:36 pm
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Just do -> F = kx -> mg = kx -> 10 = k(.03) -> k = 33.33
Post by: lzxnl on November 11, 2014, 09:59:28 pm
Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.
Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m
I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.
Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.
The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.
In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
Post by: Marrogi12 on November 11, 2014, 10:02:18 pm
Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.Gg man gg
Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m
I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.
Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.
The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.
In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
Post by: Yacoubb on November 12, 2014, 08:16:21 pm
Post by: lolaishappy on December 19, 2014, 08:00:25 pm
(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.
(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Post by: Zealous on December 20, 2014, 12:27:26 am
Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.
(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.
(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:
Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.
(http://i1282.photobucket.com/albums/a531/Ovazealous/diagram_zps1564eb19.png)
There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force. We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.
Now we can setup an equation of forces:
F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.
You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.
So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.
Hope this helps!
Post by: lolaishappy on December 20, 2014, 09:02:20 am
Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.
The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:
Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.
(http://i1282.photobucket.com/albums/a531/Ovazealous/diagram_zps1564eb19.png)
There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force. We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.
Now we can setup an equation of forces:
F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.
You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.
So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.
Hope this helps!
Thanks Zealous! Btw are applied force and Thrust the same thing? ???
Post by: FarAwaySS2 on December 31, 2014, 08:18:57 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.
How far does the police car travel before it overtakes the motorbike?
Thanks!
Post by: dankfrank420 on December 31, 2014, 08:56:36 pm
I know this question is meant to be quite easy but I can't seem to get my head around it:
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.
How far does the police car travel before it overtakes the motorbike?
Thanks!
That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.
Is the answer 612.5m? Here's how I worked it out:
d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2
Equate the two formulas, and t = 0 or t = 17.5
sub in 17.5 into the initial motorbike formula, and you get 612.5m.
Post by: AirLandBus on December 31, 2014, 09:16:24 pm
Post by: FarAwaySS2 on January 01, 2015, 01:31:05 pm
That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.
Is the answer 612.5m? Here's how I worked it out:
d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2
Equate the two formulas, and t = 0 or t = 17.5
sub in 17.5 into the initial motorbike formula, and you get 612.5m.
Post by: bts on January 05, 2015, 05:03:02 pm
Thank you
Post by: Kel9901 on January 06, 2015, 08:56:59 am
Did they first assume light was the particle model or the wave model? who?
Thank you
Newton proposed the particle (corpuscular) theory for light, but Young's double slit experiment's results could only be explained by the wave model, so the wave model became the 'only' model for light as there was not yet any evidence supporting the particle model only. The experiment that lead to wave-particle duality was the photoelectric effect, whose results could be explained only by the particle model
Post by: RedCapsicum on January 13, 2015, 11:23:05 pm
So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass
If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical
Post by: JackSonSmith on January 14, 2015, 02:56:16 pm
Post by: Kel9901 on January 14, 2015, 03:03:04 pm
Hi guys can some one help me with this question.
So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass
If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical
q1:
a. net force vertically on the mass is 0 newtons, therefore downward force due to weight=upward force due to tension.
T cos(35)=mg=2.5 * 10=25 therefore T=30.5 N
b. net force horizontally is equal to T sin(35)=17.5 N.
a=F/m=7 ms^-2
r=0.68 sin(35)=0.39 m
v^2=ar=2.73 --> v=1.65 ms^-1
q2: remember mass is still 2.5 kg and length of string is still 0.68 m
a. r=0.68 sin(θ), v=2πr/T=3.56 sin(θ)
a=v^2/r=18.6 sin(θ)
F=ma=46.6 sin(θ)
horizontal component of tension force=T sin(θ) --> T=46.6 N
b. vertical component of tension force=T cos(θ)=mg
46.6 cos(θ)=25
cos(θ)=0.54
θ=57.6 degrees
Hope this helped!
edit:
How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)53.1 degrees from the horizontal is probably best. I definitely wouldn't use bearings.
Post by: RedCapsicum on January 14, 2015, 04:48:06 pm
Post by: JackSonSmith on January 14, 2015, 05:08:05 pm
What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
Post by: AirLandBus on January 14, 2015, 05:54:11 pm
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.
What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
I have a answer, whether it is right or not is another issue. Whats the answer in the book say? Cross check with mine.
Post by: Kel9901 on January 14, 2015, 07:26:50 pm
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.
What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.
assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N
Person must exert a force of 120 N.
Post by: AirLandBus on January 14, 2015, 07:50:44 pm
Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.
assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N
Person must exert a force of 120 N.
Can confirm, also got 120 N.
Post by: JackSonSmith on January 19, 2015, 04:54:31 pm
a) What is the maximum height the ball will rise to on the rebound? Ans = 4.05m
b) What is the average size of the reaction force from the floor on the ball? Ans = 26.25N
Could someone please give me a clue as to how to work this out?
Post by: Conic on January 19, 2015, 05:26:46 pm
so the ball reaches a height of 4.05 m. Now the second question asks for the average force on the ball. We can work out the change of momentum, and use this to find the average force. Firstly, the change in momentum is given by
so the change in momentum is 3.15 kgm/s, and we are given that the contact time is 0.12 s. Now we use this in the impulse equation.
so the average reaction force is 26.25 N.
Post by: JackSonSmith on January 19, 2015, 06:02:23 pm
A box of mass 15kg is pulled 5m along a rough surface by a constant force of 80N.
The coefficient of sliding friction for the box over this surface is 0.4.
a) What is the magnitude of the friction force acting on the box? Ans = 60N
b) What is the change in kinetic energy of the box over its 5 m journey? Ans = 100J
Post by: Conic on January 19, 2015, 06:16:40 pm
so the frictional force is 60 N in the direction opposing motion. Therefore there is a net force of 20 N acting in he direction of motion. Now, we want to find the change in kinetic energy. This will be equal to the net work on the system, as no energy is being lost or converted to potential energy. The net work done is
so the box gains 100 J of kinetic energy over its journey.
Post by: FarAwaySS2 on January 19, 2015, 06:20:49 pm
A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?
Thanks!
Post by: Kel9901 on January 20, 2015, 08:31:02 am
I keep getting this question wrong.
A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?
Thanks!
u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2
Post by: AirLandBus on January 20, 2015, 12:53:30 pm
u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2
Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?
Post by: Conic on January 20, 2015, 01:23:50 pm
Post by: Kel9901 on January 20, 2015, 04:54:14 pm
Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?
Because the change in velocity is not equal to the average velocity (which you calculated to be 4 m/s).
Yes, you have to use the constant acceleration equations, as work-energy can't make use of the time figure (it needs the final velocity rather than time) and momentum-impulse can't make use of the change in displacement figure (it needs the final velocity rather than change in displacement)
Post by: alchemy on January 22, 2015, 03:26:54 pm
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.
Need help with part b and part e specifically.
Thanks.
Post by: Kel9901 on January 22, 2015, 04:05:51 pm
A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.
Need help with part b and part e specifically.
Thanks.
I'll just answer the whole Q systematically because answers to previous parts are probably needed for the later parts
a) When it was thrown KE=1/2 mv^2=32 J, at the top KE=16 J therefore GPE at the top=16 J=mgh
h=6.4 m
b) vertically:
v=0, a=-10, s=6.4, u=?
v^2=u^2+2as
0=u^2-128
u=11.3 ms^-1
c) sin(θ)=11.3/16=0.71
θ=45°
d) horizontal velocity=16 cos(45)=11.3
vertically:
u=11.3, t=1, a=-10, v=?
v=u+at=11.3-10=1.3
Using Pythagoras, speed=11.4 ms^-1
e) After 1 second:
horizontal: s=ut=11.3 m
vertically:
u=11.3, t=1, a=-10, s=?
s=ut+1/2 at^2=11.3-5=6.3 m
11.3 m to the right of (assuming it was launched to the right) and 6.3 m above the launch point
OR 13.0 m (Pythagoras) from the launch point at an angle of tan⁻¹(6.3/11.3)=29° from the horizontal
f) vertically:
u=11.3, a=-10, s=0 (as ball returns to the ground at the same height as where it was launched), t=?
s=ut+1/2 at^2
0=11.3t-5 t^2
0=t(11.3-5t)
t=0, 11.3-5t=0 --> t=2.26
t>0 therefore t=2.26 s
g) horizontal:
u=11.3, t=2.26, s=?
s=ut=25.6 m
Post by: RedCapsicum on January 23, 2015, 12:24:51 am
A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.
(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.
Post by: Kel9901 on January 23, 2015, 09:13:07 pm
Need help with this question:
A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.
(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.
I took a while to understand what this question was asking and the directions of the forces. The tension in the string counteracts the weight of the 3.0 kg and provides the centripetal force for the 0.50 kg
a) On 3.0 kg: Fnet=0=Tension-mg=Tension-30
Tension=30
On 0.50 kg: Fnet=Tension=30=ma
a=60=v^2/r=v^2/0.15
v^2=9
v=3=2πr/T
T=0.31 s
b) Let the new velocity, acceleration and net force on the 0.50 kg be v', a' and Fnet'
0.50 kg:
v'=2*3=6
a'=v'^2/r=240
Fnet'=ma'=120=T'
Other mass:
120=mg
m=12 kg
Post by: Cosec on January 27, 2015, 07:24:53 pm
1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?
2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.
3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.
Post by: odeaa on January 27, 2015, 07:59:38 pm
Our teacher has forced us to do the second type of question to death though so I'll give it a crack
A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels
In the following 5 seconds the cop travels
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)
Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to the bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
I hope this is right ahaha
Post by: Cosec on January 27, 2015, 09:24:48 pm
I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack
A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels
In the following 5 seconds the cop travelsbringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)
Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to the bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
I hope this is right ahaha
Ive done the car one to death to. But this ones weird. When i added the values for the distance initially covered by the copy im getting a different value to you.
I think its where you calculated
Shouldnt it be 22.2*5 + (27.8-22.2 *5/2)?
Take a look at the graph attached.
Imgur: http://imgur.com/wmdtr5g
And thats what i got for the second equation. Book must be wrong.
thanks buddy.
Edit: The way I got taught is a bit differnt for the first question
Where told to make two equations for each vehicle, find the distance both will cover in terms of time and then let them equal each other and solve for time.
Not sure.
Post by: odeaa on January 27, 2015, 09:47:09 pm
I'm gonna have a crack at the first one but no promises
Post by: Cosec on January 27, 2015, 09:55:27 pm
Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises
Haha no worries. The problem is when i try to solve for t im getting a negative value. Which isnt right. My dads a engineer and cant work it out either. :o :)
Post by: Cosec on January 27, 2015, 10:24:12 pm
Post by: odeaa on January 27, 2015, 10:27:46 pm
All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.(http://i.imgur.com/aUQgBHx.jpg)
Gave this one a go, have no idea if it's right but it's feasible
Post by: dankfrank420 on January 27, 2015, 10:35:27 pm
Post by: Kel9901 on January 28, 2015, 09:45:29 am
Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).
1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?
2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.
3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.
1. u=4.8, a=10sin(15)=-2.6, s=1.75, v=?
v^2=u^2+2as=23.04-9.06=13.98
v=3.74 ms^-1
You can also use energy conservation like this:
Bottom: GPE=0, KE=1/2 mv^2=0.576
1/2 way up slide: GPE=mgh=mg(1.75sin(15))=0.23
KE=0.576-0.23=0.35=1/2 mv^2
v^2=14
v=3.74 ms^-1
2. 80 km/hr=22.2 ms^-1
100 km/hr=27.8 ms^-1
In the first 15 seconds:
Car travels 22.2*15=333.3 m
Policeman travels 1/2*10*22.2 + 1/2*5*(22.2+27.8)=236.1 m
After exactly 15 seconds the policeman is behind by 97.2 m, and the difference in velocity is 27.8-22.2=5.6 ms^-1
The policeman takes a further 97.2/5.6=17.5 s to catch up, so total time taken is 15+17.5=32.5 s
3. u=0, t=2.4, s=3.2, a=?
s=ut+1/2 at^2
3.2=2.88a
a=1.11 ms^-2
For the last question, you can find out the KE just before the ball hits the ground (=GPE at top=mgh=20m). Assuming an elastic collision (I think that's what you're supposed to assume), the KE just after the ball bounces is also 20m. This means that in both cases, the speed can be calculated via:
20m=KE=1/2 mv^2
v^2=40
v=6.32 ms^-1
Post by: Kel9901 on January 29, 2015, 10:18:29 am
(http://i.imgur.com/aUQgBHx.jpg)
Gave this one a go, have no idea if it's right but it's feasible
You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculate
Post by: odeaa on January 29, 2015, 10:37:53 am
You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculateI get it now, thanks so much! I asked the class genius on facebook how to do that, I didn't even doubt his answer for one second ahaha I can't believe he was wrong
Post by: JackSonSmith on February 04, 2015, 05:32:14 pm
a) How far does the police car travel before it overtakes the motorbike?
b) At what time does the police car overtake the motorbike?
Post by: odeaa on February 04, 2015, 05:33:20 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.Almost this exact same question got answered a few days ago if you scroll back, different values but same concepts
a) How far does the police car travel before it overtakes the motorbike?
b) At what time does the police car overtake the motorbike?
Post by: JackSonSmith on February 04, 2015, 05:35:27 pm
Post by: odeaa on February 04, 2015, 05:37:42 pm
The time taken if you arrange d=vt is t=d/v
If you sub in 35m/s from the constant speed of the bike you will get an answer
I'm assuming you got the distance?
Post by: Kel9901 on February 04, 2015, 07:49:32 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.
a) How far does the police car travel before it overtakes the motorbike?
b) At what time does the police car overtake the motorbike?
a) , b)
Motorbike: s=ut=35t
Car: u=0, a=4, s=?, t=?
s=ut+1/2 at^2
s=2t^2
So the equations are s=35t and s=2t^2
35t=2t^2
t=35/2 (as t>0)=17.5 s
s=35t(=2t^2)=612.5 m
Post by: Cosec on February 04, 2015, 07:56:44 pm
I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.
Here:
Imgur: http://imgur.com/qCeuwvq
Basically, when the cop pass the car? When their displacement is the same.
Therefore, its easiest to draw a simple graph and set up two equations like shown and solve for time.
Post by: Fusuy on February 06, 2015, 05:25:03 pm
Post by: paper-back on February 08, 2015, 12:57:37 pm
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?
Would we go;
(2)(2)-(3)m2=(2+m2)(1)?
Post by: Maths Forever on February 08, 2015, 01:11:56 pm
In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?
Would we go;
(2)(2)-(3)m2=(2+m2)(1)?
Always choose a direction to be positive! Let the easterly direction be positive.
p (Before) = m1v1 + m2v2 = 2 x -2 + m2 x 3
p (After) = (m1 + m2) v = (2 + m2) x -1
p (Before) = p (After) ......conservation of momentum
Therefore, -4 + 3 x m2 = -2 - m2
-2 = -4 m2
Hence, m2 = 0.5 kg
But your way (letting westerly direction to be positive) would also work. Hope this helps!
Post by: Cosec on February 08, 2015, 01:16:04 pm
Post by: keltingmeith on February 08, 2015, 01:18:27 pm
As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.
That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.
Post by: Cosec on February 08, 2015, 01:33:39 pm
That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.
Hahha, gez cant get away from you. But yeah, thats what i was implying.
Post by: keltingmeith on February 08, 2015, 01:35:14 pm
Hahha, gez cant get away from you. But yeah, thats what i was implying.
(http://i.imgur.com/CKi9aMl.jpg)
Post by: stockstamp on February 09, 2015, 06:30:37 pm
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
The time of contact between the ball and the floor during the bounce was 0.050 s.
a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.
thanks
Post by: Maths Forever on February 09, 2015, 07:13:47 pm
Hi,
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
The time of contact between the ball and the floor during the bounce was 0.050 s.
a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.
thanks
I = Δp = F(average) Δt = m Δv
Always assign a positive direction! Let Up be positive.
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) F (av) (floor on ball) - W (weight force) = F (av) (net)
F (av) (floor on ball) = N (normal force)
Therefore, N - mg = 28.8 (from part a)
N = 28.8 + (0.080 x 10) = 28.8 + 0.8 = 29.6 N
Therefore F (av) (floor on ball) = 29.6 N in the upwards direction.
c) F (av) (floor on ball) and F (av) (ball on floor) are a Newton's 3rd Law action reaction pair.
Hence F (av) (ball on floor) = - 29.6 N
Therefore F (av) (ball on floor) = 29.6 N in the downwards direction.
This should be correct! Hope it helps!
Post by: Cosec on February 09, 2015, 07:38:17 pm
This question relates to impulse.
Since I = Δp = F(average) Δt
and I = Δp = m Δv
Always assign a positive direction! Let Up be positive.
Therefore F(average) Δt = m Δv
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)
Therefore I (floor on ball) = 1.44 N s in the upwards direction.
c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.
Hence I (ball on floor) = - 1.44 N s
Therefore I (ball on floor) = 1.44 N s in the downwards direction.
I think this is all correct! Hope it helps!
Looks good to me! Instead though i did part a with the constant acceleration motion formulae. Still got 28.8. Got a bit stumped though when i hit b and c, then i re read it and noticed "average", gets me every time. Haahha.
Post by: stockstamp on February 09, 2015, 07:47:42 pm
This question relates to impulse.
Since I = Δp = F(average) Δt
and I = Δp = m Δv
Always assign a positive direction! Let Up be positive.
Therefore F(average) Δt = m Δv
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)
Therefore I (floor on ball) = 1.44 N s in the upwards direction.
c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.
Hence I (ball on floor) = - 1.44 N s
Therefore I (ball on floor) = 1.44 N s in the downwards direction.
I think this is all correct! Hope it helps!
Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?
Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.
Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).
b) The forces acting are gravity Fg downwards and normal reaction force FN upwards.
Fg = mg = 0.78 N
totalF = Fg + FN
This gives FN = 29 N up.
c) As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.
Rendering me totally confused
thanks and sorry for my ignorance in advance
Post by: Kel9901 on February 09, 2015, 07:54:10 pm
Hi,
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
The time of contact between the ball and the floor during the bounce was 0.050 s.
a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.
thanks
regarding nature of forces: net force=both normal reaction and weight forces, forces in parts b and c are just normal reaction forces.
a) FnetΔt=mΔv
Fnet*0.050=0.08*18
Fnet=28.8 N (UP)
b) Fnet=N-W=28.8
N-0.08*10=28.8
N=29.6 N
c) According to Newton's 3rd law, this is also 29.6 N
edit: this question looks perfectly fine imo no need to hate on heinemann
Post by: Maths Forever on February 09, 2015, 08:33:26 pm
Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?
Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.
Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).
b) The forces acting are gravity Fg downwards and normal reaction force FN upwards.
Fg = mg = 0.78 N
totalF = Fg + FN
This gives FN = 29 N up.
c) As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.
Rendering me totally confused
thanks and sorry for my ignorance in advance
Sorry, I took parts b and c to mean impulse. A silly misinterpretation! Please see my edited version above, and thanks to Kel9901!
Post by: Cosec on February 09, 2015, 09:52:11 pm
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.
Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)
Can somebody explain just to clarify for me.
Post by: Maths Forever on February 09, 2015, 10:12:20 pm
Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.
Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)
Can somebody explain just to clarify for me.
Sorry, the notation you are confused about is only because of my lack of mathematical symbols.
The average was asked for in part (a). So therefore, we also take the average net force as the sum of all forces in parts (b) and (c).
We are calculating the average reaction force in part (b), or the average force that the floor exerts on the ball?
Does this help? Sorry for any confusion.
Post by: Adequace on February 10, 2015, 03:55:40 pm
Post by: Maths Forever on February 10, 2015, 04:15:14 pm
Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?
In units 1 and 2 the topics in my school were:
- Nuclear Radiation
- Electricity
- Light
- Motion
- 2 detailed studies
In units 3 and 4 all schools cover:
- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)
I hope this helps!
Post by: Kel9901 on February 10, 2015, 06:09:05 pm
In units 1 and 2 the topics in my school were:
- Nuclear Radiation
- Electricity
- Light
- Motion
- 2 detailed studies
In units 3 and 4 all schools cover:
- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)
I hope this helps!
In other words, motion+electricity are very relevant to 3/4, light is only slightly relevant (only part that is relevant is a bit of wave theory like v=fλ, radiation has 0 relevance, and the detailed studies probably aren't relevant
Post by: bobisnotmyname on February 11, 2015, 05:24:34 pm
Post by: Maths Forever on February 11, 2015, 05:54:02 pm
hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks :)
The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.
E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.
Now, using the constant acceleration formula, v = u + at:
30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s2 ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.
F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.
Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.
Post by: Kel9901 on February 11, 2015, 10:48:01 pm
The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.
E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.
Now, using the constant acceleration formula, v = u + at:
30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s2 ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.
F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.
Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.
Nope, that's for chem.
In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.
Of course, scientific notation can be used when very large numbers are present
To show what I'm saying:
Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)
Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate
Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth
g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg
Post by: Maths Forever on February 12, 2015, 08:02:13 am
Nope, that's for chem.
In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.
Of course, scientific notation can be used when very large numbers are present
To show what I'm saying:
Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)
Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate
Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth
g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg
Like I said, 2 to 3 significant figures should be fine. But it is important that the answer is not left with more figures than what the information provides. If the answer does not exceed three significant figures (e.g. 450 m), just leave it as it is.
I.e. With your question 2, 3.3 m/s or 3.33 m/s would be appropriate. Examiners are not too strict on that. Like Kel9901 said, this is more of a concern for Chemistry. Don't worry too much for Physics!
Post by: JackSonSmith on February 12, 2015, 06:31:27 pm
Post by: Cosec on February 12, 2015, 07:56:41 pm
Is work done force x displacement or force x distance?
Force x distance.
Post by: lzxnl on February 12, 2015, 10:37:31 pm
Is work done force x displacement or force x distance?
Technically it's a dot product between the force and displacement vectors; think of it as the force times the component of the displacement in the direction of the force.
Post by: Cosec on February 14, 2015, 02:48:20 pm
Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.
Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff
So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.
Or have i totally just screwed up with the theory?
Post by: Alwin on February 14, 2015, 06:25:04 pm
Can someone explain this to me.
Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.
Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff
So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.
Or have i totally just screwed up with the theory?
Hi Cosec :) Nice to see someone using VT :P
For your first equation, let's assume that we have a box resting on a table. That way we have a zero net force, and there only two forces that act on the box are the weight force, FW and FN.
As forces are vectors, we should put tilde, use arrows or use boldface to show that (see above). Now, if we don't assign a direction as being positive, then yes of course we can write:
Since we know that the net force is zero, we can let the left hand side be zero and move FN to the other side:
^ which is the set of equations you got.
Now, the alternative method is to take into account the directions first, and then work with just magnitudes (which is what's done throughout the VT vids, as working with vector notation usually confuses the hell out of ppl not doing spesh).
1) Let up be positive for the box example
2) Now, 'convert' the vectors
FW equals to a force with magnitude of FW (note no more boldface) downwards. ie:
FN equals to a force with magnitude of FN upwards. ie:
3) So our equation for the net force becomes, using only magnitudes since we have already taken into account the directions
When this is rearranged, and we sub in zero for the net force, we would get:
As for the circular motion in VT vid, let's look at the diagram on the left:
1) Choose on which direction is positive
Take up as positive
2) Look at the forces and add the sign by looking at the direction
Normal force acts up, so it is negative: - N
Weight force acts down, so it is positive: + W
Net force acts down, so it is positive: + Fnet
3) Write the equation of motion
Fnet = W - N
This is what it would look like if we used vectors
Apologies if this wasn't clear on the video, it's been quite a while haha. Hope it makes sense now :)
Post by: JackSonSmith on February 15, 2015, 11:14:58 am
Post by: Gentoo on February 15, 2015, 11:52:19 am
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?
s= 16 t=4 u=0 a=?
Sub into s=ut+1/2at^2 to find a.
Post by: lzxnl on February 15, 2015, 12:40:22 pm
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?
The important bit is not to use the right formula, but to know WHY you need to use that formula.
Here you're given the distance, time and initial velocity (0) and you want the acceleration. The only irrelevant piece of information is the final velocity, so you use the one constant acceleration equation that doesn't involve the final velocity. That's why you use x = 1/2 at^2 + ut
Post by: Floatzel98 on February 15, 2015, 08:21:29 pm
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?
Spoiler
(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009024_645555852239499_729218626_n.jpg?oh=bc99983a1c2136046e262bc25a2760b3&oe=54E32AD8&__gda__=1424148677_19c6a042e53bcc13e1049ddbd6225092)
Post by: lzxnl on February 15, 2015, 08:57:44 pm
Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?Spoiler(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009024_645555852239499_729218626_n.jpg?oh=bc99983a1c2136046e262bc25a2760b3&oe=54E32AD8&__gda__=1424148677_19c6a042e53bcc13e1049ddbd6225092)
The gravitational force depends on 1/r^2, so if the force has quartered, the radius has doubled
Post by: Floatzel98 on February 15, 2015, 09:48:20 pm
? And also how do you do 3b then?
Post by: lzxnl on February 15, 2015, 10:20:31 pm
F = k/r^2
F1/F2 = (r2/r1)^2
Force ratio is 4 as the second force F2 is a quarter of the original one, so 4 = (r2/r1)^2
r2/r1 = 2, r2 = 2 r1
As for the next one, three Moon radii above the surface = four Moon radii from the centre of the Moon => quadruple the radius => force shrinks to 1/16 of original
Post by: orgekas on February 17, 2015, 04:37:43 pm
I've got the following investigation(see photo) coming up and I have no idea how to do it.
Any ideas?
The aim is to design an experiment that will prove the formula F= mv^2/r for a body moving in a uniform circle.
I hope that you can read the attachment for more information. I f you can't, TELL ME to upload a better version.
The teacher said that we should use graphs.
Thank you.
Post by: odeaa on February 17, 2015, 09:10:19 pm
I also don't understand the whole dilation of gravity thing
Thanks
(http://i.imgur.com/gl96p42.jpg)
Post by: Stevensmay on February 18, 2015, 12:13:08 am
Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks
(http://i.imgur.com/gl96p42.jpg)
Torque is defined as a force multiplied by the perpendicular distant from that point to the force. (More technically it's the cross product of radius and force, this is why it's perpendicular). In the real world it's effectively a force that makes something want to rotate. In the picture the 20kg force wants to make the beam rotate around point P.
In our case the force from the 20kg mass is 20*9.8N (mg).
(http://i.imgur.com/HJT3vBT.png)
So we've decided that a torque is just a measure of how much something wants to rotate. Higher torque means something is wanting to rotate more. In the image above, would the left hand side 20kg mass produce a torque smaller than, equal to or greater than the torque on the right hand side?
Spoiler
Equal to, because only the perpendicular distance matters and it is the same in each diagram. Similarly, if we put the 20kg force above point P it would have a torque of 0, as it is not trying to rotate the beam.
So we now what our perpendicular distance is, it's just a matter of going T = r*F to get an answer of (hopefully, if I read the question right) 588Nm.
If this doesn't make any sense my apologies.
With the second part, I can't remember if you do summation of torques and forces.
Post by: Floatzel98 on February 20, 2015, 04:00:50 pm
Spoiler
(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009026_648111405317277_42696173_n.jpg?oh=f18f9743dda9710cc67b8e13a3037e4e&oe=54E8B477&__gda__=1424530282_c625470941f4d77c7d98f714267bf350)
Post by: Kel9901 on February 20, 2015, 06:15:07 pm
I found 7a, but don't know what to do for 7b or 8.Spoiler(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009026_648111405317277_42696173_n.jpg?oh=f18f9743dda9710cc67b8e13a3037e4e&oe=54E8B477&__gda__=1424530282_c625470941f4d77c7d98f714267bf350)
Let the distance that must be calculated be r
m is the mass of the asteroid
F(asteroid, Alpha)=GMm/r^2
F(asteroid, Beta)=G(0.01M)m)/(R-r)^2
F(asteroid, Alpha)/F(asteroid, Beta)=100(R-r)^2/r^2=8100
(R-r)^2/r^2=81
(R-r)/r=9 (remember that r and R are both positive, and that R>r)
R-r=9r
10r=R
r=0.1R
Post by: Floatzel98 on February 20, 2015, 07:21:57 pm
Post by: Kel9901 on February 20, 2015, 08:54:13 pm
GMm/r^2=G(0.01M)m/(R-r)^2
1/r^2=0.01/(R-r)^2
100/r^2=1/(R-r)^2
r^2/100=(R-r)^2
r/10=R-r
1.1r=R
r=0.91R
Post by: lolaishappy on February 21, 2015, 02:53:06 pm
Post by: Kel9901 on February 22, 2015, 08:25:18 am
Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?Yep, I did it last year and got it right (acc. to statement of marks)
a) k=F/x=mg/x pretty simple show that
b) 80 cm
c) Kinetic energy not included
d) 2 ms^-1
Post by: lolaishappy on February 22, 2015, 08:32:50 am
Yep, I did it last year and got it right (acc. to statement of marks)Thank you :)
a) k=F/x=mg/x pretty simple show that
b) 80 cm
c) Kinetic energy not included
d) 2 ms^-1
Post by: paper-back on February 25, 2015, 08:11:34 pm
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force
Do we need to go; .
9.8x14.5+sin(45)x80?
Post by: Cosec on February 25, 2015, 08:40:06 pm
For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force
Do we need to go; .
9.8x14.5+sin(45)x80?
Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).
Post by: paper-back on February 25, 2015, 08:57:04 pm
Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200
Post by: Cosec on February 25, 2015, 09:21:49 pm
Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200
Hmm, i thought of that too, but i dont see why it would be added and ive never encountered a question that takes it into consideration. But if thats the answer, i guess thats the way they expect you to work it out in which case you just find the vertical component and add it to the weight force. Someone else with a bit more experience might have to chime in.
Post by: Conic on February 25, 2015, 09:46:52 pm
What do we know in this situation? We know the gravitational force, and we can work out downward component of the pushing force. We need to relate the normal force to the things we already know. Using Newton's Second Law, we know that the net vertical force is 0, as there is no acceleration in this direction. The normal force is in the opposite direction to the weight force and the pushing force, so
This gives
Post by: lolaishappy on February 26, 2015, 05:22:11 pm
Post by: lolaishappy on February 26, 2015, 05:22:31 pm
Post by: lzxnl on February 26, 2015, 05:42:55 pm
The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused
Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.
Post by: lolaishappy on February 26, 2015, 08:17:02 pm
Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.I tried to find time as 4t=x and 8t=x however i get t=0?
Post by: lzxnl on February 26, 2015, 10:44:31 pm
Total time = t1 + t2, total distance = 2x
So average speed should be 2x / (t1 + t2)
Simplify with what you have now.
Post by: Floatzel98 on March 02, 2015, 08:21:06 pm
Post by: odeaa on March 02, 2015, 08:38:48 pm
Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.A strain percentage of 0.075% is equal to a strain of 0.00075. Also, note that while I don't convert the cm measurements to meters, the output is also in cm.
I hope that was correct ahah!
Post by: Floatzel98 on March 02, 2015, 09:22:32 pm
Post by: odeaa on March 02, 2015, 09:26:41 pm
Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.I think the examiners report tells you
But yeah definately structures, it's so straightforward.
Post by: Kel9901 on March 03, 2015, 09:22:23 am
I think the examiners report tells you
But yeah definately structures, it's so straightforward.
yeah either structure and sound (what i did) are the easiest afaik
Post by: StressedAlready on March 03, 2015, 05:53:55 pm
Post by: paper-back on March 03, 2015, 09:15:36 pm
Post by: lzxnl on March 03, 2015, 09:49:44 pm
Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?
I'm not sure if the resistance equation establishes Kirchhoff's law or the other way around...but anyway
If you have two resistors in parallel, the total resistance satisfies 1/Reff = 1/R1 + 1/R2 where R1, R2 are the resistances of each parallel part.
The total current is therefore V/Reff = V/R1 + V/R2 = I1 + I2
This shows that it's valid to say that the current after the junction is the sum of the currents before the junction. Also, I've used the same letter V for both resistors; it's clearly also valid to say that both parts of the circuit are at the same voltage.
When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?
KE = 1/2*mass*(speed)^2. Find the speed. It's not a vector.
Post by: paper-back on March 03, 2015, 10:27:16 pm
Post by: lzxnl on March 03, 2015, 11:01:10 pm
Post by: orgekas on March 04, 2015, 12:55:44 pm
So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.
Please help,
Thanks.
Post by: Stevensmay on March 04, 2015, 01:29:23 pm
Hi guys,
So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.
Please help,
Thanks.
I'm not too sure how to prove it with a graph (I've never seen anything proved like that before).
I'm happy to try writing up a derivation but it involves angular velocity/acceleration and I can't remember whether this is in the course. And a bit of calculus (basic).
Post by: lzxnl on March 04, 2015, 02:49:30 pm
Hi guys,
So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.
Please help,
Thanks.
Graph F against v^2. You should get a straight line of slope m/r
Post by: orgekas on March 04, 2015, 03:03:49 pm
Graph F against v^2. You should get a straight line of slope m/r
If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?
Post by: JackSonSmith on March 04, 2015, 04:08:44 pm
A 600g trolley was pulled by a 400g weight.
My calculation for 0 degrees, was friction = 0.95N.
I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N
I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?
Post by: Kel9901 on March 04, 2015, 05:03:58 pm
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.
A 600g trolley was pulled by a 400g weight.
My calculation for 0 degrees, was friction = 0.95N.
I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N
I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?
You need to take into account the force down the plane due to, as you said, the weight force, and that's the only difference. What you've done so far is correct. I won't do the calculations for you (since it's your prac), but expect lower friction forces
Post by: lzxnl on March 04, 2015, 09:22:06 pm
If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?
Something like that. Get Excel to display the line equation to get the slope. Right click the trendline.
Post by: chekside on March 06, 2015, 10:40:22 pm
Thanks
Post by: JackSonSmith on March 07, 2015, 10:14:54 pm
Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.
Thanks.
Post by: lzxnl on March 07, 2015, 11:47:40 pm
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks
Not part of the course I'm sure. Don't need to know about transistors
A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.1. I guess it's unlikely VCAA will test this
Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.
Thanks.
2. Any orbit must have the centre of the Earth as the centre of the orbit. If you're not over the equator, you will still orbit around the centre of the Earth so your orbit will be lopsided.
Post by: stockstamp on March 08, 2015, 12:18:00 am
If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.
I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.
Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...
Thanks
Post by: lzxnl on March 08, 2015, 10:52:12 am
Hoping someone can clarify this:
If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.
I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.
Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...
Thanks
Equal and opposite forces act on DIFFERENT OBJECTS. That's why a deceleration can result. If you had the equal but opposite forces acting on the same object, then of course no net force would result.
As for why objects stop when they hit the ground, it's because collisions with the ground are inelastic and the object tends to lose its kinetic energy thermally.
Post by: Adequace on March 08, 2015, 09:30:43 pm
I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?
It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.
Post by: silverpixeli on March 13, 2015, 10:56:26 am
Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.
I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?
It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.
luckily nuclear physics is entirely irrelevant to units 3 and 4 (except the occasional non-VCAA practice paper where they assume you know the charge on an alpha particle is +2e but yeah)
forming solid foundations for basic electrical and mechanical concepts should be your goal in year 11 because there's plenty of new stuff that you'll have to wait until year 12 to learn, both content and problem-solving skills.
to help with this, i highly recommend khanacademy's physics playlists, the first 5 of which are relevant to VCE motion and the second last (electricity and magnetism) is relevant to that. This is a resource I regularly go to for a clear and comprehensive explanation of a physics concept! it can't hurt to give it a try :)
[EDIT: I should add, I'm currently watching the thermodynamics playlist because second year thermal physics isn't taught in a very organised way at the university of melbourne, and salman khan has everything sorted. khanacademy is also great for maths and other sciences]
the other thing you can do is work at improving your algebra skills (for VCE physics, algebra is as far as you need to go, no need for calculus). I'm talking about notation, solving for x, fractions skills and mental arithmetic. Being good at maths makes physics problems waaaay more fun than if you're not a maths-y student and you kinda get the physics but you can't follow the working out. Being clear and formal in your working forces you to think methodically and improve your problem solving skills.
By improving your physics intuition and maths skills now you're forming a really really good starting point for year 12 studies!
i mean, you can work ahead if you like, but you'll have plenty of time in year 12 to learn the year 12 content. If you focus on the foundations now you'll be set when things get far busier next year.
also don't worry about test scores, seriously the only thing that matters at this point is your intuition :)
Post by: knightrider on March 16, 2015, 01:31:27 am
Post by: nerdgasm on March 16, 2015, 02:00:21 am
Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.
Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .
Post by: knightrider on March 16, 2015, 02:11:00 am
16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?
Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.
Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .
Thanks so much nerdgasm :)
Post by: lolaishappy on March 17, 2015, 06:01:51 pm
Post by: Zealous on March 17, 2015, 06:35:01 pm
When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?
This image below might help:
(http://www.ic.sunysb.edu/Class/phy141md/lib/exe/fetch.php?media=phy141:lectures:ballonstring.png)
When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have
Horizontal circular motion questions usually look like this:
(http://www.a-levelmathstutor.com/images/kinetics/kin-conpend.jpg)
The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Post by: lolaishappy on March 17, 2015, 06:41:50 pm
This image below might help:Thanks zealous, what happens in between the points of the top diagram?
(http://www.ic.sunysb.edu/Class/phy141md/lib/exe/fetch.php?media=phy141:lectures:ballonstring.png)
When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we haveat the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by
. If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.
Horizontal circular motion questions usually look like this:
(http://www.a-levelmathstutor.com/images/kinetics/kin-conpend.jpg)
The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Post by: Kel9901 on March 17, 2015, 08:01:51 pm
Thanks zealous, what happens in between the points of the top diagram?
From the study design
Quote
apply Newton's second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only
so don't worry about it
Post by: lolaishappy on March 17, 2015, 08:04:26 pm
From the study designThank you kel9901!
so don't worry about it
Post by: lolaishappy on March 22, 2015, 09:29:51 pm
In general I'm struggling to know which equations to apply to different questions :^(
Post by: Floatzel98 on March 22, 2015, 11:01:25 pm
How do we distinguish between v and u, and which to put in a equation in projectile motion?The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity if a height or time is given from this point.
In general I'm struggling to know which equations to apply to different questions :^(
Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Post by: Cosec on March 23, 2015, 06:41:23 pm
Post by: Kel9901 on March 23, 2015, 09:13:59 pm
Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.
I'm pretty sure that when the temperature increases, the resistance of thermistors decrease. Let R be the resistance of the normal resistor and R(th) that of the thermistor. Vout=Vin*R/R(th), so when R(th) decreases, the denominator only decreases and hence the voltage drop across the normal resistor increases
edit: fuck latex
Post by: lolaishappy on March 23, 2015, 10:23:31 pm
The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity if a height or time is given from this point.Thanks!
Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Post by: Ha_Nguyen on March 25, 2015, 08:50:21 pm
if an object of mass m is moving uniformly (constant speed, v) in a circle of radius r:
1. state the equation that relates the centripetal force, F to m,v,r
F= mv^2/r
2. state the equation that relates the period for 1 revolution (T) to v,r
T= (2 pi r )/v
3. from the two previous equations, write an equation relating T to m&r?
Post by: Kel9901 on March 27, 2015, 09:40:18 am
Hi, can someone help me with question 3????
if an object of mass m is moving uniformly (constant speed, v) in a circle of radius r:
1. state the equation that relates the centripetal force, F to m,v,r
F= mv^2/r
2. state the equation that relates the period for 1 revolution (T) to v,r
T= (2 pi r )/v
3. from the two previous equations, write an equation relating T to m&r?
I think you'll need F in the equation too... or does it mean T to a and r?
Post by: RedCapsicum on April 02, 2015, 11:37:26 pm
The mass of Earth is 6.0*10^24 kg and the amass of the moon is 7.4*10^22 kg. The radius of the earth is 6.4*10^6 metres and the radius of the moon is 1.7*10^6 metres. The orbital radius of the moon around the Earth taken from the centre of the Moon and the centre of the Earth is 3.8*10^8 metres.
1) Calculate the gravitational force acting between the Earth and the moon
2) Calculate to two decimal places how long it takes for the moon to orbit the Earth in days
3) Calculate the orbital speed of the moon around the earth in km/hr
Post by: JackSonSmith on April 03, 2015, 08:10:02 pm
Post by: Zealous on April 03, 2015, 08:58:36 pm
Hi guys, can someone help me with this question:
The mass of Earth is 6.0*10^24 kg and the amass of the moon is 7.4*10^22 kg. The radius of the earth is 6.4*10^6 metres and the radius of the moon is 1.7*10^6 metres. The orbital radius of the moon around the Earth taken from the centre of the Moon and the centre of the Earth is 3.8*10^8 metres.
1) Calculate the gravitational force acting between the Earth and the moon
2) Calculate to two decimal places how long it takes for the moon to orbit the Earth in days
3) Calculate the orbital speed of the moon around the earth in km/hr
1) Use
2)
3)
So with these sorts of questions, firstly write down all the variables that you are given, and then write down the ones you are trying to find. Then take the circular motion formulas and rearrange them until you find the value you are looking for. Once you've got enough practice, these questions can be done quite quickly as long as you know what value you're looking for and what formula you need to find it.
Post by: dankfrank420 on April 03, 2015, 10:36:45 pm
For electricity: does VIt equal energy or work?
Aren't they the same thing in this context?
Post by: JackSonSmith on April 04, 2015, 03:12:41 pm
Aren't they the same thing in this context?
I was a little confused about the difference between energy and work
Post by: Kel9901 on April 04, 2015, 05:32:24 pm
I was a little confused about the difference between energy and work
work is pretty much change in energy, in this case they are the same because the 'energy' means energy dissipated anyway.
Post by: dankfrank420 on April 09, 2015, 05:45:28 pm
There is a battery of 12v, a resistor of 150 ohms and a forward biased diode in circuit. The I-V characteristic of the diode says that the diode needs 0.6 v for current to flow.
Therefore, there is 11.4 volts across the resistor.
When working out the current flowing through the diode, why do we use 11.4/150 instead of 12/150? Since they're in series, shouldn't the current be constant across both the diode and the resistor? Why do we take away the 0.6v before we apply ohms law to find the current?
Post by: alchemy on April 09, 2015, 09:30:59 pm
Question about diodes.
There is a battery of 12v, a resistor of 150 ohms and a forward biased diode in circuit. The I-V characteristic of the diode says that the diode needs 0.6 v for current to flow.
Therefore, there is 11.4 volts across the resistor.
When working out the current flowing through the diode, why do we use 11.4/150 instead of 12/150? Since they're in series, shouldn't the current be constant across both the diode and the resistor? Why do we take away the 0.6v before we apply ohms law to find the current?
The diode consumes 0.6V so what's left is 11.4V. This 11.4V is for the resistor. Since the current is the same for each component, you work out the current for the resistor first by doing VR/R=11.4/150=0.076A. Hence, since this is a series circuit, that same amount of current (0.076A) passes through the diode. Hope that made sense!
Post by: dankfrank420 on April 09, 2015, 11:07:21 pm
However, why isn't the 0.6 v included? Isn't the current Vtotal/Rtotal?
So what you're saying is that the diode literally takes 0.6v away from the circuit for current to flow through it? But why isn't that voltage counted in the calculation of the total current?
Post by: lzxnl on April 09, 2015, 11:58:56 pm
Yes I understood.
However, why isn't the 0.6 v included? Isn't the current Vtotal/Rtotal?
So what you're saying is that the diode literally takes 0.6v away from the circuit for current to flow through it? But why isn't that voltage counted in the calculation of the total current?
Vtot/Rtot only works if every circuit component you have in your circuit is ohmic, or V/I is constant for every device you have.
Ohm's law isn't V=IR. Ohm's law is actually an assertion that V/I is a constant and devices that satisfy this assertion are ohmic, like normal resistors.
As you can quite clearly, diodes are NOT ohmic. Hence you can't just use normal voltage divider techniques to solve this question. You have to consider the voltage drop across each component separately.
This is what I don't like about VCE physics. Lots of concepts are not taught very well IMO.
Post by: dankfrank420 on April 10, 2015, 01:04:14 am
Vtot/Rtot only works if every circuit component you have in your circuit is ohmic, or V/I is constant for every device you have.
Ohm's law isn't V=IR. Ohm's law is actually an assertion that V/I is a constant and devices that satisfy this assertion are ohmic, like normal resistors.
As you can quite clearly, diodes are NOT ohmic. Hence you can't just use normal voltage divider techniques to solve this question. You have to consider the voltage drop across each component separately.
Thanks! Really cleared up some conceptual misconceptions (is that a phrase?) I had.
So moral of the story is don't treat a diode like another element in the circuit because it isn't ohmic.
Quote
This is what I don't like about VCE physics. Lots of concepts are not taught very well IMO.
I find this alot.
Most teachers simply teach you the way of getting the right answer instead of the concepts behind it. Most of the course can simply be ROTE learned or worst-still copied onto the cheat sheet.
Post by: Rishi97 on April 11, 2015, 04:13:29 pm
Does anyone want some physics exams ? If yes, just inbox me your email and I'll send them !
Rishi
Post by: Floatzel98 on April 11, 2015, 10:02:02 pm
Post by: Zealous on April 11, 2015, 10:19:44 pm
However, older study design exams from maybe around 2000-2006 used to have sections for a detailed study where the answers were short answer, so working out made a difference back then.
Post by: Floatzel98 on April 12, 2015, 12:51:07 am
Nope. Not at all. The detailed study consists of 13 multiple choice questions, each worth 2 marks. It's either get 0 marks or 2 marks, and you answer on a multiple choice sheet and shade bubbles, so there's no way the computer will pick out working out.Okay, thanks for the clarification :)
However, older study design exams from maybe around 2000-2006 used to have sections for a detailed study where the answers were short answer, so working out made a difference back then.
Post by: RedCapsicum on April 12, 2015, 10:35:12 am
a bullet of mass 'm' grams is fired with a velocity 'v' m/s into a wooden block of mass 'M' kg at rest on a friction less table of height 'H'. The bullet is embedded in the block which slides off the table landing at a distance of 'D'.
a) What is the velocity of the block immediately after the bullet is embedded in it?
b) How much time expires before the block lands on the floor?
Post by: Kel9901 on April 13, 2015, 11:09:11 am
Hey can someone help me with this question:
a bullet of mass 'm' grams is fired with a velocity 'v' m/s into a wooden block of mass 'M' kg at rest on a friction less table of height 'H'. The bullet is embedded in the block which slides off the table landing at a distance of 'D'.
a) What is the velocity of the block immediately after the bullet is embedded in it?
b) How much time expires before the block lands on the floor?
I think it's reasonable to assume that the block is on the edge of the table.
a) Conservation of momentum
m1v1=m2v2
mv=(m+M)v2
v2=mv/(m+M) m/s
b) u=0, a=10, s=H, t=?
s=ut+1/2 at^2
H=5t^2
t^2=H/5
t= sqrt(H/5) (time is always positive)
Post by: Floatzel98 on April 15, 2015, 09:55:39 pm
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys :)
Post by: Floatzel98 on April 17, 2015, 04:31:24 pm
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.Anyone?
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys :)
Post by: Zealous on April 17, 2015, 04:54:43 pm
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys :)
1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.
2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.
For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.
You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.
Post by: Floatzel98 on April 19, 2015, 01:18:50 pm
1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?
2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.
For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.
You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.
Post by: Adequace on April 19, 2015, 01:57:25 pm
I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.
Post by: Floatzel98 on April 19, 2015, 01:59:38 pm
For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.Yeah this is what I'm talking about. Why does that happen for?
EDIT: Another question. Is this a correct way of drawing the circuit in a more simplified way?
EDIT 2: Wait I'm pretty sure it's wrong because the 10 and the 30 should be in series with the 5 right? Not parallel. And the 5 and the second 30 should actually be in parallel i now think. So would the 3rd picture be correct now?
Post by: silverpixeli on April 19, 2015, 06:19:47 pm
Yeah this is what I'm talking about. Why does that happen for?
EDIT: Another question. Is this a correct way of drawing the circuit in a more simplified way?
EDIT 2: Wait I'm pretty sure it's wrong because the 10 and the 30 should be in series with the 5 right? Not parallel. And the 5 and the second 30 should actually be in parallel i now think. So would the 3rd picture be correct now?
Yeah picture number 2 is wrong, number 3 looks right!
You could 'clean' it a bit by drawing the 5 resistor going vertically, and then joining it with the 30-branch. Then it will look more like the circuits you are used to. You can't see the numbers but this is the shape I'm talking about:
circuit
(http://i.imgur.com/2EQNIu8.jpg)
I'm not sure if it's a technical term but I call this 'linearising' a circuit. The easiest way to do it is to start at the battery and trace yourself around the circuit, adding components in series and then branching into parallel when the circuit splits. When the splits come back together, you close those parallel bits off. This is really hard to explain by text haha.
As for the short circuit question, yeah no current will flow through a resistor if there's an alternate path with zero resistance.
here's my way of justifying it: consider a circuit with just a resistor in parallel with nothing (one of the branches is just an empty branch with only wire), and maybe another resistor somewhere else in series. The effective resistance is 0 because of the resistor-less branch, meaning the voltage across the component is also 0. Having no voltage across the resistor-less branch isnt a problem, current will still flow. But if there is zero volts across the resistor in the other branch, no current is going to be pushed through it.
Post by: Floatzel98 on April 19, 2015, 07:14:33 pm
Yeah picture number 2 is wrong, number 3 looks right!Everything makes much more sense now. Thanks so much!
You could 'clean' it a bit by drawing the 5 resistor going vertically, and then joining it with the 30-branch. Then it will look more like the circuits you are used to. You can't see the numbers but this is the shape I'm talking about:circuit(http://i.imgur.com/2EQNIu8.jpg)
I'm not sure if it's a technical term but I call this 'linearising' a circuit. The easiest way to do it is to start at the battery and trace yourself around the circuit, adding components in series and then branching into parallel when the circuit splits. When the splits come back together, you close those parallel bits off. This is really hard to explain by text haha.
As for the short circuit question, yeah no current will flow through a resistor if there's an alternate path with zero resistance.
here's my way of justifying it: consider a circuit with just a resistor in parallel with nothing (one of the branches is just an empty branch with only wire), and maybe another resistor somewhere else in series. The effective resistance is 0 because of the resistor-less branch, meaning the voltage across the component is also 0. Having no voltage across the resistor-less branch isnt a problem, current will still flow. But if there is zero volts across the resistor in the other branch, no current is going to be pushed through it.
Post by: knightrider on April 20, 2015, 08:55:17 pm
When is it like or not like this?
Post by: odeaa on April 20, 2015, 09:17:20 pm
When they say initial velocity does this always mean that time=0secs?Yeah, when t=0, v=u
When is it like or not like this?
Post by: silverpixeli on April 20, 2015, 09:29:50 pm
When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?
Depends on the context. Usually, yeah, initial velocity refers to velocity at time t=0. But t=0 is kinda arbitrary, right? Otherwise t=0 might refer to the start of the universe or whatever absolute zero point you want to define. The idea is, we define a zero point that works best for us.
In many kinds of motion, we set t=0 to the start of that motion of interest, so 'initial velocity' relates directly to that starting instant.
If we're ever talking about a change in velocity, initial velocity will refer to the start of the change.
Sometimes, in projectile motion questions, we'll deal with different values of u, (different 'initial velocities') at different points in the question. For example if you want to find the time taken to fall from the peak of the projectile path to the landing point, you'd consider the start of this motion to be the peak of the path, and hence vertical u=0. In another part of the question, you might take u to be the launch speed because that's where you're measuring from (this is far more common).
Post by: knightrider on April 21, 2015, 06:16:29 pm
Depends on the context. Usually, yeah, initial velocity refers to velocity at time t=0. But t=0 is kinda arbitrary, right? Otherwise t=0 might refer to the start of the universe or whatever absolute zero point you want to define. The idea is, we define a zero point that works best for us.
In many kinds of motion, we set t=0 to the start of that motion of interest, so 'initial velocity' relates directly to that starting instant.
If we're ever talking about a change in velocity, initial velocity will refer to the start of the change.
Sometimes, in projectile motion questions, we'll deal with different values of u, (different 'initial velocities') at different points in the question. For example if you want to find the time taken to fall from the peak of the projectile path to the landing point, you'd consider the start of this motion to be the peak of the path, and hence vertical u=0. In another part of the question, you might take u to be the launch speed because that's where you're measuring from (this is far more common).
Thanks silverpixeli :)
Post by: knightrider on April 23, 2015, 08:45:42 pm
Post by: odeaa on April 23, 2015, 09:32:36 pm
Post by: lzxnl on April 23, 2015, 10:00:26 pm
It would just increase linearly because it has no friction
The question hasn't actually said the acceleration is positive :P
Post by: odeaa on April 23, 2015, 10:01:11 pm
The question hasn't actually said the acceleration is positiveLzxnl 1 odeaa 0
Post by: knightrider on April 24, 2015, 12:01:39 am
How would you draw a velocity time graph for the following situation. A particle starts with a positive velocity and undergoes constant acceleration until the end of the time period?
How would i draw this graph?
Post by: Adequace on April 27, 2015, 08:06:31 pm
Post by: silverpixeli on April 27, 2015, 09:56:32 pm
I'm interested to know what should I be doing in units 1/2 if I'm aiming for a 40SS for 3/4? I've been putting a significant amount of effort in to electricity and will be doing the some for motion but is there anything else?
Pretty sure only motion and electricity carry forward to 3/4, sounds like you're setting yourself up with some solid foundations for these topics in year 12. Place emphasis on developing really solid problem solving skills! You can of course work on them in year 12 as well, but you should ideally be able to tackle any practice question you see without any trouble.
Post by: Kel9901 on April 28, 2015, 01:58:31 pm
Pretty sure only motion and electricity carry forward to 3/4, sounds like you're setting yourself up with some solid foundations for these topics in year 12. Place emphasis on developing really solid problem solving skills! You can of course work on them in year 12 as well, but you should ideally be able to tackle any practice question you see without any trouble.
don't forget to still put work in to the other parts of the course though, even if they aren't that relevant, because it will help you develop a good work ethic.
Post by: Floatzel98 on April 29, 2015, 08:14:57 pm
Firstly, I don't really understand the purpose of diodes. What are they used for exactly and where? How does forward and reverse biased work? Like why does it have almost no resistance and then infinite resistance? I don't understand. Why do they have a switch on voltage? Before the switch on voltage is reached, my book says that the diode does not conduct. Does this mean, if it were in series for example with another resistor, the circuit would never actually turn on/be complete? Is this phase exactly the same as what happens in reverse bias?
Secondly, in class we went over how they work, a lot of the chemistry behind them, but I really didn't understand most of it. Do we need to know this at all for any reason? Would someone be able to explain how they work anyway because I still want to understand it. I looked at a couple videos online but i still don't really understand.
Thanks :)
Post by: Adequace on April 29, 2015, 09:15:11 pm
Post by: alchemy on April 29, 2015, 09:30:13 pm
I started doing Checkpoints Physics 1/2's electric circuit chapter, I'm getting most of the questions wrong or don't understand the terminology...killed my self-esteem.
Which questions exactly? Maybe not post them on this thread, but feel free to start your own thread so we can help you! :)
Post by: silverpixeli on April 29, 2015, 11:39:23 pm
I have a couple questions about diodes:
Firstly, I don't really understand the purpose of diodes. What are they used for exactly and where? How does forward and reverse biased work? Like why does it have almost no resistance and then infinite resistance? I don't understand. Why do they have a switch on voltage? Before the switch on voltage is reached, my book says that the diode does not conduct. Does this mean, if it were in series for example with another resistor, the circuit would never actually turn on/be complete? Is this phase exactly the same as what happens in reverse bias?
Secondly, in class we went over how they work, a lot of the chemistry behind them, but I really didn't understand most of it. Do we need to know this at all for any reason? Would someone be able to explain how they work anyway because I still want to understand it. I looked at a couple videos online but i still don't really understand.
Thanks :)
Can answer some of your questions but can't explain how diodes actually work... it's not assessable so I didn't bother :)
forward or reverse bias?
first of all, forward bias refers to when conventional current in a circuit flows in the same direction that the triangle in the diode symbol points. this is opposite to the flow of electrons, but it's not like the physical diode is actually an arrow, it's just chemically a one-way thing and so we chose to line the triangle up with conventional current in our diagrams.
reverse bias is obviously the other way. if you see your conventional current hitting the point side of a triangle, your diode will not conduct because it's in reverse bias mode.
why wont it conduct below the switch on voltage?
a diode is not made of material that can inherently conduct electricity. however, because of the magical chemistry going on under the hood, the diode can --- in one direction --- 'carry' electrons across. this is different to electrons being pushed through a resistor because of potential difference from one side to the other. electrons are unable to push themselves through a diode, but they can hitch a ride through and so the diode appears to be conducting, just like a resistor does.
if there isn't enough potential across the diode, however, this 'carrying' service wont actually start and the diode wont conduct. likewise, if the circuit is pushing the other way, the diode is gonna be a dead end because electrons can't push themselves through, and the diode can only carry them in the forward direction.
i'm thinking about this 'carrying' analogy now, and it seems appropriate because of the one-way property and it's close to how diodes actually work, but theres another property of diodes that doesnt really sit well with it. When you're an electron and you push your way through a resistor, you can spend as much of your energy as you like (and arrive at the other side with less potential, hence potential 'difference' across the resistor). but in a diode, assuming we're above the switch on voltage so that the diode conducts, we'll never spend more than the switch on voltage.
because of this, maybe you should think of diodes as a one-way toll gate. you can go through as fast as you like (indeed, diodes don't influence current in a circuit -- that's decided by the elements in the rest of the circuit) but you always have to pay the same amount (the voltage across the diode will always be the switch on voltage, or less, never more). if you don't have enough potential to pay the toll, you wont even be allowed through! (this is the case where the diode has a potential difference below the switch on voltage across it) and of course, if you try to go back, you won't be allowed, because this gate is on a one way street (no conduction in reverse bias mode)
diodes dont influence current?
one final thing about the properties of a diode is the current through them. I said that diodes allow any current just like a toll gate may allow things to move through at any speed - as long as they pay their toll of potential.
think of a circuit where the battery provides the electrons with 6V of potential at the start, and they have to pass through a 0.8V diode and a 520 ohm resistor, in series. the electrons have the job of deciding how to spend their potential across the circuit before they get back to the battery - where they must have zero. if the circuit were two identical resistors, they would decide to spend their voltage evenly.
But because there's now a 'toll gate', their first priority will be to pay the toll to move through the diode. but they wont spend any more than the toll. then, they each have 6-0.8=5.2V of potential left to spend on the rest of the circuit. of course, there's only the resistor left, so they will splurge their remaining 5.2V on the 520 ohm resistor and move through it as fast as they can: I=V/R=5.2/520=0.01A. Their current through this resistor must be their current everywhere (series circuit) so they must also speed through the diode at 0.1A as well.
The point i'm getting at here is that it's the REST of the circuit that determines the current, you can't tell anything about currents just from a diode because they dont have a well defined resistance, but you can use them to tell you about voltage drops elsewhere because they have a well defined cost in crossing them.
in your example, a diode in series with a resistor, assuming the electrons have enough potential to switch on the diode, they will spend that potential on the diode and spend the rest on getting through the resistor with whatever current they can afford.
but if they weren't given enough potential to even pay the diode toll (less than switch on voltage), then none of them could get around the circuit and it wont conduct, just like if the diode was n reverse bias mode which is effectively the same as cutting one of the wires - no electrons can push through because diodes are a one-way toll gate.
non ideal diodes
in reality, diodes aren't perfectly on or off components. they will begin to conduct a little before the switch on voltage and they will charge slightly more for high current electrons - the I V graph doesn't have infinite gradient at V=switch on voltage.
also, they aren't perfectly impossible to pass in reverse bias mode. if there's enough voltage across a reverse bias diode, the materials inside will stop blocking electrons from flowing and it will begin to conduct freely (well, not freely, in this case you'd be paying a huge voltage toll to keep the gates open the wrong way)
what are they used for?
the one-way thing is actually hugely advantageous. there are many applications of regular diodes in circuitry, for example you can use them and a clever circuit layout called a bridge rectifier to turn AC current into positive-only current. (not studied in core electricity).
in photonics you'll also look at light-emitting diodes (LEDs) which I am sure you are familiar with. probably a bunch of lights in your home, school or any building are LED these days, meaning instead of a hot, glowing globe filament used in old light bulbs they just fire current through a particular type of diode and you get that blue/white light due to the emission of photons. computers, torches, toys, LOTS of things have little LED lights because it's an efficient way to illuminate stuff.
aside from looking really cool, LEDs can be used in optical applications like flashing on and off really fast sending pulsating light signals down an optic fibre for speed-of-light, highly efficient signal transmission.
there are also 'photodiodes' which you'll look at in photonics as well. These again have slightly different chemistry behind them which i dont really understand, but they basically leak a small amount of current that depends on the amount of light that illuminates their surface. In this way, they can capture changing lighting conditions and translate them into changing electrical current - we call this an opto-electric transducer, or something (i forget the technical terminology), and it's useful for doing these sorts of signal conversions
there are probably heaps more examples of applications of diodes! I've probably only scratched the surface.
Post by: Floatzel98 on April 30, 2015, 07:29:56 am
Post by: silverpixeli on April 30, 2015, 09:52:39 am
^Holy crap, thanks so much for all that information. That toll-gate analogy actually helps a lot :)
I know, I'm very proud of it, I understand diodes a lot better now after writing that :P
Post by: Adequace on April 30, 2015, 08:02:44 pm
It ranges from the different terminology used in questions and just sometimes just being confused..
Post by: Floatzel98 on April 30, 2015, 08:21:24 pm
Can someone help me with understanding questions? It seems like a broad question but I think that I understand the theory and concepts behind electronics but I can't seem to interpret the questions and understand what they want me to do.Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.
It ranges from the different terminology used in questions and just sometimes just being confused..
Try to find patterns in questions. There is only a certain number of ways they can ask you things. Maybe it involves graphs of ohmic devices. Gather a whole heap of questions and attempt them. Find patterns in them and maybe even methods or steps of what to look for first when solving them.
Maybe you do actually have some gaps in your knowledge and you don't actually know what you think you should know. If this is the case, broaden your knowledge from more than just the textbook. Look at videos online about the concepts. Maybe they will fill in the unfamiliar terminology your textbook is using or will just give you a different perspective of the topic.
I think you should start by just posting one of the questions you are having trouble with and getting help from us with it.
Post by: Adequace on April 30, 2015, 09:40:20 pm
Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.Cheers for the help, I agree I probably do have gaps in my knowledge without me knowing. I'll definitely post questions I have trouble with in the future.
Try to find patterns in questions. There is only a certain number of ways they can ask you things. Maybe it involves graphs of ohmic devices. Gather a whole heap of questions and attempt them. Find patterns in them and maybe even methods or steps of what to look for first when solving them.
Maybe you do actually have some gaps in your knowledge and you don't actually know what you think you should know. If this is the case, broaden your knowledge from more than just the textbook. Look at videos online about the concepts. Maybe they will fill in the unfamiliar terminology your textbook is using or will just give you a different perspective of the topic.
I think you should start by just posting one of the questions you are having trouble with and getting help from us with it.
Post by: StressedAlready on April 30, 2015, 10:23:26 pm
Realise this is year 11 physics. Should totally know but don't despite doing 3+4 this year. T.T
Please help?
Post by: silverpixeli on May 01, 2015, 08:26:21 am
A scientist has a 120g sample of the radioisotope polonium-218. The first three steps in the decay series of polonium-218 are an alpha emission followed by two beta particle emissions. These decays have half-lives of 3, 27 and 19 mins respectively. Calculate how many polonium-218 remains after 15 mins.
Realise this is year 11 physics. Should totally know but don't despite doing 3+4 this year. T.T
Please help?
there's some info we don't need here. The decay series it's talking about it this:
We see that after it decays for the first time, it's never going to be Polonium-218 again. So "Calculate how many polonium-218 remains after 15 mins" is just asking about the first decay, and the other halflives are meaningless to us.
The question becomes "how much of your 120g sample,half life 3min, is left after 15m?"
answer
15/3=5 half lives pass in the 15 minutes, so the sample size is halved 5 times.
120/2/2/2/2/2 = 120/32 = 3.75g
thus 3.75 grams remain after 15 mins.
120/2/2/2/2/2 = 120/32 = 3.75g
thus 3.75 grams remain after 15 mins.
Post by: lolaishappy on May 02, 2015, 05:56:37 pm
Also How do I do sketch modulation graphs? :-[
Post by: Cosec on May 02, 2015, 06:17:44 pm
When it comes to photonics, what theory is a must in knowing? Like modulation, demodulation, how about attenuation? Particularly from the Jacaranda text book.
Also How do I do sketch modulation graphs? :-[
A modulated graph, which im guessing is your just refeering to the graph of the wave itself, is a wave in which the information signal, normally one of low frequency, is super imposed on the carrier wave, one which has a much higher frequenc. producing a wave that is used to carry the information over a certain medium. Like this: http://www.sfu.ca/sonic-studio/handbook/Graphics/Amplitude_Modulation2.gif
I just had my sac, and learnt all of them which you described above. When i completed several practice sacs, a majority include a question about the modulation/demodulation process. So i would strongly advise you learnt them.
Post by: Adequace on May 02, 2015, 07:44:33 pm
This is for unit 1/2 if you guys don't mind, thanks.
Post by: Floatzel98 on May 02, 2015, 08:29:49 pm
We have our electricity SAC coming up in 2 weeks, we're allowed to bring a double sided cheat sheet. I've just finished mine but I have a lot of spare space - a whole page worth. I'm wondering if I should use that space to write questions I had trouble on or to use it for additional information about the less important things such as electric shocks and resistivity of your skin, I know it's going to be on the MCQ as well.If you know what is coming up on the test for the most part and you know what you will struggle with, definitely try to fill it up with that. Specific examples of questions you struggled with. I think simple sentence definitions could suffice for the less important information or even just try to rote learn it.
This is for unit 1/2 if you guys don't mind, thanks.
I know that we even get a cheat sheet to use in the 3/4 exam, but i would also recommend trying not to rely on it too much. It might seem counter-intuitive but the less you have to rely on a cheat sheet means that you know more of what you need to know already. You should / you usually are given a formula sheet with the standard equations you should know, so if you can try to dedicate more space on your cheat sheet to specifics of what you struggled on and not things you already know. This is not necessarily for this SAC now, but in the future as well. I know sometimes i just fill up my cheat sheets with all the equations that are already on the cheat sheet, sometimes with a few derived equations for myself. I spent time on it when i could have just been studying or i actually could have filled it up with useful information and examples that would aid me.
I know I'm just rambling on now and that my first sentence probably answers your question, but hopefully the rest helps as well if you can apply it at all. Anyway goodluck on your upcoming SAC and post any other questions you have! :)
Post by: kk.08 on May 02, 2015, 10:18:20 pm
Thanks :)
Post by: Floatzel98 on May 03, 2015, 05:41:24 pm
What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?
For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all? Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.
What exactly determines the switch on voltage of a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?
Just some question i thought of while doing my homework today. Thanks to anyone that can answer them :)
Post by: Adequace on May 03, 2015, 06:02:05 pm
In this circuit, what is voltage at X? I got the answer but I'm still confused on the method of getting the answer since I just guessed. We found the total resistance in the question above.
Post by: silverpixeli on May 03, 2015, 06:14:09 pm
Was just wondering, is it possible to find the elastic limit of a material without a stress-strain graph?
Thanks :)
I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).
If a resistor or even another diode is placed in series in front of a reversed biased diode (the current flows through the resistor before it hits the reverse biased diode), why doesn't current still flow through. For example why isn't any power dissipated? The current has to go through the resistors and such to get to the reverse biased diode?
Good question. Current in series has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.
So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.
What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?
Same explanation as first question, there is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.
For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all? Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.
Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.
Electrons actually flow the other way in a real circuit, but the convention still holds.
Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.
If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.
What exactly determines the switch on voltage of a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?
This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.
BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)
You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.
Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)
Post by: Floatzel98 on May 03, 2015, 06:19:53 pm
http://imgur.com/wPXnz5RWell first you can find the total resistance which should be
In this circuit, what is voltage at X? I got the answer but I'm still confused on the method of getting the answer since I just guessed. We found the total resistance in the question above.
Then the total current should be
The current though 12 ohm resistor should be
That leaves 2 amps to be found in the series part of the circuit and since current is constant in series they both have the 2 amps through them.
That leaves the voltage across the 2 ohm resistor to be
The voltage in the 4 ohm resistor is
You can verify this because voltage is constant in parallel and the 2 series resistors voltage add up to 12 (4 + 8 )
I'm pretty sure this is right. I'm notorious for forgetting at least something which stuffs up all the other calculations. It might be a long working but i hoped it highlighted everything for you and answered the question :)
Spoiler
I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).
Good question. Current in series has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.
So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.
Same explanation as first question, there is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.
Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.
Electrons actually flow the other way in a real circuit, but the convention still holds.
Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.
If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.
This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.
BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)
You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.
Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)
Post by: Adequace on May 03, 2015, 06:27:59 pm
Well first you can find the total resistance which should beThanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?
Then the total current should be
The current though 12 ohm resistor should besince votlage is constant in parallel.
That leaves 2 amps to be found in the series part of the circuit and since current is constant in series they both have the 2 amps through them.
That leaves the voltage across the 2 ohm resistor to be
The voltage in the 4 ohm resistor is
You can verify this because voltage is constant in parallel and the 2 series resistors voltage add up to 12 (4 + 8 )
I'm pretty sure this is right. I'm notorious for forgetting at least something which stuffs up all the other calculations. It might be a long working but i hoped it highlighted everything for you and answered the question :)
Post by: Floatzel98 on May 03, 2015, 06:43:45 pm
Thanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?If you mean find the total voltage of the series part of the circuit, then you would get a different answer to the voltage at X. You would get 6*2 = 12 V, which is the total voltage across the series arm. From there you could go and find the voltage across the 2 individual resistors, which is exactly the same as doing what i showed anyway. Otherwise as long as you get the correct value for current and you use it to find the actual voltage across X and not the entire series arm, then yes. The method would be correct.
Post by: silverpixeli on May 04, 2015, 09:15:48 am
Thank you for clearing everything up for me again. I really appreciate it :) For the part about the current flowing into a reverse biased diode: I know the answer stays the same, but does that mean that for the instant when it is turned on, current will flow up past everything to the diode? Also when you turn off the power source to a circuit, do the electrons that were currently in the circuit all leave the wire/circuit back into the power supply or can they get 'trapped' in there? Thanks again!
Actually, this is getting into the electromagnetism side, but there are always electrons distributed at all points along the wire, (conduction electrons bounce around fairly randomly when there's no potential difference affecting them)
When you 'turn on' the circuit, you're providing a 'potential difference' between one point and another. We think of this as a difference in energy, but the electrons are actually feeling a force, called 'electromotive force', because the battery is actually providing an electric field. The randomly bouncing electrons start to gradually experience a net movement with the force.
Compare this with the force of gravity, a uniform field when you're near the surface of the Earth. But sometimes we talk about potential energy (mgh). A grain of sand will fall downwards in a hollow tube, because there is a gravitational force pushing it that way. You could equivalently think of it as falling to the bottom of the tube, because then it will be at a lower gravitational potential energy (there is a 'potential difference' between the bottom and the top of the tube).
This is what's happening to electrons in your circuit. But we don't normally talk about it in terms of force in a circuit, we talk about the potential energy difference between the positive terminal and the negative terminal. In the case of the sand, we're talking about gravitational potential energy. In the case of the electron, we're talking about electrical potential energy.
'Electricity' travels at the speed of light, but this isn't the motion of the electrons, it's the speed that the field travels out and starts affecting the electrons that are further away. The electrons actually move relatively slowly :)
Post by: Floatzel98 on May 04, 2015, 03:47:44 pm
Firstly, i don't really understand how a transistor works in a circuit. From what i understood, it has a voltage input that goes through the collector and then one that comes through the base? If that's correct, are they actually 2 different power sources or do they come from the same power supply. If so, then how do you regulate how much voltage goes through it from the base input? From a picture i saw in class today, the base is apparently usually connected to a capacitor? I don't really know what a capacitor is either, but doesn't that store energy/voltage? Is that connected back to the same power supply of the circuit if that's the case. How do you regulate how much voltage goes through that anyway?
Next, from what i gathered, when no input voltage comes through the base, the resistance is infinite since it is basically 2 diodes facing each other and one of them will be in reverse bias? When there is a voltage input from the base that goes above the switch on voltage it will work and the resistance will basically be 0. How does the 'switch on voltage' allow the diode to work if it is approaching it in reverse bias? You would need like 50V to break it but it is still only about 0.7 V according to my book.
Even if that all makes sense to me, i still don't really know what it does. It amplifies the voltage but how? Is it because you are basically adding 2 voltages from 2 power sources together. Almost like as if you were adding ordinates of 2 sine graphs? I hardly understand what all the graphs are about either, but if someone can enlighten me on how all the above works, maybe i will be able to figure it out for myself.
Thanks :)
Post by: silverpixeli on May 04, 2015, 10:26:42 pm
Thanks again Silverpixeli! But alas i have more questions again. This time about voltage amplifiers. I went over it in class today and we watched a few videos and i even watched the VTextbook video on it just then, but I'm still confused.
Firstly, i don't really understand how a transistor works in a circuit. From what i understood, it has a voltage input that goes through the collector and then one that comes through the base? If that's correct, are they actually 2 different power sources or do they come from the same power supply. If so, then how do you regulate how much voltage goes through it from the base input? From a picture i saw in class today, the base is apparently usually connected to a capacitor? I don't really know what a capacitor is either, but doesn't that store energy/voltage? Is that connected back to the same power supply of the circuit if that's the case. How do you regulate how much voltage goes through that anyway?
Next, from what i gathered, when no input voltage comes through the base, the resistance is infinite since it is basically 2 diodes facing each other and one of them will be in reverse bias? When there is a voltage input from the base that goes above the switch on voltage it will work and the resistance will basically be 0. How does the 'switch on voltage' allow the diode to work if it is approaching it in reverse bias? You would need like 50V to break it but it is still only about 0.7 V according to my book.
Even if that all makes sense to me, i still don't really know what it does. It amplifies the voltage but how? Is it because you are basically adding 2 voltages from 2 power sources together. Almost like as if you were adding ordinates of 2 sine graphs? I hardly understand what all the graphs are about either, but if someone can enlighten me on how all the above works, maybe i will be able to figure it out for myself.
Thanks :)
Transistor workings are also not really part of the course. Capacitors, also, are left for the detailed study 'further electronics'.
Voltage amplification questions range from interpreting graphs, talking about cutoff/saturation, voltage gain and converting from input signals to output signals and back.
A transistor definitely has the two source things going on, but I'm not sure what you're talking about when you mention two diodes and one in reverse bias mode.
Whatever voltage applies at the base is like the voltage across a diode, in that if it's lower than some switch on value, there will be no conduction. A transistor is different in that when the diode IS switched on, the majority of the current to the emitter doesn't come through the base, it comes through the collector. Basically, the voltage across the base regulates how much current can come through the collector. Typically, you have the collector hooked up to some more powerful supply that can provide as much current to the collector as the transistor will allow through it.
This is useful for signal amplification as follows: Typically you have a signal captured as small variations in voltage/current in a circuit. If you feed this varying voltage into the base of a transistor, it will allow proportionate amounts of current through from the high-power collector. The result will be a higher power version of your original signal coming out of the emitter!
This assumes that your input signal is always within the working range of the transistor. Too low, and you won't open the gate for current from the collector. We refer to this as 'cutoff'. Too high, and you wont be able to get any more current through the transistor, and all higher signals will output that same max. We call this saturation. Both are examples of 'clipping' which result in the extremes of a signal being 'chopped off' from signals that dont lie within this range.
The complicated circuits that actually make this work are not studied and I have no understanding of them personally other than the high-level sketch above. Sorry :P
As for the amplifier graphs, Vin - Vout graphs map input voltages to output voltages. If you put in an input signal (often represented as a Vin - time graph) you can use the amplifier's characteristic graph to map it to an output signal v. time graph. This process, and the reverse, are semi-common exam questions.
Post by: TheAspiringDoc on May 06, 2015, 03:53:08 pm
Post by: silverpixeli on May 06, 2015, 04:23:18 pm
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?
Good question. afaik a 'black light' is impossible.
When we look at something black, it looks black because there are no photons (light particles) reaching our eyes from that place. If there were some red photons, it would look red. If there were some red+green+blue photons (or another combination of lots of colours) it might look white.
To make it look black, we'd have to stop it from reflecting photons towards our eyes. We can't do this with a torch.
I guess we could alternatively try to stop the photons from reaching our eyes once they have been reflected, but I don't see how this could be done either.
In short, black things look black because there is no light coming from there. There are no black photons that you could shine on something.
Post by: lzxnl on May 06, 2015, 08:52:31 pm
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?
Hmmm....
I wonder if a molecule exists that absorbs essentially completely in the visible range and has such a low fluorescent quantum yield that all of the light absorbed is re-radiated as heat. In theory, if you could make a molecule like that, said molecule would block out all light hitting it, so you would have a dark trail in the air. But I can't imagine what on earth would have such a property.
But yeah, no black photons though.
Post by: Fusuy on May 10, 2015, 02:10:32 pm
Post by: Floatzel98 on May 10, 2015, 08:15:37 pm
EDIT: Another question. There is a circuit with 3 identical resistors of 100 Ohms that are connected with one in series and then the other 2 in parallel and the max power from any one resistor is 25W.
The question was to find the max voltage the can be applied. I used
Post by: odeaa on May 10, 2015, 08:34:09 pm
I'm having a bit of trouble with the attached question. I have looked at the worked solutions for question 3 and i understand the steps but i don't understand how they thought the steps up and how they got to them. I find that a lot of these ratio questions stump me, especially when they were used in Gravity and Satellites. I can kind of see how it wants me to answer it but i can never actually go through with it and get the correct answer. Can someone show me how they went through the steps to solve this? Thanks :)
Q3) Total resistance
Using the voltage divider formula,
Alternatively, we can just use ratios- because
Q4)
Then, remembering that
b)
I hope thats right
edit: I suck at latex, I think its all good now
Post by: silverpixeli on May 10, 2015, 08:54:55 pm
Another question. There is a circuit with 3 identical resistors of 100 Ohms that are connected with one in series and then the other 2 in parallel and the max power from any one resistor is 25W.
The question was to find the max voltage the can be applied. I usedand had already found the total resistance and went
for the 3 resistors, but my answer seems to be wrong.
Why are you adding ohms and watts together? I assume typo?
Okay so just because the maximum power of one resistor is 25 Watts doesnt mean it's operating at that power. That just means that as you increase the voltage applied, you have to be careful not to take ANY of them over the max. In particular, With more resistance and twice the current, the resistor OUTSIDe of the parallel section will probably have the most power at any given time. So you should solve for the total voltage in the circuit when all you know is that the series resistor has 25W, 100Ohms.
Post by: Floatzel98 on May 10, 2015, 09:02:43 pm
Plus i have another question now. If you have a voltage divider over a resistor that is in parallel, do we take voltage divider resistors parallel voltage or it's individual voltage? From the questions I've done it's apparently the total parallel resistance but i don't think that makes sense for some reason. Is it because the voltage in parallel is constant? That's my only reasoning. If i turn the parallel resistor into a single resistor ( and put it in series with another resistor) is that still the same thing.
While typing that i think i've answered my own question. But does that make sense still? If i have a voltage divider over one resistor in parallel, i can just simplify that into a single resistor and everything will still work the same?
Thanks :)
Post by: Floatzel98 on May 10, 2015, 09:11:30 pm
Why are you adding ohms and watts together? I assume typo?Yes, that was a typo. My original workings were
Okay so just because the maximum power of one resistor is 25 Watts doesnt mean it's operating at that power. That just means that as you increase the voltage applied, you have to be careful not to take ANY of them over the max. In particular, With more resistance and twice the current, the resistor OUTSIDe of the parallel section will probably have the most power at any given time. So you should solve for the total voltage in the circuit when all you know is that the series resistor has 25W, 100Ohms.
Post by: silverpixeli on May 10, 2015, 09:23:29 pm
Yes, that was a typo. My original workings were. The total resistance is 150 ohms and if each resistor can have a max output of 25 W, then the total power output could be 75 W, so
. Is the total power across the parallel resistors not 50W? Because i'm pretty sure my answer is wrong because of my watts working out. If the total voltage is the same and the resistors are identical, the the current is exactly split across them, which would then give 12.5W + 12.5W each, changing the equation to
, but that also doesn't give the correct answer. The question was what is the total allowed voltage that can pass across them. I might still be interpreting it wrongly though. Should i try to find the voltage across the parallel and series separately and then add them together?
Actually because the resistance is down over the parallel branch, they will actually not have as much voltage as across the series resistor.
Post by: odeaa on May 10, 2015, 09:45:25 pm
Post by: lzxnl on May 10, 2015, 10:04:31 pm
What does it mean to have a negative voltage? Is it only when you have a photodiode or something in reverse bias and hence a negative current?
Firstly, what is a voltage?
A voltage is a potential difference between two points. Well that doesn't help, does it?
The electric potential is defined as the electric potential energy per unit charge. So you know how things like to go from high to low potential energy? For a positive charge, this means go from high to low potential but for a negative charge, it means it'll want to go from low to high potential.
So a negative potential just means you've measured a positive potential in the other way. This has implications for devices that are direction-specific, like diodes.
Post by: odeaa on May 10, 2015, 10:05:32 pm
Firstly, what is a voltage?Legend, thanks for clearing that up
A voltage is a potential difference between two points. Well that doesn't help, does it?
The electric potential is defined as the electric potential energy per unit charge. So you know how things like to go from high to low potential energy? For a positive charge, this means go from high to low potential but for a negative charge, it means it'll want to go from low to high potential.
So a negative potential just means you've measured a positive potential in the other way. This has implications for devices that are direction-specific, like diodes.
Post by: Floatzel98 on May 13, 2015, 09:19:50 pm
I have a physics SAC tomorrow on the first half of the Electronics and Photonics Area of Study and my teacher said today that there would be a question about Voltage RMS on it, even though we haven't actually come across it in the book yet. He explained it to us and basically told us that the question was just reading the peak AC current and finding the RMS Voltage, which is easy enough. I understand the formula and how to use it for the question we will get, but i still don't understand how they get it. In the Heinemann textbook it comes up in the further electronics AOS and quickly in Electric Power AOS, but I still wasn't able to understand it.
So basically my questions are:
Where/how do they derive it?
What is it? I gathered it was like the average voltage provided by an AC signal. I also read that it is the same the DC voltage for something? What is the relationship between them?
How does the current RMS work? Where do you ever find the peak current? Are there usually graphs of these or is it something you can solve after finding the voltage RMS?
Also i have some questions about AC and DC signals that i don't think i ever really understood, and because we are finally using them a lot more now i think my knowledge isn't really matching up with what i'm learning. And some other random questions if anybody can be bothered answering them
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
How do you actually make an alternating current? Are magnets used to do this or something else?
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.
Thanks! :)
Post by: Kel9901 on May 13, 2015, 10:27:12 pm
Hey guys. I have a couple questions today about Voltage RMS.
I have a physics SAC tomorrow on the first half of the Electronics and Photonics Area of Study and my teacher said today that there would be a question about Voltage RMS on it, even though we haven't actually come across it in the book yet. He explained it to us and basically told us that the question was just reading the peak AC current and finding the RMS Voltage, which is easy enough. I understand the formula and how to use it for the question we will get, but i still don't understand how they get it. In the Heinemann textbook it comes up in the further electronics AOS and quickly in Electric Power AOS, but I still wasn't able to understand it.
So basically my questions are:
Where/how do they derive it?
What is it? I gathered it was like the average voltage provided by an AC signal. I also read that it is the same the DC voltage for something? What is the relationship between them?
How does the current RMS work? Where do you ever find the peak current? Are there usually graphs of these or is it something you can solve after finding the voltage RMS?
Also i have some questions about AC and DC signals that i don't think i ever really understood, and because we are finally using them a lot more now i think my knowledge isn't really matching up with what i'm learning. And some other random questions if anybody can be bothered answering them
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
How do you actually make an alternating current? Are magnets used to do this or something else?
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.
Thanks! :)
i'll just answer the first part.
Try to think about it this way.
In AC, both voltage and current have a sinusoidal shape. When multiplied together to give power, it will be a graph that looks something like (sin(x))^2, ie something like this http://imgur.com/lMxj3Hf. It's pretty visually obvious that the 'average' power is half the 'peak' (top) power.
Hence, P(RMS)=P(peak)/2
V(RMS)I(RMS)=V(peak)I(peak)/2=(V(peak)/sqrt(2))(I(peak)/sqrt(2)
V(RMS)=V(peak)/sqrt(2), and I(RMS)=I(peak)/sqrt(2).
The RMS voltage of an AC signal represents the DC voltage that would provide the same average power.
current is pretty much treated the same way a voltage.
Post by: Floatzel98 on May 13, 2015, 10:44:49 pm
i'll just answer the first part.Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)
Try to think about it this way.
In AC, both voltage and current have a sinusoidal shape. When multiplied together to give power, it will be a graph that looks something like (sin(x))^2, ie something like this http://imgur.com/lMxj3Hf. It's pretty visually obvious that the 'average' power is half the 'peak' (top) power.
Hence, P(RMS)=P(peak)/2
V(RMS)I(RMS)=V(peak)I(peak)/2=(V(peak)/sqrt(2))(I(peak)/sqrt(2)
V(RMS)=V(peak)/sqrt(2), and I(RMS)=I(peak)/sqrt(2).
The RMS voltage of an AC signal represents the DC voltage that would provide the same average power.
current is pretty much treated the same way a voltage.
Post by: Kel9901 on May 14, 2015, 12:46:01 pm
Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)
If the RMS power is half the peak power, then the RMS voltage x the RMS current will be 1/2 the peak voltage x the peak current. This half is 'split' equally between the voltage and the current as 1/sqrt(2) each, because 1/sqrt(2) x 1/sqrt(2)=1/2
Post by: Peanut Butter on May 17, 2015, 04:26:15 pm
Thanks :)
Post by: Floatzel98 on May 17, 2015, 05:08:53 pm
Can someone please help me with the electricity question attached?For the first one, all we need to do is go find the voltage across the 6000 ohm resistor done by V = IR. Voltage is constant in parallel and there are no other components in the circuit, so that equals the EMF of the battery.
Thanks :)
Since voltage is the same in parallel, we can work out the current in the other resistor by going I = V/R. Use the voltage you just calculated and the resistance provided. If you wanted to check if this is right, you could find the total parallel resistance by going
With the break in the wire, nothing will flow through there. Every should just go through the 6000 ohm resistor, basically putting it in series with the battery. You should be able to work out the current of a single resistor given the total voltage and the resistance.
For the last question (and i might be wrong on this) you can think of it as putting a voltmeter across the 6000 ohm resistor in the same series scenario. X1 is on before the resistor and x2 is after it. I don't think 4000 ohm resistor will change that, but it might. If we think of it as just being in series that is right, but if we are literally testing from point x1 to x2, you would still have the 4000 ohm resistor in the way. But no current is flowing through it anyway, so it shouldn't matter.
Post by: Kel9901 on May 18, 2015, 10:37:18 am
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
No, the frequency (how quickly the current goes up and down) is so high that humans don't notice it; it appears to stay at constant intensity
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
Pretty much slows them down, yeah. The resistor pretty much makes the electrons give up some energy (potential) to get past it.
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
The same electrons run the circuit, and are 'recharged' or given back the potential they lost when going through the resistors when they reach the battery. The electrons come from the wires/resistors, as they are usually metal and hence have free moving electrons.
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
I think pretty much everything uses AC. The main exception I can think of, aside from lab experiments (lol), is trains, which use DC power.
How do you actually make an alternating current? Are magnets used to do this or something else?
You'll learn this in unit 4, but yeah, magnets.
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.
Thanks! :)
Post by: Floatzel98 on May 18, 2015, 03:26:26 pm
Thanks!
Post by: Adequace on May 18, 2015, 06:02:59 pm
Post by: Cosec on May 18, 2015, 06:43:25 pm
What's the best way to revise for a mid-year exam? Practice exams only?
Yolo, whats revision for practice exams?! Haha,
Yeah, thats what ill be doing! Smashing dem out. If i can find them. Do you know any sources that has a bunch of them?
Post by: odeaa on May 18, 2015, 07:34:29 pm
Yolo, whats revision for practice exams?! Haha,our teacher gave us a USB with all the VCAA exams back to 1997, ask your library I think they will have a CD with all the past exams
Yeah, thats what ill be doing! Smashing dem out. If i can find them. Do you know any sources that has a bunch of them?
Post by: Floatzel98 on May 19, 2015, 04:07:36 pm
Thanks!
EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.
Post by: silverpixeli on May 19, 2015, 04:17:51 pm
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?
Thanks!
quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.
The graphs for these are most interesting in the negative voltage region (whichever of photovoltaic/photoconductive this corresponds to, i cant remember) because of this light-dependent current property, and the graphs usually have multiple lines telling us what current leaks through for a certain intensity. since these lines are flat, higher reverse-bias voltages dont affect the leakage current - it is only dependent on light intensity (for an ideal photodiode anyway)
the other mode isnt as interesting and i cant remember but i think it just functions like a regular diode when in forward bias, at least that's what the graphs suggest for voltages > switch on voltage.
Post by: Kel9901 on May 19, 2015, 05:48:40 pm
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?
Thanks!
EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.
there is a lot of stuff in the electronics+photonics section of heinemann that is irrelevant. this is one of those things; phototransistors aren't on the course.
to answer your question, the formula for the energy of a single photon is E=hf, where h is Planck's constant and f is the frequency (which is also equal to speed of light/wavelength). you'll learn about it in unit 4
Post by: Floatzel98 on May 19, 2015, 06:27:23 pm
quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.So, the exact same thing happens as with a normal diode in reverse bias. Except that when light hits it it kind of decreases the resistance so that a small amount of current can go through? Or does it kind of create a current from the light that hits it (Is that possible?)? I'm still kind of confused though. If we placed it in a circuit with just another resistor, with no light hitting it, it would just stop the circuit as a normal diode does since it is in reverse bias. But what happens when you add light to it then in this example? Because there is light, it lets a current through and the circuit works as normal?
The graphs for these are most interesting in the negative voltage region (whichever of photovoltaic/photoconductive this corresponds to, i cant remember) because of this light-dependent current property, and the graphs usually have multiple lines telling us what current leaks through for a certain intensity. since these lines are flat, higher reverse-bias voltages dont affect the leakage current - it is only dependent on light intensity (for an ideal photodiode anyway)
the other mode isnt as interesting and i cant remember but i think it just functions like a regular diode when in forward bias, at least that's what the graphs suggest for voltages > switch on voltage.
Going from the IV characteristic graph of a photodiode, the positive voltage part would only ever really happen if we put the photodiode in forward bias, right? And even if you did it would just act like a normal diode. And if thats right, i'm still confused about the negative voltage region of the graph. Why are there so many different lines exactly? And they still have 'infinite' resistance like a normal reverse bias diode, but they still let current through?
there is a lot of stuff in the electronics+photonics section of heinemann that is irrelevant. this is one of those things; phototransistors aren't on the course.I think i understand the energy of a photon part, but how do they introduce the Number of photons part into P = E/t? I still don't understand it too much. I know you said phototransistors aren't in the course (thanks for that), but i would still like to know how they did this if you can explain it.
to answer your question, the formula for the energy of a single photon is E=hf, where h is Planck's constant and f is the frequency (which is also equal to speed of light/wavelength). you'll learn about it in unit 4
Thanks guys :)
Post by: silverpixeli on May 19, 2015, 06:36:42 pm
So, the exact same thing happens as with a normal diode in reverse bias. Except that when light hits it it kind of decreases the resistance so that a small amount of current can go through? Or does it kind of create a current from the light that hits it (Is that possible?)? I'm still kind of confused though. If we placed it in a circuit with just another resistor, with no light hitting it, it would just stop the circuit as a normal diode does since it is in reverse bias. But what happens when you add light to it then in this example? Because there is light, it lets a current through and the circuit works as normal?
Going from the IV characteristic graph of a photodiode, the positive voltage part would only ever really happen if we put the photodiode in forward bias, right? And even if you did it would just act like a normal diode. And if thats right, i'm still confused about the negative voltage region of the graph. Why are there so many different lines exactly? And they still have 'infinite' resistance like a normal reverse bias diode, but they still let current through?
I'm not sure on the current-being-created vs current-being-allowed-through thing, my guess would be the latter though. either way it behaves the same, the current going in that part of the circuit is the photocurrent, travelling out the back of the reverse bias photodiode (so same way as voltage is applied)
the different lines each correspond to a different level of light hitting the photodiode, that line represents the I-V behaviour of the diode under that level of light (each line is usually labelled in watts/metre^2, or 'lux' which is a unit of illumination)
Post by: Kel9901 on May 19, 2015, 06:38:13 pm
I think i understand the energy of a photon part, but how do they introduce the Number of photons part into P = E/t? I still don't understand it too much. I know you said phototransistors aren't in the course (thanks for that), but i would still like to know how they did this if you can explain it.
Thanks guys :)
Well, each photon has a certain amount of energy (Ep=hf). The total energy of the photons is the number of photons (N for the sake of this) times the individual energy of a photon- ie E=NEp. This energy is equal to the power of the light times the time- ie E=NEp=Pt. In this case, time is 1 second (since you want to find out the number of photons per second), so it becomes NEp=P, or N=P/Ep.
Post by: Floatzel98 on May 19, 2015, 07:17:29 pm
Well, each photon has a certain amount of energy (Ep=hf). The total energy of the photons is the number of photons (N for the sake of this) times the individual energy of a photon- ie E=NEp. This energy is equal to the power of the light times the time- ie E=NEp=Pt. In this case, time is 1 second (since you want to find out the number of photons per second), so it becomes NEp=P, or N=P/Ep.That makes a lot more sense now. Thanks a lot :)
I'm not sure on the current-being-created vs current-being-allowed-through thing, my guess would be the latter though. either way it behaves the same, the current going in that part of the circuit is the photocurrent, travelling out the back of the reverse bias photodiode (so same way as voltage is applied)Thanks again, but just a few more questions to clarify some stuff. So, if no light hits the photodiode, the circuit will not conduct. But the more intense the light is, more current will pass through and will work like normal diode in a circuit, except with smaller amounts of current than what comes from the power supply. It can't pull current through unless light is hitting it, and then the current level varies depending on the intensity of the light? And lastly, the obligatory, what are the used for, question. The only thing they can do is let small amounts of current through based on the intensity of the light around it. Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question. But another question kind of arises, how sensitive are photodiodes, or even LDR? Do you actually have to shine a light on them to get them to work? When you use photodiodes etc.., how is the intensity changed? Also, how long does it take. For example if i went from shining a light on a photodiode to putting my finger over the sensor (effectively not letting it conduct at all if i understand properly), how long would it take to turn off, or at least go down to a lower current? For some reason i don't think it would be very instant.
the different lines each correspond to a different level of light hitting the photodiode, that line represents the I-V behaviour of the diode under that level of light (each line is usually labelled in watts/metre^2, or 'lux' which is a unit of illumination)
Thanks again. I should pay you since you are basically tutoring me through physics ;)
Post by: silverpixeli on May 19, 2015, 09:19:53 pm
Thanks again. I should pay you since you are basically tutoring me through physics ;)
I do tutor physics but I'm full at the moment so I can't take you on as a student and you'll have to keep getting it here for free, sorry, :P
And lastly, the obligatory, what are the used for, question. The only thing they can do is let small amounts of current through based on the intensity of the light around it. Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question. But another question kind of arises, how sensitive are photodiodes, or even LDR? Do you actually have to shine a light on them to get them to work? When you use photodiodes etc.., how is the intensity changed? Also, how long does it take. For example if i went from shining a light on a photodiode to putting my finger over the sensor (effectively not letting it conduct at all if i understand properly), how long would it take to turn off, or at least go down to a lower current? For some reason i don't think it would be very instant.
First of all, you're right that it's not instant, but photodiodes are pretty fast. I think they have microsecond response times (compare with LDRs which are only millisecond-fast, so 1000 times slower.)
In many applications all you need is a tiny signal, such as in signal transmission, you can have a small current flashing off or on to transmit a code through a circuit. That small current could later be amplified by the way, by feeding it into the base of a transistor or something.
The fast response time is super useful in receiving signals that are sent through optic fibres. These light signals flash on and off really fast (created by a flashing LED or laser at the other end) and a photodiode can grab these changes and respond accordingly, at least as fast as microseconds.
Also just looking at the graphs again, when it has 0 voltage and light is hitting it, there is still current flowing through it. How and why does that work? And does the intensity of light keep getting greater over time, or does it stay constant? I know if you change the area or the power it will decrease or increase, so i think that answers my question.
So that's the 'dark current' you're hearing about. Evidently, the chemistry behind a photodiode means that it lets a few micro-amps through even when there's no light, so long as there's a reverse voltage across it.
As for the intensity thing, the unit of intensity is lux=W/m^2 (watts/metre^2) and remember that a watt is just a joule every second. So it's a rate of illumination of a certain area. So if you increase the area you absorb more energy every second (power) than before but it's still the same intensity. If you wait longer, you dont have more intensity because it's energy PER second.
Im not sure if this answers all your questions so keep asking for clarifications if you need.
Post by: Floatzel98 on May 19, 2015, 09:55:14 pm
I do tutor physics but I'm full at the moment so I can't take you on as a student and you'll have to keep getting it here for free, sorry, :PThanks for everything i really appreciate it. I didn't mean the dark current part of a diode that lets current through with no light. I mean that (if you look at the graph in the spoiler) there is still current flowing through with 0 voltage across it. Like when there is 2mW there is always just under 1mA even in the transition from postive to negative and even at 0 voltage as well. That's why i asked about the created vs being allowed through thing before. I mean now when i think about it, even if it did make the current somehow, there would still be no voltage pulling the current. Obviously the diode isnt a battery and the light doesn't charge it up. It doesn't provide its own voltage and there is no voltage being applied but there is still current going through.
First of all, you're right that it's not instant, but photodiodes are pretty fast. I think they have microsecond response times (compare with LDRs which are only millisecond-fast, so 1000 times slower.)
In many applications all you need is a tiny signal, such as in signal transmission, you can have a small current flashing off or on to transmit a code through a circuit. That small current could later be amplified by the way, by feeding it into the base of a transistor or something.
The fast response time is super useful in receiving signals that are sent through optic fibres. These light signals flash on and off really fast (created by a flashing LED or laser at the other end) and a photodiode can grab these changes and respond accordingly, at least as fast as microseconds.
So that's the 'dark current' you're hearing about. Evidently, the chemistry behind a photodiode means that it lets a few micro-amps through even when there's no light, so long as there's a reverse voltage across it.
As for the intensity thing, the unit of intensity is lux=W/m^2 (watts/metre^2) and remember that a watt is just a joule every second. So it's a rate of illumination of a certain area. So if you increase the area you absorb more energy every second (power) than before but it's still the same intensity. If you wait longer, you dont have more intensity because it's energy PER second.
Im not sure if this answers all your questions so keep asking for clarifications if you need.
Spoiler
(http://www.rp-photonics.com/img/photodiode.png)
Post by: Adequace on May 20, 2015, 05:15:26 pm
Post by: odeaa on May 20, 2015, 05:49:28 pm
I just got my results back for my 1/2 electricity SAC and managed to get 99%. I just wanted to say thanks for your help.nice man! i got like 60 or something on that ahaha, luckily managed to get my head around it this year
Post by: Kel9901 on May 20, 2015, 07:17:58 pm
I just got my results back for my 1/2 electricity SAC and managed to get 99%. I just wanted to say thanks for your help.
grats!
Post by: paper-back on May 24, 2015, 02:59:01 pm
Why and how does the resistance of the loop change the direction of an induced current?
Post by: silverpixeli on May 24, 2015, 03:03:41 pm
A central loop of wire lies inside a larger loop, which is connected to a battery. Current flows around this outer loop. The resistance of the outer loop is increasing. Determine the direction of the conventional current induced in the loop
Why and how does the resistance of the loop change the direction of an induced current?
assuming that the battery provides a fixed voltage, increasing resistance in the outer loop means that current ill be decreasing.
this does not change the direction of any induced current in the inner loop, but it does determine the direction of induced current by Lenz' law.
Post by: paper-back on May 24, 2015, 03:30:46 pm
assuming that the battery provides a fixed voltage, increasing resistance in the outer loop means that current ill be decreasing.Thanks for responding silverpixeli
this does not change the direction of any induced current in the inner loop, but it does determine the direction of induced current by Lenz' law.
How does it determine the direction of the induced current by Lenz' law? Will the direction of the new induced current always oppose the direction of the original current prior to increasing the resistance?
Post by: silverpixeli on May 24, 2015, 03:39:35 pm
Thanks for responding silverpixeli
How does it determine the direction of the induced current by Lenz' law? Will the direction of the new induced current always oppose the direction of the original current prior to increasing the resistance?
Actually Lenz' law says that it will oppose the change of field brought about by the increased R/reduced I.
Say I in the outer loop is clockwise, then by the right hand rule the field due to I on the inside of the outer loop will be into the page.
When I decreases (R increases) this field still points into the page, but it now has less magnitude. Keep in mind that magnetic field is a vector. So which was way this field changing? It's becoming less into the page so the direction of the change is out of the page (think about it. consider
So is the change in field is out of the page, the change in flux in the inner loop is out of the page. Flux is in the same direction as field, so this step of the argument is trivial and just comes about because I was talking about field before, now I'm talking about flux, which is just field scaled by area.
Lenz' law says that the induced current in the loop will oppose this change in flux. i.e. it will be directed so that it induces its own field to counter the change. Since the change is out of the page, we want a current that produces a field into the page. By the right hand rule again, that's a clockwise current.
This argument has a few steps but it's the kind of argument you go through for every single induction question, so it's worth taking the time to understand it.
Post by: odeaa on May 24, 2015, 06:42:48 pm
Post by: silverpixeli on May 24, 2015, 08:22:35 pm
Hey guys, what is the best scientific calculator for physics? I have been using my green ti from year 7,but I find it sucks for gravity questions because you can't see the whole screen
I'm a big fan of the Canon F-717SGA. There's also a Casio going around with a few extra letters you can use to store constants like G and h and stuff.
Post by: paper-back on May 24, 2015, 09:49:45 pm
Post by: alchemy on May 27, 2015, 09:47:29 pm
Hey guys, what is the best scientific calculator for physics? I have been using my green ti from year 7,but I find it sucks for gravity questions because you can't see the whole screen
#greenti4lyf.....
I find that it does most of the things I want it to for vce physics.
As for not being able to see the whole screen, that's why there's an up button :P
Post by: Adequace on June 04, 2015, 03:14:55 pm
I've heard it'll be a bonus question for our mid-year exam and we need to persuade our teacher on what it is, but I haven't come to a conclusion lol.
Post by: Adequace on June 04, 2015, 07:11:44 pm
Edit: Would like help with the second picture as well, it relates to the same circuit. I think I got them wrong.
Post by: silverpixeli on June 04, 2015, 08:25:36 pm
Legitimate question now, for Q7. I'm not sure if I got it current but would it be 0A since the diode is in reverse bias or still 0.060A?
Edit: Would like help with the second picture as well, it relates to the same circuit. I think I got them wrong.
The diode is not in reverse bias mode in the first picture. When they turn it around, then it is reverse biased and then there's no current and that's why the light goes out. This answers the last question.
When a diode's triangle is pointing the same way as conventional current, it's in forward bias mode. Conventional current always comes out of the longer side of a battery symbol, as this represents the positive terminal.
Post by: Adequace on June 04, 2015, 09:12:03 pm
The diode is not in reverse bias mode in the first picture. When they turn it around, then it is reverse biased and then there's no current and that's why the light goes out. This answers the last question.In the image the cells are placed in different directions, so it looks to be that both ends have a longer side? I'm not sure if the question is supposed to be some sort of trick question.
When a diode's triangle is pointing the same way as conventional current, it's in forward bias mode. Conventional current always comes out of the longer side of a battery symbol, as this represents the positive terminal.
Post by: Floatzel98 on June 04, 2015, 09:31:19 pm
In the image the cells are placed in different directions, so it looks to be that both ends have a longer side? I'm not sure if the question is supposed to be some sort of trick question.yeah, that cell on the left side, being the other way is a bit confusing. Is there a reason why it is like that? Because looking at the diagram alone, i might have actually said that that the direction of current was going from through ammeter 1 first (left to right) which places the diode in reverse bias. Plus you might think ammeter one and 2 would go in order, like they would place ammeter 1 where the current first comes out, and ammeter 2 after that at the end.
Also Adequace, what do you mean with your previous questions about radiation?
Post by: Adequace on June 04, 2015, 09:37:18 pm
yeah, that cell on the left side, being the other way is a bit confusing. Is there a reason why it is like that? Because looking at the diagram alone, i might have actually said that that the direction of current was going from through ammeter 1 first (left to right) which places the diode in reverse bias. Plus you might think ammeter one and 2 would go in order, like they would place ammeter 1 where the current first comes out, and ammeter 2 after that at the end.It's just about the effects of exposure to radiation, nothing too important though!
Also Adequace, what do you mean with your previous questions about radiation?
Post by: silverpixeli on June 04, 2015, 10:26:08 pm
This is an interesting electricity question relying on some nuances of Kirchoff's voltage laws that aren't really studied in year 12, so while it may be in your current course Adequace, it's certainly not something that other people should be worried about.
Long story short, one of the batteries being reversed just takes away from the total voltage that is being supplied and it's as if were were only 2 batteries there.
Post by: silverpixeli on June 04, 2015, 10:30:28 pm
It's just about the effects of exposure to radiation, nothing too important though!
From this page at gizmodo
Quote
How you would die
Large doses of ionizing radiation in a short time period lead to Acute Radiation Syndrome (ARS), aka radiation poisoning. The severity of ARS symptoms depends on the level of exposure. A radiation dose as low as 0.35 Gy could feel a bit like you have the flu—expect nausea and vomiting, headaches, fatigue, and fever. If the body is exposed to a higher dose, somewhere between 1-4 Gy, blood cells begin to die. You could still recover—treatment of this kind of radiation syndrome usually involves blood transfusions and antibiotics—but you could also suffer a weakened immune response due to a drop in white cell count, uncontrollable bleeding due to a lack of platelets, and anemia due to a reduction of red blood cells. You'll also notice a kind of odd sunburn if exposed to 2 Gy or more of ionizing radiation. Technically referred to as acute radiodermatitis, its effects include red patches, peeling skin, and sometimes blistering. Expect it to show up within 24 hours........ more
It goes on to detail the actual death... Scary stuff.
Based on this, I would say that the radiation's symptoms are death because it basically tears up the stuff that makes up your body and that causes everything to stop functioning properly and you die.
Post by: Adequace on June 04, 2015, 10:36:20 pm
Post by: silverpixeli on June 04, 2015, 10:52:12 pm
would you guys consider death an effect of radiation or the result of the symptoms of radiation?
I'm calling it a symptom of the radiation (aka an effect of the radiation) rather than an effect of its symptoms but like this is a weird distinction to make. It's definitely not like high/sudden bursts of radiation give you some other disease which then kills you, they literally tear your atoms apart. That being said, long exposure to radiation causes increased genetic mutations in new cells and that leads to cancer which then takes over and kills you too, so in that case I guess death is the result of a symptom.
But yeah this is a weird question to answer because it's a fine line in my understanding between what actually kills you.
Post by: Adequace on June 08, 2015, 08:27:50 pm
I've started unit 2 motion by myself and found myself confused with an answer/s. I've seen some answers of questions when the units have ^-2 instead of a ^-1, what does the ^-2 mean? I assume ^-1 means per 'sec/min/etc' but not sure on ^-2
Post by: Cosec on June 08, 2015, 09:02:01 pm
Sorry for asking another question.
I've started unit 2 motion by myself and found myself confused with an answer/s. I've seen some answers of questions when the units have ^-2 instead of a ^-1, what does the ^-2 mean? I assume ^-1 means per 'sec/min/etc' but not sure on ^-2
Exactly right. ^-1 means something/something. Stuff like meters/second (meters per second) can be written as ms^-1.
^-2 is similar. Most commonly used would be acceleration. In whihc case is something/something^2. Acceleration is meters/second/second (meters per second per second). And can be written as ms^-2. Or m/s^2
Post by: Floatzel98 on June 16, 2015, 05:57:18 pm
Firstly, i think my main problem is not understanding how a magnet works. Like what is flowing around a magnetic field? Electrons? What makes them move around exactly? What actually gives a magnet it's magnetism? More in the sense of a permanent magnet? How does this relate to the poles of a magnet. Is it what is inside a magnet that makes it magnetic, or does it relate more to it's magnetic field? What is different between the two poles and how can atoms/electrons(??) in a magnet form different poles. Does it stop and end at a certain point in the middle? I guess one of my questions is, what exactly is the field made out of and/or does anything flow in the field? How is this different/the same as a gravitational field?
I think this might tie into what we learn later, but my textbook says that a piece of iron will have magnetism induced in it when placed in an external magnetic field. Is this specific to Iron? Will it create it's own magnetic field or just use the other one, and how exactly does any of that work? My book explains that it happens but says it is somehow induced the piece of iron to become a magnet?
What is the difference between the grip and palm rule? I feel like they are doing the same thing. Well i understand the grip rule; your thumb shows the direction of current in a wire and your fingers will show the direction of the magnetic field. While the palm rule does the same then but then also shows the force? My book says that it gives the direction of the force on a current carrying wire placed in an external field? Does this mean it always involve >2 magnets pulling each other?
This one might be a bit hard for me to explain, but in a simple rectangular magnet, we always see that the fields always come from the two poles. And i know these are just illustrations, but how many field 'lines can actually come out of a magnet? Are there actually space in between all of them or do they fill up all the space it can? How much distance is between each field 'line'? Do all the field 'lines' come back to the south pole of the same magnet? Sometimes it looks as if there is a straight line coming out of the top of the magnet, where does this go. Does it just stop? Also, probably just bounded by book illustrations, but do all the field lines go out in all directions. Does it form an 'oval' shape around the entire magnet? And then would a magnetic field around a length of wire form a cylindrical shape around the wire? This might not make sense because i might be totally confused and wrong.
In my book there is an equation for the strength of an electric field written as
Lastly, i got up to the force on a wire carrying a charge in a magnetic field. Probably still relating to the palm rule question, where and what is this force exactly? Secondly I'm a bit confused with the equation. We have
Spoiler
(http://i.imgur.com/qP3kfgG.jpg?1)
Really lastly, my book has a part talking about how we express the relationship between the 3 vectors Force, Length and Field though vector cross multiplication. It says that 'if we write F = Il x B this is taken to mean that the magnitude of vector F is the product IlBsin(theta)'. I know this might be a bit more maths now, but how does the cross product work to get this, and is this the same way they find Torque as well?
I don't expect anyone to answer every question, but thanks to anyone that can help at all. I did try to find some of my answers elsewhere before looking here, but i didn't really find anything that helped.
Post by: odeaa on June 16, 2015, 08:18:17 pm
Hey guys, i finally started UNIT 4 AOS 1 and I am a bit confused on somethings.
Firstly, i think my main problem is not understanding how a magnet works. Like what is flowing around a magnetic field? Electrons? What makes them move around exactly? What actually gives a magnet it's magnetism? More in the sense of a permanent magnet? How does this relate to the poles of a magnet. Is it what is inside a magnet that makes it magnetic, or does it relate more to it's magnetic field? What is different between the two poles and how can atoms/electrons(??) in a magnet form different poles. Does it stop and end at a certain point in the middle? I guess one of my questions is, what exactly is the field made out of and/or does anything flow in the field? How is this different/the same as a gravitational field?
I think this might tie into what we learn later, but my textbook says that a piece of iron will have magnetism induced in it when placed in an external magnetic field. Is this specific to Iron? Will it create it's own magnetic field or just use the other one, and how exactly does any of that work? My book explains that it happens but says it is somehow induced the piece of iron to become a magnet?
What is the difference between the grip and palm rule? I feel like they are doing the same thing. Well i understand the grip rule; your thumb shows the direction of current in a wire and your fingers will show the direction of the magnetic field. While the palm rule does the same then but then also shows the force? My book says that it gives the direction of the force on a current carrying wire placed in an external field? Does this mean it always involve >2 magnets pulling each other?
This one might be a bit hard for me to explain, but in a simple rectangular magnet, we always see that the fields always come from the two poles. And i know these are just illustrations, but how many field 'lines can actually come out of a magnet? Are there actually space in between all of them or do they fill up all the space it can? How much distance is between each field 'line'? Do all the field 'lines' come back to the south pole of the same magnet? Sometimes it looks as if there is a straight line coming out of the top of the magnet, where does this go. Does it just stop? Also, probably just bounded by book illustrations, but do all the field lines go out in all directions. Does it form an 'oval' shape around the entire magnet? And then would a magnetic field around a length of wire form a cylindrical shape around the wire? This might not make sense because i might be totally confused and wrong.
In my book there is an equation for the strength of an electric field written as. What exactly is the constant K? Is it referring to, like if the magnet was is space or in air or underwater?
Lastly, i got up to the force on a wire carrying a charge in a magnetic field. Probably still relating to the palm rule question, where and what is this force exactly? Secondly I'm a bit confused with the equation. We havewhich I understand, but then we 'derive' it to
. I understand that the B force needs to be perpendicular to the current from the pictures in my textbook, but i don't understand how we find B perpendicular. I also got confused with this in Torque when i did the detailed study.
Is this picture a correct way of doing it? Can we just use cos as well? Sometimes i feel i can just use cos of an angle and it makes it easier.Spoiler(http://i.imgur.com/qP3kfgG.jpg?1)
Really lastly, my book has a part talking about how we express the relationship between the 3 vectors Force, Length and Field though vector cross multiplication. It says that 'if we write F = Il x B this is taken to mean that the magnitude of vector F is the product IlBsin(theta)'. I know this might be a bit more maths now, but how does the cross product work to get this, and is this the same way they find Torque as well?
I don't expect anyone to answer every question, but thanks to anyone that can help at all. I did try to find some of my answers elsewhere before looking here, but i didn't really find anything that helped.
I know this probably doesnt help, but unfortunately no-one really knows the answer to half those questions ahah
While we have explanantions for almost everything, these are based on assumptions, and the deeper you go in physics the more you realise how little we actually know as 100% true. For the purposes of VCE, we just take all these assumptions to be true
I'll let someone else with more knowledge (probs silverpixeli lol) answer the questions, but just thought I would chime in with my two cents
Post by: lzxnl on June 16, 2015, 08:23:10 pm
Hey guys, i finally started UNIT 4 AOS 1 and I am a bit confused on somethings.
Firstly, i think my main problem is not understanding how a magnet works. Like what is flowing around a magnetic field? Electrons? What makes them move around exactly? What actually gives a magnet it's magnetism? More in the sense of a permanent magnet? How does this relate to the poles of a magnet. Is it what is inside a magnet that makes it magnetic, or does it relate more to it's magnetic field? What is different between the two poles and how can atoms/electrons(??) in a magnet form different poles. Does it stop and end at a certain point in the middle? I guess one of my questions is, what exactly is the field made out of and/or does anything flow in the field? How is this different/the same as a gravitational field?
I think this might tie into what we learn later, but my textbook says that a piece of iron will have magnetism induced in it when placed in an external magnetic field. Is this specific to Iron? Will it create it's own magnetic field or just use the other one, and how exactly does any of that work? My book explains that it happens but says it is somehow induced the piece of iron to become a magnet?
What is the difference between the grip and palm rule? I feel like they are doing the same thing. Well i understand the grip rule; your thumb shows the direction of current in a wire and your fingers will show the direction of the magnetic field. While the palm rule does the same then but then also shows the force? My book says that it gives the direction of the force on a current carrying wire placed in an external field? Does this mean it always involve >2 magnets pulling each other?
This one might be a bit hard for me to explain, but in a simple rectangular magnet, we always see that the fields always come from the two poles. And i know these are just illustrations, but how many field 'lines can actually come out of a magnet? Are there actually space in between all of them or do they fill up all the space it can? How much distance is between each field 'line'? Do all the field 'lines' come back to the south pole of the same magnet? Sometimes it looks as if there is a straight line coming out of the top of the magnet, where does this go. Does it just stop? Also, probably just bounded by book illustrations, but do all the field lines go out in all directions. Does it form an 'oval' shape around the entire magnet? And then would a magnetic field around a length of wire form a cylindrical shape around the wire? This might not make sense because i might be totally confused and wrong.
In my book there is an equation for the strength of an electric field written as
Lastly, i got up to the force on a wire carrying a charge in a magnetic field. Probably still relating to the palm rule question, where and what is this force exactly? Secondly I'm a bit confused with the equation. We haveIs this picture a correct way of doing it? Can we just use cos as well? Sometimes i feel i can just use cos of an angle and it makes it easier.Spoiler(http://i.imgur.com/qP3kfgG.jpg?1)
Really lastly, my book has a part talking about how we express the relationship between the 3 vectors Force, Length and Field though vector cross multiplication. It says that 'if we write F = Il x B this is taken to mean that the magnitude of vector F is the product IlBsin(theta)'. I know this might be a bit more maths now, but how does the cross product work to get this, and is this the same way they find Torque as well?
I don't expect anyone to answer every question, but thanks to anyone that can help at all. I did try to find some of my answers elsewhere before looking here, but i didn't really find anything that helped.
Firstly, it's good that you're looking deeper into course material. However, you'll find that the more you look into VCE physics, the more you'll be confused as a better understanding of electromagnetic fields is...not expected of high schoolers.
Now to attempt to answer these questions. Magnetic fields are better thought of as disturbances that charges interact with. It's a bit like a gravitational field in this regard. However, the gravitational field as a vector is directed parallel to the direction of the force it would exert. A magnetic field's direction is the direction of zero torque on a magnetic dipole (i.e. the direction a compass needle would point).
Magnetic fields are, according to relativity, a different form of electric field seen by an observer moving relative to the charge. In other words, electric charges set up electric fields and anyone moving with respect to the electric charge sees a magnetic field. From our perspective, therefore, a moving charge, or a current, generates a magnetic field. Another way of setting up a magnetic field is to possess an intrinsic magnetic moment (think of this like a compass needle; has a direction). Any magnetic substance (I'm not talking about diamagnetism here) has magnetism due to their electronic structure. Paramagnetic materials have unpaired electrons. A good example would be NO2. Things like iron are magnetic because their electrons, who have their own magnetic moments, all line up so you get a macroscopic magnetic moment -> permanent magnet.
So, iron in a magnetic field does not use the other magnetic field. The external magnetic field lines up the electron magnetic moments, giving the iron its own magnetic field which will persist even after the external field is removed
The grip and palm rules are used for different things. One finds the force from a current and magnetic field. One finds the magnetic field from a current and vice versa. They are all related to vector cross products though. And yes, magnetic forces require two magnets, or one magnet in a magnetic field generated by some means.
Field lines are a mathematical construct. They are essentially just a vector field, which is a function whose inputs are x, y and z coordinates and the output is a vector. How many points are there in a parabola? That question is equally meaningless as how many field lines there can be.
As for field line shapes, they must always form closed loops. This is a fundamental law of electromagnetism known as Gauss's law for magnetism.
You are right about the cylindrical symmetry of a magnetic field around a straight wire. They form circles/cylinders (depending on how you want to think about them).
The magnetic field strength equation you have is for the magnetic field around a wire. If it is indeed the field due to a current-carrying wire, then K is equal to the permeability constant mu divided by 2pi. It may change depending on the medium though.
As for the force on a wire, this force can actually be explained in terms of relativity but I won't go into the details. F = IlB only works if the force is exactly perpendicular to the magnetic field. F = il x B (vector cross product) and the sin theta comes from the magnitude of a cross product. Torque is also defined via a cross product so it might be good to read up on those. With cross products and dot products, the angle is ALWAYS between the two vectors. No exception.
The fact that there is a sin theta in the magnitude of the cross product is a definition. You have to just live with it unfortunately :P
I know this probably doesnt help, but unfortunately no-one really knows the answer to half those questions ahah
While we have explanantions for almost everything, these are based on assumptions, and the deeper you go in physics the more you realise how little we actually know as 100% true. For the purposes of VCE, we just take all these assumptions to be true
I'll let someone else with more knowledge (probs silverpixeli lol) answer the questions, but just thought I would chime in with my two cents
Don't be TOO discouraging for the asker :P
Post by: odeaa on June 16, 2015, 08:33:29 pm
Don't be TOO discouraging for the asker :P
I feel like a dick now lol
Post by: Floatzel98 on June 16, 2015, 08:44:02 pm
Firstly, it's good that you're looking deeper into course material. However, you'll find that the more you look into VCE physics, the more you'll be confused as a better understanding of electromagnetic fields is...not expected of high schoolers.Thanks so much for that response lzxnl. It really helps. I think i will be able to just move on with the rest of the course now since all that information will tie me over and I don't need to confuse myself any further.
Now to attempt to answer these questions. Magnetic fields are better thought of as disturbances that charges interact with. It's a bit like a gravitational field in this regard. However, the gravitational field as a vector is directed parallel to the direction of the force it would exert. A magnetic field's direction is the direction of zero torque on a magnetic dipole (i.e. the direction a compass needle would point).
Magnetic fields are, according to relativity, a different form of electric field seen by an observer moving relative to the charge. In other words, electric charges set up electric fields and anyone moving with respect to the electric charge sees a magnetic field. From our perspective, therefore, a moving charge, or a current, generates a magnetic field. Another way of setting up a magnetic field is to possess an intrinsic magnetic moment (think of this like a compass needle; has a direction). Any magnetic substance (I'm not talking about diamagnetism here) has magnetism due to their electronic structure. Paramagnetic materials have unpaired electrons. A good example would be NO2. Things like iron are magnetic because their electrons, who have their own magnetic moments, all line up so you get a macroscopic magnetic moment -> permanent magnet.
So, iron in a magnetic field does not use the other magnetic field. The external magnetic field lines up the electron magnetic moments, giving the iron its own magnetic field which will persist even after the external field is removed
The grip and palm rules are used for different things. One finds the force from a current and magnetic field. One finds the magnetic field from a current and vice versa. They are all related to vector cross products though. And yes, magnetic forces require two magnets, or one magnet in a magnetic field generated by some means.
Field lines are a mathematical construct. They are essentially just a vector field, which is a function whose inputs are x, y and z coordinates and the output is a vector. How many points are there in a parabola? That question is equally meaningless as how many field lines there can be.
As for field line shapes, they must always form closed loops. This is a fundamental law of electromagnetism known as Gauss's law for magnetism.
You are right about the cylindrical symmetry of a magnetic field around a straight wire. They form circles/cylinders (depending on how you want to think about them).
The magnetic field strength equation you have is for the magnetic field around a wire. If it is indeed the field due to a current-carrying wire, then K is equal to the permeability constant mu divided by 2pi. It may change depending on the medium though.
As for the force on a wire, this force can actually be explained in terms of relativity but I won't go into the details. F = IlB only works if the force is exactly perpendicular to the magnetic field. F = il x B (vector cross product) and the sin theta comes from the magnitude of a cross product. Torque is also defined via a cross product so it might be good to read up on those. With cross products and dot products, the angle is ALWAYS between the two vectors. No exception.
The fact that there is a sin theta in the magnitude of the cross product is a definition. You have to just live with it unfortunately :P
Don't be TOO discouraging for the asker :P
Also, thanks Odeaa. Most of those questions are just ramblings in my head i feel i need answers too. I just try to connect everything for myself so I can understand it. But obviously a lot of these things are out of VCE, so your right, i shouldn't try to question everything too much. Don't feel bad :P
Thanks guys :)
Post by: paper-back on June 18, 2015, 07:46:16 pm
Post by: lzxnl on June 18, 2015, 08:20:24 pm
Why does the EMF graph get cut off when using a split-ring commutator? whereas when using an AC slip ring commutator, it goes the entire length?
The split ring commutator connects the wire loop in the magnetic field to a different battery terminal. Every half a revolution the wire loop's polarity switches. Hence the voltage/current direction changes sharply.
Post by: paper-back on June 18, 2015, 09:04:33 pm
The split ring commutator connects the wire loop in the magnetic field to a different battery terminal. Every half a revolution the wire loop's polarity switches. Hence the voltage/current direction changes sharply.
But doesn't using AC current in the loop also mean that there is a change in the direction of the current? Why isn't the graph when using slip rings also cut off?
Post by: odeaa on June 18, 2015, 09:17:10 pm
With the commutator, it is always positive, as the direction of the induced emf is always in the same direction, although it still cycles with the flux from maximum to minimum every quarter turn
hope that makes sense, its a hard topic to explain in writing
Post by: paper-back on June 21, 2015, 04:27:50 pm
E.g. for a sine flux graph, we draw a negative cosine EMF graph and,
for a cosine flux graph, we draw a positive sine EMF graph?
Post by: silverpixeli on June 21, 2015, 04:29:47 pm
When drawing EMF graphs from the magnetic flux graphs, do we draw the negative gradient function of the magnetic flux graph?yep sounds good! this isnt really assessed though (but anything goes on your sac)
E.g. for a sine flux graph, we draw a negative cosine EMF graph and,
for a cosine flux graph, we draw a positive sine EMF graph?
to determine the direction that current actually flows in a circuit (and then make sense of which way you have defined positive to be on your graph) you'll need to think through lenz' law for the situation
Post by: paper-back on June 21, 2015, 04:57:41 pm
Post by: dankfrank420 on June 21, 2015, 07:45:23 pm
Well, what exactly are we learning about in electromagnetism? Are we barely scratching the surface and not really learning anything at all?
I've been able to do many past questions without actually having a fundamental understanding of what electromagnetism exactly is. Why is this fundamental understanding not even taught or even needed?
Post by: lzxnl on June 21, 2015, 08:01:10 pm
When drawing EMF graphs from the magnetic flux graphs, do we draw the negative gradient function of the magnetic flux graph?
E.g. for a sine flux graph, we draw a negative cosine EMF graph and,
for a cosine flux graph, we draw a positive sine EMF graph?
Technically, yes. It's meant to be a derivative but welcome to VCE physics...where year 9 maths is all that's required...
But doesn't using AC current in the loop also mean that there is a change in the direction of the current? Why isn't the graph when using slip rings also cut off?
A slip ring commutator does nothing except maintain the electrical connections. A split ring commutator emf graph looks the same as a slip ring commutator graph, except every half revolution the voltage changes sign. Mathematically, this sign change corresponds to a change in sign of a sine function, meaning your sine function (the voltage) is always positive. To represent this by a function, the voltage with a split ring commutator is |a sin bt| where a and b are constants and the voltage with a slip ring commutator is just a sin bt. No mod signs.
Starting to get into electromagnetism for Physics, and the amount of times that the words "but you don't have to know about that, it's not on the course" is frustrating.
Well, what exactly are we learning about in electromagnetism? Are we barely scratching the surface and not really learning anything at all?
I've been able to do many past questions without actually having a fundamental understanding of what electromagnetism exactly is. Why is this fundamental understanding not even taught or even needed?
If you want a fully blown and completely blunt rant, PM me and I'll be more than happy to give you one on how inadequate the VCE physics course is. Yes, you essentially don't learn anything about electromagnetism in VCE physics. In a nutshell, this is essentially all you do in AoS 3. Which I've taught to my physics student in around three lessons. Maybe four.
Magnetic fields of bar magnets, right hand rules for currents/magnetic fields of wires/loops, induction, Lenz's law, transformers, magnetic forces on wires, motors/generators
Have I missed out anything that doesn't fit into one of those categories? It's not a huge list. Here is a short list of important bits of electromag that you should get but aren't getting.
1. What is magnetism
2. What is a field
3. Electric fields (if you're going to learn about magnetic fields, why not electric fields)
4. Energy of things in electric/magnetic fields (i.e. why is a compass needle stable when it points in the same direction as the magnetic field)
5. When things are NOT perpendicular or parallel
I mean VCAA cut out one of the detailed studies so I don't see why they don't put some content in so that students like you aren't completely confused about everything
Post by: dankfrank420 on June 21, 2015, 08:56:29 pm
I finally see where you and pi and all the others are coming from, VCE Physics is a joke.
Post by: odeaa on June 21, 2015, 09:07:15 pm
Post by: odeaa on June 21, 2015, 09:10:10 pm
Post by: paper-back on June 22, 2015, 08:22:56 pm
When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?
Post by: odeaa on June 22, 2015, 09:16:24 pm
Thanks lzxnl
When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?
Wouldnt the average emf of a full turn be 0, because it goes from positive to negative? that being said i think i've done a similar question and gotten it wrong ahah
Post by: lzxnl on June 22, 2015, 11:40:19 pm
Wouldnt the average emf of a full turn be 0, because it goes from positive to negative? that being said i think i've done a similar question and gotten it wrong ahah
It is zero, because of the symmetry of the sine function. If it rotates at constant speed, the emf/time relation is sinusoidal.
Thanks lzxnl
When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?
You're finding the average EMF, which is the average rate of change of the flux, which is the change in flux over the time interval.
Post by: paper-back on June 23, 2015, 05:48:08 pm
Post by: Adequace on June 28, 2015, 09:36:41 pm
(Sorry about posting this here, but I think it's best to post it here since motion overlaps)
Post by: Maths Forever on June 29, 2015, 06:35:04 pm
I've started Unit 2 motion by myself and I'm not getting the answer for this question in the textbook. Part b. of the attached image,it looks rather straight forward but I'm not entirely sure what the first step is.
(Sorry about posting this here, but I think it's best to post it here since motion overlaps)
Here is what I think:
For part (a), a quick prediction would be 90 km/h, since it is half way between 80 km/h (from Melbourne to Wodonga) and 100 km/h (back from Wodonga to Melbourne).
For part (b), find the total distance and total time covered throughout the journey.
Total Distance = 300 km + 300 km = 600 km
Now speed = distance / time, so time = distance / speed
Time covered between Melbourne to Wodonga = (300 km) / (80 km/h) = 3.75 hours
Time covered between Wodonga to Melbourne = (300 km) / (100 km/h) = 3.00 hours
Therefore, total time = 3.75 hours + 3.00 hours = 6.75 hours
Hence, average speed = total distance / total time = 600 km / 6.75 h = 89 km/h (to 2 significant figures)
Reason for difference in predicted and calculated speeds:
90 km/h would be correct if we were considering a case where the initial speed was 80 km/h and the final speed was 100 km/h (i.e. using the formula v (av) = [u + v] / 2.
However, this is not the case. We are told the average speed for one interval of the journey, and the average speed for another interval. So the formula v (av) = Δ x / Δ t must be used, which yields 89 km/h.
I hope this helps! :)
Post by: Adequace on June 29, 2015, 06:54:23 pm
Here is what I think:Thanks for the reply, greatly appreciated!
For part (a), a quick prediction would be 90 km/h, since it is half way between 80 km/h (from Melbourne to Wodonga) and 100 km/h (back from Wodonga to Melbourne).
For part (b), find the total distance and total time covered throughout the journey.
Total Distance = 300 km + 300 km = 600 km
Now speed = distance / time, so time = distance / speed
Time covered between Melbourne to Wodonga = (300 km) / (80 km/h) = 3.75 hours
Time covered between Wodonga to Melbourne = (300 km) / (100 km/h) = 3.00 hours
Therefore, total time = 3.75 hours + 3.00 hours = 6.75 hours
Hence, average speed = total distance / total time = 600 km / 6.75 h = 89 km/h (to 2 significant figures)
Reason for difference in predicted and calculated speeds:
90 km/h would be correct if we were considering a case where the initial speed was 80 km/h and the final speed was 100 km/h (i.e. using the formula v (av) = [u + v] / 2.
However, this is not the case. We are told the average speed for one interval of the journey, and the average speed for another interval. So the formula v (av) = Δ x / Δ t must be used, which yields 89 km/h.
I hope this helps! :)
Edit: your answer is correct but I don't quite understand why you need to use the total distance / the total time taken?
I interpreted the question as the return leg of the journey only, not both legs combined? Did I just interpret the question wrong?
Post by: Floatzel98 on June 29, 2015, 08:21:28 pm
Thanks for the reply, greatly appreciated!Yeah, the way they ask it does seem weird. I'm guessing they meant to write calculate the average speed for the whole return journey, like they worded it in part a. Otherwise you wouldn't need to do much calculating because they give you the average speed of the return trip haha.
Edit: your answer is correct but I don't quite understand why you need to use the total distance / the total time taken?
I interpreted the question as the return leg of the journey only, not both legs combined? Did I just interpret the question wrong?
You need to use Δx / Δt because you aren't dealing with constant acceleration. When you get up to that, hopefully a bit more will make sense. If you look at this link, and read all the way down to the 3rd box, it might help answer your question.
Post by: stockstamp on June 30, 2015, 11:14:14 pm
An F1 racing car is travelling around a banked corner. The angle of the banking is 15o and the radius of the circular corner is 300m. The mass of the car is 850 kg.
If the friction between the road and car's tyres can apply a maximum sideways acceleration of 10 ms-2 parallel to the slope of the road, how fast can the car travel around the corner now before sliding off the road?
Not sure I completely understand the wording of the question, and therefore don't know how to solve it.
Any help appreciated :)
Post by: silverpixeli on July 02, 2015, 11:38:30 am
Thanks!
Post by: odeaa on July 02, 2015, 02:47:49 pm
Hey guys, I'm taking over the physics lecture next Wednesday for Alwin and I'd love to get an idea for where you guys are up to in class. So to anyone who is coming along (or even if you're not but you feel like answering - btw there are still tickets available) please let me know where your class is at right now.
Thanks!
Electric power sac first week back next term
Post by: Floatzel98 on July 02, 2015, 02:58:47 pm
Hey guys, I'm taking over the physics lecture next Wednesday for Alwin and I'd love to get an idea for where you guys are up to in class. So to anyone who is coming along (or even if you're not but you feel like answering - btw there are still tickets available) please let me know where your class is at right now.We are also up to Electric power. We just went over magnetic flux before the holidays.
Thanks!
Post by: Floatzel98 on July 08, 2015, 06:52:20 pm
And with motors and generators, the difference is one moves with a current through it and then one moves/gets moved to make a current? How do you increase current produced from a generator and how high can it get? .Why don't cars make use of those generator principles? I might be correct to say I'm probably not the first person to suggest that though....
Also, with magnetic flux, how exactly is a current and EMF induced induced? Does the magnetic field like use the force and bring electrons up from the ground and put them into the wire? Also, since the rate of change of flux is equal to the induced EMF, how do you get there? Can you derive
With V(RMS) or I(RMS), is it just supposed to be the positive side of the graph because it is DC or is is positive and negative values of it?
Thanks :)
Post by: silverpixeli on July 08, 2015, 07:08:26 pm
Just had a question, when we are looking at pictures like these (attached), is the magnetic field one created from the current carrying wire or has a current carrying wire been placed into a magnetic field? I was just a bit confused as to how you can actually cause a magnetic field to not be perpendicular to the wire (As in the 3rd and 4th diagrams). Is it just maybe interacting with another magnetic field and that changes the direction of the original one?
And with motors and generators, the difference is one moves with a current through it and then one moves/gets moved to make a current? How do you increase current produced from a generator and how high can it get? .Why don't cars make use of those generator principles? I might be correct to say I'm probably not the first person to suggest that though....
Also, with magnetic flux, how exactly is a current and EMF induced induced? Does the magnetic field like use the force and bring electrons up from the ground and put them into the wire? Also, since the rate of change of flux is equal to the induced EMF, how do you get there? Can you deriveor something? I know it is probably more complicated but I feel it involes differentiation somehow since it is the rate of change of the flux.
With V(RMS) or I(RMS), is it just supposed to be the positive side of the graph because it is DC or is is positive and negative values of it?
Thanks :)
all good questions i am on my phone right now but i will attempt to start explaining
one, its an external field, the interaction between this external field and the one i didnt draw (caused by wire) is what causes the force
two, yes thats right, and cars only need motion and they get a lot of torque from a combustion engine that injects fuel and turns a piston turning the wheels
cars also generally charge their battery by using these pistons to turn a generator too
the production of emf is just one of the cool things about electromagnetism, youll have to consult someone like lzxnl who payed attention in first year physics and is probably studying electromag this semester to fill you in on more details and maybe the derivation
or im sure you could find it online, though maybe a bit above the reding level of a vce student, just look up faraday's law
vrms and irms are the 'root mean square' of the ac signals, they are arrived at by squaring the sine wave and finding the average value of this and then sqrt the result (like standard deviation from methods probability)
this takes care of the negatives (obviously the regular average of a plain old sine wave is just 0) because when you're talking power you dont usually care about direction
Post by: lzxnl on July 08, 2015, 10:27:41 pm
As for Faraday's law, its proper statement is indeed a derivative; the emf (aka a closed line integral of the electric field about some curve; a fancy way of defining emf) is equal to the negative total time derivative of the magnetic flux. This comes from a law that you sort of can't prove (the Maxwell-Faraday equation which relates the electric field generated to the negative partial derivative of the magnetic field) you need a foundation for every field of physics and this is one of them for electromagnetism. How do you prove F = ma in mechanics? You can't. That's essentially a definition of force.
If you want an explanation of why an EMF is induced, I'm not exactly too sure on that myself but it probably has to do with the fact that according to special relativity, electric and magnetic fields are aspects of the same phenomenon. This last point is easy to explain though.
Imagine two people, A and B. A is moving with a test particle P at velocity v, so A is in P's rest frame. B is in the lab frame, seeing P move at v. Let's put a magnetic field now in the lab frame. Assume P is moving at a constant velocity. Two observers that are both in inertial frames (non-accelerating) have to see the same forces acting. In B's frame, P is moving at speed v in a magnetic field -> there is a magnetic force on P. A must also see some force acting on P, but A is moving with P, so A sees P as stationary. The force acting on P is thus not magnetic, but electric. A must see an electric field instead of a magnetic field. This electric field is a special type of electric field for reasons I won't go into here.
But yeah, magnetism and electricity are closely intertwined.
Post by: Floatzel98 on July 09, 2015, 01:35:59 pm
Flux = BA is a definition of magnetic flux. Which only actually holds true for the very specific scenario where the magnetic field is uniform and is pointing parallel to the surface normal.Thanks guys. Do you guys have any other resources/videos on special relativity, because i have been reading up on the basics of it for a while like the detailed study in the physics textbook, I've read a Brief History of Time and a couple other books but i don't understand it too much. I know it's a hard topic to get your head around though. Thanks
As for Faraday's law, its proper statement is indeed a derivative; the emf (aka a closed line integral of the electric field about some curve; a fancy way of defining emf) is equal to the negative total time derivative of the magnetic flux. This comes from a law that you sort of can't prove (the Maxwell-Faraday equation which relates the electric field generated to the negative partial derivative of the magnetic field) you need a foundation for every field of physics and this is one of them for electromagnetism. How do you prove F = ma in mechanics? You can't. That's essentially a definition of force.
If you want an explanation of why an EMF is induced, I'm not exactly too sure on that myself but it probably has to do with the fact that according to special relativity, electric and magnetic fields are aspects of the same phenomenon. This last point is easy to explain though.
Imagine two people, A and B. A is moving with a test particle P at velocity v, so A is in P's rest frame. B is in the lab frame, seeing P move at v. Let's put a magnetic field now in the lab frame. Assume P is moving at a constant velocity. Two observers that are both in inertial frames (non-accelerating) have to see the same forces acting. In B's frame, P is moving at speed v in a magnetic field -> there is a magnetic force on P. A must also see some force acting on P, but A is moving with P, so A sees P as stationary. The force acting on P is thus not magnetic, but electric. A must see an electric field instead of a magnetic field. This electric field is a special type of electric field for reasons I won't go into here.
But yeah, magnetism and electricity are closely intertwined.
Post by: Cosec on July 09, 2015, 04:34:40 pm
1. When answers questions with regard RMS. Such as the power output when the voltage and current are given as RMS values. What is the expect answer? As in, should they be used as RMS values or converted. What happens if your given a voltage in RMS and a current in peak. To you give your answer as a RMS or peak value for say power output of a device.
2. With regards worded questions, i went to the TSFX lecture and the lecturer mentioned that in the exam one should not make refrence to the right hand grip or slap rules when answering worded questions. How should one explain them instead? Can anyone provide a perfect response to such a question as an example.
Post by: knightrider on July 10, 2015, 02:34:17 pm
How would you do these questions.
When does the bus first start gaining ground on the bicycle?
At what time does the bus overtake the bicycle?
How far has the bicycle travelled before the bus catches it?
Post by: Floatzel98 on July 10, 2015, 02:59:28 pm
Using the image attached.1) The bus will start gaining ground just after the velocities are equal. Even if it is still behind the bike, at the point when their velocities are equal i will now start lowering the distance between them because it is travelling towards it faster. Velocities are equal at t = 4s.
How would you do these questions.
When does the bus first start gaining ground on the bicycle?
At what time does the bus overtake the bicycle?
How far has the bicycle travelled before the bus catches it?
2) Well firstly you should try to remember that this is a velocity time graph and the area under it will be the displacement. You literally just have to count the squares to find when the distance is equal and this should happen at t = 10s if i counted right.
3)So you know the time when their distance is the same so you just have to find the area under the curve at that time (the actual distance for that time). So at 10s there are 10 boxes. 1 box = 2 seconds and 4 m/s so 10 x 2 x 4 = 80m
I think i did that right.
Post by: knightrider on July 10, 2015, 03:57:31 pm
1) The bus will start gaining ground just after the velocities are equal. Even if it is still behind the bike, at the point when their velocities are equal i will now start lowering the distance between them because it is travelling towards it faster. Velocities are equal at t = 4s.
2) Well firstly you should try to remember that this is a velocity time graph and the area under it will be the displacement. You literally just have to count the squares to find when the distance is equal and this should happen at t = 10s if i counted right.
3)So you know the time when their distance is the same so you just have to find the area under the curve at that time (the actual distance for that time). So at 10s there are 10 boxes. 1 box = 2 seconds and 4 m/s so 10 x 2 x 4 = 80m
I think i did that right.
Thanks so much Floatzel98 :)
Really helped and yes you are right. :D
Post by: knightrider on July 10, 2015, 05:31:18 pm
A stone is dropped vertically into a lake. Which one of the following statements best describes the motion
of the stone at the instant it enters the water?
A Its velocity and acceleration are both downwards.
B It has an upwards velocity and a downwards acceleration.
C Its velocity and acceleration are both upwards.
D It has a downwards velocity and an upwards acceleration.
Post by: Floatzel98 on July 10, 2015, 05:56:56 pm
For this multiple choice question which option is right and why?Well i think it's safe to say that it's velocity is still going down as soon as it hits the water. It hasn't stopped or floated upwards or anything because it just entered the water. So we can cancel out B and C. I think the logic here is that because the stone slows down immediately after it hits the water it decelerates which means that there is now a greater upwards force on it than it gained from falling. So the acceleration is now much greater in the opposite direction. Leaving it to be D i think.
A stone is dropped vertically into a lake. Which one of the following statements best describes the motion
of the stone at the instant it enters the water?
A Its velocity and acceleration are both downwards.
B It has an upwards velocity and a downwards acceleration.
C Its velocity and acceleration are both upwards.
D It has a downwards velocity and an upwards acceleration.
Post by: knightrider on July 10, 2015, 06:36:06 pm
Well i think it's safe to say that it's velocity is still going down as soon as it hits the water. It hasn't stopped or floated upwards or anything because it just entered the water. So we can cancel out B and C. I think the logic here is that because the stone slows down immediately after it hits the water it decelerates which means that there is now a greater upwards force on it than it gained from falling. So the acceleration is now much greater in the opposite direction. Leaving it to be D i think.
Thanks so much Floatzel98 :)
Post by: Zealous on July 25, 2015, 05:26:10 pm
Couple of questions with regards Unit 4.1. Keep it all in RMS when you are working out the power output. You can find the RMS average current and voltage then take the product of them to find the RMS power output.
1. When answers questions with regard RMS. Such as the power output when the voltage and current are given as RMS values. What is the expect answer? As in, should they be used as RMS values or converted. What happens if your given a voltage in RMS and a current in peak. To you give your answer as a RMS or peak value for say power output of a device.
2. With regards worded questions, i went to the TSFX lecture and the lecturer mentioned that in the exam one should not make refrence to the right hand grip or slap rules when answering worded questions. How should one explain them instead? Can anyone provide a perfect response to such a question as an example.
2. I personally think it's okay to refer to the rules, as long as you've properly explained why you have used the rule. So if you are using the RH grip rule, you might explain that the current in a wire causes a change in magnetic field, and therefore the magnetic field will have a certain direction as dictated by the right rule. Or if you are using the slap rule, you might explain that a current carrying wire will experience a force in a magnetic field and this force will act in a direction dictated by the slap rule. That's personally how I did things - there may be better ways of doing it but I dont think it's necessary.
Well i think it's safe to say that it's velocity is still going down as soon as it hits the water. It hasn't stopped or floated upwards or anything because it just entered the water. So we can cancel out B and C. I think the logic here is that because the stone slows down immediately after it hits the water it decelerates which means that there is now a greater upwards force on it than it gained from falling. So the acceleration is now much greater in the opposite direction. Leaving it to be D i think.Great explanation!
Post by: Floatzel98 on July 27, 2015, 03:27:37 pm
Also with question 5, is D wrong because they change the direction of the magnetic field before they actually move the coil? It's not happening while it is getting pulled out. If it was all happening simultaneously, it would induce a greater EMF though, right?
Thanks :)
Post by: odeaa on July 27, 2015, 04:19:45 pm
Initially, there is a downwards north flux. When the coil is removed, we have a change in the north flux (I just call it that, but the polarity is important) in the upwards direction (because the flux is decreasing). To oppose this, the coil induces a current that will create a downwards force, which is clockwise when viewed from above, and I think you know how to find the magnitude of the current judging from your question
As for question 5, D is incorrect because while there is a change in flux when they change the direction of the magnet, this change is only temporary and the system kinda becomes used to it (its like using a DC battery for a transformer- there is only a temporary change in flux). When the coil is removed, the same emf is generated in the coil as when the magnet was in the original orientation, just in the opposite direction.
I dont think they would ever ask you what would happen if it was being changed simueltaneously, because the only way I can think of doing it is to graph the flux and then derive it to find the emf, which is outside the scope of the course (I could be wrong though)
Post by: Floatzel98 on July 27, 2015, 04:37:36 pm
For q4 and 6, I used to struggle with those as well but I finally got my head around it (I think, havent got my SAC back yet lol)Thanks for the help. I'm still wrapping my head around it all so i have a few questions still. I understand everything up to the part in bold. I understand the direction of the changing flux, i just don't think i understand the opposing directions properly. Why is it a downwards force. i thought the magnetic field direction was the important part, it needs to oppose the changing flux?
Initially, there is a downwards north flux. When the coil is removed, we have a change in the north flux (I just call it that, but the polarity is important) in the upwards direction (because the flux is decreasing). To oppose this, the coil induces a current that will create a downwards force, which is clockwise when viewed from above, and I think you know how to find the magnitude of the current judging from your question
As for question 5, D is incorrect because while there is a change in flux when they change the direction of the magnet, this change is only temporary and the system kinda becomes used to it (its like using a DC battery for a transformer- there is only a temporary change in flux). When the coil is removed, the same emf is generated in the coil as when the magnet was in the original orientation, just in the opposite direction.
I dont think they would ever ask you what would happen if it was being changed simueltaneously, because the only way I can think of doing it is to graph the flux and then derive it to find the emf, which is outside the scope of the course (I could be wrong though)
Post by: odeaa on July 27, 2015, 04:50:46 pm
Thanks for the help. I'm still wrapping my head around it all so i have a few questions still. I understand everything up to the part in bold. I understand the direction of the changing flux, i just don't think i understand the opposing directions properly. Why is it a downwards force. i thought the magnetic field direction was the important part, it needs to oppose the changing flux?
Thats the part I was confused with as well, I'll try to explain it a bit better
Because the initial direction of the flux was downwards (assume when I'm talking about direction, I'm always talking about direction of North, because this is whats important), when we remove the loop we have a decreasing downwards flux, or an increasing upwards flux (just depends which way you think about it, but it is very important that the direction is included). To oppose this, because the system always wants to oppose the change in flux, we need an increasing downwards force to oppose the increasing upwards force (or decreasing downwards force, again just depends how you visualise it). Because of the right hand grip rule, to create a downwards force, we need a current going clockwise.
Hope that makes more sense!
Post by: Zealous on July 27, 2015, 06:57:16 pm
Thanks for the help. I'm still wrapping my head around it all so i have a few questions still. I understand everything up to the part in bold. I understand the direction of the changing flux, i just don't think i understand the opposing directions properly. Why is it a downwards force. i thought the magnetic field direction was the important part, it needs to oppose the changing flux?
I think of it as the system resisting the change and wanting to be back in its original state.
So initially, there are magnetic field lines going from N to S of the coil, or from the top to the bottom of the coil. When the coil is moved to the right, we are removing it from that initial magnetic field, and the coil wants that magnetic field back. In order to create a magnetic field going from the top to the bottom, there needs to be a clockwise current.
Conversely, for Q6, initially there is no magnetic field through the coil and we are forcing it into a position with a magnetic flux. The coil doesn't like that and wants to be back with no magnetic field, so will create a magnetic field from S to N (or bottom to top) to cancel out the increas in magnetic field. This will cause a counterclockwise current. Think of it as the system wanting to get back to where it was initially.
This guy does an amazing job of explaining it: https://www.youtube.com/watch?v=qWu82nJS42I
Post by: odeaa on July 27, 2015, 07:16:19 pm
This guy does an amazing job of explaining it: https://www.youtube.com/watch?v=qWu82nJS42I
holy shit i love that guy, ive watched legit every one of his videos
Post by: Floatzel98 on July 27, 2015, 08:41:56 pm
One very last thing. I think I'm still having trouble with using the right hand rules with coils/solenoids etc. Do you put your thumb in the direction of current or in the direction of the magnetic field, and then your fingers would wrap around the coil and be the current? I've been doing most questions the first way since that is what my teacher told me to do, but i seem to get mixed answers.
Also some other questions. Using the same question i posted above as a reference. How long does a induced EMF and current as well as the magnetic field produced last for? Would the current just stay in the loop until it got discharged somewhere?
Quote
I think of it as the system resisting the change and wanting to be back in its original state.This actually helps a lot. I'll definitely think of it like this from now on :)
Post by: odeaa on July 27, 2015, 09:04:59 pm
Thanks guys. Those explanations really helped. That video really helped as well. That guy is awesome haha.
One very last thing. I think I'm still having trouble with using the right hand rules with coils/solenoids etc. Do you put your thumb in the direction of current or in the direction of the magnetic field, and then your fingers would wrap around the coil and be the current? I've been doing most questions the first way since that is what my teacher told me to do, but i seem to get mixed answers.
Also some other questions. Using the same question i posted above as a reference. How long does a induced EMF and current as well as the magnetic field produced last for? Would the current just stay in the loop until it got discharged somewhere?
This actually helps a lot. I'll definitely think of it like this from now on :)
for normal wires, thumb in direction of current (along the wire) and fingers in the direction of field
while you can apply this to a coil or solenoid, its easier just to go fingers=current and thumb=field
as for what happens to the current in the loop, I have no idea lol. In terms of VCE, I've done all the past papers back to 1997 and havent seen anything remotely related to that, and its not in the study design afaik
Post by: Floatzel98 on July 28, 2015, 07:22:41 pm
A student builds a simple alternator consisting of a coil containing 500 turns, each of area 10cm^2, mounted on an axis that can rotate between the poles of a permanent magnet of strength 80mT. At a frequency of 50Hz, it is found that the peak voltage produced is 12.6V
a) What are the peak to peak and RMS voltages
b)If the frequency is doubled to 100hz, how will the peak and RMS voltages change?
B is what i'm having trouble with. I'm kind of having a mind blank trying to put my maths and physics skills together here, but how does the increased frequency produce a great EMF? I understand it when talking about a greater change of flux since it is now moving faster, but i don't really understand it graphically. Well basically the whole electric power generation chapter in my textbook goes over it all in like 2 pages. I understand that they differentiate the flux (
Thanks :)
Post by: Floatzel98 on July 28, 2015, 07:51:00 pm
Thanks
Post by: Kel9901 on July 29, 2015, 06:51:33 pm
Probably a simple question but I'm getting a bit confused on this:
A student builds a simple alternator consisting of a coil containing 500 turns, each of area 10cm^2, mounted on an axis that can rotate between the poles of a permanent magnet of strength 80mT. At a frequency of 50Hz, it is found that the peak voltage produced is 12.6V
a) What are the peak to peak and RMS voltages
b)If the frequency is doubled to 100hz, how will the peak and RMS voltages change?
B is what i'm having trouble with. I'm kind of having a mind blank trying to put my maths and physics skills together here, but how does the increased frequency produce a great EMF? I understand it when talking about a greater change of flux since it is now moving faster, but i don't really understand it graphically. Well basically the whole electric power generation chapter in my textbook goes over it all in like 2 pages. I understand that they differentiate the flux (and that it equals the induced EMF, but i don't understand the angular velocity part of the equation (
). The book also does some stuff to be able to use frequency but I'm just not understanding it at all.
Thanks :)
With doubled frequency comes halved period (T=1/f), so the denominator is halved and hence the EMF (both rms and peak) is doubled.
Post by: Adequace on July 29, 2015, 08:29:25 pm
With doubled frequency comes halved period (T=1/f), so the denominator is halved and hence the EMF (both rms and peak) is doubled.This might be a dumb comment, but is that because f + T has to equal 1 or something like that?
Anyway, I have a Unit 2 SAC tomorrow. I've not really been doing much tonight apart from revising my notes and cheat sheet which makes me feel unproductive. I completed the checkpoints questions, textbook questions and questions I could find online. Is there anything else I could've done?
I want to develop a strong work ethic towards test and know what to do when I have nothing to do especially when units 3/4 come along, if that makes sense.
Post by: odeaa on July 29, 2015, 08:37:20 pm
This might be a dumb comment, but is that because f + T has to equal 1 or something like that?
Anyway, I have a Unit 2 SAC tomorrow. I've not really been doing much tonight apart from revising my notes and cheat sheet which makes me feel unproductive. I completed the checkpoints questions, textbook questions and questions I could find online. Is there anything else I could've done?
I want to develop a strong work ethic towards test and know what to do when I have nothing to do especially when units 3/4 come along, if that makes sense.
Imo, theres really not that much point doing any more for a 1/2 if you have done all that already. thats what I did for all my sacs in year 11 and it still set me up with a good work ethic in year 12. I wouldnt waste time on something like physics if you dont have any practise questions to do (really theres not you can do), and just do an english essay or something like that, in the long run its time better spent
if you really want to study but have literally no questions to do, write your own exam, making it as hard as possible but still covering as much of the course as you can, and then write detailed solutions (probs wont have time but if you really want something to do)
also, i dunno if you have it, but buy the neap questions book. Really good quality questions, better than checkpoints imo (didnt even know they had checkpoints for 1/2)
Post by: Floatzel98 on July 30, 2015, 07:36:35 pm
Post by: paper-back on July 30, 2015, 07:59:32 pm
Just checking, can we add N, number of turns, into the equation
Yeah, you add N(number of turns) to the equation
Post by: Floatzel98 on August 01, 2015, 04:09:44 pm
Thanks :)
Post by: Kel9901 on August 01, 2015, 04:39:12 pm
I just saw a questions where we had to find an estimate for the magnetic force on a 'vertical lightning conductor' that was perpendicular to the Earths magnetic field. They gave us the current and they estimated a length for it. But they estimated Earths magnetic field to be. My textbook (this question was in checkpoints) usually uses
and online it says it ranges from
to
. Would we actually ever get a question on the exam like this anyway that i should't worry about it?
Thanks :)
I can state with certainty that there won't be a question that requires you to use estimates that aren't given.
Post by: timton on August 01, 2015, 10:12:43 pm
Context: Generator for a small town.
Why would you make the turn ratio for a step up and step down generator the same?
Is it to lower power loss?
Post by: silverpixeli on August 02, 2015, 12:35:20 pm
QUESTION
Context: Generator for a small town.
Why would you make the turn ratio for a step up and step down generator the same?
Is it to lower power loss?
Not exactly sure what you're asking, I think you mean a step up/down transformer which is connected to a generator. Also, by 'the same' do you mean e.g. 1:100 for the step up and 100:1 for the step down?
If that's the case, then there's not really any advantage to this, it just means that when you step up the power coming out of the generator to transmit it, and then step it down later, the stepping up and the stepping down exactly cancel each other out and the ratio of voltage to current after transmission is the same as the ratio coming out of the generator, before transmission.
Post by: timton on August 02, 2015, 05:05:08 pm
Not exactly sure what you're asking, I think you mean a step up/down transformer which is connected to a generator. Also, by 'the same' do you mean e.g. 1:100 for the step up and 100:1 for the step down?
If that's the case, then there's not really any advantage to this, it just means that when you step up the power coming out of the generator to transmit it, and then step it down later, the stepping up and the stepping down exactly cancel each other out and the ratio of voltage to current after transmission is the same as the ratio coming out of the generator, before transmission.
My bad, was meant to write transformer instead of generator.
The question specifically asks "why would the turn ratio be chosen to be the same for both the step up and step down transformer pair?"
You are not given additional information about the number of turn in the coil (question later on) but you are asked why they are the same.
Post by: odeaa on August 02, 2015, 05:19:04 pm
My bad, was meant to write transformer instead of generator.
The question specifically asks "why would the turn ratio be chosen to be the same for both the step up and step down transformer pair?"
You are not given additional information about the number of turn in the coil (question later on) but you are asked why they are the same.
It's a bit of a weirdly worded question if you ask me. I can't see any reason why they would be the same ratio, usually a question like that would ask you why they need to step the voltage up/down. The only thing I can think of is for safety (if not to minimise power loss) because you aren't directly connected to the power source
Post by: silverpixeli on August 02, 2015, 07:17:46 pm
My bad, was meant to write transformer instead of generator.
The question specifically asks "why would the turn ratio be chosen to be the same for both the step up and step down transformer pair?"
You are not given additional information about the number of turn in the coil (question later on) but you are asked why they are the same.
yeah i'm not sure on 'why' you would do that specifically, maybe they are looking for something along the lines of 'even though stepping up the voltage and then stepping it down by the same amount causes no change between the generator voltage and the voltage at the house, it allows the transmission to take place at a lower current, which means the power lost in transmission is lower (according to P = I^2 R)'
or something like that.
Post by: dankfrank420 on August 02, 2015, 07:45:08 pm
As I understand it, you increase the voltage by increasing the turns on the secondary coil (to "step it up"). Since P = V*I and P is constant, therefore the current is decreased as voltage is increased. That's all good.
But isn't power also defined as v^2/r as well as defined as I^2*r? Wouldn't increasing the voltage also increase the power loss?
Post by: silverpixeli on August 02, 2015, 08:13:27 pm
Re: transformers
As I understand it, you increase the voltage by increasing the turns on the secondary coil (to "step it up"). Since P = V*I and P is constant, therefore the current is decreased as voltage is increased. That's all good.
But isn't power also defined as v^2/r as well as defined as I^2*r? Wouldn't increasing the voltage also increase the power loss?
Yes, P = V^2/R is a valid formula, but to calculate the power loss across the wires using this you would need to use the voltage that is across the wires (which is probably not the voltage in the second part of the transformer because some is also across the houses or the transformer at the other end)
The current in the secondary coil, the wires and in the houses is the same because they are all in series, so this formula is usually easier to use
Post by: dankfrank420 on August 02, 2015, 09:20:34 pm
Yes, P = V^2/R is a valid formula, but to calculate the power loss across the wires using this you would need to use the voltage that is across the wires (which is probably not the voltage in the second part of the transformer because some is also across the houses or the transformer at the other end)
The current in the secondary coil, the wires and in the houses is the same because they are all in series, so this formula is usually easier to use
Ah, cheers.
Post by: Kel9901 on August 03, 2015, 11:47:00 am
My bad, was meant to write transformer instead of generator.
The question specifically asks "why would the turn ratio be chosen to be the same for both the step up and step down transformer pair?"
You are not given additional information about the number of turn in the coil (question later on) but you are asked why they are the same.
I'm guessing because the voltage produced by the generator is itself the desired voltage, and keeping the step up/down ratios the same also keeps the voltage (nearly) the same.
Post by: Floatzel98 on August 04, 2015, 07:56:17 pm
Also i don't understand the paragraph on Extent of diffraction from my book. It says that the extent of diffraction
One more question that I didn't understand: When white light is passed through double slits, colored fringes can be observed. What is responsible for this phenomenon?
Post by: Zealous on August 04, 2015, 08:27:13 pm
Hey guys. I'm just having a bit of trouble understanding how fringe spacing can be increased/decreased. How does increasing the wavelength and distance to the screen increase the spacing and reducing the slits increase it? I just can't seem to get my head around how it happens.
Also i don't understand the paragraph on Extent of diffraction from my book. It says that the extent of diffractionwavelength / size of slit =
One more question that I didn't understand: When white light is passed through double slits, colored fringes can be observed. What is responsible for this phenomenon?
These are the ways I like to think about it. May or may not be truly accurate but it worked for me. ;D
Wavelength: Imagine an 800nm wave of light and a 300nm wave of light. If they are both trying to fit through a slit which is 400nm, which one is going to have more trouble getting through the gap? The 800nm will have much more trouble, and therefore will experience greater diffraction which will spread out the band spacing.
Distance: This one is trickier to understand. Have a look at this diagram here:
(http://i1282.photobucket.com/albums/a531/Ovazealous/screenshot.319_zps9nqprtvj.png)
If you imagine one arrow being a maximum (bright fringe), moving the rectangle further back will result in only 3 of the fringes appearing on the rectangle, so they will spread apart compared to 5 fringes over the same area when the rectangle is much closer.
Reducing Slit: Well, it's the same as the wavelength scenario, but instead of different wavelengths changing, its different slit sizes. If we have light of 500nm wave length, is it going to have an easier time passing through a gap which is 600nm or 400nm? The smaller the slit size, the more impedence it will have with the slit and more diffraction will occur.
Extent of Diffraction: They're taking the ratio of wavelength over slit size. If the wavelength and the slit size are exactly the same, the fraction will equal 1. If the wavelength is bigger than the size of the slit, the fraction will be > 1 and if the wavelength is smaller than the slit the fraction will be < 1. You can use this to get a rough idea as to the extent of the diffraction.
White Light: White light is made up of light of many different wavelengths. When it passes through the slits, the different wavelengths that make up the white light will be diffracted to different degrees, which will cause the red light to spread differently from the green light, which leads to colored fringes.
Post by: Floatzel98 on August 04, 2015, 08:59:25 pm
Also with the photoelectric effect quickly, is the photo current actually the 'photons' of light from the light or is it just electrons from the metal plate that create that current? I'm also kind of lost on how this predicts light as a particle and not a wave. I can't seem to understand why a wave doesn't work in this situation. Is the fact that there isnt a delay as a particle one of the points? Why isnt there a delay for?
Post by: Zealous on August 04, 2015, 10:28:02 pm
Also with the photoelectric effect quickly, is the photo current actually the 'photons' of light from the light or is it just electrons from the metal plate that create that current? I'm also kind of lost on how this predicts light as a particle and not a wave. I can't seem to understand why a wave doesn't work in this situation. Is the fact that there isnt a delay as a particle one of the points? Why isnt there a delay for?
It's not the photons which create the current - it's the electrons in the metal. The photons provide enough energy to the electrons so that they can "escape" the metal.
The particle model explains this because if the energy of the photon (E=hf) is less than the work function of the metal, each electron will not have enough energy to escape the metal. If we used the wave model, we'd expect energy to accumulate in the metal as a continuous flow, then the electrons will eject when they have the energy.
Think of the particle model as a one to one transaction, the photon will either transfer enough energy or the electron won't leave. Think of the wave model as a continuous transfer of energy, even if it has an extremely low energy, eventually the electron will gain enough energy over time to leave.
Because the electron flow is instantaneous, it supports the particle model the best. There are other things I could talk about such as the amplitude and intensity but I won't go into that since you've just started.
This video explains it really well: https://www.youtube.com/watch?v=GiM21YCvWh4
Post by: lzxnl on August 05, 2015, 12:10:09 am
Hey guys. I'm just having a bit of trouble understanding how fringe spacing can be increased/decreased. How does increasing the wavelength and distance to the screen increase the spacing and reducing the slits increase it? I just can't seem to get my head around how it happens.
Also i don't understand the paragraph on Extent of diffraction from my book. It says that the extent of diffraction
One more question that I didn't understand: When white light is passed through double slits, colored fringes can be observed. What is responsible for this phenomenon?
Welcome to VCE physics, where concepts are presented in a hand-wavy manner and you're expected to understand everything perfectly for the exam. Kidding. You have a cheat sheet for that. The concepts are still hand-wavy though.
A much better interpretation of 'extent of diffraction' is 'width of the central maximum'. You see, diffraction is actually just interference (and Huygen's principle but forget that for now), so you have a central bright band and other less bright bands. It just so happens that the central bright band is gazillions brighter than the other bands (or more intense for the case of sound) so we normally consider the extent of diffraction to be the width of the central bright band.
If we define the width of the central bright band as the distance between the two perfectly dark spots (aka total destructive interference), it can be shown, by splitting the slit into two separate slits and considering this now as a two-slit interference experiment, that the angle made between a line drawn from the slit to the dark spot and a line drawn from the slit perpendicular to the slit, theta, satisfies the equation d sin theta = wavelength where d is the slit width
AKA sin theta = wavelength / slit width
If this quantity approaches 1, theta approaches 90 degrees, and 90 degrees corresponds to essentially no dark band (draw out a diagram). If the quantity gets bigger than 1, you NEVER have destructive interference and that's what VCAA means by 'complete diffraction'. It's complete in that the diffracted wave has significant amplitude everywhere.
Post by: Cosec on August 05, 2015, 06:49:18 pm
In a quesiton like this, do you just assume the value they provided is RMS?
Post by: Floatzel98 on August 05, 2015, 07:01:28 pm
Post by: knightrider on August 05, 2015, 08:37:33 pm
A rope is allowed to move freely over a ‘frictionless’ pulley backstage of a theatre. A 30 kg sandbag,
which is at rest on the ground, is attached at one end. A 50 kg work-experience student, standing on
a ladder, grabs onto the other end of the rope to lower himself.
What is the tension in the rope?
Post by: Kel9901 on August 06, 2015, 01:41:50 pm
How would you do this question?
A rope is allowed to move freely over a ‘frictionless’ pulley backstage of a theatre. A 30 kg sandbag,
which is at rest on the ground, is attached at one end. A 50 kg work-experience student, standing on
a ladder, grabs onto the other end of the rope to lower himself.
What is the tension in the rope?
Clearly the sandbag moves up and the person moves down
On the sandbag:
Fnet=T-mg=ma
T-300=30a
T/30-10=a
On the person:
Fnet=mg-T=ma
500-T=50a
10-T/50=a
Equating accelerations: (in a connected bodies question the accelerations are the same)
T/30-10=10-T/50
4T/75=20
T=375 N
Post by: knightrider on August 06, 2015, 04:07:57 pm
Clearly the sandbag moves up and the person moves down
On the sandbag:
Fnet=T-mg=ma
T-300=30a
T/30-10=a
On the person:
Fnet=mg-T=ma
500-T=50a
10-T/50=a
Equating accelerations: (in a connected bodies question the accelerations are the same)
T/30-10=10-T/50
4T/75=20
T=375 N
Thanks Kel9901 :)
but the book has the answer FT = 367.5 N ?
Post by: silverpixeli on August 06, 2015, 04:46:15 pm
Thanks Kel9901 :)
but the book has the answer FT = 367.5 N ?
using g=9.8 instead of g=10 should give you what the book has, though ofc 10 makes the working a little simpler
Post by: dankfrank420 on August 06, 2015, 07:12:54 pm
Post by: Floatzel98 on August 06, 2015, 08:44:48 pm
Can someone explain how current/fields are induced when a bar magnet is pushed into a coil of wire? Like how do we know what direction they will be?I think the current is induced by a changing flux because the magnetic force applies a force on the free electrons in the coil which causes them to move. That probably isn't correct, but i don't think you actually need to know how a current is created. Maybe lzxnl or someone else could help you there.
As for the direction of the current produced, which we do need to know, that can be predicted using Lenz's Law, which states that 'If an induced current flows, its direction is always such that it will oppose the change which produced it'. We need Lenz's Law so everything we do obeys Newton's Third Law and the Conservation of Energy.
I probably can't explain it too well without maybe an example to work through, so maybe check out Brightstorm's video on it.
Post by: Floatzel98 on August 07, 2015, 05:18:58 pm
And in this situation where there is an open switch on the secondary load, is there still actually voltage induced in the coil? Like it will induce the current, but it will have no where to follow since there is an open switch.
Thanks :)
Post by: Floatzel98 on August 11, 2015, 10:01:02 pm
I'm kind of confused as to what happens when there is no load connected to the secondary coil of a generator. From what I do understand, no current will be able to be generated in the secondary coil, so the magnetic field will loop around the transformer and induce a voltage back in the primary coil that will oppose the change of flux and the current will basically cancel out. How exactly does the current cancel each other out? Why doesn't this Back EMF happen all the time? For every transformer, shouldn't the changing flux after it has made a revolution of the loop just cancel out the primary voltage? Or does a magnetic field cease to exist after it has induced a current (I don't know how else to word that. Why isn't there always a Back EMF that cancels out the current? Does the magnetic field get weaker after it has induced a current once?)?A bump for this if anyone can answer it.
And in this situation where there is an open switch on the secondary load, is there still actually voltage induced in the coil? Like it will induce the current, but it will have no where to follow since there is an open switch.
Thanks :)
Also I have another question. Can someone explain 'standing waves' to me. I don't really understand the electrons having wavelengths around their orbits and what exactly is happening when they destructively interfere. Also my book is drawing parallels between them and violin strings. Is this a common analogy for standing waves? Could someone explain what they mean by it.
Thanks :)
Post by: lzxnl on August 11, 2015, 10:48:32 pm
A bump for this if anyone can answer it.
Also I have another question. Can someone explain 'standing waves' to me. I don't really understand the electrons having wavelengths around their orbits and what exactly is happening when they destructively interfere. Also my book is drawing parallels between them and violin strings. Is this a common analogy for standing waves? Could someone explain what they mean by it.
Thanks :)
Don't get too caught up in what is actually waving when they say an electron waves. What they really mean is that the electron has a probability of being in a particular place at a particular time due to the uncertainty principle and something connected to this probability distribution in space is waving. The VCE interpretation of energy levels coming from electron standing waves is wrong but I won't explain why because it's pretty confusing.
Standing waves are waves that don't travel. Regular sound waves travel through the air; standing waves only have an amplitude oscillation over time but don't move. Violin strings are a good example (as is any form of string or pipe) because all of the pitches you can hear are due to standing waves. Look up videos; they'll explain better than words.
Post by: knightrider on August 15, 2015, 06:39:42 pm
Calculate the momentum of an object: that experiences a net force of magnitude 45 N, if
the net force is applied for 3.5 s.
Post by: lzxnl on August 15, 2015, 06:47:20 pm
How would you do this question ?
Calculate the momentum of an object: that experiences a net force of magnitude 45 N, if
the net force is applied for 3.5 s.
You can't. We don't know the initial momentum. The change in momentum, however, is given by p = F*t
Post by: Floatzel98 on August 15, 2015, 10:57:00 pm
Describe how the wave-particle duality of electrons can be used to explain the quantised energy levels in atoms.
I may be missing something crucially simple here, but I don't really know how to answer this question.
Which one or more of the following phenomena can be modeled by a pure wave model of light?
A - The Photoelectric Effect
B - Refraction
C - Double - source interference of light
D - Reflection
E - Diffraction
F - The Compton Effect
Maybe I'm just getting caught up on 'pure wave model' for this question, but my i keep reading that reflection and refraction can be modeled by either a wave or particle. So going off that i said the answers were C and E, but apparently the answers are B , C, D ,E. Is that correct?
Post by: lzxnl on August 16, 2015, 10:59:33 am
Can someone explain how current/fields are induced when a bar magnet is pushed into a coil of wire? Like how do we know what direction they will be?
A little bit of insight into how magnetic fields can be induced is related to the fact that electric and magnetic fields are actually the same thing. Here's an example of what I mean. You'll need to understand the result that two people moving at a constant velocity with respect to each other in inertial frames (i.e. obeys Newton's first law, so that they're not accelerating) measure the same forces.
Suppose you have two people, A and B. A is 'stationary' and B moves at a constant velocity v with respect to A. Let's suppose there is a proton moving with B, so that B sees the proton as stationary. Suppose A sets up a magnetic field. As the proton is a charged moving particle, it is affected by the magnetic field (assuming it's not moving parallel to the magnetic field) and experiences a force. B must therefore also record a force on the proton, somehow. But according to B, the proton is stationary and we know magnetic fields don't operate on stationary charges. Hence this force must be due to something else, and this something else is an electric field.
I can't definitively answer your question as to how induction comes about, but it'll be something to do with that.
Couple questions:
Describe how the wave-particle duality of electrons can be used to explain the quantised energy levels in atoms.
I may be missing something crucially simple here, but I don't really know how to answer this question.
Which one or more of the following phenomena can be modeled by a pure wave model of light?
A - The Photoelectric Effect
B - Refraction
C - Double - source interference of light
D - Reflection
E - Diffraction
F - The Compton Effect
Maybe I'm just getting caught up on 'pure wave model' for this question, but my i keep reading that reflection and refraction can be modeled by either a wave or particle. So going off that i said the answers were C and E, but apparently the answers are B , C, D ,E. Is that correct?
Refraction can't be modelled by a particle theory. I remember in year twelve seeing a presentation on how Newton's particle theory of light predicts an increase in the speed of light in water as opposed to air, which is clearly false.
The question says what 'CAN' be modelled by a wave model. So anything that is explained by a wave model is a valid answer.
Post by: knightrider on August 16, 2015, 07:13:41 pm
should this be W= force*distance instead?
Because say you moved an object 500 metres forwards and then back 500 metres you have done work to move the object.
But using W=force*displacement this would mean you have done no work?
Whereas using W= force*distance this does suggest you have done work?
Post by: lzxnl on August 16, 2015, 08:48:29 pm
The work formula W=force*displacement
should this be W= force*distance instead?
Because say you moved an object 500 metres forwards and then back 500 metres you have done work to move the object.
But using W=force*displacement this would mean you have done no work?
Whereas using W= force*distance this does suggest you have done work?
Another failing of VCE physics. Sigh.
Work is actually defined as a vector dot product of the force vector and displacement vectors (this will make more sense if you do spesh and I didn't check to see if you did). Essentially what it means is that the work done is dependent on the angle between the force and the displacement. For your question, if you push on an object for 500 m to the left with a constant force F, then apply the exact same constant force over 500 m to the right afterwards, you WILL have done no net work on the object. This means the object will be moving at the same speed as it was in the beginning.
Force * distance is quite dangerous to use.
Post by: Adequace on August 16, 2015, 08:50:52 pm
Post by: knightrider on August 16, 2015, 10:20:00 pm
Another failing of VCE physics. Sigh.
Work is actually defined as a vector dot product of the force vector and displacement vectors (this will make more sense if you do spesh and I didn't check to see if you did). Essentially what it means is that the work done is dependent on the angle between the force and the displacement. For your question, if you push on an object for 500 m to the left with a constant force F, then apply the exact same constant force over 500 m to the right afterwards, you WILL have done no net work on the object. This means the object will be moving at the same speed as it was in the beginning.
Force * distance is quite dangerous to use.
Thanks so much lzxnl :)
so we must use The work formula W=force*displacement ?
Post by: Kel9901 on August 17, 2015, 10:16:37 am
Thanks so much lzxnl :)force*change in displacement*cos (angle between force and direction of motion)
so we must use The work formula W=force*displacement ?
Post by: Floatzel98 on August 19, 2015, 10:16:25 pm
In a laboratory class at school, Lee is given a spring with a stiffness of 20 N/m and an unstretched length of 0.40 m. He hangs it vertically, and attaches a mass of 0.40kg to it. The new length is 0.60 m. Lee pulls the mass down a further distance of 0.10 m. By how much has the potential energy stored in the spring changed?
I thought I would just go
Also with the attached I just can't seem to do question 103 properly. I don't even know if I'm doing it correctly. Are we supposed to use the graph or something?
Post by: silverpixeli on August 20, 2015, 09:09:34 am
Hey guys. I'm going over the checkpoints books now for motion because that was the AOS I did the worst in so far this year. And i just have a few questions I need help with.
In a laboratory class at school, Lee is given a spring with a stiffness of 20 N/m and an unstretched length of 0.40 m. He hangs it vertically, and attaches a mass of 0.40kg to it. The new length is 0.60 m. Lee pulls the mass down a further distance of 0.10 m. By how much has the potential energy stored in the spring changed?
I thought I would just go, but the answer says to go
and I'm a bit confused why they do that. Why are they using the total length of the spring at those points for? Shouldn't they be using the stretched length?
Also with the attached I just can't seem to do question 103 properly. I don't even know if I'm doing it correctly. Are we supposed to use the graph or something?
You're right of course they should be using Δx rather than total length, must be an error
I would do 103 by solving for x in: 6000 = 1/2 k x^2 with k the gradient of the graph, in the right units, from 102
Alternatively I you can find 'at what x the area under the F-x graph is 6000J' and you may realise that this is the exact same calculation as the first method
Post by: Floatzel98 on August 23, 2015, 01:16:23 pm
Another question as well. How does the impulse that the airbag exerts on the driver's head compare with the impulse that the drivers head exerts on the airbag? The Impulse is 56 Ns from the previous part of the question, but the answer for this question says they will be exactly the same. Should it not be equal and opposite? Impulse is a vector quantity isn't it?
Post by: odeaa on August 23, 2015, 02:04:46 pm
If a question is asking us to find magnitude of an impulse and we give the units in kg m/s instead of Ns, would we be incorrect? I know on the VCAA exams they usually have a box for the answer with the units already there, but if we stated different units in our workings, would that be incorrect?
You should get a mark, its not an incorrect unit for momentum at all. No way they can ping you if they didnt specify it, and on the exam it won't be an issue anyway
Another question as well. How does the impulse that the airbag exerts on the driver's head compare with the impulse that the drivers head exerts on the airbag? The Impulse is 56 Ns from the previous part of the question, but the answer for this question says they will be exactly the same. Should it not be equal and opposite? Impulse is a vector quantity isn't it?I think that's just a badly worded question, it should say 'how does the magnitude compare' if they want that answer
You're correct, they are opposite and equal because
Post by: Orson on August 24, 2015, 12:02:51 pm
a) Calculate the number of photons that are incident on the reflecting surface each second.
b) What is the momentum of each photon in the beam?
I got the correct answer to part a) #ph =
I am struggling with the second part. De Broglie?
Post by: Floatzel98 on August 24, 2015, 03:20:18 pm
A beam of blue light,Wellhz, is incident normally on a perfect reflecting surface. The beam power is
.
a) Calculate the number of photons that are incident on the reflecting surface each second.
b) What is the momentum of each photon in the beam?
I got the correct answer to part a) #ph =photons per second.
I am struggling with the second part. De Broglie?
Post by: odeaa on August 31, 2015, 06:06:38 pm
Is this even possible? All we have is the radius and mass of the planet, which is useless. Don't you need the mass of the sun (not given in data book afaik) to solve it? Note: its a 5 mark question, which is more than you would get for a substitution usually, so I'm a bit thrown off. Thanks
Post by: lzxnl on September 01, 2015, 03:06:37 pm
A comet is in orbit around the sun. The mass of the comet is 3.7x10^22 kg, and the radius of the comets orbit around the sun is 7.2x10^9m. Calculate the period of the comets orbit.
Is this even possible? All we have is the radius and mass of the planet, which is useless. Don't you need the mass of the sun (not given in data book afaik) to solve it? Note: its a 5 mark question, which is more than you would get for a substitution usually, so I'm a bit thrown off. Thanks
You can look up the sun's mass in your own data book for physics. For reference, it's about 1.99 x 10^30 kg (that number will change over time for reasons you don't need to worry about). Then, you have enough information.
Alternatively, you can compare the comet's orbit to Earth's orbit, which has a mean radius of 150 million km and a period of 365.26 days (I know, you're not expected to know the former).
Post by: knightrider on September 05, 2015, 06:04:18 pm
Post by: odeaa on September 06, 2015, 01:30:39 pm
Post by: lzxnl on September 06, 2015, 01:56:11 pm
For a loop in a field, it has to be a complete loop for there to a flux doesnt it? Textbook explanation is a bit iffy
Erm...no. A magnetic flux is defined by essentially how much of a magnetic field goes through an area. This area does not have to physically exist as an object; it can be air, for all you know. You can have a voltage generated with a changing magnetic flux with an incomplete loop; you just won't get a current as the resistance of the air between the loop ends is huge.
knightrider, your question doesn't make a whole lot of sense as you can't really disturb a spring like that; you can't disturb part of a spring only
Post by: odeaa on September 06, 2015, 01:59:13 pm
Erm...no. A magnetic flux is defined by essentially how much of a magnetic field goes through an area. This area does not have to physically exist as an object; it can be air, for all you know. You can have a voltage generated with a changing magnetic flux with an incomplete loop; you just won't get a current as the resistance of the air between the loop ends is huge.Ah I see, I was a bit confused there. Thanks for clearing that up
knightrider, your question doesn't make a whole lot of sense as you can't really disturb a spring like that; you can't disturb part of a spring only
Post by: Orson on September 06, 2015, 02:56:39 pm
Question [NEAP 2012 U3]:
A 4WD of mass 1800kg pulls a boat on a trailer at a constant speed up a boat ramp with an incline of 18 degrees. The combined mass of the boat and trailer is 700kg.
A frictional force of 300N acts between the boat trailer and the ramp, and a frictional force of 850N acts between the 4WD and the ramp.
a) Calculate the force exerted by the boat ramp on the 4WD (I'm good with this one). 1.7*10^4 N
b) Calculate the magnitude of the tension in the coupling joining the trailer and the 4WD. 2.5*10^3 N
c) Calculate the power developed by the 4WD if it moves the boat 15 meters up the ramp in 8 seconds. 1.7*10^4 W
Thanks heaps!
Post by: knightrider on September 06, 2015, 06:00:49 pm
What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?
Post by: Rishi97 on September 06, 2015, 10:28:24 pm
How would you do this question?
What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?
Hey Knightrider :)
These sort of questions are best answered by finding a relationship between the variables asked. In this case, since the question is asking for speed and frequency, we can use the formula speed = frequency x wavelength. So, if the frequency is halved (decreased), the speed must also decrease since they are proportional to each other.
Hope that makes sense :-)
Post by: knightrider on September 07, 2015, 06:44:29 am
Hey Knightrider :)
These sort of questions are best answered by finding a relationship between the variables asked. In this case, since the question is asking for speed and frequency, we can use the formula speed = frequency x wavelength. So, if the frequency is halved (decreased), the speed must also decrease since they are proportional to each other.
Hope that makes sense :-)
Thanks Rishi97 :)
but the answer says there is no change in speed?
Post by: Rishi97 on September 07, 2015, 11:28:12 am
Thanks Rishi97 :)
but the answer says there is no change in speed?
Oh really?
I'm not sure then sorry :(
Post by: Floatzel98 on September 07, 2015, 08:09:37 pm
For part b I'm really lost with what to do. 0.9, 1.5 and 2.2 eV are all between the first excited state. I don't know how to use them at all to find out the unknown excited state.
Thanks
Post by: Orbit on September 07, 2015, 10:52:20 pm
How would you do this question?
What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?
One of the most important things to remember about waves is their speed depends only on the medium they travel though. If you have red and green light passing through a vacuum, for example, they still have the same speed:
I'm having a bit of trouble with VCAA 2014 Exam Question 22 (If anyone has time to help). With part a) is it's first excited state at 4.9 eV? It can absorb a 1.8eV photon because that is the exact difference between the 1st and 2nd excited states but it can't emit a photon from the first excited state because there is no energy level at (4.9eV - 1.8eV) 3.1 eV? I got really confused on this when i did it today. I read the diagram from top to bottom instead of bottom to top. Just wanted to clear it all up.
For part b I'm really lost with what to do. 0.9, 1.5 and 2.2 eV are all between the first excited state. I don't know how to use them at all to find out the unknown excited state.
Thanks
You are essentially correct for part a. Remember that in general atoms can only take on specific discrete energy levels. So absorbing a
For part b, we just need to use some simple observation. Since a
Post by: Floatzel98 on September 10, 2015, 04:42:41 pm
a) Calculate how many photons leave the searchlight each second.
b) What is the momentum of a single 520nm photon.
c) All these photons are focused onto a perfectly reflecting mirror of area 0.1 m^2. Calculate the average force on the mirror.
d) If the mirror in the previous question was replaced by a perfectly absorbing surface, then the answer would:
A - increase by a factor 2
B - decrease by a factor 2
C - remain the same
D - change, but not by a factor of 2
I'm having a bit of trouble on parts c and d. I don't understand how to incorporate the area into the answer. I mean there is no pressure given so that we can. I tried using F =
Any help would be great. Thanks
Post by: Floatzel98 on September 10, 2015, 08:25:31 pm
Post by: odeaa on September 10, 2015, 09:08:26 pm
If you were asked to find a value for the gradient of a stopping voltage against frequency graph, and just say you didn't know how to do it, would you get marks for just writing down the actual value of Planck's Constant or not?
I dont think so unfortunately. I havent seen a question that explicitly asks for the gradient, but I have seen several (VCAA) ones that ask to calculate h from the graph and they all say students were given 0 marks for just saying the value from the data sheet.
If it was worded as asking specifically for the gradient and not plancks constant, they might give you a mark because that shows you know the gradient is h, but i doubt it
Post by: Floatzel98 on September 10, 2015, 10:17:36 pm
I dont think so unfortunately. I havent seen a question that explicitly asks for the gradient, but I have seen several (VCAA) ones that ask to calculate h from the graph and they all say students were given 0 marks for just saying the value from the data sheet.Yeah, didn't think you would get a mark.
If it was worded as asking specifically for the gradient and not plancks constant, they might give you a mark because that shows you know the gradient is h, but i doubt it
This is a bit of a specific question again, but if you were finding the value of h from a graph/table with multiple points and you chose a set of points that gave you a value 'outside' the preferable range, would you still get full marks for that? A couple of VCAA questions I did said that a preferable range would be between 4.7 to 5.3 x 10^-15 eV but I ended up using points that gave me a value closer to 4.14x10^-15 eV
Post by: odeaa on September 10, 2015, 10:20:59 pm
Yeah, didn't think you would get a mark.I've had that exact problem, and I didn't give myself any marks. Usually, they don't want you to use the values from the table, but the graph itself (that's what the question asks for after all) because the line of best fit will be slightly different to the gradient of two random points. You're usually safe with the x and y intercepts (threshold freq and work function respectively) because they are actually on the graph and not just values from the table
This is a bit of a specific question again, but if you were finding the value of h from a graph/table with multiple points and you chose a set of points that gave you a value 'outside' the preferable range, would you still get full marks for that? A couple of VCAA questions I did said that a preferable range would be between 4.7 to 5.3 x 10^-15 eV but I ended up using points that gave me a value closer to 4.14x10^-15 eV
Post by: Kel9901 on September 11, 2015, 02:02:00 pm
A searchlight is beaming out green light. The wavelength of the light is 520nm and the power of the search light is 5.0kW.
a) Calculate how many photons leave the searchlight each second.
b) What is the momentum of a single 520nm photon.
c) All these photons are focused onto a perfectly reflecting mirror of area 0.1 m^2. Calculate the average force on the mirror.
d) If the mirror in the previous question was replaced by a perfectly absorbing surface, then the answer would:
A - increase by a factor 2
B - decrease by a factor 2
C - remain the same
D - change, but not by a factor of 2
I'm having a bit of trouble on parts c and d. I don't understand how to incorporate the area into the answer. I mean there is no pressure given so that we can. I tried using F =p/
Any help would be great. Thanks
I think that the answer for c) would be multiplying the change in momentum of one photon by the number of photons per second (which has a unit of s-1 anyway). The change in momentum that one photon experiences would be double the answer in b), since the photon REFLECTS off the surface (the mirror provides momentum for it to 'stop' AND 'turn around'). So probably 2*what you did
For d), it's probably B, because the photons only 'stop' and don't 'turn around', hence experiencing only half the change in momentum.
Post by: Floatzel98 on September 13, 2015, 04:09:27 pm
Post by: silverpixeli on September 13, 2015, 11:44:20 pm
For the attached question, how could an electron return to the ground state with 13.6 eV if that is equal to the ionization energy? I know now that n = 3 obviously isn't the highest excited state, but I thought that 0 eV or n = infinity was the ionisation energy, not actually an excitation state.
I guess it's a technicality, i think there are lots of excited states up there at not-quite-ionized, and 13.6 is like an asymptote, in that it actually takes you until n=infinity before you get there. also, if an electron was hanging around just outside the electron with no energy i suppose it could 'fall in' emitting a photon on the way down. (waves hands)
Post by: odeaa on September 14, 2015, 02:27:53 pm
Post by: lzxnl on September 14, 2015, 06:20:04 pm
How come when finding the work done using a F/d graph, you use the area under the graph and not simply the force*distance (from the graph)?
Because work = force * distance when the force is constant in magnitude and direction only. If it changes in magnitude, you have to do an integral to sum up all the differential work elements dW = F dx and this integral is the area under the aforementioned graph.
Post by: Floatzel98 on September 15, 2015, 06:18:25 pm
Post by: silverpixeli on September 17, 2015, 12:01:26 pm
In the Light and Matter area of study, is it possible for a question to come up asking us to find the photo-current produced by the photoelectric effect experiment? Because a question like that came up on my SAC which stumped me for a while because I hadn't seen a question like it in any of the practice I had done. Just wondering if this is something that is likely to come up in the exam at all?
never seen it :) what info did it give you?
Post by: Floatzel98 on September 17, 2015, 03:55:49 pm
never seen it :) what info did it give you?It was worded like ' If all the light of the laser was absorbed/used (something like that) then what would the photocurrent be'.
We were given the Power output of the laser and the wavelength of the light.
I tried finding the number of electrons emitted per second (
Firstly I tried finding the Kinetic energy of the light (we also had the work function of the metal) and then finding a voltage from it by
Also is there an allowed value of hc (planck's constant x speed of light) we can use for workings and to put in our calculators because it takes a lot of time typing everything in your calculator constantly.
Post by: silverpixeli on September 17, 2015, 05:36:41 pm
It was worded like ' If all the light of the laser was absorbed/used (something like that) then what would the photocurrent be'.
We were given the Power output of the laser and the wavelength of the light.
I tried finding the number of electrons emitted per second () and then using
to find the current.
Firstly I tried finding the Kinetic energy of the light (we also had the work function of the metal) and then finding a voltage from it by, but what I was doing didn't really seem to make much sense.
Also is there an allowed value of hc (planck's constant x speed of light) we can use for workings and to put in our calculators because it takes a lot of time typing everything in your calculator constantly.
For the constants I would try to store h and c individually in your scientific calculator (h in both units, along with charge and mass of electron, and G)
Type the number, find the 'sto' (store) feature on your calc, and then press the button with whatever capital letter you want to store it under (or google storing constants for your specific calculator model to find instructions)
for ideas, I think i had:
A: gravitational constant
B: charge on electron
C: speed of light
D: mass of electron
X: planck's constant (eVs)
Y: planck's constant (Js)
M: nothing
after you do that, you can just press alpha>letter and put these letters in your formulas rather than typing out the big numbers
ANYWAY in terms of the actual question, i wouldn't expect it on a real exam but you were on the right track
number of photons that hit the metal = number of photoelectrons emitted (assuming everything is absorbed, and that individual photons are above the threshhold frequency)
number of photons:
is equal to number of photoelectrons:
if i havent made mistakes that should work, use whatever value of h and e is consistent with the units P is given in to make sure it works out
Post by: dankfrank420 on September 17, 2015, 09:30:02 pm
Post by: Floatzel98 on September 17, 2015, 10:02:58 pm
How did we know which of plancks' constants to use (eV or J/s)Depends on what you need to find and what is given. You can use either as long as the units are constant with everything else you are using. It's the same as having your speed in m/s or km/h. If you are given speeds in km/h and need an answer in m/s you can opt to convert everything first or leave everything in km/h till the end. You don't want to have an energy in eV while using plancks' constant in J/s.
Post by: silverpixeli on September 18, 2015, 09:29:48 am
Depends on what you need to find and what is given. You can use either as long as the units are constant with everything else you are using. It's the same as having your speed in m/s or km/h. If you are given speeds in km/h and need an answer in m/s you can opt to convert everything first or leave everything in km/h till the end. You don't want to have an energy in eV while using plancks' constant in J/s.
Also if the formula has momentum in it you should put everything in SI units (planck's constant in Js and energy in J if it's there too) because momentum in units like eV is meaningless for our purposes.
Post by: Floatzel98 on September 18, 2015, 07:04:48 pm
Any kind of explanation would be great. Thanks :)
Post by: dankfrank420 on September 18, 2015, 07:21:39 pm
Depends on what you need to find and what is given. You can use either as long as the units are constant with everything else you are using. It's the same as having your speed in m/s or km/h. If you are given speeds in km/h and need an answer in m/s you can opt to convert everything first or leave everything in km/h till the end. You don't want to have an energy in eV while using plancks' constant in J/s.
Ah ok
I read this after I did my exam, luckily I used your logic. Thanks guys!
Post by: Floatzel98 on September 21, 2015, 07:40:41 pm
So the only thing I still don't really understand in the course is modulation and transmission of waves (in electronics). I don't really understand amplitude modulation and frequency modulation (which I don't think is in the course though?). A lot of the questions about carrier waves transferring analog information talk about the carrier wave being in time variation of the intensity or something. I'm just not really understanding it at all.Any help? Tried my best to understand but still a bit unsure.
Any kind of explanation would be great. Thanks :)
Post by: odeaa on September 21, 2015, 10:13:19 pm
Any help? Tried my best to understand but still a bit unsure.I'll give you a quick run down tomorrow, as my computer has decided to shit itself (I'm on my phone atm)
Post by: zsteve on September 22, 2015, 10:52:41 pm
A spring hangs from a ceiling, and a mass is attached to it, causing it to extend.
Say we have a 1 kg mass, and k = 10 N/m so the extension is 1 m.
Thus spring potential energy = 0.5kx^2 = 5 J
However change in GPE = mg(delta h) = 10 J
So some of the GPE has been lost from the system? Where has it gone?
Post by: silverpixeli on September 22, 2015, 11:31:17 pm
Just found out I'm a bit confused with this scenario:
A spring hangs from a ceiling, and a mass is attached to it, causing it to extend.
Say we have a 1 kg mass, and k = 10 N/m so the extension is 1 m.
Thus spring potential energy = 0.5kx^2 = 5 J
However change in GPE = mg(delta h) = 10 J
So some of the GPE has been lost from the system? Where has it gone?
When you attach a mass to a spring and lower it until it balances at equilibrium, you need to take that energy out yourself! (it comes out into your hand as you lower the mass slowly)
If you didn't take this extra 5J of energy out, and you just attached the mass and let it go as a closed system, the extra energy would stay there and the mass would bounce up and down, oscillating.
Post by: Floatzel98 on September 25, 2015, 10:07:04 pm
Thanks
EDIT: Another question, do magnetic field lines ever cross if they come from the same magnet. Like for example, do the magnetic field lines of a simple bar magnet ever cross each other?
EDIT 2: For the attached image, can someone answer the question. The solutions aren't specific. They just talk about Lenz's law, not specifically why the current flows from A to B. How do we know when we can just be vague when answering the question? Are we not actually able to find out the direction of the current on our own?
Post by: Zealous on September 25, 2015, 10:19:56 pm
I know this is super simple, but I'm getting really confused looking at these solutions for a prac exam. If you are in an elevator going down at a constant speed and then you slow down to a stop, what is the direction of the acceleration while you are stopping? Hopefully that is worded okay enough so it's not confusing.
Thanks
EDIT: Another question, do magnetic field lines ever cross if they come from the same magnet. Like for example, do the magnetic field lines of a simple bar magnet ever cross each other?
EDIT 2: For the attached image, can someone answer the question. The solutions aren't specific. They just talk about Lenz's law, not specifically why the current flows from A to B. How do we know when we can just be vague when answering the question? Are we not actually able to find out the direction of the current on our own?
1. Upwards. The normal force from the ground will push up on you more than gravity pushes down on you, so you will accelerate upwards to counter your downwards velocity.
2. Nope. Field lines shouldn't touch at all.
3. The solutions are probably not very good. If the question asks to explain why it flows in that direction, I'd be looking to apply Lenz's law to the situation, not just state it.
Post by: Floatzel98 on September 26, 2015, 11:03:36 am
EDIT: Sorry for so many questions, these exams I'm doing don't have very good solutions. If we needed to draw the normal reaction force for a car do we need to draw two normal forces, for each wheel? I attached the picture.
Post by: odeaa on September 26, 2015, 02:24:34 pm
How do we do question 2 of the attached. I understand we have to use, but I have no idea how to find the force with this information.The solutions just state that the force can be found as 0.005N but I have no idea where they get it from
EDIT: Sorry for so many questions, these exams I'm doing don't have very good solutions. If we needed to draw the normal reaction force for a car do we need to draw two normal forces, for each wheel? I attached the picture.
Dude I was stuck on that exact same question yesterday, 2012 Insight yeah?
Makes absolutely no sense where they got the force from, I have asked multiple people and I think it might just be a mistake
Post by: Floatzel98 on September 26, 2015, 03:14:27 pm
Dude I was stuck on that exact same question yesterday, 2012 Insight yeah?Yeah, 2012 Insight. Hopefully it was just a mistake. I have been sitting here for ages wondering how they got it. I guess I'll just write it off as a mistake for now then.
Makes absolutely no sense where they got the force from, I have asked multiple people and I think it might just be a mistake
Post by: Floatzel98 on September 29, 2015, 03:07:43 pm
Also on the same thought, can a transformer have input voltages on either side? I know that doesn't really make sense, but from the solutions to my above problem it's making me think that power can be transmitted back through the lines and go back to the initial step up transformer.
One more question, when we have to draw de broglie wavelengths around atoms what is the difference between saying an electron in the third excited state or the n = 4 quantum number?
Thanks :)
Post by: zsteve on September 29, 2015, 04:07:31 pm
Quote
It flows across the first wire until it reaches the transformer and then will never need to come back through the transformer again.Current WILL have to flow through the transformer in the reverse direction... it will induce the current in the reverse direction that is. Current does flow in both directions over the transformers (otherwise you violate conservation of charge)
A transformer can take input voltages from either side. There is nothing in its mechanization which precludes this, that is, for the fundamental transformer concept.
3rd excited state <=> n=4 level, so 4 wavelengths around the atomic orbital. No difference, just n=1 is ground state.
Post by: Floatzel98 on September 29, 2015, 05:25:53 pm
@FloatZel98: You need to double the resistance. In a way, you're right in saying that the voltage drop due to the return wire occurs 'on the way back'. However, note that the resistance for the return wires DOES change the voltage drop across EVERY COMPONENT IN THE CIRCUIT. This is because it affects total resistance and hence reduces the current flowing through what is effectively a series circuit. So in response to "The voltage drop over both resistors doesn't matter until you get up to it, which is the case here", yes that is right, but the voltage drop over everything is different because the extra resistance at the 'end' of the current's path.Current WILL have to flow through the transformer in the reverse direction... it will induce the current in the reverse direction that is. Current does flow in both directions over the transformers (otherwise you violate conservation of charge)Thanks so much! Clears everything up :)
A transformer can take input voltages from either side. There is nothing in its mechanization which precludes this, that is, for the fundamental transformer concept.
3rd excited state <=> n=4 level, so 4 wavelengths around the atomic orbital. No difference, just n=1 is ground state.
Post by: dankfrank420 on September 29, 2015, 06:45:24 pm
It's not that they're difficult, its just that the lengthy calculations mean I always mess up.
Post by: Floatzel98 on September 29, 2015, 07:35:19 pm
Anyone got any strategies for those astrophysics questions?Storing values of some of the constants can be very helpful. That way you make sure you don't input any numbers incorrectly. There isn't much else you can do besides double checking what you are writing out and putting in your calculator.
It's not that they're difficult, its just that the lengthy calculations mean I always mess up.
I have a question again. With questions about thermistors (or LDR's) that ask about whether you should increase/decrease the resistance of the resistor in the circuit (or any other variation), is it okay to use a calculation as part of your answer.
Post by: odeaa on September 29, 2015, 07:45:49 pm
I have a question again. With questions about thermistors (or LDR's) that ask about whether you should increase/decrease the resistance of the resistor in the circuit (or any other variation), is it okay to use a calculation as part of your answer.They can't penalise you for it if it's correct
I usually don't bother but for 3 marks might be better of using a calculation just to be on the safe side
Post by: odeaa on September 30, 2015, 04:54:32 pm
So the only thing I still don't really understand in the course is modulation and transmission of waves (in electronics). I don't really understand amplitude modulation and frequency modulation (which I don't think is in the course though?). A lot of the questions about carrier waves transferring analog information talk about the carrier wave being in time variation of the intensity or something. I'm just not really understanding it at all.
Any kind of explanation would be great. Thanks :)
sorry it took so long, here ya go
Post by: dankfrank420 on September 30, 2015, 06:15:53 pm
Post by: Zealous on October 01, 2015, 11:28:26 am
When using the radius of orbit, we include the radius of the mass we're orbiting right? (eg. r = radius of earth + height of satellite above Earths surface).Yep! Make sure to include the altitude in all your calculations.
Post by: Adequace on October 01, 2015, 01:25:23 pm
Post by: zsteve on October 01, 2015, 05:35:26 pm
If they give you an actual diagram (i.e. car) -
- FOUR normal forces, from the GROUND/PLANE on the TYRES
- ONE weight force from CENTRE OF MASS.
If you get a block
- Draw everything from centre of mass, this is a free body diagram.
Unlike Spesh
Post by: odeaa on October 01, 2015, 06:37:44 pm
Just came back from a revision lecture today. Yes, you do need to draw from point of application (something I've been avoiding all year LOL).So the normal reaction for a block is drawn from the middle?
If they give you an actual diagram (i.e. car) -
- FOUR normal forces, from the GROUND/PLANE on the TYRES
- ONE weight force from CENTRE OF MASS.
If you get a block
- Draw everything from centre of mass, this is a free body diagram.
Unlike Spesh
Post by: zsteve on October 01, 2015, 08:47:03 pm
My mistake
Post by: odeaa on October 01, 2015, 09:01:17 pm
Oops sorry: normal force for block still drawn from surface IF it is a force diagram. For free-body diagram, from centre of mass.
My mistake
Allg, cheers for clearing that up
Post by: stockstamp on October 01, 2015, 11:01:57 pm
If you are asked to find the Max Kinetic Energy, Stopping Voltage, Stopping potential, energy of the electron - do you consider this the same thing?
And if so are the units always the same or can they change?
Post by: odeaa on October 02, 2015, 07:40:18 am
In Light, when you get questions on the photoelectric effectEnergy is in joules or eV, stopping voltage/potential is in volts
If you are asked to find the Max Kinetic Energy, Stopping Voltage, Stopping potential, energy of the electron - do you consider this the same thing?
And if so are the units always the same or can they change?
But yes they will be the same value
Post by: zsteve on October 03, 2015, 03:27:30 pm
Also would they reflect VCAA's opinion
Post by: zsteve on October 03, 2015, 04:20:08 pm
Post by: knightrider on October 04, 2015, 12:49:02 am
Angle a is equal to angle b
Post by: TheAspiringDoc on October 04, 2015, 12:55:17 am
Can anyone please explain why in this image attached.Because they are alternating angles, within two parallel lines.
Angle a is equal to angle b
I'd recommend googling ;D
Post by: odeaa on October 04, 2015, 12:33:15 pm
Also do we need to know the different merits/disadvantages of photonic devices, e.g. extremely fast response time for photodiode, etc. for the exam? SD is not too clear on thisYes, I've had a question on why you would use a photo diode and the solutions said because a fast response time was needed. Doesn't hurt to chuck it on your cheatsheet anyway
Post by: odeaa on October 04, 2015, 02:06:44 pm
cheers
Post by: Calculus on October 07, 2015, 06:28:53 pm
Question 15
Students are experimenting with an ideal transformer. The circuit is shown in Figure 19.
The primary coil has 1000 turns; the secondary coil has 6000 turns. There is a 1200 Ω resistor in the
secondary circuit. A 3.0 VRMS AC power supply is connected across the primary coil.
The students now modify the circuit, and connect a 3.0 V DC battery and a switch in the primary circuit,
as shown in Figure 20.
d. The students have been asked to observe the current in the resistor as the switch is closed. Before the
switch is closed, there is no current in the resistor. This does not surprise them. When the switch is
closed, there is a very short pulse of current in the resistor. When the switch remains closed, there is no
current in the resistor.
Explain why there is a short pulse of current as the switch is closed and why there is no current in the
resistor as the switch remains closed. No numbers are required in your answer, but you should refer to
the relevant law of physics. 3 marks
Examiner's Report Comment:
When the switch was closed there was a sudden change (increase) in the current. This resulted in a change in the flux. By applying Faraday’s law this flux change induced a voltage and thus current in the secondary coil. When the switch remained closed there was no further change in the current and thus no change in the flux. This resulted in no more voltage or current in the secondary coil.
Some students referred to the Physics principle involved as Lenz’s law instead of Faraday’s law. It was important to refer to a change in flux rather than a change in magnetic field. Many students described how a transformer worked with AC, but this did not address the question.
Figures 19 and 20 are attached.
Thanks very much in advance. :)
Post by: odeaa on October 07, 2015, 06:35:58 pm
Hi, I was wondering in the question below why it is incorrect to state Lenz's law instead than Faraday's law. This question is from 2013 VCAA, question 15. I don't really understand the statement given in the examiner's report.Lenz law only deals with the direction of the induced current, while faradays law is related to the idea of having a current induced to oppose the change in flux.
Question 15
Students are experimenting with an ideal transformer. The circuit is shown in Figure 19.
The primary coil has 1000 turns; the secondary coil has 6000 turns. There is a 1200 Ω resistor in the
secondary circuit. A 3.0 VRMS AC power supply is connected across the primary coil.
The students now modify the circuit, and connect a 3.0 V DC battery and a switch in the primary circuit,
as shown in Figure 20.
d. The students have been asked to observe the current in the resistor as the switch is closed. Before the
switch is closed, there is no current in the resistor. This does not surprise them. When the switch is
closed, there is a very short pulse of current in the resistor. When the switch remains closed, there is no
current in the resistor.
Explain why there is a short pulse of current as the switch is closed and why there is no current in the
resistor as the switch remains closed. No numbers are required in your answer, but you should refer to
the relevant law of physics. 3 marks
Examiner's Report Comment:
When the switch was closed there was a sudden change (increase) in the current. This resulted in a change in the flux. By applying Faraday’s law this flux change induced a voltage and thus current in the secondary coil. When the switch remained closed there was no further change in the current and thus no change in the flux. This resulted in no more voltage or current in the secondary coil.
Some students referred to the Physics principle involved as Lenz’s law instead of Faraday’s law. It was important to refer to a change in flux rather than a change in magnetic field. Many students described how a transformer worked with AC, but this did not address the question.
Figures 19 and 20 are attached.
Thanks very much in advance. :)
So it would be incorrect to cite lenz law by itself because that would only tell you the direction of the emf, whereas Faraday actually explains why there would be no emf without a change in flux
Post by: Calculus on October 07, 2015, 06:40:16 pm
Lenz law only deals with the direction of the induced current, while faradays law is related to the idea of having a current induced to oppose the change in flux.
So it would be incorrect to cite lenz law by itself because that would only tell you the direction of the emf, whereas Faraday actually explains why there would be no emf without a change in flux
Thanks Odeaa. :)
Post by: zsteve on October 08, 2015, 08:01:17 pm
Also, what is the difference between an AC generator and an alternator? From what I've read before, alternator = AC generator, 'generator' = DC generator. However, the TSSM exam in front of me says differently ...
Post by: zsteve on October 08, 2015, 08:05:31 pm
Post by: lzxnl on October 09, 2015, 03:40:14 pm
Also, is it possible for an electron to de-excite fromlevel - I don't think so, but just confirming?
That's basically capturing an electron by a cation, and yes that happens.
If you somehow managed to make He+, like in a mass spectrometer, you'd reform neutral He pretty easily.
Post by: Adequace on October 09, 2015, 08:46:14 pm
Post by: zsteve on October 09, 2015, 08:50:33 pm
Post by: Adequace on October 09, 2015, 09:03:50 pm
The arrow might be the air resistance force, not net force? Otherwise possibly an error.Ah, appears like it. Cheers.
Post by: paper-back on October 11, 2015, 12:22:48 pm
Or do we just leave it as a sine graph, regardless of whether the flux graph is sine or cosine?
Post by: zsteve on October 11, 2015, 09:11:37 pm
Post by: lzxnl on October 12, 2015, 11:39:14 am
You need to follow the rate of change. This means a sine graph -> negative cosine graph due to Lenz's law. You can simply read points and gradients off the graph and draw out the graph.
To be honest, I don't even know if VCAA expects you to know that a + sine flux graph gives you a - cosine EMF graph (that isn't Lenz's law; that's Faraday's law of induction). I mean, they get you to count squares to work out gravitational potential energy changes so they clearly don't expect knowledge of rudimentary calculus.
Post by: Floatzel98 on October 12, 2015, 07:37:06 pm
We have probably (obviously) made some assumptions that don't make sense (maybe we just haven't understood the course properly yet). We just don't understand why it is such a small speed. Is this all incorrect because a single person can't be treated as single particle (point of mass)? We only did this for fun, but I'm confused about our answer and what it means exactly / why it doesn't make sense.
Also, when doing a double slit experiment, why doesn't air effect any part of the experiment. The light obviously still has to pass through air, so why doesn't this effect the interference pattern at all? Is it just because of the relative size of light compared to these atoms/molecules?
Another thing, I know I've only been doing real physics (that's probably a stretch for VCE) for two years now, but I honestly don't feel I can explain anything that happens in the real world. Like I can't seem to ever really apply any of what I've learn't to real life scenarios. Is this normal? I know I can't expect much out of VCE, but do you actually end up 'learning' things when you get to university?
Thanks guys, any help would be appreciated :)
Post by: lzxnl on October 13, 2015, 07:21:41 pm
Me and a friend were trying to calculate how fast a person would have to travel to be able to 'diffract' through a door. We took the width of a door to be 1m. For diffraction to be observable(?) the wavelength / slit width has to be approximately 1 (). So to find the velocity of a 60kg person to produce a wavelength of approx 1m we went
m/s
We have probably (obviously) made some assumptions that don't make sense (maybe we just haven't understood the course properly yet). We just don't understand why it is such a small speed. Is this all incorrect because a single person can't be treated as single particle (point of mass)? We only did this for fun, but I'm confused about our answer and what it means exactly / why it doesn't make sense.
Also, when doing a double slit experiment, why doesn't air effect any part of the experiment. The light obviously still has to pass through air, so why doesn't this effect the interference pattern at all? Is it just because of the relative size of light compared to these atoms/molecules?
Another thing, I know I've only been doing real physics (that's probably a stretch for VCE) for two years now, but I honestly don't feel I can explain anything that happens in the real world. Like I can't seem to ever really apply any of what I've learn't to real life scenarios. Is this normal? I know I can't expect much out of VCE, but do you actually end up 'learning' things when you get to university?
Thanks guys, any help would be appreciated :)
Actually what you've done looks fine to me. It's such a tiny speed because we're such large objects. Diffraction essentially means there's a macroscopic uncertainty in where your position is. Use Heisenberg's uncertainty principle and you'll find that the uncertainty in momentum required for a person to have a positional uncertainty of 1 m is tiny.
Past VCE, you'll find that VCE physics was absolutely nothing compared to what's out there in uni. Then, when you leave uni, you find that what you did in uni was nothing compared to the real world.
You do learn things at uni; it's just that you learn fundamentals at uni that you'd have to try and then apply in real life. You can't apply VCE stuff to real life because...high school sucks, full stop. There's only so much stuff you CAN learn in two years of high school with such a limited background. Plus, remember high school curricula have to accommodate for all types of students, so incredibly fast-paced accelerated courses aren't allowed.
Post by: Orson on October 14, 2015, 06:27:22 pm
Post by: odeaa on October 14, 2015, 07:34:34 pm
Hey everyone...I was wondering what kind of stuff you are all putting on your cheat sheets...would I need definitions for Work function and stuff? (I remember this coming on an exam, but can't remember if it was VCAA or not...I'm just putting equations, if you have too much theory you won't think intuitively and will just copy down irrelevant answers (well that's my theory)
Post by: Adequace on October 14, 2015, 07:37:18 pm
I'm just putting equations, if you have too much theory you won't think intuitively and will just copy down irrelevant answers (well that's my theory)Isnt there an equation sheet attached to the exam?
Post by: odeaa on October 14, 2015, 07:49:18 pm
Isnt there an equation sheet attached to the exam?Yeah but it's useless as it only has the basics
In 3/4 there a quite a few 'shortcut' formulas that aren't even really shortcuts because literally everyone uses them, you'd probably struggle to finish the exam if you derived all your formulas from their sheet
Also it's nice to have them laid out in a specific format
Post by: Adequace on October 14, 2015, 08:16:06 pm
Yeah but it's useless as it only has the basicsAh, yeah true.
In 3/4 there a quite a few 'shortcut' formulas that aren't even really shortcuts because literally everyone uses them, you'd probably struggle to finish the exam if you derived all your formulas from their sheet
Also it's nice to have them laid out in a specific format
Would assssors care if you just use sqrt(height/5) to find the time in projectile questions? I've heard people say it's not a 'valid' formula to use in the end of year exam.
Post by: Orson on October 14, 2015, 08:20:40 pm
Yeah but it's useless as it only has the basics
In 3/4 there a quite a few 'shortcut' formulas that aren't even really shortcuts because literally everyone uses them, you'd probably struggle to finish the exam if you derived all your formulas from their sheet
Also it's nice to have them laid out in a specific format
I totally agree...it's pretty much useless unless you have memorised the derivations or something. However, I disagree with the theory thing. I hope you put the explanation to Young's Double Slit Experiment at least...otherwise you'll kinda be wasting time trying to phrase it. Just my thoughts...not dissin' yours or anything mate...
Can someone their definition/explanation of YDSE please?
Post by: odeaa on October 14, 2015, 08:35:12 pm
Ah, yeah true.Nah in the assessors report they often say "some students used a pretransposed formula from their cheatsheet which was generally effective if copied correctly" or something along those lines
Would assssors care if you just use sqrt(height/5) to find the time in projectile questions? I've heard people say it's not a 'valid' formula to use in the end of year exam.
Post by: zsteve on October 15, 2015, 05:51:30 pm
Could someone just confirm that this convention is simply due to assigning directions to the sign of flux change? (otherwise it would be dodgy :P)
Post by: odeaa on October 15, 2015, 06:53:08 pm
I'm aware of the convention of describing flux like a vector, e.g. 'flux change upwards' or 'flux change downwards' but flux is a scalar.I usually say increasing/decreasing flux and it's polarity rather than give a direction which I think is more accurate. At the very least, I haven't been penalised for it so I don't think it's inaccurate (for vce anyway).
Could someone just confirm that this convention is simply due to assigning directions to the sign of flux change? (otherwise it would be dodgy)
Lzxnl or someone else may be able to clarify but I don't think you would lose marks in vce for describing it like that
Post by: lzxnl on October 17, 2015, 08:26:52 am
In common-sense terms, imagine you have a door and you're shooting water with a toy gun through it. You can consider the water flow as positive going into the door, or negative going out of the door on the other side. The sign in the magnetic flux is analogous to the sign of your 'water flux'.
You can't really learn electromagnetism properly in high school. Or even in first year uni. You just don't have the mathematical tools to do so.
Post by: GeniDoi on October 17, 2015, 02:52:17 pm
Nah in the assessors report they often say "some students used a pretransposed formula from their cheatsheet which was generally effective if copied correctly" or something along those lines
It's not a cheat sheet if it's endorsed by the examiners...
Post by: schooliskool on October 17, 2015, 04:25:59 pm
I don't do it but does it matter?
Post by: Mc47 on October 17, 2015, 05:35:56 pm
(http://i.imgur.com/43y7oBC.png)
Cheers
Post by: Orson on October 17, 2015, 06:55:35 pm
Could anyone help me out with this question?
Sure! Could you please tell me what paper it was? I've seen it (and done it). I just want to double check!
The answer is: C A D C
C: The signal wave. This is the initial medium, in this case it is sound.
A: The carrier wave. (I'm not sure how to explain this...can someone jump in?)
D: The modulated signal. This is the signal piggybacking, or being superimposed on the medium of transmission. (This could be through a Laser Diode sending, and a Photodiode receiving).
C: The signal wave. This is the output, again sound, only it has been amplified.
I'm sure someone can explain it in more detail. Remember to put this on your summary sheet, along with a definition for modulation.
Post by: odeaa on October 17, 2015, 07:41:27 pm
Post by: Mc47 on October 17, 2015, 08:15:30 pm
Sure! Could you please tell me what paper it was? I've seen it (and done it). I just want to double check!
The answer is: C A D C
C: The signal wave. This is the initial medium, in this case it is sound.
A: The carrier wave. (I'm not sure how to explain this...can someone jump in?)
D: The modulated signal. This is the signal piggybacking, or being superimposed on the medium of transmission. (This could be through a Laser Diode sending, and a Photodiode receiving).
C: The signal wave. This is the output, again sound, only it has been amplified.
I'm sure someone can explain it in more detail. Remember to put this on your summary sheet, along with a definition for modulation.
Ahh okay cheers. It was NEAP 2014
When you say put it on your summary sheet, are the answers always going to be the same?
Post by: Orson on October 17, 2015, 08:45:07 pm
Ahh okay cheers. It was NEAP 2014
When you say put it on your summary sheet, are the answers always going to be the same?
In my experience, these have a pattern. I'll find another example and post it. Stay posted mate. When I say put it on your summary sheet, I mean put the question on it. Then if you get a similar question, you can use it to work backwards or something. The answers won't be the same...I can't guarantee that.
Post by: Orson on October 17, 2015, 08:46:05 pm
Anyone here done the exampro challenge exam? Mind sharing what you got?
Haha...thanks for reminding me to do that. I'll let you know when I do it...
Post by: zsteve on October 17, 2015, 08:48:23 pm
Just a quick one, if you get an answer of 41666.7 etc, should I round it to 4.2 x 10^4?
I don't do it but does it matter?
Yeah I would, definitely wouldn't leave the raw value. The num of sit figs you leave depends somewhat on your input, but not strictly as in chem
Post by: Adequace on October 17, 2015, 09:48:25 pm
Ek: work done by force - W=1/2mv^2 * x
Ug: work done by gravity - W=mgh * x
Thermal Energy: work done by friction - W=Fr * x
Cheers
Post by: GeniDoi on October 17, 2015, 10:08:15 pm
Can someone confirm if these energy/work formulas are legitimate, they were in our notes from my teacher but people have told me they're wrong.
Ek: work done by force - W=1/2mv^2 * x
Ug: work done by gravity - W=mgh * x
Thermal Energy: work done by friction - W=Fr * x
Cheers
Well work = force * distance, but the first two of those formulae aren't in terms of a force, so they are wrong.
The first equation, whoever made it was probably going for W = ΔKE which leads to v^2 = u^2 + 2ax.
The second equation, close but not quite... Work = Force * distance = GMm/r^2 * distance for gravity. Since for most situations the Δr is very small it's possible to approximate GMm/r^2 to mg, so Work = mg * x = mgx = mgh.
The third one is correct but usually friction is W(friction) < 0 and friction opposes the motion and hence removes energy from a system.
Post by: Adequace on October 17, 2015, 10:21:08 pm
Well work = force * distance, but the first two of those formulae aren't in terms of a force, so they are wrong.Ah cheers.
The first equation, whoever made it was probably going for W = ΔKE which leads to v^2 = u^2 + 2ax.
The second equation, close but not quite... Work = Force * distance = GMm/r^2 * distance for gravity. Since for most situations the Δr is very small it's possible to approximate GMm/r^2 to mg, so Work = mg * x = mgx = mgh.
The third one is correct but usually friction is W(friction) < 0 and friction opposes the motion and hence removes energy from a system.
Hopefully I just copied the notes down incorrectly, lmao..
Post by: Orson on October 17, 2015, 11:16:01 pm
Can someone confirm if these energy/work formulas are legitimate, they were in our notes from my teacher but people have told me they're wrong.
Ek: work done by force - W=1/2mv^2 * x
Ug: work done by gravity - W=mgh * x
Thermal Energy: work done by friction - W=Fr * x
Cheers
I've never seen these before... :o I think your better off not using these derivations...Just a thought...
Post by: paper-back on October 18, 2015, 11:03:47 am
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?
Post by: Orson on October 18, 2015, 11:12:20 am
How do you prove that maximum speed occurs at the center of oscillation?
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?
SPOILERS! :o
I haven't seen the question yet, but I assume it is an energy question...Sounds like a Spring/Energy situation....
Post by: odeaa on October 18, 2015, 11:24:50 am
How do you prove that maximum speed occurs at the center of oscillation?Make a table with the gpe, ke and spe at the top, middle and bottom of the oscillation
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?
Because the ke is 0 at both top and bottom, it makes sense that it would be the highest in the centre of oscillation
Post by: paper-back on October 18, 2015, 11:39:06 am
Make a table with the gpe, ke and spe at the top, middle and bottom of the oscillation
Because the ke is 0 at both top and bottom, it makes sense that it would be the highest in the centre of oscillation
Oh I see
Thanks!
Post by: odeaa on October 19, 2015, 10:41:12 am
Post by: Orson on October 19, 2015, 12:17:55 pm
How come when a plane is flying in a vertical circle, it has an apparent weight at the top? Isn't there no normal reaction there? Simple question but it's got me confused
I think it's because there is a reaction force to the centripetal force exerted on the plane. I don't really understand it...but I've come to accept it.
Post by: lzxnl on October 19, 2015, 01:05:48 pm
Post by: Orson on October 19, 2015, 04:28:42 pm
This depends actually on the radius of the circle and speed of the plane. At the top of the circle, there is a 'normal reaction' force provided by the air which pushes upwards on the plane. This may partially balance the downwards gravitational force on the plane.
Ohhhh! So is this called 'lift'? Is this the force created by the aerodynamics (vents and shapes) and whatnot?
Thanks!
Post by: lzxnl on October 19, 2015, 04:50:27 pm
Post by: Cosec on October 19, 2015, 06:03:11 pm
Ohhhh! So is this called 'lift'? Is this the force created by the aerodynamics (vents and shapes) and whatnot?
Thanks!
I could go in to quite some detail, but ill spare it. Basically. Lift is based on a variety of factors, including speed, angle of attack, area of the wing, and some coefficients based on air density, etc. The lift force itself is like the normal for a car on the air. Its the force that keeps them vertically stationary. If a plane is to ascend, it has to have a greater amount of lift than its weight. Vice versa for descending. So while it isnt a normal force, the reaction force is whats keeping the plane flying and thus because their is no 'normal' force but their is a reaction force present, the plane is not going to be apparent weightlessness.
Post by: odeaa on October 21, 2015, 12:51:15 pm
Post by: zsteve on October 21, 2015, 03:31:23 pm
'Net force' is the sum of all vector forces, i.e.
Acceleration is only calculated from net force, i.e.
Post by: odeaa on October 21, 2015, 05:00:38 pm
Looks like you're using ready made formulas - there's no set formula for 'driving force', you derive that.
'Net force' is the sum of all vector forces, i.e.. on a system. It is not the 'driving force'.
Acceleration is only calculated from net force, i.e.. So if you know the acceleration of a system and its mass, you essentially have its net force.
so friction isnt taken into account for net force if you have the acceleration?
Post by: schooliskool on October 21, 2015, 05:07:45 pm
so friction isnt taken into account for net force if you have the acceleration?
Yeah, if you have mass and acceleration given, the net force is calculated ignoring friction. If the driving force is asked, you use net force= driving force - friction
Post by: odeaa on October 21, 2015, 05:11:33 pm
Yeah, if you have mass and acceleration given, the net force is calculated ignoring friction. If the driving force is asked, you use net force= driving force - friction
awesome, cheers
Post by: zsteve on October 22, 2015, 06:55:00 pm
Make a table with the gpe, ke and spe at the top, middle and bottom of the oscillation
Because the ke is 0 at both top and bottom, it makes sense that it would be the highest in the centre of oscillation
Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.
Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.
Post by: GeniDoi on October 24, 2015, 10:44:20 am
Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.
Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.
You would probably be penalized. It's an explain style question where there are dot points you need to hit to get marks. Your solution might be valid in the methods scope but you needed to explain it in terms of physics 3/4 language and ideas.
Post by: lzxnl on October 25, 2015, 09:13:35 am
How do you prove that maximum speed occurs at the center of oscillation?
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?
Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.
Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.
Let me mathematically convince you that the centre of oscillation necessarily is the point of maximum KE.
I'll do this two ways.
First way: using forces.
The centre of oscillation is the point of zero net force, so you have F = ma = 0. This means that you have zero acceleration here, so the velocity must be stationary. Now, F = -kx - mg. As F is linear in x, it changes sign at the stationary point. That means on one side of the centre of oscillation the velocity is increasing and on the other side it's decreasing.
If you increase x, F decreases so when x gets higher, the force is negative. If you decrease x, F increases so when x gets lower, the force is positive. The net effect is that as the force opposes any displacement from equilibrium, your speed will decrease either side of the centre of oscillation. So the centre of oscillation gives the maximum KE.
Second way: using energy
You have 1/2 mv^2 + 1/2 kx^2 + mgx = constant
Let's find the centre of oscillation explicitly. From above. 0 = mg + kx -> x = -mg/k. Remember this result.
Now, let's complete the square on the above.
1/2 mv^2 + 1/2 k(x^2 + 2mg/k x) = 1/2 mv^2 + k/2 ((x+mg/k)^2 - some constant) = constant
I haven't written out the constant because the exact values of these constants aren't important.
1/2 mv^2 + 1/2 k(x+mg/k)^2 = constant
So this is an ellipse. Can you see that the speed is greatest when x = -mg/k, aka the centre of oscillation?
Post by: odeaa on October 25, 2015, 09:39:44 am
Let me mathematically convince you that the centre of oscillation necessarily is the point of maximum KE.You are a king, that helped heaps
I'll do this two ways.
First way: using forces.
The centre of oscillation is the point of zero net force, so you have F = ma = 0. This means that you have zero acceleration here, so the velocity must be stationary. Now, F = -kx - mg. As F is linear in x, it changes sign at the stationary point. That means on one side of the centre of oscillation the velocity is increasing and on the other side it's decreasing.
If you increase x, F decreases so when x gets higher, the force is negative. If you decrease x, F increases so when x gets lower, the force is positive. The net effect is that as the force opposes any displacement from equilibrium, your speed will decrease either side of the centre of oscillation. So the centre of oscillation gives the maximum KE.
Second way: using energy
You have 1/2 mv^2 + 1/2 kx^2 + mgx = constant
Let's find the centre of oscillation explicitly. From above. 0 = mg + kx -> x = -mg/k. Remember this result.
Now, let's complete the square on the above.
1/2 mv^2 + 1/2 k(x^2 + 2mg/k x) = 1/2 mv^2 + k/2 ((x+mg/k)^2 - some constant) = constant
I haven't written out the constant because the exact values of these constants aren't important.
1/2 mv^2 + 1/2 k(x+mg/k)^2 = constant
So this is an ellipse. Can you see that the speed is greatest when x = -mg/k, aka the centre of oscillation?
Post by: Orson on October 25, 2015, 06:45:04 pm
Post by: jyce on October 25, 2015, 08:39:40 pm
Does anyone remember doing any questions that asked you to find the 'slowest photoelectron'? Do these even exist?
There are definitely the 'slowest photoelectrons' - they would have the greatest ionisation energy. However, I have never come across a question asking you to find the slowest ones; only the fastest ones, using the smallest ionisation energy (i.e. the work function).
Post by: Orson on October 25, 2015, 09:01:50 pm
There are definitely the 'slowest photoelectrons' - they would have the greatest ionisation energy. However, I have never come across a question asking you to find the slowest ones; only the fastest ones, using the smallest ionisation energy (i.e. the work function).
Ohk thanks. Do you think you can make one up? I can't find any at the moment. :o
Post by: jyce on October 25, 2015, 09:23:52 pm
Ohk thanks. Do you think you can make one up? I can't find any at the moment. :o
I doubt you'll find one, so don't stress about it. And even if VCAA did decide, on the off chance, to do such a question, it'd be the same kind of thing as working out the greatest kinetic energy / speed.
Ek max. = hf - Eionisation min.
Ek min. = hf - Eionisation max.
Post by: odeaa on October 25, 2015, 10:29:50 pm
Does anyone remember doing any questions that asked you to find the 'slowest photoelectron'? Do these even exist?
they would only have just enough energy to make it past the work function, so their energy would be the same as the work function (roughly)
you could use the threshold frequency to figure this out if they dont give the WF
Post by: jyce on October 25, 2015, 10:41:36 pm
they would only have just enough energy to make it past the work function, so their energy would be the same as the work function (roughly)
you could use the threshold frequency to figure this out if they dont give the WF
Not too sure about your statement that the energy of these slowest photoelectrons "would be the same as the work function"; isn't it the case that those photoelectrons who receive the threshold frequency would have (roughly) NO kinetic energy? I thought what was instead being asked was regarding a case where some frequency, above the threshold, was selected and you would have a range of different kinetic energies - with a minimum and a maximum. In your situation, there would not be a minimum or maximum kinetic energy; rather, there would simply be no kinetic energy?
Post by: lzxnl on October 26, 2015, 04:13:47 pm
More specifically, electron energies follow some kind of distribution after being excited by the light (I'm not too sure what sort of distribution tbh as you have to take into account the light absorption and interactions between electrons) and as the only requirement for a photoelectron is for its total energy to be greater than the energy required to break free, the lowest possible energy of a photoelectron is at this requirement. The kinetic energy would then be 0.
Post by: Floatzel98 on October 30, 2015, 04:12:29 pm
Thanks :)
Post by: yeahm8 on October 31, 2015, 12:00:38 am
Post by: jyce on October 31, 2015, 12:20:33 am
Does your cheatsheet have to be taped together if its 2 A4 pages?
Well mine was taped and that was okay. I'm not sure if there's any other acceptable ways of attaching the two pages, but they do definitely have to be attached.
Post by: paper-back on November 03, 2015, 12:20:35 pm
Post by: Orson on November 03, 2015, 02:29:56 pm
Post by: jyce on November 03, 2015, 03:03:13 pm
Do we need to know about the Doppler effect for the Sound detailed study?
No.
Post by: odeaa on November 03, 2015, 09:08:31 pm
Post by: jyce on November 03, 2015, 10:08:40 pm
Anyone done the 2013 sample exam? Do you reckon its a good indicator or a bit easy?
A bit easy, in my opinion.
EDIT: Well, actually, I just looked at it and I wouldn't say it's that easy. After all, all the questions are from past VCAA exams. However, I'd definitely say it's easier than both the 2013 and 2014 VCAA exams.
Post by: Adequace on November 03, 2015, 11:18:18 pm
Post by: Mc47 on November 03, 2015, 11:37:25 pm
Did you guys learn about impulse in Units 1&2 or 3&4? We've pretty much finished the 1&2 course at school but in the 1&2 checkpoints book it has an energy chapter which requires knowledge of impulse for most of the momentum questions.
We learnt it in 1&2
Post by: Adequace on November 04, 2015, 08:58:58 pm
What are you guys aims? Exam score and SS...Even though I'm finishing physics next year, I'm aiming for around 45. I'll be happy with anything over 40 though but obviously want to get more than just a bare 40.
Post by: sheeep on November 05, 2015, 02:14:01 pm
Post by: jyce on November 05, 2015, 05:26:23 pm
Which grip/slap rule do we use when finding an induced current? I always got confused on when to use which rule
You use the right-hand-grip rule when applying Lenz's law to determine the direction of an induced current. The right-hand-slap rule is used when you are considering magnetic force.
Post by: bedigursimran on November 05, 2015, 07:35:54 pm
A 4WD (2,500 kg) is towing a caravan up in an incline of 110. The 4WD and the caravan encounter a constant
frictional force of 1,250 N and 800 N respectively. The driving force of 13,250 N provided by the 4WD is enough to
accelerate the pair at 1.5 m s-2
Calculate the mass of the caravan.
.
Post by: Floatzel98 on November 05, 2015, 07:49:55 pm
Hey guys. Can you help me with this question? Thanks!!I don't know what the actual incline angle is supposed to be, because it shouldn't be 110 degrees (unless I just don't understand the question), but I'll try to go through a general method to the question. It probably helps to draw up a diagram and label everything btw.
A 4WD (2,500 kg) is towing a caravan up in an incline of 110. The 4WD and the caravan encounter a constant
frictional force of 1,250 N and 800 N respectively. The driving force of 13,250 N provided by the 4WD is enough to
accelerate the pair at 1.5 m s-2
Calculate the mass of the caravan.
.
Firstly set up your equation of motion:
F_(net) = ma = (2500 + m)*1.5
Then pick a direction for positive motion (up the ramp), then:
(2500 + m)*1.5 = 13250 - 1250 - 800 - m_(4WD)gsin(theta) - m_(caravan)gsin(theta)
Just sub in your values for the angle and solve for m that way. Should be correct...
Post by: bedigursimran on November 05, 2015, 08:11:02 pm
I don't know what the actual incline angle is supposed to be, because it shouldn't be 110 degrees (unless I just don't understand the question), but I'll try to go through a general method to the question. It probably helps to draw up a diagram and label everything btw.
Firstly set up your equation of motion:
F_(net) = ma = (2500 + m)*1.5
Then pick a direction for positive motion (up the ramp), then:
(2500 + m)*1.5 = 13250 - 1250 - 800 - m_(4WD)gsin(theta) - m_(caravan)gsin(theta)
Just sub in your values for the angle and solve for m that way. Should be correct...
The incline is 11 degrees, sorry. And cute floatzel. I'll do it now and see if I get it :D
Post by: sheeep on November 06, 2015, 12:36:05 pm
Post by: mtse on November 06, 2015, 12:44:48 pm
Post by: Muchos Help on November 06, 2015, 11:56:04 pm
Good luck in the physics exam boys.
Post by: Floatzel98 on November 07, 2015, 10:29:49 am
Question 5 & 7: For question 5, is this just the net force? And how do we do question 7 exactly? A bit lost on how they got B
Post by: jyce on November 07, 2015, 11:11:51 am
Simple questions here, all from VCAA 2005 Exam 1 (Motion).
Question 5 & 7: For question 5, is this just the net force? And how do we do question 7 exactly? A bit lost on how they got B
Yes, Question 5 is just the net force and therefore is towards the centre of the circle.
For Question 7, when the train applies its breaks there are now two forces acting on the train: the force of the rails on the wheels, which acts toward the centre of the circle, and the breaking force, which acts along a tangent directly backwards from Q. On the diagram, A is the frictional force and C is the breaking force and they add together to give a force somewhere in the middle, which is B.
Post by: Floatzel98 on November 07, 2015, 11:15:50 am
Yes, Question 5 is just the net force and therefore is towards the centre of the circle.That makes sense now. Thank you!
For Question 7, when the train applies its breaks there are now two forces acting on the train: the force of the rails on the wheels, which acts toward the centre of the circle, and the breaking force, which acts along a tangent directly backwards from Q. On the diagram, A is the frictional force and C is the breaking force and they add together to give a force somewhere in the middle, which is B.
Post by: Floatzel98 on November 07, 2015, 11:32:02 am
Any help would be great!
Post by: paper-back on November 07, 2015, 12:54:52 pm
Post by: jyce on November 07, 2015, 01:11:12 pm
When a light bulb does not receive the required voltage, does it stop working entirely or does it just dim down?
Unless it's receiving no where near the correct voltage, I believe it would just dim down. Voltage drop, after all, is a measure of how much electrical energy is being transformed. If the voltage drop across a bulb is lower than desirable, it would just be transforming less electrical energy into light, and consequently it would be shining less brightly.
Post by: Orson on November 07, 2015, 01:20:32 pm
I keep getting 250N, but the answers say 1000N.
Also, what did you all get in VCAA 2013 and VCAA 2014?
Thanks!
Post by: Floatzel98 on November 07, 2015, 01:23:23 pm
What did you guys get for VCAA 2011 Exam 1 Question 2?F = ma = 2000*0.5 = T - 0 (There are no frictional forces)
I keep getting 250N, but the answers say 1000N.
Also, what did you all get in VCAA 2013 and VCAA 2014?
Thanks!
Therefore T = 1000 N
How did you get 250N exactly?
Post by: Orson on November 07, 2015, 01:39:13 pm
F = ma = 2000*0.5 = T - 0 (There are no frictional forces)
Therefore T = 1000 N
How did you get 250N exactly?
T = 500* 0.5 = 250N
Post by: Floatzel98 on November 07, 2015, 01:59:56 pm
T = 500* 0.5 = 250NIf you wanted to find the tension through the tractor you would need to set up an equation of motion of the tractor which would need to include the driving force of it. Your equation finds the net force acting on the tractor alone.
Post by: dankfrank420 on November 07, 2015, 02:47:23 pm
Motion Question 2d)
It states that the "height" of the mass at the max height is 0.8m. However, I thought that considering the max extension of the spring to have a height of 0m, then the max height would therefore be 0.4m as the extension is 0.4m?
If this doesn't make sense sorry, I managed to get the entire methodology right but I just used the wrong value of h. Pretty confused tbh.
Post by: GeniDoi on November 07, 2015, 04:32:34 pm
For VCAA 2014
Motion Question 2d)
It states that the "height" of the mass at the max height is 0.8m. However, I thought that considering the max extension of the spring to have a height of 0m, then the max height would therefore be 0.4m as the extension is 0.4m?
If this doesn't make sense sorry, I managed to get the entire methodology right but I just used the wrong value of h. Pretty confused tbh.
I know the question your talking about. "Extension" refers to whatever the displacement of the end of the spring is after you've done something to it, like attaching a mass to it. If the spring was 0.4m long and you added a mass that made it's total length 1m, it's extension is 1-0.4 = 0.6m
Post by: Adequace on November 07, 2015, 04:34:29 pm
I think I'm putting too many expectations on myself already and expect myself to understand U3&4 exams with my U1&2 knowledge, sigh.
Post by: Floatzel98 on November 07, 2015, 05:00:51 pm
Might be off topic but I'm kinda getting anxious for physics next year...I looked at VCAA 2014 a couple hours ago and literally couldn't do any of it even the motion and electronics section which made me scared lmao.I wouldn't worry at all. You are definitely more prepared than I was. I doubt I even looked at a VCAA exam when I was doing 1/2. I had the exact same problem in specialist this year. At the start of the year when our teacher gave us past exam questions I didn't know how to do almost any of them. If it makes you feel better, I honestly didn't understand a lot of motion physics, conceptually and otherwise, until we did dynamics in spesh haha.
I think I'm putting too many expectations on myself already and expect myself to understand U3&4 exams with my U1&2 knowledge, sigh.
My best advice would be just forget about looking at exams, you aren't expected to know everything already. Contrary to that, maybe try to use it as motivation; keep learning. Make a checklist of things you didn't know from the 2014 exam and tick things off as you do learn them. Just some ideas, but in the end don't worry about it., you'll be fine :)
Post by: dankfrank420 on November 07, 2015, 05:04:02 pm
I know the question your talking about. "Extension" refers to whatever the displacement of the end of the spring is after you've done something to it, like attaching a mass to it. If the spring was 0.4m long and you added a mass that made it's total length 1m, it's extension is 1-0.4 = 0.6m
Exactly, that's why I'm so confused.
The VCAA examiners report contradicts what you and I are agreeing on :o
Post by: Adequace on November 07, 2015, 05:16:53 pm
I wouldn't worry at all. You are definitely more prepared than I was. I doubt I even looked at a VCAA exam when I was doing 1/2. I had the exact same problem in specialist this year. At the start of the year when our teacher gave us past exam questions I didn't know how to do almost any of them. If it makes you feel better, I honestly didn't understand a lot of motion physics, conceptually and otherwise, until we did dynamics in spesh haha.Ah, thanks for the reassurance.
My best advice would be just forget about looking at exams, you aren't expected to know everything already. Contrary to that, maybe try to use it as motivation; keep learning. Make a checklist of things you didn't know from the 2014 exam and tick things off as you do learn them. Just some ideas, but in the end don't worry about it., you'll be fine :)
Good luck on the physics exam everyone.
Post by: zsteve on November 07, 2015, 05:20:40 pm
Only thing I'm a bit worried about for the Physics exam are the spring questions that turned up in 2013 and 2014 - they were highly conceptual. Does anyone have ideas of how to prepare for those? (i.e. any questions in Checkpoints, etc) (Other than that, I'm bracing myself for Spesh and then Chem :P)
Post by: odeaa on November 07, 2015, 05:49:41 pm
Hi,Those spring questions will be the death of me man, praying that they give em a miss this year. It's just the wording that got me, made absolutely no sense considering it was a vcaa which are usually crystal clear
Only thing I'm a bit worried about for the Physics exam are the spring questions that turned up in 2013 and 2014 - they were highly conceptual. Does anyone have ideas of how to prepare for those? (i.e. any questions in Checkpoints, etc) (Other than that, I'm bracing myself for Spesh and then Chem
Post by: zsteve on November 07, 2015, 05:52:35 pm
Those spring questions will be the death of me man, praying that they give em a miss this year. It's just the wording that got me, made absolutely no sense considering it was a vcaa which are usually crystal clearWell hopefully, since Tasmania Jones is nowhere to be seen :)
I mean, I coped with those questions, but I couldn't see the proof (2014) that maximal Ek occurred in the middle of the oscillation. It made logical/practical sense, but I wasn't satisfied until I had shown that it was so using a mathematical method. Not entirely sure how VCAA wanted it to be argued, although I suspect they just wanted people to go by 'instinct'
Post by: dankfrank420 on November 07, 2015, 06:07:57 pm
Well hopefully, since Tasmania Jones is nowhere to be seen :)
I mean, I coped with those questions, but I couldn't see the proof (2014) that maximal Ek occurred in the middle of the oscillation. It made logical/practical sense, but I wasn't satisfied until I had shown that it was so using a mathematical method. Not entirely sure how VCAA wanted it to be argued, although I suspect they just wanted people to go by 'instinct'
The answer to this question confused the fuck outta me.
Did you manage to get it correct?
Post by: dankfrank420 on November 07, 2015, 06:08:44 pm
Those spring questions will be the death of me man, praying that they give em a miss this year. It's just the wording that got me, made absolutely no sense considering it was a vcaa which are usually crystal clear
3 years in a row? I wouldn't think so.
Post by: odeaa on November 07, 2015, 06:09:54 pm
3 years in a row? I wouldn't think so.That's what I'm hoping, although each year they've been poorly done. And 90% of physics is exact same as the preceding year ahaha it's not as though they're scared of repetition
Post by: Orson on November 07, 2015, 06:14:07 pm
Post by: zsteve on November 07, 2015, 06:22:11 pm
Post by: zsteve on November 07, 2015, 07:23:22 pm
Attempted doing a general proof but messed that up twice (misconceptions abounded :( )
However, here's my working for the VCAA 2014 question again. It completely avoids the assumption that max speed occurs in middle of oscillation (and is thus the mathematically rigorous method of proof). The examiners' report simply stated that max speed = middle of oscillation. This is probably a handy fact to remember then ;) But for those who are curious or don't like assumptions:
Spoiler
https://onedrive.live.com/redir?resid=2692D5EC8060E581!6587&authkey=!AHb8GmRybJWZUUo&ithint=file%2cpdf
Post by: odeaa on November 07, 2015, 07:27:16 pm
Hi guys,
Attempted doing a general proof but messed that up twice (misconceptions abounded :( )
However, here's my working for the VCAA 2014 question again. It completely avoids the assumption that max speed occurs in middle of oscillation (and is thus the mathematically rigorous method of proof). The examiners' report simply stated that max speed = middle of oscillation. This is probably a handy fact to remember then ;) But for those who are curious or don't like assumptions:Does use methods knowledge which is beyond the Physics SD, but who cares. My physics teacher said it was acceptable.Spoilerhttps://onedrive.live.com/redir?resid=2692D5EC8060E581!6587&authkey=!AHb8GmRybJWZUUo&ithint=file%2cpdf
link doesnt seem to work for me, keen to see this
Post by: Orson on November 07, 2015, 07:55:20 pm
link doesnt seem to work for me, keen to see this
Can someone please explain why the 'wrong' way of doing VCAA 2013 Question 1 is wrong? itute did it that way, but VCAA said it was wrong.
Thanks so much everyone!
Post by: odeaa on November 07, 2015, 08:14:13 pm
Can someone please explain why the 'wrong' way of doing VCAA 2013 Question 1 is wrong? itute did it that way, but VCAA said it was wrong.
Thanks so much everyone!
Which part of the question? part a?
Post by: zsteve on November 07, 2015, 08:17:52 pm
link doesnt seem to work for me, keen to see thisHi, you need to copy the entire link (only part of it is hyperlinked for some reason)
Post by: odeaa on November 07, 2015, 08:38:32 pm
Hi, you need to copy the entire link (only part of it is hyperlinked for some reason)cheers, that makes sense
surely if you wrote that in an exam you wouldnt be penalised? shows greater knowledge than just guessing that max is in the middle
Post by: GeniDoi on November 08, 2015, 11:11:39 am
There was a really good PBS space time challenge episode about which particle would win the race to travel to the opposite side of the Earth (assuming it is uniform density):
One that is thrown into orbit versus one that is "dropped" down a straight hole that goes through the center of the Earth and to the other side.
Link for those who are interested (if you like physics you 100% will be): https://www.youtube.com/watch?v=MUThGpp6ze4
The solution turned out to be that they tie and meet on the opposite end of the Earth at the exact same time, and the way you solve the problem is you assume the particle that goes through the middle of the Earth behaves kinda of like a spring, so it obey's hooks law, so if you never touched the particle again and there were no frictional forces it would oscillate between the two ends of the Earth forever.
My point is, if you treat the particle as a spring (a completely valid assumption), it becomes very obvious that the maximum speed will be at the center of the oscillation (at the exact centre of the Earth) because the only type of energy it has at the center of the Earth is kinetic energy. It has no gravitational potential energy whatsoever because at the center of the Earth your height = 0. Moreover, due to the symmetry of the problem (conservation of energy + a circular earth) the particle will have no speed at the surface of the earth, so you can (somewhat naively, but still validly) say that the only logical place that it can have a maximum speed is at the midpoint, the centre of the Earth.
If this has confused anyone I'm sorry. Just trying to put in different terms.
Post by: Floatzel98 on November 08, 2015, 11:50:52 am
Post by: GeniDoi on November 08, 2015, 11:53:13 am
In the solutions for the orbiting particle how exactly do they express the period in terms of density (Equation (6))? Do they just take the volume for a sphere and substitute v = m / p to get the equation they did? Why exactly do they only want it in terms of density for?
Here are the solutions. The Newtonian one is completely accessible to a physics 3/4 understanding.
http://bit.ly/particle_challenge
Post by: Floatzel98 on November 08, 2015, 12:00:01 pm
Here are the solutions. The Newtonian one is completely accessible to a physics 3/4 understanding.Yeah, that's what I was referring to. They make sense, I just don't understand why they need to express it in terms of the density? Why nothing else? Is it just because the planet is supposed to be 'uniformly dense'?
http://bit.ly/particle_challenge
Post by: GeniDoi on November 08, 2015, 12:31:12 pm
Yeah, that's what I was referring to. They make sense, I just don't understand why they need to express it in terms of the density? Why nothing else? Is it just because the planet is supposed to be 'uniformly dense'?
Because a varying density planet changes the problem, making it significantly more complex to solve algebraically (if not downright impossible, I'm not even sure if there is an algebraic mathematical representation of the density of the earth vs depth).
As a result of that, the assumption that the falling particle only experiences a force towards the inner sphere becomes invalid and you ruin the symmetry of the problem. The point is that by making these assumptions elegant algebraic expressions can be found. There are tonnes of things which would prevent this from working in real life, from the Coriolis effect to the atmosphere to the gravitational perturbations by the sun and planets, and of course the down right impossibility of drilling a hole through the center of the Earth. If you take real life factors into account it really doesn't work anymore. Abstraction is necessary to arrive to these nice expressions.
If all that wasn't enough, it's actually wrong since it uses Newtonian mechanics but I have no idea about how the Einstein solution works.
EDIT: For anyone stumbling here worrying what this discussion is about; it's not part of the course.
Post by: Orson on November 08, 2015, 12:43:22 pm
There are 2 ways, using SUVAT, and using forces.
Quote from: VCAA
There were two methods of approaching this question. One involved using the constant acceleration
formula
x = ut + 1/2at2. Alternatively, because the there was no friction, the net force on the trolley was
the component of the weight down the plane (5 sin10). Substituting this into Newton’s second law (5
sin10 = 0.5 × a) gave an acceleration of
1.74 m s–2.
A common error was to determine the average speed (3.5/2 = 1.75), incorrectly assume this was the
final speed at the bottom of the ramp and calculate the acceleration from v = u + at.
Post by: GeniDoi on November 08, 2015, 12:49:14 pm
F(net) = ma
=>
=> gsin(10) = a
=> a = 10sin(10) = 1.74m/s^2
(note that in general for inclined planes with gravity being the only force, a = gsin(theta).
The good thing about doing it this way is you don't need to know the time or distance it's moved, which VCAA might not always tell you.
Post by: Orson on November 08, 2015, 12:54:45 pm
The good thing about doing it this way is you don't need to know the time or distance it's moved, which VCAA might not always tell you.
Cheers. I did it this way in the exam:
u = 1, t = 2, s = 3.5, a = ?
s = ut + at^2
3.5 = 0 + (1/2)*a*(2)^2
3.5 = (1/2)*a*4
3.5 = 2a
a = 1.75m/s^2
Is this correct? They did it this way for part b)
Thanks mate!
Post by: GeniDoi on November 08, 2015, 12:57:22 pm
Cheers. I did it this way in the exam:
u = 1, t = 2, s = 3.5, a = ?
s = ut + at^2
3.5 = 0 + (1/2)*a*(2)^2
3.5 = (1/2)*a*4
3.5 = 2a
a = 1.75m/s^2
Is this correct? They did it this way for part b)
Thanks mate!
Completely fine and had VCAA not given us the value of the angle, your method would be the only way to do it. However if the time/distance wasn't supplied but the angle was, the net force way would be the only way to solve it.
Post by: Orson on November 08, 2015, 01:04:56 pm
Completely fine and had VCAA not given us the value of the angle, your method would be the only way to do it. However if the time/distance wasn't supplied but the angle was, the net force way would be the only way to solve it.
Okay. Cheers heaps mate for the help! I wasn't sure whether SUVAT works on slopes...
Post by: lzxnl on November 08, 2015, 07:07:51 pm
Because a varying density planet changes the problem, making it significantly more complex to solve algebraically (if not downright impossible, I'm not even sure if there is an algebraic mathematical representation of the density of the earth vs depth).
As a result of that, the assumption that the falling particle only experiences a force towards the inner sphere becomes invalid and you ruin the symmetry of the problem. The point is that by making these assumptions elegant algebraic expressions can be found. There are tonnes of things which would prevent this from working in real life, from the Coriolis effect to the atmosphere to the gravitational perturbations by the sun and planets, and of course the down right impossibility of drilling a hole through the center of the Earth. If you take real life factors into account it really doesn't work anymore. Abstraction is necessary to arrive to these nice expressions.
If all that wasn't enough, it's actually wrong since it uses Newtonian mechanics but I have no idea about how the Einstein solution works.
EDIT: For anyone stumbling here worrying what this discussion is about; it's not part of the course.
Yeah, you can actually still represent the gravitational force if you don't have a constant density. The solution would then depend on the nature of the density. If you naturally assume spherical symmetry (i.e. the density only depends on the distance to the centre of the Earth and that rotation of the Earth doesn't change the density), then you can use Gauss's law for gravity to work out the gravitational field (and hence force) at any point within the Earth.
Essentially, Gauss's law for gravity would say that g * 4pi*r^2 = -4pi GM(enc) = -4pi GM(total) * (r/R)^3. It still does this for a spherically symmetric mass distribution; a uniform density is just a special case of this.
g = -GM(total)r/R^3
It's the same as if you had a uniform density, so GeniDoi's assumption that it behaves as a spring would be justified if and only if you dropped the spring on the surface of the Earth but that condition is met here.
Then you'd just use R^3/T^2 = GM/4pi^2
Find that T^2 = 4pi^2R^3/GM
From the spring equation above, g = a = -GMr/R^3 = -kr/m where k = GmM/R^3
Period of a spring with spring constant k is 2pi * sqrt(m/k) = 2pi * sqrt(R^3/GM) = T
If the periods are equal, so are the half periods
The two equations are equivalent
So the half periods are the same
Essentially, this proof works for any spherically symmetric body and that's the most general case possible.
Post by: lzxnl on November 09, 2015, 05:32:17 am
You'd.have to replace it with an integral of the volume over the desired volume instead and that would get annoying.
Post by: odeaa on November 09, 2015, 02:56:39 pm
Hi guys,
Attempted doing a general proof but messed that up twice (misconceptions abounded :( )
However, here's my working for the VCAA 2014 question again. It completely avoids the assumption that max speed occurs in middle of oscillation (and is thus the mathematically rigorous method of proof). The examiners' report simply stated that max speed = middle of oscillation. This is probably a handy fact to remember then ;) But for those who are curious or don't like assumptions:Does use methods knowledge which is beyond the Physics SD, but who cares. My physics teacher said it was acceptable.Spoilerhttps://onedrive.live.com/redir?resid=2692D5EC8060E581!6587&authkey=!AHb8GmRybJWZUUo&ithint=file%2cpdf
yeah if anyone was wondering, I asked my teacher who used to be chief assessor and actually helped write a few of the older study designs, and he said that if you use correct physics they will never penalise you. This includes using integration for the area under graphs- although I think this is only really useful if you know the equation of the graph (eg. kinetic energy, gravity equations etc)
Post by: Orson on November 10, 2015, 04:44:45 pm
Thanks for all the help over the year, and especially these last few weeks. See you all on the other side...
Post by: Orson on November 10, 2015, 05:04:46 pm
g = (G*M)/(R^2), g = 10m/s^2
Why is this wrong?
Thanks!
Post by: byCrypt on November 10, 2015, 05:06:26 pm
Post by: zsteve on November 10, 2015, 05:19:56 pm
Post by: Adequace on November 10, 2015, 05:31:21 pm
Is it too late to start studying for physics?Dunno, just ask pi.
(He apparently studied the night before the physics exam only :P).
Post by: odeaa on November 10, 2015, 05:37:18 pm
Is it too late to start studying for physics?
Dunno, just ask pi.
(He apparently studied the night before the physics exam only :P).
beat me to it ahah
Post by: Orson on November 10, 2015, 05:51:29 pm
deltaX = 0.4, not 0.8
Post by: byCrypt on November 10, 2015, 07:11:57 pm
deltaX is the change of length of a spring right? Could someone have a look at VCAA 2014 2b? They ask for the extension, but the answer is the extension + length.
deltaX = 0.4, not 0.8
I think the answer is extension, as they equated the gravitational potential energy at the top (bottom of spring at unstretched length) to the spring potential energy at the bottom. The h in mgh is equal to the x in 1/2kx^2. So I believe the extension is 0.8m.
Post by: cooldude123 on November 10, 2015, 07:35:21 pm
I was doing VCAA 2014 Q5a. I used this:
g = (G*M)/(R^2), g = 10m/s^2
g=10m/s^2 only applies to gravity at the Earth's surface (not to a planet orbiting a star). The value for g would be different.
You need to use Kepler's Laws and transpose for m.
Post by: Orson on November 10, 2015, 07:43:58 pm
g=10m/s^2 only applies to gravity at the Earth's surface (not to a planet orbiting a star). The value for g would be different.
You need to use Kepler's Laws and transpose for m.
Cheers dude. You're pretty cool, you know that?
Post by: SayJay on November 10, 2015, 08:32:08 pm
In section A, question 8 (b+c) I've calculated the period value to be 1.56 * 10^4 s, and yet itute solutions say 1.53 * 10^4 s? I've used that value from 8(b) in 8(c) when finding the velocity, so I got 5.44 * 10^3 m/s (itute says 5.54 * 10^3 m/s).
Post by: SayJay on November 10, 2015, 09:13:40 pm
Just checking this: for AC calculations (power, energy, etc) always use RMS values - yes?
Yes, unless stated otherwise.
Post by: paper-back on November 11, 2015, 09:16:16 am
Post by: Orson on November 11, 2015, 09:55:01 am
What is RMS exactly?
Root Mean Square...
Post by: schooliskool on November 11, 2015, 10:19:47 am
Is it too late to start studying for physics?Is it too late now or na
Post by: Orson on November 11, 2015, 10:43:01 am
Is it too late now or na
No...you can probably learn something...
Post by: paper-back on November 11, 2015, 11:11:51 am
Root Mean Square...
Sorry, I meant; what does RMS voltage/current describe?
Is it the AC equivalent of DC? So DC/(root 2) provides the AC voltage that is equivalent to the DC voltage?
Post by: Orson on November 11, 2015, 12:22:01 pm
Sorry, I meant; what does RMS voltage/current describe?
Is it the AC equivalent of DC? So DC/(root 2) provides the AC voltage that is equivalent to the DC voltage?
I'm not sure...sorry. I think its ACs voltage respective to DC...not sure...all I know is that you use the RMS when doing equations and finding current, voltage drop and power and stuff, and you only use Vpeak-peak when graphing graphs...
Post by: jumcakes on November 11, 2015, 12:29:58 pm
Sorry, I meant; what does RMS voltage/current describe?
Is it the AC equivalent of DC? So DC/(root 2) provides the AC voltage that is equivalent to the DC voltage?
If DC voltage is the same as AC(peak)/root (2) = rms of AC, then both the AC and the DC provide the same amounts of power, I believe.
Post by: ClivePalmer on November 12, 2015, 10:50:46 am
(http://i.imgur.com/rMvzwbs.jpg)
Cheers just wondering how the marking works
Post by: Fusuy on November 12, 2015, 05:30:53 pm
Sacs: Rank 1
Exam 147/150
Thanks!
Post by: Davos on November 12, 2015, 05:40:16 pm
Hey Guys, was just wondering if someone could give me a study prediction.
Sacs: Rank 1
Exam 147/150
Thanks!
I'm no pro at predictions but that is certainly at least a 49, likely 50
Post by: Zealous on November 12, 2015, 05:50:38 pm
Hey Guys, was just wondering if someone could give me a study prediction.
Sacs: Rank 1
Exam 147/150
Thanks!
Yeah, you'll get yourself another 50 with those exam scores. :)
Post by: Davos on November 12, 2015, 05:53:47 pm
Yeah, you'll get yourself another 50 with those exam scores. :)
Zealous do you know what you got on the Physics exam if I may ask?
Post by: Fusuy on November 12, 2015, 06:17:54 pm
Post by: dave101 on November 12, 2015, 08:15:31 pm
GA1: Low A
GA2: Low A
GA3: 128/150
Rank: 20/65
I did physics 3/4 in year 11 this year
Cheers :)
Post by: Mc47 on November 12, 2015, 10:41:31 pm
Hey guys what score do you reckon this would be?
GA1: Low A
GA2: Low A
GA3: 128/150
Rank: 20/65
I did physics 3/4 in year 11 this year
Cheers :)
38-40
Post by: Maz on December 08, 2015, 05:53:12 pm
just wandering if anyone could answer these questions for me please?
1. Explain the consequences of the total reaction force of a roadway on a car being at an angle to the surface on a horizontal road.When would this occur?
2. At the end of many steep declines there are emergency exit points for trucks that have sand as a base rather than a solid railway. Explain why these are necessary.
Thankyou in advance :)
Post by: lzxnl on December 08, 2015, 06:49:07 pm
Hey guys
just wandering if anyone could answer these questions for me please?
1. Explain the consequences of the total reaction force of a roadway on a car being at an angle to the surface on a horizontal road.When would this occur?
2. At the end of many steep declines there are emergency exit points for trucks that have sand as a base rather than a solid railway. Explain why these are necessary.
Thankyou in advance :)
Well, if the reaction force is at an angle, the reaction can actually contribute to the centripetal force required to make the car go in a circle. This way, you're less likely to have a car skidding out.
As for the emergency exit points, I'd assume it's to quickly slow the wheels down by (rolling) friction as it's not easy for wheels to travel in sand.
Post by: Muchos Help on December 10, 2015, 04:24:56 pm
Post by: odeaa on December 10, 2015, 05:36:15 pm
For the last question of the attached. How likely will these sort of motion questions be in the VCAA end of year exams? Struggled with this question since I didn't expect it to get this mathsy in VCE physics, ended up getting it by looking at the worked solution though lmao.Yeah I've seen a few like that, those graphs are pretty common
Post by: lzxnl on December 10, 2015, 06:38:53 pm
For the last question of the attached. How likely will these sort of motion questions be in the VCAA end of year exams? Struggled with this question since I didn't expect it to get this mathsy in VCE physics, ended up getting it by looking at the worked solution though lmao.
Questions 9 and 10 are fairly typical year 12 questions.
Question 11 is a little trickier in that you have to equate the areas under two graphs and while that's assessable in VCE physics, I don't recall seeing a question like this in a while. VCE physics motion questions now tend to be fairly simple number crunching questions. A lot has happened in 35 years.
Post by: Muchos Help on December 11, 2015, 09:07:08 pm
Questions 9 and 10 are fairly typical year 12 questions.Thanks.
Question 11 is a little trickier in that you have to equate the areas under two graphs and while that's assessable in VCE physics, I don't recall seeing a question like this in a while. VCE physics motion questions now tend to be fairly simple number crunching questions. A lot has happened in 35 years.
I have another tough question I couldn't do, it's the last question of the attached. In the answer it states that the motorcycle travels 100m in every 5secs after t=15 which I don't quite understand how they got that.
Also, they then calculate the time the motorcycle takes to travel the 350m but isn't the car also moving at the same time?
Post by: Adequace on December 16, 2015, 03:44:26 pm
I got 7m/s^2 but they got 7.2m/s^2 for the question prior. Although their answer is closer, I'm not exactly sure how you can guess the correct value.
Post by: wyzard on December 18, 2015, 02:30:51 pm
For the attached on Q6. The answer just said "this ratio must always equal 1", why is that and does this mean anything?
I got 7m/s^2 but they got 7.2m/s^2 for the question prior. Although their answer is closer, I'm not exactly sure how you can guess the correct value.
The value of 7.2 is gotten by reading the graph more accurate with the use of a ruler to measure the length. FOr the second part, the ratio of 1 means that the gravitational force acting always stays the same.
Post by: Adequace on December 18, 2015, 03:01:16 pm
The value of 7.2 is gotten by reading the graph more accurate with the use of a ruler to measure the length. FOr the second part, the ratio of 1 means that the gravitational force acting always stays the same.Thanks.
So, would the gravitational force always equal 1? Would there ever be a situation where it doesn't equal 1?
Post by: wyzard on December 18, 2015, 05:30:47 pm
Thanks.
So, would the gravitational force always equal 1? Would there ever be a situation where it doesn't equal 1?
Gravitational force don't equal to one, it's the ratio of the forces at different time that equal to one, meaning they're the same.
Of course there are situations when the ratio will not equal to one. When we talk about large distances away from Earth where gravity is no longer uniform, like satellites that orbit the Earth. From there we cannot use
Post by: Adequace on December 18, 2015, 05:34:01 pm
Gravitational force don't equal to one, it's the ratio of the forces at different time that equal to one, meaning they're the same.Many thanks.
Of course there are situations when the ratio will not equal to one. When we talk about large distances away from Earth where gravity is no longer uniform, like satellites that orbit the Earth. From there we cannot useanymore, but we'll need another way to calculate gravitational force via Newton's Law of Universal Gravitation which is:
Post by: Maz on December 21, 2015, 06:19:32 pm
1. How long will it take a ball thrown at 12ms^1 at an angle of 70 degrees above the horizontal to reach a height of 4m above its launch position?
2. Sketch separate speed versus time graphs for the horizontal and vertical motions of a projectile that it launched directly upwards and lands below the launch position
Thankyou in advance :)
Post by: Muchos Help on December 21, 2015, 09:38:03 pm
hello...can someone please help me on these...its projectile motion :)For 1, you probably just have to use the equations of motion with the variables that the question gives you. (a,x u,v)
1. How long will it take a ball thrown at 12ms^1 at an angle of 70 degrees above the horizontal to reach a height of 4m above its launch position?
2. Sketch separate speed versus time graphs for the horizontal and vertical motions of a projectile that it launched directly upwards and lands below the launch position
Thankyou in advance :)
2. For the horizontal graph, it would just be a horizontal line at +12m/s since horizontal a=0. I'm not 100% on the vertical graph but I'd say it's a negative slope from +12m/s
Post by: Peanut Butter on December 22, 2015, 02:54:36 pm
Tam throws a ball across a field a horizontal distance of 100m. She throws it at an angle of 30 degrees. It lands at the same height (ignore air resistance). At what speed does the ball leave her hand?
The answer is in the spoiler below :D
Spoiler
The answer is 34 m/s
Thank you!
Post by: lzxnl on December 22, 2015, 03:21:30 pm
Could someone please help me with the following projectiles question? :)
Tam throws a ball across a field a horizontal distance of 100m. She throws it at an angle of 30 degrees. It lands at the same height (ignore air resistance). At what speed does the ball leave her hand?
The answer is in the spoiler below :DSpoilerThe answer is 34 m/s
Thank you!
Let's break this down completely. I'm assuming that you recognise this is a constant acceleration in 2D problem (vertical acceleration =g downwards, horizontal acceleration = 0. This part is important; please be sure of this first)
What do we need for a horizontal distance? Well, because the acceleration is zero in that direction (the vertical direction doesn't affect the horizontal direction to a very good approximation; this has been verified by experiment and it follows from the fact that the gravitational force is vertical. I say approximation because the Earth is a rotating sphere and that complicates things but let's disregard that), the distance is just the horizontal velocity * time. The horizontal velocity is related to the speed and the angle, which you have.
So we have 100 = v(x)t = u*cos 30*t
The problem is, this is one equation in two unknowns, so we need another piece of information. This is where we use the vertical motion.
If the ball is thrown across a horizontal field, its total vertical displacement is identically zero. This is also important. In these questions, you need to recognise what words mean what mathematically. That's the art of problem solving: converting given information into useful information.
Anyway, you now have the vertical displacement and the acceleration. Remember how we still don't know the speed at the time? Let's write out another equation that uses them so that we can solve for them. The relevant constant acceleration equation that we want now is the one that involves the initial vertical velocity (to include the speed), the time and the other two known quantities, so our equation becomes 0 = v(y)t - 1/2 at^2.
t isn't zero because we're not interested in the instant the ball is thrown.
So v(y) = 1/2 at -> u sin 30 = 1/2 at, u = at
Plug this into the other equation and solve for the speed.
Of course, you can do this in general and with a bit of algebra, you can find that the general formula for the range for a projectile with no gain in height is R = v^2 sin (2*theta) / g. I'll leave that up to you. Hint: you'll need a double angle trig identity somewhere for this.
Post by: Peanut Butter on December 22, 2015, 03:24:40 pm
Post by: JI2015 on December 22, 2015, 03:34:59 pm
I know someone has already answered the question for you, but you may like to see how I approached this question.
http://jmp.sh/v9IEs96
All the best!
Also, just some advice for physics calculations: When I had trouble trying to start the question, I always wrote down the information the question provided me with and then what information I was trying to find. It makes it much easier to see where you are going and how you can relate each piece of information together to find the answer. If still stuck, my cheat sheet had formulas grouped for each question type (e.g. projectile motion) and I would quickly look and see which one's I would have to use given the specific information provided by the question.
Post by: Peanut Butter on December 22, 2015, 03:49:54 pm
The formulas at the bottom and the way you set out your solution was really helpful! :)
Post by: knightrider on December 28, 2015, 07:12:01 pm
Like how does the rounding work ?(when giving answers)
Post by: JI2015 on December 28, 2015, 07:55:50 pm
Post by: knightrider on December 28, 2015, 08:04:17 pm
In VCE Physics (Units 3 + 4 2016), you are not required to have answers correct to the appropriate amount of significant figures (someone correct me if I'm wrong). You will cover basic data analysis in physics and will be taught how sig. figs. work. You will also learn about sig. figs. in VCE Chemistry where they have a question every year where they randomly check your sig. figs. and marks can be lost, but in physics as long as your answer is reasonable (i.e. 2-4 sig. figs.) then you should be fine. Your answers will usually have 2 decimal places at most, anything more is inappropriate considering the uncertainty present. If you are studying Units 3 + 4 Physics in 2017, then there might be more weight placed on accuracy of answers if I recall correctly.
Thanks so much JI2015 for clarifying :)
Post by: knightrider on December 28, 2015, 08:12:32 pm
8a and b are 6.26ms^-1
8c is 5.42ms^-1
Now for these following questions can someone check if my working out is right?
d Which object had the greater change in speed as it landed? Calculate the speed change of this object.
golf ball-
tomato-
e Which of these objects experienced the greater change in velocity as it landed? Calculate the velocity change of this object.
Let down be positive.
golf ball-
tomato-
Post by: JI2015 on December 28, 2015, 08:24:21 pm
Post by: knightrider on December 30, 2015, 01:58:48 pm
When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )
Like refer to the image attached.
How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)
Post by: knightrider on December 30, 2015, 02:12:06 pm
All the book says is "The forces acting on a bouncing ball."
Post by: JI2015 on December 30, 2015, 02:54:48 pm
When resolving vectors into a triangle.
When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )
Like refer to the image attached.
How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)
Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.
Can anyone explain what is happening in this diagram attached.
All the book says is "The forces acting on a bouncing ball."
Ok, let's start with Picture 1 (noting that air resistance is being neglected):
The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.
Picture 2:
When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.
Picture 3:
The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.
Picture 4:
The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.
Picture 5:
Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.
I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.
Post by: Syndicate on December 30, 2015, 03:09:51 pm
Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.
Ok, let's start with Picture 1 (noting that air resistance is being neglected):
The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.
Picture 2:
When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.
Picture 3:
The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.
Picture 4:
The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.
Picture 5:
Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.
I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.
Just want to extend on JI2015's response.
If you also have a look at the force G (gravity), it stays the same, as you may remember, that the gravity on Earth is 9.8 m/s/s.
Due to the acceleration of gravity, all objects have the same rate of acceleration, when they free fall (no matter what their mass is). So if you divide the Netforce by the weight, you will always get 9.8 m/s/s.
I have attached a picture below, hopefuly that helps :)
Post by: Syndicate on December 30, 2015, 03:16:36 pm
When resolving vectors into a triangle.
When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )
Like refer to the image attached.
How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)
If you have a look at Vectors, you see that the theta is placed from the tail of a vector to another tail of another vector. If you have a look at the attached picture below, the angle will be placed in between the vectors a and c (a + b). [look at the second triangle]
Post by: knightrider on December 30, 2015, 04:55:51 pm
Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.
Ok, let's start with Picture 1 (noting that air resistance is being neglected):
The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.
Picture 2:
When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.
Picture 3:
The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.
Picture 4:
The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.
Picture 5:
Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.
I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.
Just want to extend on JI2015's response.
If you also have a look at the force G (gravity), it stays the same, as you may remember, that the gravity on Earth is 9.8 m/s/s.
Due to the acceleration of gravity, all objects have the same rate of acceleration, when they free fall (no matter what their mass is). So if you divide the Netforce by the weight, you will always get 9.8 m/s/s.
I have attached a picture below, hopefuly that helps :)
If you have a look at Vectors, you see that the theta is placed from the tail of a vector to another tail of another vector. If you have a look at the attached picture below, the angle will be placed in between the vectors a and c (a + b). [look at the second triangle]
Thanks so much JI2015 :) and Syndicate :). Really helped !!
Post by: Adequace on December 30, 2015, 07:39:26 pm
http://imgur.com/a/v8lrM (All uploaded here)
For Ex43, Q22: The answer is B but I wrote C. Could someone clarify why it's B and not C?
For Ex45, Q34: I got the correct answer in the end but I initially just used W=Fx which was wrong. Is there a reason why W=Fx doesn't work? Is it because the force isn't constant?
Ex47, Q36 is the last part of the question above. The answer just said "since the elastic is 2x as long, therefore the mass will extend the original length by twice as much" which is 0.04x2=0.08m. Can someone explain the theory behind this?
For Ex50, Q37: the answer states "0J, because 8N will not alter the compression". I used W=Fx then subtracted 0.25J which gave me an answer >0, wouldn't using 8N to compress the spring require more work regardless?
Cheers, sorry about all the questions. (If you need more context/answers of the previous questions just ask :))
Post by: JI2015 on December 30, 2015, 08:47:38 pm
I've prepared some solutions for you in the pdf. Please let me know if you need further explanation.
Some points:
Q34: As you suggested force is not constant so cannot use W=Fx but remember that W=change in Energy and we can use the Energy formulas to solve.
Q36: I used information gained from the structures and materials detailed study, if there were easier ways to explain hopefully someone points it out. However, I'm confident that you will understand my explanation.
Q37: I didn't prepare anything on this in the pdf so I'll comment here. Pay attention to the graph, the maximum extension occurs at a force of 5N. 8N of force will not change the extension and so no work is done. Also some theory, the area under the Force vs Extension graph gives the Work. There is no area between F=5 and F=8 as the extension remains 0.1m and hence no work is done.
Hope this helps you! Let me know if you understand what I have written!
http://jmp.sh/VA5TX1k
Post by: Adequace on December 30, 2015, 09:49:13 pm
Hi Adequace,You legend, thanks so much!
I've prepared some solutions for you in the pdf. Please let me know if you need further explanation.
Some points:
Q34: As you suggested force is not constant so cannot use W=Fx but remember that W=change in Energy and we can use the Energy formulas to solve.
Q36: I used information gained from the structures and materials detailed study, if there were easier ways to explain hopefully someone points it out. However, I'm confident that you will understand my explanation.
Q37: I didn't prepare anything on this in the pdf so I'll comment here. Pay attention to the graph, the maximum extension occurs at a force of 5N. 8N of force will not change the extension and so no work is done. Also some theory, the area under the Force vs Extension graph gives the Work. There is no area between F=5 and F=8 as the extension remains 0.1m and hence no work is done.
Hope this helps you! Let me know if you understand what I have written!
http://jmp.sh/VA5TX1k
But yeah with Q36, if anyone could provide an answer with motion theory that would be great since.(regardless thanks for the working/insight JI2015)
Post by: Syndicate on December 31, 2015, 12:55:33 am
You legend, thanks so much!
But yeah with Q36, if anyone could provide an answer with motion theory that would be great since.(regardless thanks for the working/insight JI2015)
From my knowledge and a little bit of research, the question simply states that the length of the same elastic band is no 0.4 metres with the same mass hung on it. Last time, the length was .200, so it stretched 0.04 metres, however this time, only the length has doubled, thus the extension would double too. If the mass had been halved, then the extension would have been same as the last one (0.04 metres).
I couldn't exactly find the exact theory behind this, but, you can have a look at Hooke's Law (it's about springs, however, sometimes elasticity applies too).
I also wanted to extend a little on question 34 from JL2015's description (which is amazing :) )
As the body is moving towards the spring (compressing it), the kinetic energy is being transformed into potential energy. However, it's not constant as it would require more energy to compress the spring even more. As the body slows down, the kinetic energy is slowly being lost into potential energy, thus, it can't be constant (so it can't be Answer A). Then you may think, about Answer C. It is wrong, due to the slope of the graph in Answer C, as it displays that the amount of kinetic energy being lost accelerates (if you look at the end part of it), however, it should slow down, as it compresses the spring.
Hope the explanation helps [too] ;D
Post by: Adequace on December 31, 2015, 02:01:34 pm
From my knowledge and a little bit of research, the question simply states that the length of the same elastic band is no 0.4 metres with the same mass hung on it. Last time, the length was .200, so it stretched 0.04 metres, however this time, only the length has doubled, thus the extension would double too. If the mass had been halved, then the extension would have been same as the last one (0.04 metres).Thanks
I couldn't exactly find the exact theory behind this, but, you can have a look at Hooke's Law (it's about springs, however, sometimes elasticity applies too).
I also wanted to extend a little on question 34 from JL2015's description (which is amazing :) )
As the body is moving towards the spring (compressing it), the kinetic energy is being transformed into potential energy. However, it's not constant as it would require more energy to compress the spring even more. As the body slows down, the kinetic energy is slowly being lost into potential energy, thus, it can't be constant (so it can't be Answer A). Then you may think, about Answer C. It is wrong, due to the slope of the graph in Answer C, as it displays that the amount of kinetic energy being lost accelerates (if you look at the end part of it), however, it should slow down, as it compresses the spring.
Hope the explanation helps [too] ;D
Edit: In regards to Ex36, I'm still not entirely sure on how doubling the length of the same elastic = a doubled extension length. I can't really apply Hooke's law to this doubled elastic length situation either(since x=compression length).
Post by: lzxnl on December 31, 2015, 03:56:18 pm
Here's my take on things.
Q22: 1/2 mv^2 + 1/2 kx^2 = constant
So the relationship between KE and x should be quadratic in x; negative parabola.
Q34: as above
Q37: work, by definition, is a force acting on a distance. If your force doesn't move anything, there is no work being done. You can think of work as a change in kinetic energy. If there is no change in displacement, then your object isn't moving any faster or smaller -> no work done.
How to deal with Q36?
What is the origin of this spring force? You have bonds between atoms that make up the spring force. By using Hooke's law, we're approximating this force as being linear in the displacement, which is generally a pretty good approximation. Now, if your spring is twice as long with the same material, you have twice as many of the same identical atoms that pull on each other. The quadratic nature of the energy means that the most stable configuration is when all of the atoms are the same distance from each other (this can be proved using some version of the AM-QM inequality I think but I'm not going to go into the details here).
Now, let's look at the atomic cross section that is attached to the mass. This layer of atoms has to bear the entire weight; by Newton's third law, if it is pulling the mass up, the mass pulls this layer of atoms down. The next layer of atoms therefore has to exert a force equal to the weight force on the first layer of atoms to hold them up. Apply this logic to all of the layers of atoms there and you'll find that the extension = number of layers * extension between layers
Double the length of the spring -> double the number of layers of atoms -> double the total extension
Post by: Adequace on December 31, 2015, 04:53:43 pm
Post by: knightrider on January 02, 2016, 09:28:44 pm
is it fine to just show that the initial kinetic energy does not equal the final kinetic energy to receive full marks ?
Post by: odeaa on January 02, 2016, 09:43:41 pm
In order to demonstrate an inelastic collision
is it fine to just show that the initial kinetic energy does not equal the final kinetic energy to receive full marks ?
yeah, just calculate the initial and final energies (must be correct values, usually a mark here) and then make a quick statement saying that because initial and final are not equal it is inelastic
Post by: knightrider on January 03, 2016, 12:19:41 am
yeah, just calculate the initial and final energies (must be correct values, usually a mark here) and then make a quick statement saying that because initial and final are not equal it is inelastic
Thanks for clarifying odeaa :)
Post by: knightrider on January 03, 2016, 04:42:03 pm
Post by: Syndicate on January 03, 2016, 04:55:48 pm
How would you do this question attached?
hey knighrider,
formula:
m1 = 10kg
m2 = 10kg
v= 5m/s
50/20 = 2.5m/s
Edited^
hope this helps :)
Post by: knightrider on January 03, 2016, 05:13:26 pm
hey knighrider,
formula:= the subsequent speed
m1 = 10kg
m2 = 10kg
v= 5m/s
50/20 = 2.5m/s
Edited^
hope this helps :)
Thanks Syndicate :)
I also found a way of doing it (refer to below :) )
Post by: Syndicate on January 03, 2016, 06:38:49 pm
Thanks Syndicate :)
I also found a way of doing it (refer to below :) )speed is
yes, that is certainly a way of writing the formula down in a different state, but its a good idea, remembering both equations, as you may never know, what comes on the SAC.
Post by: knightrider on January 03, 2016, 07:59:22 pm
Calculate the speed of each ball after the collision.
Post by: Syndicate on January 03, 2016, 09:01:03 pm
How would you do this question relating to image attached?
Calculate the speed of each ball after the collision.
That questions seems a little inaccurate, as KE = 1/2 x mv^2, which is this case is 18 J and if 20 J is lost, thus, while calculating the final amount of Kinetic Energy, we would get -2 J, which is not possible (as there is nothing like negative kinetic energy)
Post by: nerdgasm on January 03, 2016, 10:39:54 pm
Post by: Adequace on January 03, 2016, 11:11:50 pm
Hmm, I think the question is saying that the 20J is the total amount of kinetic energy lost, not the energy lost from each ball. So, I think you're meant to work out the amount of kinetic energy 'remaining', then work out how much kinetic energy each ball has after the collision (it should be the same amount for both balls), and use that to work out the final speed of each ball.I did the same as nerdgasm, I got 2m/s as my final answer. If it's correct I can post my working if you want but you're pretty much going E initial = E final as your first line.
Post by: knightrider on January 03, 2016, 11:39:04 pm
I did the same as nerdgasm, I got 2m/s as my final answer. If it's correct I can post my working if you want but you're pretty much going E initial = E final as your first line.
Yep thats the right answer :)
Could you please post your working. :D
Post by: Syndicate on January 03, 2016, 11:59:55 pm
Hmm, I think the question is saying that the 20J is the total amount of kinetic energy lost, not the energy lost from each ball. So, I think you're meant to work out the amount of kinetic energy 'remaining', then work out how much kinetic energy each ball has after the collision (it should be the same amount for both balls), and use that to work out the final speed of each ball.
I am still a little short of this
I did the same as nerdgasm, I got 2m/s as my final answer. If it's correct I can post my working if you want but you're pretty much going E initial = E final as your first line.
Can you kindly show the essential parts of your working out?
Thanks :)
Post by: Adequace on January 04, 2016, 02:27:58 pm
EKi = EKf + Mechanical Energy(sound)
0.5mu^2 + 0.5mu^2 = 0.5mv^2 + 0.5mv^2 + 20
Sub in values then you'll get: 36 = 4v^2 + 20
Solve for v: =2m/s
Post by: knightrider on January 04, 2016, 04:11:13 pm
How is the normal force down ? Shouldn't the normal force be up as it opposes gravity?
I have written out my working below shouldn't this be correct ?
let down be positive
Post by: Syndicate on January 04, 2016, 04:28:47 pm
For this question and solution attached.
How is the normal force down ? Shouldn't the normal force be up as it opposes gravity?
I have written out my working below shouldn't this be correct ?
let down be positive
Normal force doesn't always oppose gravitational force. In this case, the toy truck will basically be weightless, thus normal force will be in the same direction as the gravitation force.
Look at this website and scroll down: http://www.ux1.eiu.edu/~cfadd/1350/06CirMtn/VertCircle.html
If you don't understand it, don't hesitate to reply back :)
EDIT: you gained a negative normal force, so your working out is fine. (negative force = same direction as the gravity)
Post by: knightrider on January 04, 2016, 04:42:32 pm
Normal force doesn't always oppose gravitational force. In this case, the toy truck will basically be weightless, thus normal force will be in the same direction as the gravitation force.
Look at this website and scroll down: http://www.ux1.eiu.edu/~cfadd/1350/06CirMtn/VertCircle.html
If you don't understand it, don't hesitate to reply back :)
EDIT: you gained a negative normal force, so your working out is fine. (negative force = same direction as the gravity)
Thanks Syndicate :)
why will the toy truck be weightless ?
also in my working out my final answer had the normal force as up what did i do wrong (working out is below) ?
let down be positive
Post by: Syndicate on January 04, 2016, 04:52:48 pm
Thanks Syndicate :)
why will the toy truck be weightless ?
also in my working out my final answer had the normal force as up what did i do wrong (working out is below) ?
let down be positive
You did everything right, except the last line. You already got a negative normal force (-7.31 N), which means that the normal force will face in the direction of gravity (which will always be downwards, so it should be 7.31 Ndown)
As for the toy truck being weightless.
As both forces (gravity and normal force) will be facing downwards, the toy truck will seem to be weightless.
Think it about this way:
When you freefall, you would feel weightless, as there is no force impacting in the opposite direction of gravity ( I think, there is no normal force, when an object is free falling).
Post by: knightrider on January 04, 2016, 06:08:24 pm
You did everything right, except the last line. You already got a negative normal force (-7.31 N), which means that the normal force will face in the direction of gravity (which will always be downwards, so it should be 7.31 Ndown)
As for the toy truck being weightless.
As both forces (gravity and normal force) will be facing downwards, the toy truck will seem to be weightless.
Think it about this way:
When you freefall, you would feel weightless, as there is no force impacting in the opposite direction of gravity ( I think, there is no normal force, when an object is free falling).
Thanks Syndicate :)
Post by: knightrider on January 04, 2016, 06:09:07 pm
Post by: Syndicate on January 04, 2016, 07:05:17 pm
How would you do this question attached.
Hey Knightrider,
You can workout the angle by simply using the formula:
where v = velocity (18 m/s)
r = radius (80)
g = gravity (9.8 m/s)
so if you sub in the quantities
=
=
which can be rounded to a whole number of 22 degrees
hopefully this helps :)
Post by: knightrider on January 04, 2016, 07:08:09 pm
Hey Knightrider,
You can workout the angle by simply using the formula:
so if you sub in the quantities
=
=
=
which can be rounded to a whole number of 22 degrees
hopefully this helps :)
Thanks Syndicate :)
Where did you get this formula from
Post by: Syndicate on January 04, 2016, 07:13:33 pm
Thanks Syndicate :)
Where did you get this formula from?
I discovered it last year, while reading over my cousin's Physics notes, however, I am sure, it can be found on a webpage :)
Post by: JI2015 on January 04, 2016, 07:27:58 pm
Post by: lzxnl on January 04, 2016, 07:42:35 pm
Object in a banked curve is not moving vertically (horizontal plane) so the vertical acceleration has to be zero. Vertical component of net force is zero. Through geometry, the incline angle happens to also be the angle between the normal and the vertical. It makes sense as the incline and normal are at 90 degrees and the horizontal/vertical are at 90 degrees.
Hence, resolving the forces vertically gives N cos theta = mg and N sin theta = mv^2/r (centripetal acceleration requirement as the net force is directed radially towards the centre, much like the horizontal force component)
Post by: knightrider on January 04, 2016, 09:09:51 pm
These are the equations from which the banked corner formula is derived from. Fc is the centripetal force (net force). I'll try to find a sheet I have that explains this in detail but here are the equations.
The idea is:
Object in a banked curve is not moving vertically (horizontal plane) so the vertical acceleration has to be zero. Vertical component of net force is zero. Through geometry, the incline angle happens to also be the angle between the normal and the vertical. It makes sense as the incline and normal are at 90 degrees and the horizontal/vertical are at 90 degrees.
Hence, resolving the forces vertically gives N cos theta = mg and N sin theta = mv^2/r (centripetal acceleration requirement as the net force is directed radially towards the centre, much like the horizontal force component)
Thanks JI2015 and lzxnl. Really helped :)
Post by: knightrider on January 05, 2016, 12:01:59 am
Post by: JI2015 on January 05, 2016, 12:06:41 am
Post by: knightrider on January 05, 2016, 12:34:32 am
Draw your forces on the mass on the diagram. There is a tension force vertically upwards acting on the mass and a vertically downwards force of gravity acting on the mass. Since the mass is at rest (pay attention to the wording), it means the net force at this point is zero. So, the magnitude of the tension in the rope is equal to the magnitude of the weight force which in this case is 40N. Also, specify that tension is in the upwards direction.
Thanks JI2015 :)
just wondering is my working out below correct in relation to this situation?
let up be positive
Post by: JI2015 on January 05, 2016, 12:46:48 am
F(net) = F(centripetal) = Tension - Weight = 0
Tension - Weight = 0
therefore, Tension = Weight
Remember to refer to tension rather than F(N) as it is more specific. Also with the net force equation, make sure that there is a negative before the Fg/ Weight as it is acting in the opposite direction to the tension force.
What I also want to point out is that you can treat the equation with the variables as vectors or scalars. I prefer scalars (you account for direction with negative signs). Your net force equation is fine if you keep vectors in mind, but since you specified up as positive then treating them as scalars is the way to go. This also means that when referring to the forces in the equations, you are only concerned with their magnitudes when solving. If you do this correctly for all force equations, you will only get positive values for magnitudes of forces when solving as you have accounted for the direction (whether positive or negative) already in the equation.
If you would like me to show you both methods (vector and scalar) and how the solutions present themselves when you solve, Iet me know and I'll prepare something tomorrow.
Post by: knightrider on January 05, 2016, 01:03:52 am
Treating upwards as postive:
F(net) = F(centripetal) = Tension - Weight = 0
Tension - Weight = 0
therefore, Tension = Weight
Remember to refer to tension rather than F(N) as it is more specific. Also with the net force equation, make sure that there is a negative before the Fg/ Weight as it is acting in the opposite direction to the tension force.
What I also want to point out is that you can treat the equation with the variables as vectors or scalars. I prefer scalars (you account for direction with negative signs). Your net force equation is fine if you keep vectors in mind, but since you specified up as positive then treating them as scalars is the way to go. This also means that when referring to the forces in the equations, you are only concerned with their magnitudes when solving. If you do this correctly for all force equations, you will only get positive values for magnitudes of forces when solving as you have accounted for the direction (whether positive or negative) already in the equation.
If you would like me to show you both methods (vector and scalar) and how the solutions present themselves when you solve, Iet me know and I'll prepare something tomorrow.
Thanks so much JI2015 :) Really clarified things :D
It would be great if you could show me both methods (vector and scalar) and how the solutions present themselves when you solve. :)
Thanks so much for your help :)
Post by: JI2015 on January 05, 2016, 01:11:55 am
Some quick points:
When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.
Good Night!
And also well done with your commitment to physics as working hard and understanding this subject is really important.
Post by: knightrider on January 05, 2016, 01:19:27 am
Look for a post later today (since it is past midnight) about this as I want you to understand fully.
Some quick points:
When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.
Good Night!
And also well done with your commitment to physics as working hard and understanding this subject is really important.
Nice tips thanks again !!
Also it is with the help of such wonderful people like you and the many others on this site , that my skills and understanding in physics improves .!! :)
Good night :D
Post by: lzxnl on January 05, 2016, 12:05:30 pm
Look for a post later today (since it is past midnight) about this as I want you to understand fully.
Some quick points:
When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.
Good Night!
And also well done with your commitment to physics as working hard and understanding this subject is really important.
Technically, those aren't scalars. Scalars are actually quantities that are invariant (don't change) under coordinate transformations like rotation or reflection. So, rotating an object doesn't change its mass or any distances so those are scalars. Dealing with vector quantities in 1D and treating their direction as a scalar quantity is strictly not correct. Your 'scalar' is not invariant under rotation as rotating your axis 180 degrees changes the sign. So technically you're still dealing with a vector, just representing it as a real number.
Post by: JI2015 on January 05, 2016, 01:25:25 pm
As promised I have prepared a more thorough explanation for you. Hope this clears up any hesitation.
http://jmp.sh/bnRFytP
Post by: JI2015 on January 05, 2016, 01:55:37 pm
More of a 1/2 question, but I'll ask anyway since it's relevant to 3/4. When adding or subtracting numbers, do we get our answer to the least amount of sig figures after the decimal point? For example, is 6.9536 (4 sig figs after decimal point) + 3772.650 (3 sig figs after d.p) equal to 3779.604 (3 sig figs after d.p)?
When multiplying or dividing, is our answer to least amount of sig figures (regardless of decimal place)? For example, is 63.84 (4 sig.figs) / 2.5010 (5 sig figs) = 25.53 (4 sig figs)
Your reasoning is spot on!
Adding or Subtracting: Decimal places of answer is to the lower amount of decimal places that one of the terms has.
Multiplying or Dividing: Answer has the same significant figures as the number with the lower amount of significant figures of the two used in the calculation.
Usually people only have trouble with the Addition/Subtraction but you seem to understand the rules well!
Post by: knightrider on January 05, 2016, 07:17:13 pm
Hi knightrider,
As promised I have prepared a more thorough explanation for you. Hope this clears up any hesitation.
http://jmp.sh/bnRFytP
Wow awesome explanation :) Thanks so much JI2015 :)
Post by: Adequace on January 05, 2016, 08:12:23 pm
Post by: JI2015 on January 05, 2016, 08:17:48 pm
For projectile motion are we allowed to use pre-derived formulas? My teacher encourages us to use a pre-derived formula for time which is convenient. I've also heard that assessors don't mind it.
You can use the 'shortcut' formulas but just make sure you know how to use them properly (i.e.when the object lands at the same height that it starts at). Maybe even learning how they are derived may benefit your understanding.
Also, just to add, I used some of these formulas in practice exams and also the actual exam and had no problems, so yeah assessors do not mind at all as long as your answer is correct.
Post by: Syndicate on January 05, 2016, 08:18:43 pm
For projectile motion are we allowed to use pre-derived formulas? My teacher encourages us to use a pre-derived formula for time which is convenient. I've also heard that assessors don't mind it.if the assessors don't mind it, then why not?
Post by: lzxnl on January 06, 2016, 01:19:59 pm
if the assessors don't mind it, then why not?
Simply because it's not good for your understanding in general. It's sooo easy for students to forget where a formula is applicable. For instance, range = v^2 sin 2 theta / g only for landing at the height of launch.
Post by: Adequace on January 06, 2016, 01:49:07 pm
Post by: knightrider on January 06, 2016, 01:59:44 pm
lets take the situation of a projectile motion.
Say you use a shortcut formula and then get the answer wrong , is it true that you will not get method marks too?
Also say if you do things from first principles and then get the answer wrong, is it true that you will still have a chance to get method marks ?
Post by: JI2015 on January 06, 2016, 02:23:53 pm
In physics, the examiners don't clearly tell us how marks are awarded and teachers often judge answers based on their own expectations. The exam reports are not in-depth and usually only present one interpretation especially for worded questions.
My advice would be to show as much working out as possible so that the assessor can see if you made a careless error, and therefore be more inclined to award you marks for demonstration of the correct physics principles.
So for projectile motion, I would write the 'shortcut' formula, then write it again with the known information substituted in and then solve and write the answer. Assessors are familiar with these formulas, and if you use them in the correct situation but just enter something wrong in the calculator, you may only lose 1 mark as they are aware that you knew what you were doing.
However, if you use the 'shortcut' formulas when they are not applicable and as a result get the wrong answer, you will not get marks.
Lastly, to address your point on the traditional method of dealing with projectile motion: It may take too long in an exam to set all your equations up and will not be advantageous in dealing with a question that the 'shortcut' formulas can be used for. Some questions, you have to set it out the long way, and method marks will be awarded if you demonstrate correct physics principles.
As long as you demonstrate an understanding of the situation presented to you, and apply relevant and correct physics ideas, then you will most likely gain some marks with or without a correct answer!
Post by: knightrider on January 06, 2016, 03:35:39 pm
Hi knightrider,
In physics, the examiners don't clearly tell us how marks are awarded and teachers often judge answers based on their own expectations. The exam reports are not in-depth and usually only present one interpretation especially for worded questions.
My advice would be to show as much working out as possible so that the assessor can see if you made a careless error, and therefore be more inclined to award you marks for demonstration of the correct physics principles.
So for projectile motion, I would write the 'shortcut' formula, then write it again with the known information substituted in and then solve and write the answer. Assessors are familiar with these formulas, and if you use them in the correct situation but just enter something wrong in the calculator, you may only lose 1 mark as they are aware that you knew what you were doing.
However, if you use the 'shortcut' formulas when they are not applicable and as a result get the wrong answer, you will not get marks.
Lastly, to address your point on the traditional method of dealing with projectile motion: It may take too long in an exam to set all your equations up and will not be advantageous in dealing with a question that the 'shortcut' formulas can be used for. Some questions, you have to set it out the long way, and method marks will be awarded if you demonstrate correct physics principles.
As long as you demonstrate an understanding of the situation presented to you, and apply relevant and correct physics ideas, then you will most likely gain some marks with or without a correct answer!
Thanks so much for clarifying JI2015 :)
Post by: JI2015 on January 06, 2016, 04:05:37 pm
Great job learning data analysis principles as they are often neglected and not taken seriously!
EDIT:
The percentage error for (10.5 +/- 0.5) is supposed to have 2 sig. figs. and therefore it become 4.8%
4.8%+0.9%=5.7% total percentage error
Post by: Syndicate on January 06, 2016, 04:05:47 pm
Hi, I was wondering which of these working outs are correct:
(21.4 +/- 0.2) * (10.5 +/- 0.5)
1)
0.2 / 21.4 = 0.0093457943925234
= 0.009 (1 s.f)
= 0.9%
0.5 / 10.5 = 0.0476190476190476
= 0.05 (1 s.f)
= 5%
0.9% + 5% = 5.9%
21.4 * 10.5 = 224.7
= 225 (3 s.f)
Final answer = 225 +/- 5.9%
2)
0.2 / 21.4 = 0.0093457943925234
0.5 / 10.5 = 0.0476190476190476
0.0093457943925234 + 0.0476190476190476 = 0.05696484201
= 5.7%
21.4 * 10.5 = 224.7
= 225 (3 s.f)
Final answer = 225 +/- 5.7%
the second seems more accurate, as you have added all the decimals points together, while in the first one, you rounded 0.0476190.... to 0.05 (see the difference).
Post by: JI2015 on January 06, 2016, 04:09:38 pm
the second seems more accurate, as you have added all the decimals points together, while in the first one, you rounded 0.476190.... to 0.05 (see the difference).
In my notes I had written down that you add the individual percentage errors together and those are supposed to be to 1 or 2 sig. figs. and therefore the top calculation seems right. I'm not completely certain though.
Post by: Syndicate on January 06, 2016, 04:13:49 pm
In my notes I had written down that you add the individual percentage errors together and those are supposed to be to 1 or 2 sig. figs. and therefore the top calculation seems right. I'm not completely certain though.
hey Jl2015
What I did was, that I added those two answers together (0.09... + 0.47...), and get an answer closer to 0.56 , which seems much closer to 5.7% than 5.9%
Post by: JI2015 on January 06, 2016, 04:15:38 pm
hey Jl2015
What I did was, that if you added those two answers together (0.09... + 0.47...), i get a answer higher to 0.56 , which seems much closer to 5.7%
Take a look at my notes screenshot, that is the correct way to deal with these calculations. I modified my initial response as I found an error in the working out.
Post by: JI2015 on January 06, 2016, 05:05:46 pm
Sorry to be pedantic, but I thought when we divide 10.5 by 0.5 we take our answer to the least amount of sig figs, which is 1. Or is there a special exception with percentage errors?
dividing 0.5 by 10.5 yields 0.0476, to convert it into a percentage we multiply it by 100 and we get 4.76%
Since 4.76% is > 1% then it will have two sig figs and it will become 4.8%
This is just how it's done.
Post by: JI2015 on January 06, 2016, 05:09:52 pm
Spoiler
http://pegsnet.pegs.vic.edu.au/studentdownloads/Physics/Students/UNIT%201/Data%20analysis/DataAnalysisBooklet.pdf
Spoiler
Heaps of physics booklets are also available on my school's website so just take what you need, but let's agree that I did not send you, haha: http://pegsnet.pegs.vic.edu.au/studentdownloads/Physics/Students /
Post by: YellowTongue on January 07, 2016, 02:29:03 pm
A ball with a mass of 100g is thrown at a wall horizontally with a speed of 4.0m/s. Calculate the force that the wall exerts on the ball during impact.
Thanks ;)
Post by: JI2015 on January 07, 2016, 02:35:21 pm
Can someone please help me with this question. I just can't figure it out... :'(
A ball with a mass of 100g is thrown at a wall horizontally with a speed of 4.0m/s. Calculate the force that the wall exerts on the ball during impact.
Thanks ;)
Where did you find this question? You need more information.
If you are given a time of contact with the wall and a rebound velocity then you can use the change of momentum/ impulse formulas.
Otherwise there isn't much you can do with the information you have provided.
Post by: YellowTongue on January 07, 2016, 02:51:17 pm
Post by: Muchos Help on January 08, 2016, 02:09:12 pm
My variables were x=?, u=5, *v=5, a=0 and t=0.6. For v, is 5 incorrect? I thought that horizontal velocity was constant throughout the entire flight, but I tried using v2 = u2 +2ax but it was wrong, then used x = ut + 0.5at2 which I got the correct answer from.
http://imgur.com/yr7WRUb, similarly for this question. Why can't you use the horizontal direction's variables then use v=u+at?
The correct answer for this Q is considering the vertical direction and I used x=vt-0.5at2 but I encountered a problem, why does x= -5 if I take down to be positive? Whereas when you use x=ut+0.5at2, x needs to = 5 for the answer to be correct (1second).
Sorry if my writing is confusing, I'll clarify if needed.
Post by: Syndicate on January 08, 2016, 05:05:41 pm
http://m.imgur.com/ZjeJkM2 - In the attached for Ex15,Q7: I initially got this question wrong but got it correct in the end by using a different equation of motion equation.
My variables were x=?, u=5, *v=5, a=0 and t=0.6. For v, is 5 incorrect? I thought that horizontal velocity was constant throughout the entire flight, but I tried using v2 = u2 +2ax but it was wrong, then used x = ut + 0.5at2 which I got the correct answer from.
http://imgur.com/yr7WRUb, similarly for this question. Why can't you use the horizontal direction's variables then use v=u+at?
The correct answer for this Q is considering the vertical direction and I used x=vt-0.5at2 but I encountered a problem, why does x= -5 if I take down to be positive? Whereas when you use x=ut+0.5at2, x needs to = 5 for the answer to be correct (1second).
Sorry if my writing is confusing, I'll clarify if needed.
The second rule you have used is right, as it is to find the horizontal displacement. If you re-read the question, it says that your intial horizontal velocity is 5 m/s , therefore, you are required to use this formula to find the horizontal displacemnt. It can also be used to find the vertical displacement, by simply subbing in your intial vertical veloctiy, instead of intial horizontal velocity.
As for your first formula. It used to find the final vertical/ horizontal velocity{squared} (velocity = speed in given direction, think of vectors), while, you had to find the displacment (distanced covered from the beginning and concluding points)
Displacement formula : d =
where
d = displacment
v = intial vertical/horizontal velocity
t = time
a = acceleration
Do note: whenever you are working out with any of the vertical formulas, acceleration will always = 9.8 m/s/s
Hope this helps :)
Post by: Muchos Help on January 08, 2016, 05:41:58 pm
Post by: Syndicate on January 08, 2016, 06:02:11 pm
I'm wondering why v^2=u^2+2ax doesn't work, just to clarify..
As I stated, the formula
Therefore, it is relevant to use the formula
Post by: Muchos Help on January 08, 2016, 06:22:50 pm
As I stated, the formulaNormally you could just rearrange the first formula for x, but since a=0 you can't. I just need some who knows the theory on why it isn't possible.will get you the velocity of the projectile, while you question clearly states that, it wants the distance.
Therefore, it is relevant to use the formulain this situation
Thanks though.
Post by: lzxnl on January 08, 2016, 06:38:24 pm
If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error
However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.
To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)
Post by: Syndicate on January 08, 2016, 06:51:43 pm
OK. I see your problem now.
If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error
However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.
To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)
never knew cheat sheets were allowed in VCE exams
Post by: Muchos Help on January 08, 2016, 06:57:56 pm
OK. I see your problem now.Ah okay, thanks.
If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error
However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.
To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)
Post by: lzxnl on January 08, 2016, 07:17:48 pm
never knew cheat sheets were allowed in VCE exams
They shouldn't be allowed. But ask any maths or physics student. It's silly how there's a formula sheet in physics AND you get to bring in 2 A4 double-sided. At least, that was way back in 2013.
Post by: Syndicate on January 08, 2016, 07:38:08 pm
They shouldn't be allowed. But ask any maths or physics student. It's silly how there's a formula sheet in physics AND you get to bring in 2 A4 double-sided. At least, that was way back in 2013.
is the formula sheet like the one on a maths exam?
EXAMPLE:
thermodynamics
q = ml
q = mc*change in temperature
velocity
v(final) = v(initial) + at
??
Post by: lzxnl on January 08, 2016, 08:37:45 pm
is the formula sheet like the one on a maths exam?
EXAMPLE:
thermodynamics
q = ml
q = mc*change in temperature
velocity
v(final) = v(initial) + at
??
Difference is, in maths there's no limit on how much stuff you bring as long as it's bound. Otherwise...it's the same really.
Post by: Adequace on January 11, 2016, 05:01:48 pm
Both F and x is also positive so I'm not sure how you could cancel out the negative of k?
Post by: JI2015 on January 11, 2016, 05:13:56 pm
Quick question when you're using F=-kx, I see multiple/all of the time that the questions' worked solution just uses this formula as F=kx, so when the spring constant will always be positive. Are you allowed to do this?
Both F and x is also positive so I'm not sure how you could cancel out the negative of k?
A quick online search:
https://answers.yahoo.com/question/index?qid=20101231173352AAfckwX
I used F=kx, it doesn't really matter which you use as the magnitude of the force is the same. Negative just indicates direction which you can find out from a diagram so it's not the most useful to have a negative in the equation, especially if you haven't set a positive direction. Bottom line, draw a diagram and label the direction the force and use F=kx to find the magnitude.
Post by: lzxnl on January 11, 2016, 06:12:11 pm
Quick question when you're using F=-kx, I see multiple/all of the time that the questions' worked solution just uses this formula as F=kx, so when the spring constant will always be positive. Are you allowed to do this?
Both F and x is also positive so I'm not sure how you could cancel out the negative of k?
It matters when the sign of the force matters. Otherwise, as mentioned, you can just relate the force to magnitudes.
When does it matter? When you're actually solving for the equation of motion or the energies.
The potential energy is given by -int F dx, so the negative sign for the force gives rise to U = 1/2 kx^2. Without the negative sign, the potential energy would be -1/2 kx^2
Similarly, if you're trying to solve for the shape of the motion, you would have F = -kx -> ma = -kx
a = -kx/m = d^2x/dt^2
The general solution to this differential equation is a sinusoid, x = A cos(sqrt(k/m)t + constant)
Without the negative sign, the solutions would be sums of exponentials and there'd be no oscillation (try it yourself if you want xD)
Post by: Adequace on January 11, 2016, 06:15:49 pm
Post by: Maz on January 11, 2016, 06:49:48 pm
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou :)
Post by: lzxnl on January 11, 2016, 07:09:48 pm
hey- just a few questions related to tension? please...i don't really understand this...
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou :)
Tension is the force on the string pulling on the object. So, if it's a vertical circle, at the bottom of the circle the tension is pointing up, and at the top it's pointing down.
To do this question, you would treat the tension force like a normal reaction force and just solve net force = mv^2/r
Then assign the direction based on the direction requirement for circular motion
Post by: Maz on January 11, 2016, 08:05:30 pm
Tension is the force on the string pulling on the object. So, if it's a vertical circle, at the bottom of the circle the tension is pointing up, and at the top it's pointing down.okay cool- thank you :)
To do this question, you would treat the tension force like a normal reaction force and just solve net force = mv^2/r
Then assign the direction based on the direction requirement for circular motion
Post by: Syndicate on January 12, 2016, 10:46:04 am
hey- just a few questions related to tension? please...i don't really understand this...
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou :)
By "vertical of tension", your book is simply try to state that it is gravity, as simply all the forces going downwards will equal to gravity(9.8 m/s/s). If you think about it, vertical is a straight line (from top to bottom), therefore, it must equal the gravitational force.
To do the following question, you must use the formula
where:
T = Tension (what you want to calculate)
W = Work (mass x gravity) [0.2 x 9.8 = 1.96 J]
m = mass [0.2 kg]
a = acceleration of the moving object [2m/s/s]
therefore ma = 0.2 x 2 = 0.4 kg.m/s/s
Now we can work out tension at the bottom of the vertical circle, by simply having a negative acceleration quanity, and knowing that, the speed is constant throughout the course, you will have an acceleration of 2m/s/s downwards => -2m/s/s
Therefore,
Hope this helps :)
Post by: Maz on January 12, 2016, 03:33:13 pm
By "vertical of tension", your book is simply try to state that it is gravity, as simply all the forces going downwards will equal to gravity(9.8 m/s/s). If you think about it, vertical is a straight line (from top to bottom), therefore, it must equal the gravitational force.thankyou :)
To do the following question, you must use the formula
where:
T = Tension (what you want to calculate)
W = Work (mass x gravity) [0.2 x 9.8 = 1.96 J]
m = mass [0.2 kg]
a = acceleration of the moving object [2m/s/s]
therefore ma = 0.2 x 2 = 0.4 kg.m/s/s(at the top)
Now we can work out tension at the bottom of the vertical circle, by simply having a negative acceleration quanity, and knowing that, the speed is constant throughout the course, you will have an acceleration of 2m/s/s downwards => -2m/s/s
Therefore,(at the bottom)
Hope this helps :)
Post by: Maz on January 12, 2016, 03:37:13 pm
why does a train track for a very fast train have to have curves of a big radius? I'm supposed to use an equation to demonstrate...
is it something to do with having to overcome the centrifugal force and is the equation mv^2/r?
thankyou in advance :)
Post by: Syndicate on January 12, 2016, 03:50:13 pm
hello...quick question please can someone help?
why does a train track for a very fast train have to have curves of a big radius? I'm supposed to use an equation to demonstrate...
is it something to do with having to overcome the centrifugal force and is the equation mv^2/r?
thankyou in advance :)
yes, you can have a look at centipetal force, which keeps an object inside the turn. Usually train tracks have banked curves, so the train doesnt move away from the center of the curvature. The formula seems right.
more imformation is available below:
http://www.diffen.com/difference/Centrifugal_Force_vs_Centripetal_Force
Post by: Maz on January 12, 2016, 04:06:54 pm
yes, you can have a look at centipetal force, which keeps an object inside the turn. Usually train tracks have banked curves, so the train doesnt move away from the center of the curvature. The formula seems right.thankyou :)
more imformation is available below:
http://www.diffen.com/difference/Centrifugal_Force_vs_Centripetal_Force
Post by: Adequace on January 12, 2016, 05:25:24 pm
For the last question of the attached, the answer states "X is exerting the same force on Y as Y is exerting on X" then they equated ma=ma and ended up with (a of y)/(a of x) = 0.2
I don't understand how they exert the same force on eachother? Is Newton's third law applied here, if it is I don't understand how? What even is the force that is being exerted on each other?
Thanks, if you could clarify these questions that would be great.
Post by: Syndicate on January 12, 2016, 05:47:04 pm
http://imgur.com/kMVtqqkHey Adequance,
For the last question of the attached, the answer states "X is exerting the same force on Y as Y is exerting on X" then they equated ma=ma and ended up with (a of y)/(a of x) = 0.2
I don't understand how they exert the same force on eachother? Is Newton's third law applied here, if it is I don't understand how? What even is the force that is being exerted on each other?
Thanks, if you could clarify these questions that would be great.
The last question can be worked by simply solving for the accelaration of these masses. So the right formula, we must use in this situation is:
As there is no force, other than gravity, we can state that your Net Force is equal to 10 N (as requested above, by the book)
So, your acceleration of y would be
As the book has asked you to workout the ratio
If you need anymore clarification, dont hesitate to ask :)
Post by: Adequace on January 12, 2016, 07:12:11 pm
Thanks though
Post by: Syndicate on January 12, 2016, 10:00:03 pm
Wouldn't the tension in the string balance out the weight force of both of the masses combined?
Thanks though
Yes, it would, due to the masses being equally distributed on both sides. However, the question is asking for the acceleration of Y and X, while both masses are rotating with the frequency of 2 revolutions per second (2 complete rotations in 1 second), and because acceleration doesnt equal to 0 (in this case), the rod isnt balanced.
Post by: Maz on January 14, 2016, 03:18:57 pm
can someone please help me with this question?
you are having difficulty undoing a screw...should you use a screwdriver with a thicker handle or a longer handle?
I'm thinking the answer is thicker as torque= r*F*sin(theta)...and thinker would give a bigger r value? how do you discard the longer handle...
thankyou in advance :)
Post by: Syndicate on January 14, 2016, 03:37:54 pm
hello humans... :)
can someone please help me with this question?
you are having difficulty undoing a screw...should you use a screwdriver with a thicker handle or a longer handle?
I'm thinking the answer is thicker as torque= r*F*sin(theta)...and thinker would give a bigger r value? how do you discard the longer handle...
thankyou in advance :)
hey mq123,
Yes, you are correct, the handle of the screwdriver must be thicker to get a larger torque value. As for the longer handle screwdriver, it wouldnt directly increase/ decrease the torque and would stay exactly the same, as with the original screwdriver. It would only matter, if the situation was different. Example: Try to open a the lid from can.
What did you exactly mean by discard?
Post by: Maz on January 14, 2016, 04:03:04 pm
hey mq123,thankyou so much...yeah by discard i was basically just asking why the longer handle one wouldn't work- which you answered so thank you :)
Yes, you are correct, the handle of the screwdriver must be thicker to get a larger torque value. As for the longer handle screwdriver, it wouldnt directly increase/ decrease the torque and would stay exactly the same, as with the original screwdriver. It would only matter, if the situation was different. Example: Try to open a the lid from can.
What did you exactly mean by discard?
Post by: Maz on January 14, 2016, 05:09:41 pm
a person stands next to a table. explain how they can apply forces to the table to:
1.increse the normal force of the floor acting on them
2. decrease the normal force of the floor acting on them
thankyou in advance :)
Post by: Syndicate on January 14, 2016, 07:32:57 pm
An increase in normal force can simply be achieved, if that person pushes down on the table.
Theory?
Well, as the person is pushing down the table, there would be an increase in static friction. The increase in static friction would also cause the normal force to increase, as
A decrease in the normal force can be achieved, if the person puts the table on an incline.
Theory?
Well, as if you put any object on an incline, the normal force would decrease. This can be proved, if you use the formula
For example:
Lets say the person puts the table on an incline of 45 degrees and the table's mass is 2 kg
however, if the table was not on an incline, you would use the formula:
Hopefully this helps :)
Post by: lzxnl on January 14, 2016, 07:38:05 pm
hey...another question please :)
a person stands next to a table. explain how they can apply forces to the table to:
1.increse the normal force of the floor acting on them
2. decrease the normal force of the floor acting on them
thankyou in advance :)
Hey mq123,
An increase in normal force can simply be achieved, if that person pushes down on the table.
Theory?
Well, as the person is pushing down the table, there would be an increase in static friction. The increase in static friction would also cause the normal force to increase, as. Lets say the amount of increase in static friction equals to a, then
, which proves that an increase in Static Friction would also cause an increase in normal force.
A decrease in the normal force can be achieved, if the person puts the table on an incline.
Theory?
Well, as if you put any object on an incline, the normal force would decrease. This can be proved, if you use the formula, and as you may know that any object on an incline cant equal to 0 dgrees or 180 degrees, the value of cos would also decrease, as cos 0/360 and cos 180 = 1, and any lower/higher that 180 and higher than 0 degrees, would end with a lower value of 1.
For example:
Lets say the person puts the table on an incline of 45 degrees and the table's mass is 2 kg
however, if the table was not on an incline, you would use the formula:
Hopefully this helps :)
Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.
2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.
2. Decrease normal force could be done by partially leaning on the table; table now helps support your body
Post by: Syndicate on January 14, 2016, 07:42:11 pm
Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.
2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.
2. Decrease normal force could be done by partially leaning on the table; table now helps support your body
doesn't it ask for normal force increase/decrease of floor on the table? as it said, that how can they apply forces to the table
*I was a little confused with this one, sorry if I am wrong :P*
Post by: natdogg on January 14, 2016, 07:53:18 pm
a person stands next to a table. explain how they can apply forces to the table to:
1.increse the normal force of the floor acting on them
I believe 'them' suggests that they are talking about the 'person'.
ie. increase the normal force of the floor acting on the person
Post by: Syndicate on January 14, 2016, 07:56:12 pm
I believe 'them' suggests that they are talking about the 'person'.(them) but what if suggested both? because how can the use of person, be they, later? xD
ie. increase the normal force of the floor acting on the person
Books need to clarify their questions more fluently O.o
Post by: Maz on January 14, 2016, 09:47:58 pm
Hey mq123,
An increase in normal force can simply be achieved, if that person pushes down on the table.
Theory?
Well, as the person is pushing down the table, there would be an increase in static friction. The increase in static friction would also cause the normal force to increase, as. Lets say the amount of increase in static friction equals to a, then
, which proves that an increase in Static Friction would also cause an increase in normal force.
A decrease in the normal force can be achieved, if the person puts the table on an incline.
Theory?
Well, as if you put any object on an incline, the normal force would decrease. This can be proved, if you use the formula
For example:
Lets say the person puts the table on an incline of 45 degrees and the table's mass is 2 kg
however, if the table was not on an incline, you would use the formula:
Hopefully this helps :)
Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.
2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.
2. Decrease normal force could be done by partially leaning on the table; table now helps support your body
doesn't it ask for normal force increase/decrease of floor on the table? as it said, that how can they apply forces to the table
*I was a little confused with this one, sorry if I am wrong :P*
I believe 'them' suggests that they are talking about the 'person'.
ie. increase the normal force of the floor acting on the person
(them) but what if suggested both? because how can the use of person, be they, later? xD
Books need to clarify their questions more fluently O.o
thankyou all of you...yeah books do need to write their questions better ::)
Post by: knightrider on January 15, 2016, 10:20:58 pm
Post by: lzxnl on January 15, 2016, 11:35:42 pm
This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?
The net force is thus the horizontal component of the normal reaction force. This must therefore be the centripetal force requirement for circular motion.
Post by: anastasiaaa on January 15, 2016, 11:51:42 pm
Post by: knightrider on January 16, 2016, 12:44:42 am
Note that the vertical force components must cancel and sum to zero.
This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?
The net force is thus the horizontal component of the normal reaction force. This must therefore be the centripetal force requirement for circular motion.
Thanks lzxnl :)
"This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?"
Would this be the force they have labelled R
" vertical force components must cancel and sum to zero."
Why is this?
Also if you were to draw a triangle to find the centripetal force how would it look like and where would theta be placed?
Post by: Syndicate on January 16, 2016, 12:57:14 am
Thanks lzxnl :)
"This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?"
Would this be the force they have labelled R
" vertical force components must cancel and sum to zero."
Why is this?
Also if you were to draw a triangle to find the centripetal force how would it look like and where would theta be placed?
Hey knightrider,
Normal force is the force, which is points [vertically] upwards, thus it is R (generally, but not in all cases)
As Iznxl said that the vertical components must cancel [Subract] from one another, so you can get the net force.
Do remeber that the Net force = centripetal force =
As this is a banked curve, the centripetal force would face towards the centre of the radius (so in the same direction as Friction), which is facing [horizontally] right.
Therefore, the correct answer would be Rcos(theta), as cos corresponds to the horizontal value of a graph.
EDIT: what do you mean by your last question? You already have a triangle made up in the question, then why make another one?
Hopefully this helps :)
Post by: knightrider on January 16, 2016, 07:29:05 am
Hey knightrider,
Normal force is the force, which is points [vertically] upwards, thus it is R (generally, but not in all cases)
As Iznxl said that the vertical components must cancel [Subract] from one another, so you can get the net force.
Do remeber that the Net force = centripetal force ==
As this is a banked curve, the centripetal force would face towards the centre of the radius (so in the same direction as Friction), which is facing [horizontally] right.
Therefore, the correct answer would be Rcos(theta), as cos corresponds to the horizontal value of a graph.
EDIT: what do you mean by your last question? You already have a triangle made up in the question, then why make another one?
Hopefully this helps :)
Thanks Syndicate :)
But the answers say Rsin(theta) how did they get this ?
Post by: lzxnl on January 16, 2016, 09:12:49 am
Hence, the horizontal component of the normal reaction force is equal to R sin theta
The reason why the vertical force components must cancel is simple. Is your object moving up and down? No. So the net force must not have a vertical component.
Post by: Syndicate on January 16, 2016, 10:52:16 am
Sorry about the wrong answer :-[ (I clearly should have worked out it, before giving out an answer, carelessly)
Luckily, I worked it out on a paper, look at the attachment below!
http://imgur.com/FkC2T4s
Post by: knightrider on January 16, 2016, 09:28:12 pm
With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical.
Hence, the horizontal component of the normal reaction force is equal to R sin theta
The reason why the vertical force components must cancel is simple. Is your object moving up and down? No. So the net force must not have a vertical component.
Thanks so much lzxnl :)
"With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical."
What geometry do you use for this ?
Hey,
Sorry about the wrong answer :-[ (I clearly should have worked out it, before giving out an answer, carelessly)
Luckily, I worked it out on a paper, look at the attachment below!
http://imgur.com/FkC2T4s
no worries thanks so much Syndicate :)
Post by: lzxnl on January 16, 2016, 11:09:30 pm
Thanks so much lzxnl :)
"With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical."
What geometry do you use for this ?
no worries thanks so much Syndicate :)
You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.
Post by: Syndicate on January 16, 2016, 11:26:50 pm
You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.
I attached a screenshot with my last post, is that what you mean? But I didn't add the angle between the horizontal (net/ centripetal force) and the incline.
Post by: knightrider on January 17, 2016, 12:22:31 am
You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.
Thanks for clarifying !! :)
Post by: lzxnl on January 17, 2016, 10:09:56 am
I attached a screenshot with my last post, is that what you mean? But I didn't add the angle between the horizontal (net/ centripetal force) and the incline.
Yeah, but you need to relate the two angles in your diagram.
Post by: sweetiepi on January 18, 2016, 03:02:45 pm
Post by: lzxnl on January 18, 2016, 03:16:07 pm
Can somebody please explain how to do the question attached? 8)
In 1 second there are 2 revolutions.
A period is the time for 1 revolution.
How many seconds needed?
Post by: sweetiepi on January 18, 2016, 03:20:40 pm
Post by: Maz on January 27, 2016, 01:40:18 am
the question: Explain why curved railway tracks are usually banked towards the inside of the curve? and I'm supposed to use a force vector diagram to illustrate
some help would be much appreciated
Thankyou sooooo much in advance :)
Post by: Syndicate on January 27, 2016, 01:11:59 pm
hey guys...i kinda somewhere in my brain know how to do this question and it has to do with centripetal and centrifugal forces...Hey mq123,
the question: Explain why curved railway tracks are usually banked towards the inside of the curve? and I'm supposed to use a force vector diagram to illustrate
some help would be much appreciated
Thankyou sooooo much in advance :)
The train tracks are usually banked (on a curve), as fast moving trains has a great amount of Inertia, it tends to move tangentially off the track (contrifugal force), however, when the tracks are banked, it produces the necessary amount of centripetal force to keep the train moving in the curvature at the same speed. Friction can't provide the required amount of Centripetal force, and thus, the tracks must be banked for the reason being.
The vertical forces (Normal and Gravitation force) balances out the weight, while the Horizontal forces will provide the necessary centripetal force.
Your vector diagram would look like:
Net force (centripetal force) facing towards the center of the curvature.
Normal force facing directly vertical from the banked curve (on a tilt)
Gravitation force facing directly downwards
Post by: lzxnl on January 27, 2016, 01:30:36 pm
Hey mq123,
The train tracks are usually banked (on a curve), as fast moving trains has a great amount of Inertia, it tends to move tangentially off the track (contrifugal force), however, when the tracks are banked, it produces the necessary amount of centripetal force to keep the train moving in the curvature at the same speed. Friction can't provide the required amount of Centripetal force, and thus, the tracks must be banked for the reason being.
The vertical forces (Normal and Gravitation force) balances out the weight, while the Horizontal forces will provide the necessary centripetal force.
Your vector diagram would look like:
Net force (centripetal force) facing towards the center of the curvature.
Normal force facing directly vertical from the banked curve (on a tilt)
Gravitation force facing directly downwards
Centrifugal forces are a little problematic because they're fictitious and don't actually exist. A centrifugal force only appears if you're considering the forces felt by an accelerating object. In other words, they come about because the object doesn't have the right force to move with that acceleration.
Your explanation hasn't mentioned where this centripetal force comes from. Its origin is the tilt of the banked curve; as now the normal to the banked curve has a horizontal component, the normal reaction force will partially push vertically and partially push horizontally. The vertical component balances its weight force (it's a horizontal circle, no vertical motion) and the horizontal component is equal to the centripetal force.
Post by: TheAspiringDoc on January 27, 2016, 03:56:52 pm
I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O
Post by: Syndicate on January 27, 2016, 04:09:44 pm
Hey, here's a real life physics application for you guys :)
I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O
well offcourse... xD (unless it's fully scratched :P)
Your explanation hasn't mentioned where this centripetal force comes from. Its origin is the tilt of the banked curve; as now the normal to the banked curve has a horizontal component, the normal reaction force will partially push vertically and partially push horizontally. The vertical component balances its weight force (it's a horizontal circle, no vertical motion) and the horizontal component is equal to the centripetal force.
yea, but he only asked if why the train tracks are banked (when going through a curve)
Centrifugal forces are a little problematic because they're fictitious and don't actually exist. A centrifugal force only appears if you're considering the forces felt by an accelerating object. In other words, they come about because the object doesn't have the right force to move with that acceleration.
It's good to learn something new everyday :)
Post by: Maz on January 27, 2016, 04:15:03 pm
Hey, here's a real life physics application for you guys :)I'm still confused on how a situation like that happenes -LOL
I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O
it will work- water doesn't do anything to a DVD...the actual data burned on the dvd is a couple of layers deep- not on the surface...which is why scratches sometimes harm and sometimes don't harm the disc...but the ice may make a few marks (possibly)
Good Luck :P
Post by: bts on January 27, 2016, 06:03:50 pm
is someone able to help me here please
A bullet is fired from a gun with a barrel of 50cm, towards a target exactly 100m away. To fire a bullet, gunpowder is ignited and the expanding gasses that result from the burning gunpowder applies a constant force to the bullet while it travels down the barrel of the gun. The force that is applied to the bullet is 100N, for each cm2 of surface area of the bullet, per gram of gunpowder used. The bullet is 9mm in diameter and has a mass of 5g. The bullet leaves the gun at a velocity of 500m/s and experiences negligible air friction.
6. How much gunpowder, rounded to the nearest gram, was used to fire the bullet?
thank you in advance
Post by: lzxnl on January 27, 2016, 08:45:18 pm
Hello!
is someone able to help me here please
A bullet is fired from a gun with a barrel of 50cm, towards a target exactly 100m away. To fire a bullet, gunpowder is ignited and the expanding gasses that result from the burning gunpowder applies a constant force to the bullet while it travels down the barrel of the gun. The force that is applied to the bullet is 100N, for each cm2 of surface area of the bullet, per gram of gunpowder used. The bullet is 9mm in diameter and has a mass of 5g. The bullet leaves the gun at a velocity of 500m/s and experiences negligible air friction.
6. How much gunpowder, rounded to the nearest gram, was used to fire the bullet?
thank you in advance
I'll simplify the wording of the question.
1. Bullet experiences constant force from gas.
2. This gas acts over 50 cm
3. Constant force is given by 100 N/cm^2 * area of bullet (calculate assuming bullet is a sphere) * mass of gunpowder (need to work this out)
4. Assume bullet is initially at rest, and after the force acts on it 50 cm later it's moving at 500 m/s
5. What is the force acting on the bullet, and consequently what is the mass of gunpowder used? (calculate using point 3)
Post by: bts on January 27, 2016, 08:51:55 pm
where m is 5g, v is 500m/s and the r is 50cm?
the answer is apparently around twenty but i cant seem to get it using this
(0.005x500^2)/0.5 = 2500
2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?
the answer wouldve been 9.8 if that was the case... what am i doing wrong?
Post by: Syndicate on January 27, 2016, 10:38:31 pm
To calculate force do I use F=(mv^2)/r
where m is 5g, v is 500m/s and the r is 50cm?
the answer is apparently around twenty but i cant seem to get it using this
(0.005x500^2)/0.5 = 2500
2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?
the answer wouldve been 9.8 if that was the case... what am i doing wrong?
You are not trying to find the centripetal force (there is none in this case)
As Iznxl said, you need to use the formula 100/aw
To workout a, you must use the formula (believing it's a sphere): pi x r^2 (1.41 cm)
Post by: lzxnl on January 27, 2016, 11:01:00 pm
well offcourse... xD (unless it's fully scratched :P)
yea, but he only asked if why the train tracks are banked (when going through a curve)
It's good to learn something new everyday :)
My point was that you didn't connect banked curves with centripetal force. A non-banked curve has to wholly rely on friction for the centripetal force, whereas with a banked curve, you don't need as much friction.
To calculate force do I use F=(mv^2)/r
where m is 5g, v is 500m/s and the r is 50cm?
the answer is apparently around twenty but i cant seem to get it using this
(0.005x500^2)/0.5 = 2500
2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?
the answer wouldve been 9.8 if that was the case... what am i doing wrong?
Think about what information the question is giving you. You have constant acceleration (don't know magnitude), known initial and final velocities and a known distance. That's enough to give you the value of the acceleration. As you know the force is some number divided by a known area and unknown mass of gunpowder and you know the force, you can work out the gunpowder mass.
Post by: knightrider on January 28, 2016, 01:21:11 am
mass of Earth = 6.0 × 10^24 kg
radius of Earth = 6.4 × 10^6 m
mass of the Moon = 7.3 × 10^22 kg
mean radius of Moon’s orbit = 3.8 × 10^8 m
Calculate the gravitational force of attraction that
exists between the following objects.
d ) The Moon and the Earth
solution is as follows
Gravitational force F
= (6.67 x 10^-11 × 7.3 x 10^22 × 6.0 x 10^24 )/(3.8 x 10^8)^2
= 2.0 x 10^20 N
shouldnt R be 6.4x10^6+3.8x10^8=3.9x10^8 m instead of 3.8 x 10^8 ?
Post by: lzxnl on January 28, 2016, 01:22:55 am
For this question, assume that:
mass of Earth = 6.0 × 10^24 kg
radius of Earth = 6.4 × 10^6 m
mass of the Moon = 7.3 × 10^22 kg
mean radius of Moon’s orbit = 3.8 × 10^8 m
Calculate the gravitational force of attraction that
exists between the following objects.
d ) The Moon and the Earth
solution is as follows
Gravitational force F
= (6.67 x 10^-11 × 7.3 x 10^22 × 6.0 x 10^24 )/(3.8 x 10^8)^2
= 2.0 x 10^20 N
shouldnt R be 6.4x10^6+3.8x10^8=3.9x10^8 m instead of 3.8 x 10^8 ?
Radius of Moon's orbit IS the distance between the centres of the two planets
You don't need to add the Earth's radius to that
Post by: knightrider on January 28, 2016, 02:24:31 am
Radius of Moon's orbit IS the distance between the centres of the two planets
You don't need to add the Earth's radius to that
Thanks for clarifying !!
Post by: Maz on January 28, 2016, 11:08:18 am
but thanks guys :)
Post by: Maz on January 28, 2016, 11:52:46 am
Can someone please tell me how to do this one? please- would greatly appreciate it :)
A girl is swinging on a maypole
The girl has mass of 36kg and when she is moving with a speed of 2ms^1 the light rope makes an angle of 20 degrees with the vertical. Consider the motion of the centre of mass of the girl, which moves in a horizontal circle of radius r.
1. what are the vertical and horizontal components of tension? i think vertical is Tcos30=mv^2/r....(36*2)/r= Tcos30
which will mean 76.62=Tr...and just T would be (76.62/r)...but i don't know if thats right...or how to do the horizontal tension
2. What is the tension in the rope
3. What is the net force acting on the girl and the radius of the circle
Thankyou so much in advance :)
Post by: Shinkaze on January 30, 2016, 11:22:54 am
A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10.0 s and reaching a constant speed of 100 km/h after a further 5.0 s. At what time will the policeman catch up with the car?
Post by: TheAspiringDoc on January 30, 2016, 12:44:13 pm
And if so, why does this process release 'pure energy'?
Thanks! :D
Post by: happymeal on January 30, 2016, 12:45:14 pm
Can anyone help me with one of these super basic physics questions that I still don't know how to do .-. thanks!
A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10.0 s and reaching a constant speed of 100 km/h after a further 5.0 s. At what time will the policeman catch up with the car?
To do this, the first step you need to do is construct a velocity-time graph.
You want to depict both the policeman and the car on it (remember to convert km/h to m/s, you can do this by dividing km/h by 3.6)
Your graph should show the policeman accelerating to the same speed as the car in 10seconds, and then rising above it in the next five seconds.
Next, you want to tackle the distance that the police car has moved first. You do this by calculating the area of the triangle under the graph (10 seconds) and the trapezium under the graph (from 10 to 15 seconds). The police car now continues at a constant speed, so the formula you should have is x=(distance triangle)+(distance trapezium)+(speed times t-15). The t-15 part takes into account the distance travelled over the past 15 seconds.
Next, is the distance that the car travels. Throughout all this time, it travels at a constant velocity, so you can use the formula x=vt, subbing in your value for v.
Now to find when they meet, you equate the two equations together, giving you (distance triangle)+(distance trapezium)+(speed times t-15)=vt. The only unknown variable you should have here is t, which you will find using algebra.
Let me know if you have any questions, I tried to go through each step and it got pretty long in the end.
Post by: lzxnl on January 30, 2016, 02:32:19 pm
Does the sun convert hydrogen nuclei into helium nuclei?
And if so, why does this process release 'pure energy'?
Thanks! :D
Yes. It's called (not surprisingly) the 'proton-proton process'
I won't go into the details, but this process releases energy because of the strong nuclear force between protons and neutrons (if you ever wondered how positive charges could be so close together in the nucleus, it's because of this strong force that is REALLY strong at small distances but drops off really fast at higer distances)
This attraction releases energy. It's a bit like putting two opposite-pole magnets near each other; their opposite attraction will lead to an increase in kinetic energy.
Post by: TheAspiringDoc on January 30, 2016, 05:01:05 pm
Yes. It's called (not surprisingly) the 'proton-proton process'thank you!
I won't go into the details, but this process releases energy because of the strong nuclear force between protons and neutrons (if you ever wondered how positive charges could be so close together in the nucleus, it's because of this strong force that is REALLY strong at small distances but drops off really fast at higer distances)
This attraction releases energy. It's a bit like putting two opposite-pole magnets near each other; their opposite attraction will lead to an increase in kinetic energy.
I'm a little unsure about Red Giant formation - the contraction of a main sequence star (due to gravity) heats up the MSS (I get that). This causes Helium atoms to fuse together, whilst H fusion continues elsewhere. (I get all that too). But then, how does the He fusion (and H fusion) cause the main sequence star to expand (resultin in a red giant)?? ???
Post by: mohakmalhotra on January 30, 2016, 09:22:52 pm
i am not sure whether i have got the right answer for this question. It would be very nice if I could get your answers for this question. Thanks in advance :)
Question : Theon is firing arrows from the ground towards targets mounted on a wall 6 m above the ground. He fires his arrows at a speed of 70 m/s and at an angle of 60°.
a. What is the maximum height above the ground the arrows reach? (2 marks)
b. How long does it take the arrow to reach the target? (3 marks)
c. What horizontal distance is the target away from Theon? (2 marks)
Post by: Syndicate on February 01, 2016, 05:25:31 pm
Hi there,
i am not sure whether i have got the right answer for this question. It would be very nice if I could get your answers for this question. Thanks in advance :)
Question : Theon is firing arrows from the ground towards targets mounted on a wall 6 m above the ground. He fires his arrows at a speed of 70 m/s and at an angle of 60°.
a. What is the maximum height above the ground the arrows reach? (2 marks)
b. How long does it take the arrow to reach the target? (3 marks)
c. What horizontal distance is the target away from Theon? (2 marks)
a. 193.5 m
b. 112.37 s
c. 433.01 m
If you want an explanation with how I derived with these answers, I can surely help :)
Post by: TheAspiringDoc on February 02, 2016, 06:16:59 pm
few more astronomy questions!
What is the nature of cosmic microwave background radiation?
A)faint uniform x-ray glow
B)bright uniform x-ray glow
C)faint uniform radio signal
D)weak and patchy glow at visible wavelengths
The major cause of a supernova is:
A)gravitational collapse of the core of a massive star
B)exhaustion of nuclear fuel
Any help would be really appreciated as I have a test tomorrow! :o
Post by: Syndicate on February 02, 2016, 08:20:38 pm
Hi :)
few more astronomy questions!
What is the nature of cosmic microwave background radiation?
A)faint uniform x-ray glow
B)bright uniform x-ray glow
C)faint uniform radio signal
D)weak and patchy glow at visible wavelengths
The major cause of a supernova is:
A)gravitational collapse of the core of a massive star
B)exhaustion of nuclear fuel
Any help would be really appreciated as I have a test tomorrow! :o
Question 1: C
Question 2: A
Post by: TheAspiringDoc on February 02, 2016, 08:58:38 pm
Question 1: Creasoning?
Question 2: A
And for the first Q, it's not really uniform is it?!
Post by: sweetiepi on February 04, 2016, 05:13:04 pm
Post by: JellyBeanz on February 04, 2016, 06:07:38 pm
How would I be able to solve this? :)
Fc = Fnet = W − N
As N = 0:
Fc = W = mg
Fc = 1200 × 10
Fc = 12000
Fc=mv^2/r
v^2= Fc*r/m
v= √12000*10/1200
Post by: sweetiepi on February 04, 2016, 06:58:26 pm
Fc = Fnet = W − N
As N = 0:
Fc = W = mg
Fc = 1200 × 10
Fc = 12000
Fc=mv^2/r
v^2= Fc*r/m
v= √12000*10/1200
Thanks so much! :)
Post by: strawberry7898 on February 06, 2016, 11:31:05 am
Post by: Maz on February 06, 2016, 07:25:29 pm
Find the mass of an electron for the alpha decay (4.003u) of Po-214 (213.995u) to Pb-210 (209.984u). is this consistent with the alpha particle having an energy of 7.68 MeV....
the answer is 7.46 MeV
Thankyou so much in advance :)
Post by: Peanut Butter on February 07, 2016, 04:13:07 pm
A car enters a roundabout at 24 kmh-1 travelling in an easterly direction and leaves the roundabout at 18 kmh-1 travelling in a southerly direction. Calculate the change in velocity of the car (include the direction).
Thanks :)
Post by: Adequace on February 07, 2016, 04:37:04 pm
Can someone please help me with this question?I'd assume you could use change in v = v(f) - v(i), afterwards using vector addition which would require pythag.
A car enters a roundabout at 24 kmh-1 travelling in an easterly direction and leaves the roundabout at 18 kmh-1 travelling in a southerly direction. Calculate the change in velocity of the car (include the direction).
Thanks :)
Edit: I got 30m/s with direction pointing southwest.
Post by: alanyin1 on February 07, 2016, 09:13:52 pm
If I have 2 diodes arranged in series, both having a switch-on voltage of 0.7V but my supply voltage is strictly within the range of (0.7, 1.4) Volts (exclusive), what happens?
I remember my teacher from last year saying that electrons are already present in the wire and that they do not "begin flowing" from the negative terminal of the battery to the positive terminal. If this is the case, the closest one to the battery isn't necessarily the answer. He also explained P-N junctions in semi-conductors to us at one stage, which may be involved in the answer.
I asked my teacher last year during a Prac Period and he replied simply with a "you should try and find out" but I never actually got around to seeing what happens. I have a couple of speculations as to what happens:
1. Either Diode 1 or Diode 2 turn on, but the other remains off. The battery supplies its total voltage, but since not enough is supplied to turn both on, one takes 0.7 and the other takes the remaining and thus does not reach the required switch-on voltage. Perhaps each time this happens, it is a random process as to which one turns on?
2. Neither turn on, but this is probably the least logical answer. Imagine a Diode and Ohmic Resistor arranged in series; the Diode will always use 0.7 V, regardless of how the series circuit is arranged, given that the supply voltage >0.7V. Therefore, shouldn't speculation 1. be more feasible?
3. When I attended the ATARNotes Physics Headstart Lecture at RMIT during the holidays, I asked Alwin the presenter this question. He explained that both would be on, but would be dim. Both would receive a voltage in-between 0 and the switch-on voltage of 0.7. Since in reality, the "boundary" between "on" and "off" is not as clear-cut as 0.7V, there is a "zone" where they are "on." Therefore, both are on but are dim.
However, if the 3rd option is true, if I were to supply 0.6 V to a circuit containing only 1 diode, shouldn't it be "on"? Shouldn't each coulomb of charge expend 0.6 Joules of energy to cross the P-N junction?
I tried to find an answer to this question on Google, but the circumstance might have been too specific to type into Google and couldn't find a suitable solution
Post by: Swagadaktal on February 07, 2016, 09:45:53 pm
Can someone please help me with this question?Correct me if I'm wrong, but isn't this a question of pythag?
A car enters a roundabout at 24 kmh-1 travelling in an easterly direction and leaves the roundabout at 18 kmh-1 travelling in a southerly direction. Calculate the change in velocity of the car (include the direction).
Thanks :)
So you get 24 km/h east and 18 km/h south, soo 24^2 + 18^2 and then sq root that answer
You should get 30 km/h
So the change in velocity is 30 km/h south east --- IS THIS CORRECT?
Post by: alanyin1 on February 07, 2016, 10:02:50 pm
If we were to further scrutinise the question (probably not necessary), because we are travelling in a circle, wouldn't there be horizontal and vertical components to the respective velocities?
Post by: Swagadaktal on February 07, 2016, 10:07:50 pm
Regarding the velocity question above^ wouldn't the direction be [arctan(18/24)+90 degrees from North] ASSUMING that the same amount of time was used to travel in the respective velocities? Or do we just assume that the journey would be exactly SE due to the roundabout entrance/exits? If the same amount of time is used to travel at the velocities, it wouldn't be an isosceles triangle would it?erm can you get that spesh outta here :P
If we were to further scrutinise the question (probably not necessary), because we are travelling in a circle, wouldn't there be horizontal and vertical components to the respective velocities?
Nah jks - I don't personally think VCE Physics goes into that much detail - I may be wrong but I find it highly unlikely that they have questions of that kind of depth.
Can someone concur?
Post by: alanyin1 on February 07, 2016, 10:14:42 pm
Post by: JellyBeanz on February 09, 2016, 07:40:42 pm
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?
I've done the questions, but i don't have the solutions and i want to know if i am correct, could someone please help me with explanations for each answer.
sorry for the long question.
Post by: Maz on February 09, 2016, 10:29:30 pm
could someone please help me with this question?
why does a photon have energy but not obey Einstein's equation E=mc^2
Post by: lzxnl on February 09, 2016, 11:58:44 pm
Rather, you have to use the total energy, which is given by E^2 = (pc)^2 + (mc^2)^2
Post by: Maz on February 10, 2016, 06:48:31 pm
That is the rest mass energy of a particle. Photons have no rest mass, so E = mc^2 doesn't work for them. Rest mass energy means the energy the particle has in a stationary reference frame.
Rather, you have to use the total energy, which is given by E^2 = (pc)^2 + (mc^2)^2
Thankyou :)
Post by: Maz on February 12, 2016, 10:22:08 pm
i was wandering if someone could please help me with this relativity question
while you are on earth you notice a space ship travel past you at 0.99c in 30ns. this spacecraft is normally stored in a hanger. what minimum length would this hanger need to be?
i tried doing it but my answer came out as 0.9785m and well that doesn't seem very realistic for a space ship :)
id very much appreciate any help :)
Post by: lach3087 on February 13, 2016, 04:18:12 pm
Post by: Gogo14 on February 16, 2016, 10:27:12 pm
Post by: lzxnl on February 17, 2016, 12:38:39 am
hello humans...again :)
i was wandering if someone could please help me with this relativity question
while you are on earth you notice a space ship travel past you at 0.99c in 30ns. this spacecraft is normally stored in a hanger. what minimum length would this hanger need to be?
i tried doing it but my answer came out as 0.9785m and well that doesn't seem very realistic for a space ship :)
id very much appreciate any help :)
Well, you measure a distorted length; the length you measure is contracted by a factor of 1/(sqrt(1-0.99^2)) which turns out to be 7.1
The length you measure is 0.99c * 30 ns = 8.91 m
The actual length therefore is around 63.3 m
I was doing a prac, but there is a variable resistor ( like the dial one) with 3 wires. 1 red, black and white. What does the white wire do and where do you connect it.
How many places to plug wires are there in the resistor?
Post by: Gogo14 on February 17, 2016, 01:57:58 pm
Post by: Maz on February 17, 2016, 07:49:46 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: lzxnl on February 17, 2016, 09:49:41 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
First question: no. If two trains aren't moving at the same velocity in one frame, you can't possibly find a reference frame in which they're moving at the same velocity. This is partly common sense (if one person thinks two things are moving at different speeds/directions, surely everyone else would think so too), but it can be proven quite easily because Lorentz transformations (i.e. changing from one reference frame to another) are reversible. If the transformation is reversible, then if you have two different velocities transforming into the same velocity, the reverse transformation isn't uniquely defined, which physically makes no sense.
Second question: yes. Although if they're both non-inertial, that means they're both accelerating. Inertial reference frame means obeys Newton's first law (i.e. if you let go of something in that frame of reference, does it appear to not accelerate). Curiously, this means that if you have someone in free-fall, the above test of whether a frame is inertial will tell you that this free-fall person is in an inertial frame because anything they drop will 'accelerate at the same rate' (because gravity is independent of mass) and thus they won't perceive any acceleration. This is the basis for Einstein's theory of general relativity which I'm hoping to look at next year :P
Post by: Maz on February 17, 2016, 10:11:47 pm
First question: no. If two trains aren't moving at the same velocity in one frame, you can't possibly find a reference frame in which they're moving at the same velocity. This is partly common sense (if one person thinks two things are moving at different speeds/directions, surely everyone else would think so too), but it can be proven quite easily because Lorentz transformations (i.e. changing from one reference frame to another) are reversible. If the transformation is reversible, then if you have two different velocities transforming into the same velocity, the reverse transformation isn't uniquely defined, which physically makes no sense.Thankyou so much
Second question: yes. Although if they're both non-inertial, that means they're both accelerating. Inertial reference frame means obeys Newton's first law (i.e. if you let go of something in that frame of reference, does it appear to not accelerate). Curiously, this means that if you have someone in free-fall, the above test of whether a frame is inertial will tell you that this free-fall person is in an inertial frame because anything they drop will 'accelerate at the same rate' (because gravity is independent of mass) and thus they won't perceive any acceleration. This is the basis for Einstein's theory of general relativity which I'm hoping to look at next year :P
Post by: Maz on February 17, 2016, 11:04:13 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: soNasty on February 18, 2016, 05:56:05 pm
'a terrorist is cornered down the end of a dead end street with a nuclear power plant behind him. a tank is advancing on him but he knows it will not fire and endanger the power plant. he tries to defend himself with a flame thrower. given that he can get maximum range from the flame thrower when the angle of projection is (theta=45 deg) and that, when fresh, the flame thrower can project liquid at a speed of 40root2 m/s, which of the following is closest to the maximum distance the tank can be from him for the weapon to be at all effective?
A. 50m
B. 100m
C. 200m
D. 300m
Answer is D.
Help pls
Post by: JellyBeanz on February 18, 2016, 06:58:53 pm
Idk if this is the right place to post this but im in need of help with a gamsat physics question
'a terrorist is cornered down the end of a dead end street with a nuclear power plant behind him. a tank is advancing on him but he knows it will not fire and endanger the power plant. he tries to defend himself with a flame thrower. given that he can get maximum range from the flame thrower when the angle of projection is (theta=45 deg) and that, when fresh, the flame thrower can project liquid at a speed of 40root2 m/s, which of the following is closest to the maximum distance the tank can be from him for the weapon to be at all effective?
A. 50m
B. 100m
C. 200m
D. 300m
Answer is D.
Help pls
So this is a projectile question.
We are given two things, the angle at which the liquid was projected (45) degrees and the initial velocity of the liquid. 40root2 m/s
1st thing you should do is find the vertical and horizontal components of the velocity.
Initial vertical velocity = initial velocity * sin (45)
= 40root2 * 1/root2
= 40 m/s
Initial horizontal velocity = initial velocity * cos (45)
= 40root2 * 1/root2, *Note you already know this because it is a right angled isosceles triangle
Since you have both components of velocity first thing you have to find is the total time it takes to reach x (distance).
using the initial vertical velocity (40m/s) use newton's Constant acceleration formula, v = u+at, substituting -10 m/s^2 for acceleration, v = 0 at top of the flight and u = 40 m/s, and solving for time, you get t = 4 seconds for half of the flight.
Therefore, total time it takes for the flight = 2* t
= 8 seconds for the total time.
Now what we do is use the horizontal component to find the distance the liquid will reach. we know that for the horizontal component x (distance) = (horizontal velocity) * (time)
using this distance = 40 m/s * 8
= 320 meters
hence the maximum distance the tank can be from him for the weapon to be effective is 320m, answer is D as 320m is closest to 300m
I hope it can get through to you somehow, i'm sure there may be easier ways to do this, and my explanation is not the best, a better explanation can be done with the aid of a diagram, i'm not great at diagrams XD
Post by: soNasty on February 18, 2016, 07:07:11 pm
So this is a projectile question.
We are given two things, the angle at which the liquid was projected (45) degrees and the initial velocity of the liquid. 40root2 m/s
1st thing you should do is find the vertical and horizontal components of the velocity.
Initial vertical velocity = initial velocity * sin (45)
= 40root2 * 1/root2
= 40 m/s
Initial horizontal velocity = initial velocity * cos (45)
= 40root2 * 1/root2, *Note you already know this because it is a right angled isosceles triangle
Since you have both components of velocity first thing you have to find is the total time it takes to reach x (distance).
using the initial vertical velocity (40m/s) use newton's Constant acceleration formula, v = u+at, substituting -10 m/s^2 for acceleration, v = 0 at top of the flight and u = 40 m/s, and solving for time, you get t = 4 seconds for half of the flight.
Therefore, total time it takes for the flight = 2* t
= 8 seconds for the total time.
Now what we do is use the horizontal component to find the distance the liquid will reach. we know that for the horizontal component x (distance) = (horizontal velocity) * (time)
using this distance = 40 m/s * 8
= 320 meters
hence the maximum distance the tank can be from him for the weapon to be effective is 320m, answer is D as 320m is closest to 300m
I hope it can get through to you somehow, i'm sure there may be easier ways to do this, and my explanation is not the best, a better explanation can be done with the aid of a diagram, i'm not great at diagrams XD
thanks! i get it all except how is the total time distance equal to 2t?
Post by: JellyBeanz on February 18, 2016, 07:16:58 pm
thanks! i get it all except how is the total time distance equal to 2t?
The total time distance is equal to 2t because (ignoring air resistance) and the only force acting on the liquid is the force of gravity, the motion of the liquid is identical whether it is coming up or it's coming down. since it takes 4 seconds for the liquid to reach it's maximum height, it must take twice as much time to reach the ground.
EDIT: the quickest way to find the time in this case, when giving the angle and initial velocity is through the formula t=2usin(45)/g
notice the 2, it's basically a derivation of newton's formula v=u+at but multiplied by two.
Post by: lzxnl on February 18, 2016, 09:52:47 pm
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
Assume all lengths given are proper lengths.
Post by: Maz on February 18, 2016, 10:10:57 pm
Assume all lengths given are proper lengths.right thanks- so l0 would be 600
Post by: Gogo14 on February 20, 2016, 01:30:41 pm
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.
So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help
Post by: lzxnl on February 20, 2016, 09:16:42 pm
For 1/2 physics, the q is
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.
So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help
Currents have directions given by the sign of the moving charge. An electron moving to the right at 10000 m/s is mathematically the same current as a proton moving to the left at 10000 m/s. Hence you add the currents there.
Post by: Gogo14 on February 20, 2016, 09:46:39 pm
Currents have directions given by the sign of the moving charge. An electron moving to the right at 10000 m/s is mathematically the same current as a proton moving to the left at 10000 m/s. Hence you add the currents there.But if they are travelling in opposite directions, why would you add them together?But even if you add them, how do you know which way the current will flow. In the q, it has more current flowing left than right, but the final total current is flowing right anyway according to the answers
Post by: lzxnl on February 20, 2016, 10:07:48 pm
But if they are travelling in opposite directions, why would you add them together?But even if you add them, how do you know which way the current will flow. In the q, it has more current flowing left than right, but the final total current is flowing right anyway according to the answers
They're opposite charges. That's why you'd add them even though they're in opposite directions
Post by: Alwin on February 21, 2016, 08:29:09 pm
For 1/2 physics, the q is
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.
So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help
Another way to think of it is: "a total negative charge of 30 C moved to the left" is equivalent to +30 C moving to the right as lzxnl stated in his post.
So then we would have +15 C moving to the right and another +30 C moving towards the right which will give a total of 45 C moving towards the right. Then we could find current by dividing by time and get 9A.
Also: "So I got that there is a current of 6A to the left and 3A to the right" ... This doesn't really sit well with me (sorry) because conventional current is defined as the movement of positive charge. That's why we had to consider -30 C moving left as equal to +30 C. Hope it makes sense :)
Post by: lach3087 on February 27, 2016, 03:54:06 pm
Post by: Syndicate on February 27, 2016, 06:53:52 pm
Hi, i have a question about projectile motion that i was wondering if i could get some help with. I need to know why the angle at which a projectile is fired at affects its range and how to relate that answer to the formula R = (v^2sin(2theta))/g. I know the optimum angle is 45 degrees but apart from it meaning that sin(2theta) will be equal to the highest value (1), i am unsure why.
*haven't been here since a long time :P*
well sin(2x45) = sin(90)
sin(90) = 1
Post by: lzxnl on February 27, 2016, 11:30:39 pm
Hi, i have a question about projectile motion that i was wondering if i could get some help with. I need to know why the angle at which a projectile is fired at affects its range and how to relate that answer to the formula R = (v^2sin(2theta))/g. I know the optimum angle is 45 degrees but apart from it meaning that sin(2theta) will be equal to the highest value (1), i am unsure why.
On the one hand, you need to throw your projectile high so that it has enough time to travel far.
On the other hand, you need to throw it flat enough so that its horizontal velocity is decent.
Post by: lach3087 on February 27, 2016, 11:51:54 pm
On the one hand, you need to throw your projectile high so that it has enough time to travel far.
On the other hand, you need to throw it flat enough so that its horizontal velocity is decent.
Can any of that be explained using W=mg?
Post by: Maz on March 03, 2016, 07:26:32 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: lzxnl on March 03, 2016, 09:10:05 pm
Can any of that be explained using W=mg?
Yes, you can.
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 :)
Consider the S frame as the lab frame. Then the velocity of one of the particles is 0.75c.
Pretend the S' frame is the other particle, which is moving at -0.75c relative to the S frame. This -0.75c is your v.
Then the velocity u' = (u - v)/(1 - uv/c^2) = (0.75c + 0.75c)/(1.5625) = 1.5c/1.5625 = 0.9375c
I'm assuming, of course, that you mean both particles are moving at 0.75c towards each other. If that's not the question, please say so.
If they're moving in the same direction, then both particles are moving at the same velocity in the S frame and they must necessarily be perceived to be moving at the same velocity in any other reference frame.
Post by: Maz on March 03, 2016, 10:55:20 pm
thankyou! :D
Consider the S frame as the lab frame. Then the velocity of one of the particles is 0.75c.
Pretend the S' frame is the other particle, which is moving at -0.75c relative to the S frame. This -0.75c is your v.
Then the velocity u' = (u - v)/(1 - uv/c^2) = (0.75c + 0.75c)/(1.5625) = 1.5c/1.5625 = 0.9375c
I'm assuming, of course, that you mean both particles are moving at 0.75c towards each other. If that's not the question, please say so.
If they're moving in the same direction, then both particles are moving at the same velocity in the S frame and they must necessarily be perceived to be moving at the same velocity in any other reference frame.
Post by: NerdyPi on March 05, 2016, 03:28:30 pm
I have a question regarding air resistance in projectile motion;
If you were to describe the air resistance (to quote the VCAA study design "a qualitative description of the effects of air resistance"), would it be correct to say that the force of air resistance in the vertical direction is increasing as the object accelerates downwards (assuming this is before it reaches terminal velocity), but that the air resistance in the horizontal direction is constant, as there is no acceleration?
If not, could someone please clear by my misunderstanding...
Thanks :)
Post by: Syndicate on March 05, 2016, 04:17:58 pm
Hi guys :)
I have a question regarding air resistance in projectile motion;
If you were to describe the air resistance (to quote the VCAA study design "a qualitative description of the effects of air resistance"), would it be correct to say that the force of air resistance in the vertical direction is increasing as the object accelerates downwards (assuming this is before it reaches terminal velocity), but that the air resistance in the horizontal direction is constant, as there is no acceleration?
If not, could someone please clear by my misunderstanding...
Thanks :)
1) You are absolutely correct about your first point (due to the acceleration of gravity, the air resistance would increase vertically). The gravitational force is "stronger" than air resistance, which causes an increase in air resistance, as the object tends to accelerate faster vertically (until reaching its terminal velocity).
2) Air resistance would decrease horizontally, as there would be acceleration (as the projectile tends to slow down). If you think about it... when you throw a ball, it would start with a maximum force, and then gradually start decreasing as it goes further away from you. As it slows down, there would be a decrease in air resistance.
Post by: NerdyPi on March 05, 2016, 05:16:03 pm
1) You are absolutely correct about your first point (due to the acceleration of gravity, the air resistance would increase vertically). The gravitational force is "stronger" than air resistance, which causes an increase in air resistance, as the object tends to accelerate faster vertically (until reaching its terminal velocity).
2) Air resistance would decrease horizontally, as there would be acceleration (as the projectile tends to slow down). If you think about it... when you throw a ball, it would start with a maximum force, and then gradually start decreasing as it goes further away from you. As it slows down, there would be a decrease in air resistance.
Ah, makes sense. Thanks you again :)
Post by: lzxnl on March 06, 2016, 05:16:16 pm
1. Its direction is against the direction of motion of the object in consideration.
2. Its magnitude increases with object speed (exact dependence varies; can be linear, quadratic in speed)
So, as you throw something, initially there'll be a horizontal air resistance. However, over time the horizontal air resistance will reduce the horizontal velocity to zero and the velocity will be purely vertical. Also, as the air resistance depends on velocity, as the speed increases, so does the air resistance.
Post by: Maz on March 12, 2016, 02:46:40 pm
I'm trying to calculate the gravitational and electrostatic force between two electrons neutrons that
are 1 x 10-15m apart...however my answer isn't matching up with the book...can someone please help me?
the answer in th book is 0 for electrostatic and 2x10-34 N for gravitational...
thankyou sooo much in advance :)
Post by: Syndicate on March 12, 2016, 07:24:48 pm
hey,
I'm trying to calculate the gravitational and electrostatic force between two electrons neutrons that
are 1 x 10-15m apart...however my answer isn't matching up with the book...can someone please help me?
the answer in th book is 0 for electrostatic and 2x10-34 N for gravitational...
thankyou sooo much in advance :)
well, the electrostatic force between the neutrons is 0, as they don't have any charge, therefore, no attraction/repulsion between them. To work out the gravitational force between the two neutrons, you can easily use the formula: F = Gm^2/r^2
where:
G = 6.67*10-11
m = 1.6749 X 10-24 (mass of a neutron)
r = 1x 10-15
I believe you can do the rest
Post by: Maz on March 13, 2016, 04:17:54 pm
well, the electrostatic force between the neutrons is 0, as they don't have any charge, therefore, no attraction/repulsion between them. To work out the gravitational force between the two neutrons, you can easily use the formula: F = Gm^2/r^2thanks...thats what i thought you did...turns out the answers in my book are wrong :)
where:
G = 6.67*10-11
m = 1.6749 X 10-24 (mass of a neutron)
r = 1x 10-15
I believe you can do the rest
thank you again :)
Post by: Maz on March 14, 2016, 09:56:17 pm
It's on the standard model
An electron and a positron undergo pair annihilation. if they initially had no kinetic energy, what is the energy of each gamma produced by the annihilation? why must there be 2 gamma rays produced?
thankyou so much in advance :)
Post by: lzxnl on March 14, 2016, 10:13:59 pm
Hi humans, can someone please help me on this?
It's on the standard model
An electron and a positron undergo pair annihilation. if they initially had no kinetic energy, what is the energy of each gamma produced by the annihilation? why must there be 2 gamma rays produced?
thankyou so much in advance :)
Two gamma rays to conserve momentum. If you only have one gamma ray, you have momentum coming from a system of two stationary particles.
Energy of each gamma ray must be the same (so that their momentums cancel). You can work this out from total energy = rest energy of essentially two electrons
Post by: Maz on March 15, 2016, 06:51:30 am
Post by: Maz on March 15, 2016, 06:51:59 am
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: lzxnl on March 15, 2016, 11:11:12 pm
hi,
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
a. Find mass of a muon. Convert total mass to an energy. Divide this energy amongst two photons evenly.
b. Find wavelength from E = hc/wavelength
c. Answered above
d. Opposite each other, conserve momentum
e. Depends on wavelength. Look up anything online that has a table of these
Post by: shlblk on March 16, 2016, 10:14:01 pm
Can someone please explain to me (preferably with formulas) why a ball thrown at a 45 degree angle will travel further than one thrown horizontally (0 degrees) if launch speed is the same? I want to know this for my physics investigation.
I was trying to see how it works with the range formula, but since sin0 is 0, I get no solution.
Thanks! :)
Post by: lzxnl on March 16, 2016, 10:45:30 pm
Hi,
Can someone please explain to me (preferably with formulas) why a ball thrown at a 45 degree angle will travel further than one thrown horizontally (0 degrees) if launch speed is the same? I want to know this for my physics investigation.
I was trying to see how it works with the range formula, but since sin0 is 0, I get no solution.
Thanks! :)
Actually that is a solution. As sin 0 is 0, that means it won't travel anywhere.
Remember how the range formula works. You're starting and finishing at the same height. Which means you're tossing something parallel to the ground...while you're on the ground. So as you never leave the ground, your motion never starts.
Post by: shlblk on March 16, 2016, 10:51:02 pm
Actually that is a solution. As sin 0 is 0, that means it won't travel anywhere.Ok, that makes sense. Could you then please explain to me why from a particular height, a ball thrown at a 45 angle will go further than one thrown horizontally, and why a horizontally thrown ball travels further than one thrown at say 30 or 60? Thanks.
Remember how the range formula works. You're starting and finishing at the same height. Which means you're tossing something parallel to the ground...while you're on the ground. So as you never leave the ground, your motion never starts.
Post by: Syndicate on March 17, 2016, 04:32:30 pm
Ok, that makes sense. Could you then please explain to me why from a particular height, a ball thrown at a 45 angle will go further than one thrown horizontally, and why a horizontally thrown ball travels further than one thrown at say 30 or 60? Thanks.
Think about the formula:
Post by: Redoxify on March 17, 2016, 06:14:13 pm
Post by: Syndicate on March 19, 2016, 03:24:23 pm
Help plz </3
(a) use the formula of horizontal displacement (to find the time it takes for it to fall 1.46 m):
Therefore:
1.46 = 0+4.9t^2
1.46/4.9 = t^2
t = 0.545856376
Now to find the velocity of the mug falling down the distance 2.2 m, we must divide 2.2 by t (because v = distance/ time)
2.2/t = 4.03 m/s (approximately)
(b) This can be solved by finding the vertical value, and subbing it in the formula:
verticle velocity = sqrt(2g x h) (derived formula)
= 5.35 m/s
Post by: Syndicate on March 23, 2016, 10:20:17 pm
In class, we're doing this weird identifying errors worksheet as our holiday homework to get us in the right mindset to scrutinise our pracs and better identify errors or whatever. I'm stuck on this question though (found here because the photo was apparently too big of a file: http://uploadpie.com/2eKGy). Basically, we have to find an error and suggest potential ways of resolving it.
We have to come up with five potential sources of error and so far, all I've got is
1. Friction between the board and the trolley rolling down the board. - How would I solve this error? Especially considering we can't exactly tell friction to go get lost so we can do our physics experiment in peace...
2. An inaccurate reading of the distance travelled (because I'm assuming they're measuring the distance with like a ruler)
Errrr.... be more precise about readings?
3. Reaction times? Shorter travel times = shorter time for the person with a stopwatch to react. Could be solved by using some of that fancy gadgetry where the moment the trolley starts rolling, a sensor detects it and starts timing and when it passes the end point, another sensor detects it and stops timing and you get a super exact time. These things do exist, right?
Someone please help me find 2 more errors?
Just saying... errors are not the mistakes you have made in the prac (so errors are the things that are wrong in your method). Your points 2 and 3 are mistakes (the things that can be fixed, if you do the Parc again), and I don't believe they would considered as errors.
Post by: lzxnl on March 23, 2016, 11:52:38 pm
Post by: Syndicate on March 25, 2016, 07:25:10 pm
Bump. Anyone? Pretty pleaseeee?A few things I came up with:
- stability of the track over different angles
- the force applied when the trolley was released (by the person)- this can cause inaccuracies
- At the end, the trolley was slowed down due to a sudden bump
Post by: Syndicate on March 26, 2016, 02:50:29 pm
What do you mean by the first point?
I mean that was the track stable enough to allow the trolley car to accelerate down it easily. Was the track moving? shaking? etc.. (I have done this experiment before, and we used wooden planks to make the track, so it was a little shaky at a higher angle).
Post by: knightrider on March 31, 2016, 04:53:25 pm
say your variable was angles . What formulas could you use to work out the angles?
Also what would be some other good variables to investigate relating to the pendulum EPI ?
Post by: Syndicate on March 31, 2016, 07:00:14 pm
Ohhh. Thanks Syndicate. I only just saw this. :)
For that inclined plane experiment with the cart rolling down the plane, is there any way we can account for friction? Is there any way to figure out the amount of friction acting?
Generally, most of the motion formulas used in VCE Physics tend to ignore friction. I don't think you can calculate friction, as there isn't enough information given, however, you can say that the friction was not 'stronger' than the trolley car's acceleration, as it was able to move (whereas, if it wasn't able to move, then it would have meant that friction was stronger than the force applied to the trolley car). So yeah... you shouldn't really worry about it.
For a pendulum EPI.
say your variable was angles . What formulas could you use to work out the angles?
I am not quite sure about the formula, however, I did a little research and found this great website (it has some good information)- https://www.boundless.com/physics/textbooks/boundless-physics-textbook/waves-and-vibrations-15/periodic-motion-123/the-simple-pendulum-431-8324/
hopefully others will be able to help you here :)
Also what would be some other good variables to investigate relating to the pendulum EPI ?
- length of the pendullum (does it increases/decreases it's velocity?)- l (length variable)
- mass of bob (again, what effects does it have on the velocity)- m (mass of bob)
Post by: Syndicate on March 31, 2016, 08:32:18 pm
Then how would we solve the source of error that is friction for this experiment? All of our equations assume ideal-ness but obviously there is no such thing as an ideal situation so how would you be able to calculate an acceleration similar to the theoretical acceleration if the friction was present? Wouldnt you end up with a huge experimental error because of friction which we did not take into consideration?
Friction is an error because it's causing a change in the trolley's speed, and we know that friction is not accounted for when solving for speed. You are absolutely correct about your last point, as if you were to measure the trolley's final speed with hi-tech equipment, you would end up with something completely different compared to what you worked out on your piece of paper. We cannot completely remove friction, so maybe your answer to this question can be using a track with a surface producing minimal friction.
Post by: knightrider on April 01, 2016, 01:45:55 pm
I am not quite sure about the formula, however, I did a little research and found this great website (it has some good information)- https://www.boundless.com/physics/textbooks/boundless-physics-textbook/waves-and-vibrations-15/periodic-motion-123/the-simple-pendulum-431-8324/
hopefully others will be able to help you here :)
- length of the pendullum (does it increases/decreases it's velocity?)- l (length variable)
- mass of bob (again, what effects does it have on the velocity)- m (mass of bob)
Thanks Syndicate :)
Post by: YellowTongue on April 04, 2016, 08:31:14 am
Post by: Peanut Butter on April 04, 2016, 09:58:56 am
If a mass follows a circular trajectory in a vertical plane, will increasing the size of the mass or radius of motion change the period of motion or the minimum speed that the mass can travel at?
minimum speed = root(r x g)
Hence, changing the radius will affect the minimum speed however changing the mass will not make a difference 😊
Post by: Edge_of_Chaos on April 13, 2016, 08:43:46 pm
A) What is the velocity of the basketballs centre of mass when it returns to its original height above the ground?
B) For how long was the basketballs centre of mass above its original height?
Post by: lzxnl on April 13, 2016, 09:35:57 pm
A basketball player jumps directly upwards so that his centre of mass reaches a maximum displacement of 50 cm.
A) What is the velocity of the basketballs centre of mass when it returns to its original height above the ground?
B) For how long was the basketballs centre of mass above its original height?
This is a 1 D gravitational motion question. The centre of mass bit is a more accurate way of saying the basketball player moved up 50 cm (because then you have the problem of what part of their body moved up 50 cm). Does that help?
Post by: Maz on April 19, 2016, 11:56:56 am
can someone please help me with this question?
A canoeist can paddle a kayak for short bursts at a speed of 2.7 ms-1 in still water. he wants to cross a stream in which a 2ms-1 current flows. At what angle to the current must he point his canoe if he wants to land on the bank directly opposite where he started.?
Thankyou so much in advance :)
Post by: bedigursimran on April 20, 2016, 05:03:34 pm
Post by: Syndicate on April 20, 2016, 05:11:19 pm
hey
can someone please help me with this question?
A canoeist can paddle a kayak for short bursts at a speed of 2.7 ms-1 in still water. he wants to cross a stream in which a 2ms-1 current flows. At what angle to the current must he point his canoe if he wants to land on the bank directly opposite where he started.?
Thankyou so much in advance :)
Is it all the information given? I believe the question is missing some valuable information (how far is the bank?)
Post by: Elisha913 on April 25, 2016, 02:36:13 pm
Post by: YellowTongue on April 25, 2016, 02:48:58 pm
Post by: Elisha913 on April 27, 2016, 06:37:15 am
What actually is a data analysis SAC? I've never done one before...
I have mine today so I'll tell you later :')
Post by: Maz on May 01, 2016, 11:55:07 am
I'm doing circular motion...could someone please help me with this question?
A pilot flies her aeroplane in a vertical loop of diameter 1.6km
Ignoreing air resistance, what is the speed of the aeroplane as it emerges from the bottom of the loop?
i'd really appreciate the help
thankyou so much in advance :)
Post by: JI2015 on May 01, 2016, 12:00:17 pm
mgh=0.5mv^2
gh=0.5v^2
v=(2gh)^0.5
In this case: v=(2*9.8*1.6*1000)^0.5= 177m/s
Post by: Maz on May 01, 2016, 12:15:18 pm
could you please possibly help me with another?
An aeroplane flies in a vertical loop of radius 650m. at the top of the loop the pilot experiences a downward reaction force, from her seat equal to one fifth of her weight. Estimate the aeroplane's speed at this instant.
thankyou :)
Post by: JI2015 on May 01, 2016, 12:24:31 pm
The Fnet=W+R= mg+1/5(mg)=6/5(mg)
Fnet=F(centripetal)=mv^2/r
so 1.2mg=mv^2/r 1.2gr=v^2 Hence v=(1.2gr)^0.5
Sub the known values into v=(1.2gr)^0.5, we get v=(1.2*9.8*650)^0.5= 87m/s
Post by: HopefulLawStudent on May 14, 2016, 10:55:25 pm
1. In the study design, in the photonics and electronics section, they say we need to calculate the effective resistance for circuits comprising of unloaded voltage dividers. What's an unloaded voltage divider?
2. The study design says we need to know stuff re:"energy transfers and transformations in opto-electric devices". Is that that stuff about how electrical energy is transferred to thermal/light/whatever energy?
Post by: Swagadaktal on May 14, 2016, 11:08:46 pm
I have two urgent questions. SAC is on Monday... Please help. Will be forever grateful.1) A loaded voltage divider is where the resistance of a load in parallel is far too great for it to impact the effective resistance, so I'm going to assume that unloaded is an object that is attached in parallel where the resistance is not too high so that it actually impacts the total effective resistance in that parallel section of the circuit (hope this makes sense)
1. In the study design, in the photonics and electronics section, they say we need to calculate the effective resistance for circuits comprising of unloaded voltage dividers. What's an unloaded voltage divider?
2. The study design says we need to know stuff re:"energy transfers and transformations in opto-electric devices". Is that that stuff about how electrical energy is transferred to thermal/light/whatever energy?
2) Yeah energy transfer: Using source-> input transducer -> modulator -> demodulator -> output transducer -- used in the transmission of a signal/data (such as audio)
Opto-electric devices specifically talks about LED's and LDRs, you could potentially be asked about the benefits of using an LED over an LDR in the process of transmission through modulation. (it's more immediate so the signal would be more accurate)
Hope this helps
Post by: HopefulLawStudent on May 15, 2016, 09:40:32 am
So basically for the first point, we just need to know circuits like this one:
(https://upload.wikimedia.org/wikipedia/commons/8/8f/Voltage_divider.svg)
Or circuits that can be simplified into something like that?
Basically stuff that we can use the Vout = Vin (R1/(R1 + R2)) formula for?
Post by: Swagadaktal on May 15, 2016, 10:49:45 am
Thanks Swag!OOPs wait a sec I think I go my definitions mixed.
So basically for the first point, we just need to know circuits like this one:
(https://upload.wikimedia.org/wikipedia/commons/8/8f/Voltage_divider.svg)
Or circuits that can be simplified into something like that?
Basically stuff that we can use the Vout = Vin (R1/(R1 + R2)) formula for?
Unloaded follows the laws of Vout = Vin*r1/r1+r2 with no potential for the effective resistance to be changed.
Loaded is where the resistance attached can either be too high for it to make an impact or it can change the effective resistance if its low enough - generally speaking though you can approach the questions the same way you'll just have to add in an extra step. I doubt they'd ask the differences between.
And yes that's a kind of circuit you'll encounter - your practice qs in book or practice sacs should have plenty of example circuits with voltage dividers
Post by: HopefulLawStudent on May 15, 2016, 11:08:35 am
Post by: Swagadaktal on June 03, 2016, 04:39:33 pm
Does sound change air density?
Like with compressions and rarefactions does the air density change at those points?
Post by: jyce on June 03, 2016, 04:42:32 pm
Hey guys
Does sound change air density?
Like with compressions and rarefactions does the air density change at those points?
Yep, compressions of sound = areas of high air pressure, rarefactions = areas of low air pressure.
Post by: Swagadaktal on June 03, 2016, 04:48:44 pm
Yep, compressions of sound = areas of high air pressure, rarefactions = areas of low air pressure.wew thank God omg u make me so happy jyce <3
Post by: @#035;3 on June 06, 2016, 10:12:27 pm
Cheers.
Post by: Syndicate on June 07, 2016, 01:01:18 pm
Can someone pls show me how to do VCAA 2006 EXAM 2 Q6 structures and materials.
Cheers.
What page number is it on the exam?
Post by: @#035;3 on June 07, 2016, 10:40:15 pm
here's the link: http://www.vcaa.vic.edu.au/Documents/exams/physics/physics22004.pdf it's on page18 CHEERS
EDIT: DW figured it out :)
Post by: CarrymetoUni on June 08, 2016, 02:09:17 pm
I'm unsure about this question, please help me out. Is the ratio 1/2 since 2 resistors are in parallel and one is in series. I have attached a picture of the circuit.
Post by: Swagadaktal on June 08, 2016, 02:23:23 pm
If all three resistors have the same value, what is the ratio Vout/Vin?if u have 2 resistances in parallel of same resistance, then the effective resistance halves.
I'm unsure about this question, please help me out. Is the ratio 1/2 since 2 resistors are in parallel and one is in series. I have attached a picture of the circuit.
So the Voltage dissipated of the parallel will be half the voltage on the single R.
Therefore the ratio it 2:1
Post by: CarrymetoUni on June 08, 2016, 02:47:49 pm
An LED and a light dependent resistor (LDR) are being used in the circuit shown below to transmit data from a computer to a printer. The circuits are as shown below. The control block (data) adds the zero referenced data signal (one that varies plus and minus around zero) to the DC 10V supply. The control block has a max, peak to peak output of 20V (I.e. from -10V to +10V)
What will the minimum and maximum voltages at A be?
Post by: bedigursimran on June 15, 2016, 09:48:13 pm
I don't understand why the answers are figuring out the length y to find torque and thus tension in cable. I may be missing something in my understanding. Thanks so much!
Post by: StupidProdigy on June 17, 2016, 11:51:17 am
Hi guys, I have a materials and structures question (and answer) that I have attached below. Hope you guys can have a look at and help me.torque=f*d (note that d must be the perpendicular distance to the force! 'y' is the perpendicular distance from point O to the tension force). Alternatively, to do this question without finding the distance 'y', you can resolve the tension force in horizontal and vertical components and find the torques which these components create (the horizontal component will create zero torque as the force's line of action will pass through point O). Finally, they have chosen point O to take the torque about since there is a reaction force here, but it becomes neglected in the analysis since it passes through point O. Did that help? :)
I don't understand why the answers are figuring out the length y to find torque and thus tension in cable. I may be missing something in my understanding. Thanks so much!
Post by: HopefulLawStudent on July 03, 2016, 12:18:09 pm
Quote
There are two conductors, m and n that are placed next to each other a distance of 2R apart from one another on the east-west axis. If m carries a current into the page and n carries a current out of the page, at which of the following points could a third conductor experience zero magnetic force?
A. a point between m and n
B. a point west of m
C. a point east of n
Also attempted a diagram of what they're saying (please excuse the dodginess of my drawing tho :P )
Please. Someone help.
Post by: wyzard on July 03, 2016, 03:46:48 pm
Also attempted a diagram of what they're saying (please excuse the dodginess of my drawing tho :P )
Please. Someone help.
The answer's A, as the magnetic field from both wires cancel each other out in the middle.Just trace out the magnetic field produced by each wire using the right-hand grip rule and you'll see why.
Post by: HopefulLawStudent on July 03, 2016, 04:53:14 pm
Post by: wyzard on July 03, 2016, 06:05:54 pm
Nope. The answer's B and C, apparently. I thought the answer would be A too but it's not and I don't get why.
Oh wait a minute, made a silly mistake earlier as I was in a hurry. :(
The magnetic fields in the middle are pointing in the same direction. Along the west and east of conductor m and n, the magnetic fields cancel out partially, however it will never be zero. It does however approach to zero as you get further and further away, so I think that is what the question meant.
On a side note, you can actually make A the correct answer by orientating the third wire perpendicular m and n, making it parallel to the field produced.
Post by: Swagadaktal on July 03, 2016, 10:13:12 pm
Oh wait a minute, made a silly mistake earlier as I was in a hurry. :(When you say perpendicular you mean have the wire going into our out of the page right (the current)?
The magnetic fields in the middle are pointing in the same direction. Along the west and east of conductor m and n, the magnetic fields cancel out partially, however it will never be zero. It does however approach to zero as you get further and further away, so I think that is what the question meant.
On a side note, you can actually make A the correct answer by orientating the third wire perpendicular m and n, making it parallel to the field produced.
But in the question they assume it's just going flat across the page and hence the magnetic fields are added at that point?
Just trying to clarify my understanding here - will read this AOS during the upcoming week :)
thanks for help
EDIT: Added in question marks - coz grammar right?
Post by: wyzard on July 03, 2016, 10:41:07 pm
Post by: Swagadaktal on July 03, 2016, 10:47:40 pm
What I meant was having the wire lying flat on the paper, as shown in the attached image ;D It's pretty difficult to describe 3D objects in space with words.OOH yeah that's what I meant too soz, forgot that the wires were going in and out but yeah that's what I meant.
How would you diagramatically demonstrate the magnetic field? It's going in a circle around it right? so would you just draw a circle around the circle and put an arrow going anticlockwise for the one going outta page and a circle with an arrow going clockwise with the one going into the page?
Wanna get a perfect understanding of the fundamentals coz if I dont it's gonna bite me in the ass later on :P
Post by: wyzard on July 04, 2016, 11:04:23 am
OOH yeah that's what I meant too soz, forgot that the wires were going in and out but yeah that's what I meant.
How would you diagramatically demonstrate the magnetic field? It's going in a circle around it right? so would you just draw a circle around the circle and put an arrow going anticlockwise for the one going outta page and a circle with an arrow going clockwise with the one going into the page?
Wanna get a perfect understanding of the fundamentals coz if I dont it's gonna bite me in the ass later on :P
Sure thing ;D I've attached the sketch of the magnetic field line. They're not really circles as the fields produced by each wires are interfering with each other. Also notice how the middle field line is exactly straight? If you place a wire exactly on the field line, it will not experience any magnetic force.
A more detailed explanation can be found here: http://www.schoolphysics.co.uk/age16-19/Electricity%2520and%2520magnetism/Electromagnetism/text/Forces_between_currents/index.html
Post by: Swagadaktal on July 04, 2016, 11:50:29 am
Sure thing ;D I've attached the sketch of the magnetic field line. They're not really circles as the fields produced by each wires are interfering with each other. Also notice how the middle field line is exactly straight? If you place a wire exactly on the field line, it will not experience any magnetic force.thanks :)
A more detailed explanation can be found here: http://www.schoolphysics.co.uk/age16-19/Electricity%2520and%2520magnetism/Electromagnetism/text/Forces_between_currents/index.html
Post by: HopefulLawStudent on July 06, 2016, 09:37:02 am
Along the west and east of conductor m and n, the magnetic fields cancel out partially, however it will never be zero.
How do they cancel out partially?
Post by: lzxnl on July 08, 2016, 09:55:14 pm
What wyzard means is that if you have m on the left, n on the right and the currents are in opposite direction, the net magnetic field to the far left of m will have a contribution from current m and a contribution from current n in the opposite direction, so they'll somewhat cancel.
Post by: HopefulLawStudent on July 11, 2016, 11:34:12 am
Post by: zsteve on July 11, 2016, 12:34:53 pm
Someone please help. How do I find the direction of the induced current?
Hey HLS:
Your wire is moving out of the page in a magnetic field. B is pointing to the left within the vicinity of the wire.
Using the RH palm rule and considering the positive charges in the wire:
- Point fingers in direction of B (i.e. to the left)
- Point thumb in direction of movement of +ve charges (i.e. out of the page)
- Palm gives direction in which +ve charges are 'pushed' (i.e. X -> Y)
Note that using the RH palm rule for negative charges (e.g. electrons) gives the opposite result. (as expected)
Hence current (flow of +ve charge) is from X to Y.
Post by: HopefulLawStudent on July 11, 2016, 09:18:19 pm
Post by: HopefulLawStudent on July 12, 2016, 08:36:51 pm
Post by: Swagadaktal on July 12, 2016, 08:53:30 pm
HALP PLS. Just for the flux-time graph question.Yo just gonna write from my understanding here - some physics legend is probably gonna comment underneath me and outdo me 1000 times but I still want the satisfaction hehe :P
No idea how to sketch but wouldnt the flux go in a U shape -
So
\----/ (it's negative on the axis coz the magnetic field lines are going into the page so gonna be negative right? Can someone confirm here... soz about my shitty explanation lmao)
The induced voltage follows the equation
induced voltage (E) = -dflux/dt
How do the physics people put those fancy equations in damn man im so underwhelming compared to them
So the induced voltage would look like
+ Horizontal line
0 gradient
-ve horizontal line
If the flux is positive (I.e i got it wrong b4 hehe) then the signs are reversed...
tbh i think this helps me more than it helps you soz lmao :P <3
can someone teach me how to input those formulas for future purposes?
Post by: Adequace on July 12, 2016, 09:04:43 pm
Correct me if I'm wrong, I don't want to look like a bum and not a physics legend either :)
Post by: Swagadaktal on July 12, 2016, 09:13:46 pm
Yeah, it's pretty much just a trapezium looking shape as Swag said. It doesn't have to be negative though because the direction that's being taken isn't specified.Yeah I forgot what determines the sign - its just decided by the person right? What does the axis look like hls? Is it just a single quadrant? I think my U should be positive.. .but then again no reason why it cant be negative if i wanna define it to be negative right?
Correct me if I'm wrong, I don't want to look like a bum and not a physics legend either :)
Post by: Adequace on July 12, 2016, 09:27:46 pm
I accidently misread your comment. It isn't a U shape, it's literally a trapezium. It'll first be increasing at a constant rate, then it'll be constant (entire loop is in the magnetic field), then it'll decrease at a constant rate with the same magnitude as it did when it was increasing.
It shouldn't matter if the shape is positive or negative, it just shouldn't be in both from my understanding.
Post by: HopefulLawStudent on July 12, 2016, 09:40:47 pm
So I went digging around on VCAA (this is a past exam question; probably shoulda checked there first but didn't even think to; it turns out the answer given in the teacher-compiled handout I got was completely and utterly wrong). It turned out to be a 2015 question and omg. This question was done incredibly poorly (11% got full marks for part a and 79% got 0 marks).
For anyone who's interested answer is attached.
And while I was typing this message (was originally gonna ask y'all to explain it cuz VCAA didn't offer an explanation in the exam report beyond "y'all screwed up. do better" and then something clicked and I got it so everything's a-okay again. Thanks guys! :D
Post by: Swagadaktal on July 12, 2016, 09:48:07 pm
Yeah it's just decided by the person.Yeah I meant trapezium. Lmao my explanations. Yeah defs not a U shape soz if I misled you there.
I accidently misread your comment. It isn't a U shape, it's literally a trapezium. It'll first be increasing at a constant rate, then it'll be constant (entire loop is in the magnetic field), then it'll decrease at a constant rate with the same magnitude as it did when it was increasing.
It shouldn't matter if the shape is positive or negative, it just shouldn't be in both from my understanding.
Edit: I refuse to acknowledge the existence of trapeziums after methods exam 1 q10d.
But yeah trapezium - soz about that it would've been really confusing thinking it was a U shape. In my mind I think of it as a linear U shape (not as a trapezium for some reason lmao)
Post by: Swagadaktal on July 12, 2016, 09:52:09 pm
Soooo... y'all confused me. (sorry... be rest assured tho, it's definitely me, not you. My brain just can't comprehend physics in general.)Pwoah 11% only? Like by the end of the year you'd expect more wouldn't you?
So I went digging around on VCAA (this is a past exam question; probably shoulda checked there first but didn't even think to; it turns out the answer given in the teacher-compiled handout I got was completely and utterly wrong). It turned out to be a 2015 question and omg. This question was done incredibly poorly (11% got full marks for part a and 79% got 0 marks).
For anyone who's interested answer is attached.
And while I was typing this message (was originally gonna ask y'all to explain it cuz VCAA didn't offer an explanation in the exam report beyond "y'all screwed up. do better" and then something clicked and I got it so everything's a-okay again. Thanks guys! :D
Is this as hard as the questions get in physics?
Glad to see everything is A-OK :D. Cant believe that VCAA physics explanations are aligned with the explanations in the crappy textbooks. Give us a break physics :'(.
EDIT: Damn teacher giving you different answers? That would give me trust issues tbh..
Btw gonna do this to save some time for the mod
Mod note: Merged double post. You can edit previous posts.
yw fam ;)
Post by: HopefulLawStudent on July 12, 2016, 10:06:56 pm
Btw gonna do this to save some time for the mod
Mod note: Merged double post. You can edit previous posts.
#SwagForMod2k17 :P
Post by: HopefulLawStudent on July 12, 2016, 10:11:27 pm
Post by: zsteve on July 13, 2016, 08:57:35 am
Question: What's the formula for maximum EMF? (Is it that 2pi*fNBA*sin(2pi*ft) thing or is it 2pi*fNBA?) + Is that in the study design?
EMF is given by Faraday's Law of Electromagnetic Induction:
As
Hence
The other formula with sine is the emf at any time during the coil's rotation. Letting sin(...)=1 gives the max emf.
Post by: HopefulLawStudent on July 13, 2016, 11:17:01 pm
EMF is given by Faraday's Law of Electromagnetic Induction:
Aswhere
is the angular velocity (basically how many radians/sec your coil spins at)
Hence.
The other formula with sine is the emf at any time during the coil's rotation. Letting sin(...)=1 gives the max emf.
Gotcha. Thank you! Just wondering: Is the max emf thing in the study design? My teacher seems a bit vague about whether or not it is... :/
Post by: Adequace on July 13, 2016, 11:52:00 pm
Post by: Swagadaktal on July 14, 2016, 12:33:57 am
Just wondering have you guys been taught your detailed study before starting unit 4?Yeah
Post by: HopefulLawStudent on July 14, 2016, 08:54:01 am
Just wondering have you guys been taught your detailed study before starting unit 4?
Nope.
Post by: sweetiepi on July 14, 2016, 10:10:40 am
Just wondering have you guys been taught your detailed study before starting unit 4?
Yeah. My class actually did it before moving onto electricity. :)
Post by: lzxnl on July 15, 2016, 01:11:32 am
Gotcha. Thank you! Just wondering: Is the max emf thing in the study design? My teacher seems a bit vague about whether or not it is... :/
You'd be pretty stiff to get a question on that. I never saw it on SACs or on the exam.
Then again, I never understood why VCAA decided to make it possible to take physics without methods because it really makes no sense teaching it without maths. It's a bit like teaching someone addition without teaching them to count. Sure, you can remember 5+5=10, but good luck knowing what 5 means.
Just wondering have you guys been taught your detailed study before starting unit 4?
Depends on detailed study. Further electrons/structures and materials are related to unit 3 content, so they might be taught before unit 4. Sound is related to unit 4 so it'd be taught during/at the end of unit 4.
Post by: solution on July 19, 2016, 07:43:53 pm
Post by: NerdyPi on July 25, 2016, 08:00:19 pm
Explain why an alternator results in an AC output voltage? ...
My response was :
Faraday’s law states that the induced emf is equal to the rate of change of flux. As a coil initially rotates, it is on one particular side (take as positive), so as the maximum initial flux decreases to zero, (negative rate of change), a positive (according to Lenz’s law) voltage in induced.
After this, the coil is on the opposite side, and rotates from 0 flux to a maximum flux (in opposite direction, so is negative); this is a hence still decreasing flux, so a positive emf is induced.
Then, as the coil rotates from –max flux to 0, the rate of change of flux is increasing, hence inducing a negative voltage.
Therefore, as coil rotates from a positive to negative flux (and continues this cycle), flux both increases and decreases, so both a positive and negative flux is induced.
But I have been told this is wrong as flux is a scalar, and that it has something to do with the current? Can someone please clarify this for me? Thanks heaps :)
Post by: lzxnl on July 25, 2016, 11:34:41 pm
Mathematically, flux is a scalar because it arises from a dot product of two vectors: the magnetic field vector and the surface normal.
Post by: zsteve on July 26, 2016, 07:16:44 am
Flux IS a scalar, but it's a scalar that takes positive and negative values. If defined properly, it has to be defined with a consistent choice of surface direction. What I mean is, you need to measure flux in a way such that if the surface flips 180 degrees, the flux changes sign, obviously. So in this way, you do have to specify a direction for the flux, but this direction is only a convention for how your flux was defined.
Mathematically, flux is a scalar because it arises from a dot product of two vectors: the magnetic field vector and the surface normal.
I thought I would add - 'flux' through a surface is pretty much the 'flow' of the B-field through your given surface. In this sense, you can have flux in two directions - out of surface, and into surface (you define what directions 'out' and 'in' represent).
Post by: HopefulLawStudent on July 27, 2016, 02:06:12 pm
Quote
A coin will rest on a long-playing record rotating at 45 rpm provided that it is not more than 10cm from the centre of the record. How far away from the centre may it be placed if it is to remain on the record when rotated at 33.3 rpm?
The answer is apparently
Spoiler
18cm.
Post by: zsteve on July 27, 2016, 08:45:52 pm
How do I approach this question?
The answer is apparentlyHow do they get to this answer???Spoiler18cm.
Let the coin be of mass m, the record rotating at 45rpm, so one revolution is made in 1/45 min = 60/45 = 1.33s.
Hence:
is the centripetal force (supplied by maximal friction).
Now, if it's rotated at 33.3 rpm (i.e. one rotation each 1.802s), we've still got the same friction, so
Hence r = 0.18m
Post by: wyzard on July 27, 2016, 11:41:49 pm
How do I approach this question?
The answer is apparentlyHow do they get to this answer???Spoiler18cm.
Zsteve answered this really well, but I'll present an alternative method which doesn't involve the calculation of the force, which is awkward because you don't know the coin's mass.
The centripetal force acting against the friction on the coin to prevent it from spinning away from the record is:
Since mass remains unchanged, centripetal force only depends on the distances squared over the rotational period square. So for the centripetal force to stay the same, we arrive at the following equation:
Which can be used to to find the second radius, which is the distance from the records since we know its RPM.
Post by: YellowTongue on July 31, 2016, 07:07:58 am
I just can't figure out how I would go about answering this question. Thank you for your assistance :)
Post by: HopefulLawStudent on July 31, 2016, 01:41:33 pm
Please help.
Just the one question (even though there are three attachments + the fourth attachment which is the supplied answer).
I know that work per unit mass is the area under the graph but I don't quite understand where they're getting their numbers from.
PS: Realised I never thanked y'all for the help you gave me for the last question so thank you (better late than never, right?)
Post by: wyzard on August 01, 2016, 11:32:22 pm
Hey guys:
Please help.
Just the one question (even though there are three attachments + the fourth attachment which is the supplied answer).
I know that work per unit mass is the area under the graph but I don't quite understand where they're getting their numbers from.
PS: Realised I never thanked y'all for the help you gave me for the last question so thank you (better late than never, right?)
The reason why the area underneath the gravitational field strength graph with respect to distance is the work done per unit mass is the mass times gravitational field strength is the gravitational force, and recall that area underneath the force-distance graph is the work done to counteract the gravitational force to get further away from Earth ;D
Think the gravitational field strength as force per unit mass, hence the you're finding work done per unit mass when you're evaluating the area.
As to where the graph comes from, it's nothing more than the formula for gravitational field strength:
Post by: HopefulLawStudent on August 18, 2016, 09:14:27 am
Two quick questions:
1) Why is interference not seen when the beams of a car’s headlights overlap?
2) To demonstrate the interference of light waves, Young allowed light from an illuminated single slit to fall on a pair of narrow closely-spaced slits. Why does light from two separate lamp bulbs never give an interference pattern?
I asked my teacher and his answer was something along the lines of "it's because x, y, z, oh wait... maybe it's because of l, m, n. Oh I don't know. I'll get back to you on those." And when I brought it back up in the following class, he blinked at me and was basically just like "I'm busy haven't thought about it yet, I'll get back to you."
Post by: jamonwindeyer on August 18, 2016, 10:08:58 am
Haaaaai
Two quick questions:
1) Why is interference not seen when the beams of a car’s headlights overlap?
2) To demonstrate the interference of light waves, Young allowed light from an illuminated single slit to fall on a pair of narrow closely-spaced slits. Why does light from two separate lamp bulbs never give an interference pattern?
I asked my teacher and his answer was something along the lines of "it's because x, y, z, oh wait... maybe it's because of l, m, n. Oh I don't know. I'll get back to you on those." And when I brought it back up in the following class, he blinked at me and was basically just like "I'm busy haven't thought about it yet, I'll get back to you."
I know the first one would be because headlights are a wide spectrum light source! They aren't coherent and they don't have a single frequency, instead spread over a wide spectrum shifted towards the infra-red end of the spectrum. Light bulbs are like mini, not quite proper, black bodies, that emit a spectrum of radiation rather than a specific frequency. No interference occurs for separate sources like those ;D
Quick Definition: A coherent light source is a source of light that produces waves that have a close to constant phase difference between them. Another way to think about it is that the light waves more in sync with one another, even if they aren't in phase.
This also answers the second question. Essentially, two separate lamps will not be able to produce light waves that have a consistent phase difference, there is too much variation in frequency and too much fluctuation from the source. That means no noticeable interference. I mean, it could occur, but it's never consistent enough to actually be noticeable ;D
Post by: MB_ on August 20, 2016, 12:44:30 pm
Post by: HopefulLawStudent on August 21, 2016, 01:15:19 pm
Could someone please tell me how realistic this question is as a potential VCAA exam question?
Quote
Would a proton microscope be able to achieve a higher degree of resolution than an electron microscope with the same accelerating potential difference of 10kV?
Post by: zsteve on August 21, 2016, 07:15:39 pm
Hi.
Could someone please tell me how realistic this question is as a potential VCAA exam question?
Yeah, that sounds like a legit VCAA question, sounds somewhat familiar to me actually (might have encountered that in some ancient pre-1990s VCAA questions). Just compare the wavelengths.
Post by: HopefulLawStudent on August 22, 2016, 04:46:52 pm
Post by: MB_ on August 27, 2016, 08:39:10 pm
Post by: wyzard on August 30, 2016, 10:07:57 am
What is the purpose of the armature in a DC motor?
The simple design of the DC motor learned in VCE physics has a major flaw of having a very uneven rotation (or in more technical term uneven torque), due to the fact that torque is maximum when the coil is parallel to the field, and zero when it is perpendicular; so the rotation is very rough and shakes around a lot.
An armature is basically an improvement in the design, where the torque applied to the coil not only more smoothed out, the motor is a lot more compact as well 8)
Post by: CarrymetoUni on September 13, 2016, 05:31:00 pm
Robert flips a coin at a velocity of 3m/s and as it comes back, he fails to catch it resulting in it falling down an 8m deep well. How much time does it take for the coin to fall down the well.
Post by: jakesilove on September 13, 2016, 05:36:36 pm
I've been trying to figure this question our for a while but for some reason I keep getting some stupid quadratic.
Robert flips a coin at a velocity of 3m/s and as it comes back, he fails to catch it resulting in it falling down an 8m deep well. How much time does it take for the coin to fall down the well.
Hey! So, if we ignore air resistance, the coin should move up and down in a perfect parabola. This means that, when it reaches the point at which it was originally flipped, it will be travelling at exactly 3m/s, but in the opposite direction! You can use formulas to prove this, but there really isn't any need; it's sort of intuitive.
Now, we can use the equation
to find the time! Just sub in 8m for delta y, 9.8 for a and 3m/s (initial velocity) for u. Solve the quadratic, and you're done!
Jake
Post by: lzxnl on September 14, 2016, 11:21:28 am
I've been trying to figure this question our for a while but for some reason I keep getting some stupid quadratic.
Robert flips a coin at a velocity of 3m/s and as it comes back, he fails to catch it resulting in it falling down an 8m deep well. How much time does it take for the coin to fall down the well.
You're going to get a quadratic, like it or not. Try not to fear them.
Post by: bedigursimran on September 14, 2016, 09:34:43 pm
Thanks :)
Post by: sweetiepi on September 14, 2016, 10:02:48 pm
Hey guys! In Young's double slit experiment, what does decreasing the slit width do to pattern(as in the spacing) when compared to decreasing slit separation?
Thanks :)
Hey! Decreasing slit width will cause the fringe spacing/pattern to come closer, whereas increasing the slit width will bring the fringe spacing/pattern out (or more spaced out). :)
Post by: bedigursimran on September 14, 2016, 10:04:50 pm
Hey! Decreasing slit width will cause the fringe spacing/pattern to come closer, whereas increasing the slit width will bring the fringe spacing/pattern out (or more spaced out). :)
Consise and helpful! Thanks ;D
Post by: lzxnl on September 15, 2016, 11:02:20 pm
Hey! Decreasing slit width will cause the fringe spacing/pattern to come closer, whereas increasing the slit width will bring the fringe spacing/pattern out (or more spaced out). :)
You have it the wrong way around. If your slit width decreases, then for a given spacing, your path difference decreases. To maintain the same path, and thus phase, difference as well between the two slits, the pattern needs to widen.
Post by: sweetiepi on September 15, 2016, 11:04:54 pm
)
Post by: HopefulLawStudent on September 24, 2016, 01:40:47 pm
Post by: HopefulLawStudent on September 24, 2016, 01:49:40 pm
Post by: FallingStar on September 24, 2016, 04:31:10 pm
Also, could someone please help me with this question (follow on from the last question)? I get that we use F=nBIL but the answers have 0.004 as their value for B... where do they get that value from??
Sorry HLS, I can only answer this part of your questions.
1. Find Area
2. Using the flux formula, find B
Mods: I'm still kinda new to Latex, so can you please correct anything I haven't done properly.
Post by: Adequace on September 24, 2016, 09:03:07 pm
Post by: HopefulLawStudent on September 25, 2016, 12:44:28 pm
Sorry HLS, I can only answer this part of your questions.
1. Find Area
2. Using the flux formula, find B
Mods: I'm still kinda new to Latex, so can you please correct anything I haven't done properly.
Gah. I'm so silly. Thank you.
Post by: HopefulLawStudent on September 30, 2016, 01:02:05 pm
Why is the answer clockwise? The current is going from A to B, right? So using our RH slap rule, with my thumb pointing into the page and my fingers pointing toward south, I get down so shouldn’t the answer be anticlockwise?
We’re later told that the turbine’s coil is connected to an output circuit via slip rings and asked whether it is the set up of an alternator or a generator. Why is is it a generator? If slip rings are used, isn’t it an alternator?
Post by: MB_ on October 03, 2016, 06:32:09 pm
Post by: FallingStar on October 03, 2016, 08:02:12 pm
Would someone be able to summarise the key ideas and differences between DC and AC generators?
AC generator:
-Uses a slip ring (no gaps).
-Current reverses every half turn, as shown in the graph.
-Sinusoidal wave (trig function)
-Graph of current produced:
(http://www.smartlearner.mobi/science/VideoPastPapers/Electrodynamics/Images/Electrodynamics_papers_58.jpg)
-Diagram:
(http://www.everythingmaths.co.za/science/grade-12/11-electrodynamics/images/3a3c2ea57b313384441dfcd64f812817.png)
DC generator:
-Uses split rings commmutators (there is a gap between the rings)
-Current does not reverse: Modulus/absolute function
-Sinusoidal wave, but in absolute values.
-Graph of current produced:
(http://www.smartlearner.mobi/science/VideoPastPapers/Electrodynamics/Images/Electrodynamics_papers_32.jpg)
-Diagram:
(http://www.everythingmaths.co.za/science/grade-12/11-electrodynamics/images/03920d7581e52a9bb724e75fdcfe5447.png)
And just remember:
Slip ring = AC
Split ring commutator = DC
Edit: Added commutator to the "split rings" since my teachers said that VCAA is going to be pedantic about that
Post by: MB_ on October 03, 2016, 09:33:15 pm
AC generator:
-Uses a slip ring (no gaps).
-Current reverses every half turn, as shown in the graph.
-Sinusoidal wave (trig function)
-Graph of current produced:
(http://www.smartlearner.mobi/science/VideoPastPapers/Electrodynamics/Images/Electrodynamics_papers_58.jpg)
-Diagram:
(http://www.everythingmaths.co.za/science/grade-12/11-electrodynamics/images/3a3c2ea57b313384441dfcd64f812817.png)
DC generator:
-Uses split rings (there is a gap between the rings)
-Current does not reverse: Modulus/absolute function
-Sinusoidal wave, but in absolute values.
-Graph of current produced:
(http://www.smartlearner.mobi/science/VideoPastPapers/Electrodynamics/Images/Electrodynamics_papers_32.jpg)
-Diagram:
(http://www.everythingmaths.co.za/science/grade-12/11-electrodynamics/images/03920d7581e52a9bb724e75fdcfe5447.png)
And just remember:
Slip ring = AC
Split ring = DC
Would you also be able to explain the purpose of the slip rings in the AC generator and the purpose of the split rings in the DC generator. I've read quite a few different things but I'm not entirely sure of their main purpose. Thanks for the help
Post by: sweetiepi on October 03, 2016, 09:46:14 pm
Would you also be able to explain the purpose of the slip rings in the AC generator and the purpose of the split rings in the DC generator. I've read quite a few different things but I'm not entirely sure of their main purpose. Thanks for the help
Hey MB_! :)
A split ring commutator changes the current's direction every half-rotation so that the torque is in the same direction throughout the rotation.
This makes the current/time graph look like this
On the otherhand, slip rings allow the current to alternate, so the torque is changing direction every half turn.
This makes the current time graph look like this
Edit: I couldn't get the pics to work- will try later! :)
Post by: wyzard on October 25, 2016, 10:04:28 pm
Would you also be able to explain the purpose of the slip rings in the AC generator and the purpose of the split rings in the DC generator. I've read quite a few different things but I'm not entirely sure of their main purpose. Thanks for the help
A slip ring ensures the sides of the coil always remain in contact with same side the wire. Say the left side of the coil, which we will call A for now. After rotating 180 degrees, A will be on the right side, but the slip ring ensures A remains in contact with the same side of the wire.
A split ring on the other hand reverses this and A is now connected to the other terminal of the wire.
Due to this, a slip ring generates AC, while split ring generates DC.
This will take some thinking and visualisation to get it, but once you get it, it's actually a really smart design ;D
Post by: HopefulLawStudent on October 30, 2016, 02:44:57 pm
Post by: Swagadaktal on October 30, 2016, 02:58:10 pm
Would you be able to please explain why the light source for the photoelectric experiment needs to be coherent? Checked several textbooks and they just say that it has to be coherent, no explanation why. :(Yoyo hls,
It's gotta be coherent coz you want only 1 frequency wave length hitting the plates, coz if there were other wavelengths with diff energies they could cause some photoelectrons to be emitted - makes no sense to have 8 different frequency lights to test the photoelectric effect coz how are you gonna gauge the work function etc
Is that what you're asking?
Post by: wyzard on October 30, 2016, 03:51:17 pm
Would you be able to please explain why the light source for the photoelectric experiment needs to be coherent? Checked several textbooks and they just say that it has to be coherent, no explanation why. :(
Pretty sure monochromatic's the right word. I think in the context in photoelectric effect it means light of a single frequency/wavelength striking the surface of the metal. Reason for this is that if there were multiple frequencies, the energy of the ejected electrons will be more chaotic and difficult to study.
The term coherent light source is more applicable to the double slit experiment, where the light diffracted from the two slits are coherent. In this context, it means that the electromagnetic wave leaving the slits are in phase.
You can think of light waves oscillating side by side in the slits, if one goes up, the other goes up as well. The crest and the trough will leave the slits exactly at the same time. The coherence of light diffracted from the double slit is necessary to produce the double slit interference pattern observed.
Post by: Swagadaktal on October 30, 2016, 03:56:18 pm
Pretty sure monochromatic's the right word. I think in the context in photoelectric effect it means light of a single frequency/wavelength striking the surface of the metal. Reason for this is that if there were multiple frequencies, the energy of the ejected electrons will be more chaotic and difficult to study.A google search defined coherent light source as being a light source of a singular wavelength and not multiple wave lengths in the context of physics- is this incorrect?
The term coherent light source is more applicable to the double slit experiment, where the light diffracted from the two slits are coherent. In this context, it means that the electromagnetic wave leaving the slits are in phase.
You can think of light waves oscillating side by side in the slits, if one goes up, the other goes up as well. The crest and the trough will leave the slits exactly at the same time. The coherence of light diffracted from the double slit is necessary to produce the double slit interference pattern observed.
Post by: wyzard on October 30, 2016, 04:01:02 pm
A google search defined coherent light source as being a light source of a singular wavelength and not multiple wave lengths in the context of physics- is this incorrect?
Yeap you're right, having a single wavelength is one of the criteria for the light source to be coherent.The other is that they are in phase.
Here's the full definition I've found:
"Two sources of light are said to be coherent if the waves emitted from them have the same frequency and are 'phase-linked'; that is, they have a zero or constant phase difference."
From: http://schools.matter.org.uk/Content/Interference/coherent.html
They also provide a nice gif to illustrate this 8)
Post by: Swagadaktal on October 30, 2016, 04:06:31 pm
Yeap you're right, having a single wavelength is one of the criteria for the light source to be coherent.The other is that they are in phase.+1 for gif
Here's the full definition I've found:
"Two sources of light are said to be coherent if the waves emitted from them have the same frequency and are 'phase-linked'; that is, they have a zero or constant phase difference."
From: http://schools.matter.org.uk/Content/Interference/coherent.html
They also provide a nice gif to illustrate this 8)
Post by: FutureDoctor2k16 on October 30, 2016, 10:22:51 pm
Would my explanation have been sufficient?
Gradient = Planck's constant = a constant
Therefore, the magnesium and selenium graphs should be parallel
Therefore not C
The two lines cannot be collinear either otherwise magnesium and selenium graphs would have the same work function.
Therefore not D
Magnesium's Work function < Selenium's Work function.
Therefore, the answer is A.
Post by: JellyBeanz on October 30, 2016, 10:28:19 pm
2010 Exam 2, Question 6
Would my explanation have been sufficient?
Gradient = Planck's constant = a constant
Therefore, the magnesium and selenium graphs should be parallel
Therefore not C
The two lines cannot be collinear either otherwise magnesium and selenium graphs would have the same work function.
Therefore not D
Magnesium's Work function < Selenium's Work function.
Therefore, the answer is A.
Yeah that seems solid enough. Although might want to clarify why < work function results in the graph in this case. (value of y intercept smaller than Selenium's and same gradient)
Post by: Gogo14 on November 03, 2016, 02:51:40 pm
a Determine the time taken by the platinum sphere to strike the ground.
b Calculate the time taken by the lead sphere to strike the ground.
c Determine the average velocity of each sphere over their respective distances.
how do you do b and c?
I got b=4.42s and c= 31.67ms for lead
Answer for b is 4.08s and for c for lead is 29.9ms.
Post by: trystan.paderno on November 04, 2016, 09:23:45 pm
also how do i know if my calculator is truncating answers and will this cause problems in the exam if it is? and how do i fix it if so? im using the casio fx-82AU plus
Post by: Swagadaktal on November 04, 2016, 09:55:10 pm
In Young's double slit experiment, why does the intensity of the bright fringes decrease as the path difference increases?Can someone confirm whether this is within the scope of the SD?
But basically intensity = power/ area, both waves are travelling a greater distance so their intensity decreases
Post by: skobaby on November 05, 2016, 10:41:33 am
LINK: https://www.youtube.com/playlist?list=PLs46E_EDggXrJFhmwe_fMDXgCu_bY99kd
Post by: HopefulLawStudent on November 05, 2016, 11:07:47 am
Can someone confirm whether this is within the scope of the SD?
But basically intensity = power/ area, both waves are travelling a greater distance so their intensity decreases
I'd say that it is; largely cos I think there was a VCAA exam that went into why the central band had the largest intensity (and that isn't a big jump from what ^ was asking...)??? Well... I think it was VCAA??
Post by: Swagadaktal on November 05, 2016, 12:01:02 pm
I'd say that it is; largely cos I think there was a VCAA exam that went into why the central band had the largest intensity (and that isn't a big jump from what ^ was asking...)??? Well... I think it was VCAA??Yeah I think I know which q you were talking about but that was pre 2013 SD - I don't think they can ask about that in the new SD?
And any questions about intensity post 2013 simply states "bright band" or "dark band" and nothing more...
Nothing about that in SD soo
Post by: trystan.paderno on November 05, 2016, 12:17:02 pm
how many decimal places should i correct to on the physics exam? they dont specify
also how do i know if my calculator is truncating answers and will this cause problems in the exam if it is? and how do i fix it if so? im using the casio fx-82AU plus
anyone ? just need to put my mind at ease, for example i get a radius or 109325730.6 and the vcaa report says 1.1x10^8 whitch i understand is to 2 significant figures but would i be penalised for my answer? and do i need to correct to the 2 sig figs
Post by: HopefulLawStudent on November 05, 2016, 01:15:36 pm
anyone ? just need to put my mind at ease, for example i get a radius or 109325730.6 and the vcaa report says 1.1x10^8 whitch i understand is to 2 significant figures but would i be penalised for my answer? and do i need to correct to the 2 sig figs
Significant figures doesn't tend to be that big of a deal in Physics... unlike chem where sig figs is life. Maybe don't write "109325730.6" as an answer, purely because it's a waste of time; generally speaking 2-3 should be fine imo. But don't stress too much about it. :)
Post by: trystan.paderno on November 05, 2016, 01:59:04 pm
Significant figures doesn't tend to be that big of a deal in Physics... unlike chem where sig figs is life. Maybe don't write "109325730.6" as an answer, purely because it's a waste of time; generally speaking 2-3 should be fine imo. But don't stress too much about it. :)
so 1.09x10^8 would be fine
thanks for the reply
Post by: HopefulLawStudent on November 06, 2016, 07:51:21 am
Quote
Compare the types of energy transfers occurring in the LED with the energy transfers in the resistor in this circuit. - 2013, VCAA
Am I allowed to say that in the LED, electrical energy was converted to light energy and heat energy. Or is that not true??
Post by: Calebark on November 06, 2016, 04:25:25 pm
Quick Q:
Am I allowed to say that in the LED, electrical energy was converted to light energy and heat energy. Or is that not true??
This is true. I think you could get away with that. Alternatively, you could call it an ideal LED, and say it converts to light energy.
Post by: kateedelaney on November 07, 2016, 12:36:27 pm
– qualitative effect of wavelength, distance of screen and slit separation on interference patterns
• Interpret electron diffraction patterns as evidence for the wave-like nature of matter
• explain the production of atomic absorption and emission spectra, including those from metal vapour lamps
• analyse the absorption of photons by atoms, not including their bombardment by electrons, in terms of:
– the change in energy levels of the atom due to electrons changing state
– the frequency and wavelength of emitted photons, E = hf = hc/λ
• describe the quantised states of the atom in terms of electrons forming standing waves, recognising this as evidence of the dual nature of matter
Post by: plants on November 07, 2016, 02:52:05 pm
Post by: FallingStar on November 07, 2016, 03:09:55 pm
Equation:
- Increasing the wavelength increases fringe spacing.
- Decreasing the separation between the two slits increases fringe spacing.
- Increasing the distance to the screen from the source increases fringe spacing.
Interpret electron diffraction patterns as evidence for the wave-like nature of matter
If electrons were a particle, we would expect the pattern to look like this, if they are passed through a double slit:
(https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQWceuVPM1zs115VIh9XPdG2jSGEXuwyNA6KTlWzTCzYNZVfcUUXw)
But instead, we get this:
(https://logicaluniverseblog.files.wordpress.com/2015/11/double-slit-electrons.jpg)
Isn't that similar to what happened in Young's double slit experiment? Well, waves causes interference patterns, which explains the multiple light and dark screens of electrons.
Although I don't have any good ways to explain the latter of your question (can someone else please explain).
Post by: HiggsBShouldB140GeV on November 07, 2016, 04:48:02 pm
Hey dudes!! In the 2015 exam, question 13, it asks you to identify the direction of current as the loop enters the magnetic field. I'm getting Y to X, but their answers say X to Y. I don't understand why - even with their explanation in the assessment report.
Hey dude! I'm not sure if you've picked up on it yet, but the question states for the direction of the current THROUGH the voltmeter. So we from the perspective of the current flowing through the voltmeter first we can see that it flows from X to Y.
Soz if its already solved prior to this response. :)
Post by: plants on November 07, 2016, 05:23:53 pm
Hey dude! I'm not sure if you've picked up on it yet, but the question states for the direction of the current THROUGH the voltmeter. So we from the perspective of the current flowing through the voltmeter first we can see that it flows from X to Y.
Soz if its already solved prior to this response. :)
Ohh lol. Thanks my man. What a cheeky bastard VCAA is
Post by: Willba99 on November 07, 2016, 09:54:41 pm
Thanks
Post by: plants on November 07, 2016, 10:06:56 pm
Hey, so i recently did the sample exam at http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf which is vcaa's sample exam for this study design. Problem is, i can't find the solutions for them. Does anyone know where to find them? The link to the exam is on the vcaa site at http://www.vcaa.vic.edu.au/Pages/vce/studies/physics/physicsindex.aspxSolutions can be found on itute :^):
Thanks
http://www.itute.com/2010/10/11/free-download-2013-2016-sample-physics-examination-solutions/
Post by: nkemobi13 on November 13, 2016, 07:25:21 pm
Thanks
Post by: robibl476 on November 16, 2016, 10:39:48 am
Post by: Gogo14 on November 22, 2016, 10:43:24 pm
Post by: Buddster on November 23, 2016, 02:00:19 pm
Hey guys this is urgent. How do you calculate uncertainties for sine and arcsine functions. This is for my practical investigation about calculating refractive angles. thannnnnkkkkkkkssssss!!!!!!!!
I found the following on some random forum, need verification:
If you have sin (1.5 ± 0.2) radians
The uncertainty is 0.2 Cos 1.5
(This must be in radians)
Post by: wyzard on November 24, 2016, 07:09:09 pm
Hey guys this is urgent. How do you calculate uncertainties for sine and arcsine functions. This is for my practical investigation about calculating refractive angles. thannnnnkkkkkkkssssss!!!!!!!!
A way to work out the uncertainty of sin and arcsin functions is by using the linear approximation:
Where delta x and y represent the uncertainty in the quantities x and y.
For example, suppose you measured x to be 1.0 rad with uncertainty of 0.1 rad, the uncertainty of sin x is then cos(1.0) * 0.1.
Post by: dylang99 on November 26, 2016, 11:50:33 am
I am having a little bit of trouble understanding how to work out this question. Any help would be much appreciated! :D
A child is holding a garden hose at ground level and the water stream from the hose is travelling at 15 ms-1. which angle to the horizontal will result in the water stream travelling the greatest horizontal distance through the air?
Post by: Buddster on November 26, 2016, 12:02:05 pm
Hi All,long way:
I am having a little bit of trouble understanding how to work out this question. Any help would be much appreciated! :D
A child is holding a garden hose at ground level and the water stream from the hose is travelling at 15 ms-1. which angle to the horizontal will result in the water stream travelling the greatest horizontal distance through the air?
Use these two formulae for the x and y axis motion:
x=v*t=15t
y=u*t-a/2*t2=15t-5t^2
rearrange y for t in terms of y. Derive x in terms of y. Use d/dx=0 to find the maximum value for x, and for what y value that is. angle=arctan(y/x)
short way:
range formula: u2*sin(2*angle)/g
the max value of the function is where sin(2*a)=1, look at a sine graph ;)
Post by: Nicko912 on November 26, 2016, 01:02:07 pm
Thanks
(http://puu.sh/suD3b/39551af676.jpg)
Post by: Buddster on November 26, 2016, 02:07:03 pm
Does anyone know how to approach this question?
They're in equilibrium so the weight force and the electromagnetic force is equal.
The electromagnetic force is 3.55×10^4 multiplied by the charge in Columbus as the unit is N/C
MG=-3.55×10^4*c
Once you find c, divide by the charge of an electron to find the amount of electrons
Post by: Nicko912 on November 26, 2016, 03:01:53 pm
They're in equilibrium so the weight force and the electromagnetic force is equal.
The electromagnetic force is 3.55×10^4 multiplied by the charge in Columbus as the unit is N/C
MG=-3.55×10^4*c
Once you find c, divide by the charge of an electron to find the amount of electrons
Thanks for the response, but how exactly do I find 'c'?
Post by: Buddster on November 26, 2016, 03:05:29 pm
Thanks for the response, but how exactly do I find 'c'?
You have the values for the mass an gravity (1.161×10^-14, 9.8) and rearrange
Post by: peanut on December 09, 2016, 12:25:20 pm
When working through a question, do I round to appropriate sig figs at each step, or just for the final answer? In most cases, it shouldn't matter, but what is the correct way of doing it? Similarly, do I use a rounded value for the next step in my calculation, or just use the full (as many decimal places as possible) value on my calculator?
Thanks!
Post by: Buddster on December 09, 2016, 12:29:45 pm
I asked this on the Chemistry thread, but I wanted to see if the same applies to Physics.
When working through a question, do I round to appropriate sig figs at each step, or just for the final answer? In most cases, it shouldn't matter, but what is the correct way of doing it? Similarly, do I use a rounded value for the next step in my calculation, or just use the full (as many decimal places as possible) value on my calculator?
Thanks!
Sig figs don't matter in physics! Best to keep within realms of the correct number though for readability and accuracy. For example: if the correct number of figures is 5, don't round to 1.
Post by: Jakeybaby on December 09, 2016, 09:02:45 pm
Sig figs don't matter in physics! Best to keep within realms of the correct number though for readability and accuracy. For example: if the correct number of figures is 5, don't round to 1.Are you sure?
They certainly do here in SA, I'd assume that they would be assessed throughout Australia as they are extremely relevant to the accuracy of an answer.
I'd suggest the exact same method as what I suggested in the Chemistry thread.
Post by: Syndicate on December 09, 2016, 09:07:45 pm
Are you sure?
They certainly do here in SA, I'd assume that they would be assessed throughout Australia as they are extremely relevant to the accuracy of an answer.
I'd suggest the exact same method as what I suggested in the Chemistry thread.
Significant figures are not assessed VCE physics, which is a shame I guess.
To OP: like Buddster asserted, only round your answer to a sensible amount of numbers.
Post by: Swagadaktal on December 09, 2016, 09:09:12 pm
Are you sure?Yeah for physics it isnt like chemistry, they honestly dont care about sig figs. The sig figs are mostly incorrect in the examiners reports :) (recently they've been better but pre 2012 there's like 0 care for sig figs)
They certainly do here in SA, I'd assume that they would be assessed throughout Australia as they are extremely relevant to the accuracy of an answer.
I'd suggest the exact same method as what I suggested in the Chemistry thread.
as long as u dont round too much, (i.e answer is 2.6 and you write 3)
Post by: Jakeybaby on December 09, 2016, 09:20:05 pm
Significant figures are not assessed VCE physics, which is a shame I guess.
To OP: like Buddster asserted, only round your answer to a sensible amount of numbers.
Yeah for physics it isnt like chemistry, they honestly dont care about sig figs. The sig figs are mostly incorrect in the examiners reports :) (recently they've been better but pre 2012 there's like 0 care for sig figs)Fair enough, wasn't expecting that haha
as long as u dont round too much, (i.e answer is 2.6 and you write 3)
Post by: Gogo14 on January 03, 2017, 03:58:25 pm
2. In the photo, why is the reaction force acting on the cyclist at an angle. I thought the reaction force was always perpendicular?
Post by: Syndicate on January 03, 2017, 04:11:23 pm
1. When an object is spinning, why does the object continue to move along the circumference if it is accelerating towards the centre. I.e why does centripedal acceleration not cause it to reach the centre?Because its accelerating toward the centre, shouldnt it eventuall reach the centre?
2. In the photo, why is the reaction force acting on the cyclist at an angle. I thought the reaction force was always perpendicular?
1. Centrifugal force (it is an apparent force) works opposite to the centripetal force, which draws the rotating body out the centre.
2. I think it is a mistake because normal force is always perpendicular to the ground (the book would only be right, if they are trying to show the bike on a banked track).
Post by: homosapien on January 05, 2017, 12:07:14 am
Thanks :)
Post by: Syndicate on January 05, 2017, 10:20:01 am
Could someone please explain the trend in size of the weight force (Fg) and normal force (FN) as a ball bounces / goes up and down?
Thanks :)
As the:
- ball is bouncing down, only gravity exists (since it is free falling).
- ball touches the ground: Fg = Fn
- ball compresses: Fn > Fg
- ball exits it's compression stage, however, is still touching the ground: Fg = Fn
- ball going back up: Fn = 0, and only Fg exists
Post by: Xandorious on January 11, 2017, 11:53:25 am
Post by: wyzard on January 11, 2017, 12:20:10 pm
A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?
That is a pretty intense question 8) I'll give you some hints and pointers.
The gravitational force by the Earth and Moon is at opposite directions, and the closer you get to one planet the stronger the gravitational pull. So there must be a point somewhere in between that will have the forces cancelling each other out.
To find the distance, let x be the distance from the center of Earth. Use this x, set up the equation of forces from the Earth and Moon using Newton's law of gravity, then solve for x.
Post by: Rathin on January 11, 2017, 01:27:52 pm
A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?
Fnet (from moon) = GM(moon)m(satellite)/r(moon)^2
Fnet (from earth = GM(earth)m(satellite)/r(earth)^2
Where r(moon) is the distance from moon centre to the satellite
and r(earth) is the distance from the earth centre to the satellite
Hence at the point where net force on satellite = 0
We will equate the two formulas;
m(moon)/r(moon)^2 = m(earth)/r(earth)^2
and since r(moon) = 3.84 x 10^8 - r(earth)
m(moon)/ (3.84 x 10^8 - r(earth))^2 = m(earth)/r(earth)^2
sub in relevant numbers
Answer = 3.46x10^8m
Post by: Xandorious on January 12, 2017, 03:16:46 pm
Hi, i'm glad that you got the answer that I got but the answer in my book states 3.02 * 10^8 m
Post by: Gogo14 on January 19, 2017, 01:00:13 pm
http://imgur.com/6lYLwa2
Post by: Shadowxo on January 19, 2017, 11:24:00 pm
@RathinUnless we both made the same mistake, I got that answer too so it may be a book error
Hi, i'm glad that you got the answer that I got but the answer in my book states 3.02 * 10^8 m
How do you do quesiton 9, graph relates to question
http://imgur.com/6lYLwa2
Hi, I'm a bit rusty on this, so someone correct me if I'm wrong.
I'm assuming at the start of the question it means 6000km, not 600km as that's what's shown on the graph.
J = Nm
Change in the energy is the change in kinetic energy, final kinetic energy is zero, so the initial kinetic energy is the change in kinetic energy.
Change in kinetic energy is the area under the graph, so just the number of squares * value of squares (in this case 2*(1/2)*10^6 = 10^6)
16 squares * 10^6
1.6*10^7 Joules
Or you could multiply each of the values and calculate the difference aka change in energy
Initial energy = 11*6.0*10^6 = 6.6*10^7
Final energy = 6.3*8.0*10^6 = 5.0*10^7
Difference = 1.6*10^7 Joules
Hope this helps
Post by: Gogo14 on January 23, 2017, 03:19:18 pm
Unless we both made the same mistake, I got that answer too so it may be a book errorThe answer says 1.7*10^9J
Hi, I'm a bit rusty on this, so someone correct me if I'm wrong.
I'm assuming at the start of the question it means 6000km, not 600km as that's what's shown on the graph.
J = Nm
Change in the energy is the change in kinetic energy, final kinetic energy is zero, so the initial kinetic energy is the change in kinetic energy.
Change in kinetic energy is the area under the graph, so just the number of squares * value of squares (in this case 2*(1/2)*10^6 = 10^6)
16 squares * 10^6
1.6*10^7 Joules
Or you could multiply each of the values and calculate the difference aka change in energy
Initial energy = 11*6.0*10^6 = 6.6*10^7
Final energy = 6.3*8.0*10^6 = 5.0*10^7
Difference = 1.6*10^7 Joules
Hope this helps
Post by: Shadowxo on January 23, 2017, 04:13:45 pm
The answer says 1.7*10^9J
Ah I see, haven't done many questions on this and don't think I've seen any on an exam (so very rusty, sorry)
First of all I misread the question partially :P as it's in orbit 600km from earth's surface, and the radius of earth is approx 6.4*10^3km, the satellite is approx 7*10^3km or 7*10^6m from the centre of the earth.
So the energy per kg of the satellite is 7 squares *10^6 (value of each of the squares) = 7*10^6J
Total change in energy for the satellite is 7*10^6 J * 240kg = 1.7*10^9J
:)
Post by: Gogo14 on January 23, 2017, 04:44:14 pm
Ah I see, haven't done many questions on this and don't think I've seen any on an exam (so very rusty, sorry)But why do you multiply 7*10^6 by 240 kg? Isnt the graph a force vs distance graph not a gravitational field strength vs distance graph
First of all I misread the question partially :P as it's in orbit 600km from earth's surface, and the radius of earth is approx 6.4*10^3km, the satellite is approx 7*10^3km or 7*10^6m from the centre of the earth.
So the energy per kg of the satellite is 7 squares *10^6 (value of each of the squares) = 7*10^6J
Total change in energy for the satellite is 7*10^6 J * 240kg = 1.7*10^9J
:)
Post by: Shadowxo on January 23, 2017, 05:00:27 pm
But why do you multiply 7*10^6 by 240 kg? Isnt the graph a force vs distance graph not a gravitational field strength vs distance graph
It says on the graph that it's the force acting on 1kg, aka the gravitational field strength, which is why you have to multiply the weight.
Post by: Gogo14 on January 23, 2017, 05:18:31 pm
It says on the graph that it's the force acting on 1kg, aka the gravitational field strength, which is why you have to multiply the weight.*repeatedly slams head on desk*
Post by: Shadowxo on January 23, 2017, 05:29:30 pm
*repeatedly slams head on desk*
Haha dw, it's the little things that get you (and me)
Post by: Gogo14 on January 27, 2017, 12:07:56 pm
For question 9, how do you do it?
Post by: Gogo14 on January 27, 2017, 01:40:55 pm
How do you question 8, really lost
Thanks heaps!
http://imgur.com/uZccyZh
Post by: Shadowxo on January 29, 2017, 01:55:39 am
For question 1, the answer is A, but isnt the travelator doing work?
For question 9, how do you do it?
1. For the travelator, there is no acceleration as everything is going at the same speed, therefore F = 0 therefore work = 0
9. You can use Work = Fd
F = ma, a = 9.8, F = .8*9.8 = 7.84
W = Fd = 7.84*1.9 = 14.90
Alternative Way:
Net work done = change in kinetic energy
You should resolve the speed into vertical and horizontal components.
v = 30 ms-1
Horizontal speed = 30cosx = 21.21 ms-1
Vertical speed = 30sinx = 21.21 ms-1
Horizontal speed is unchanging (no force acting on it)
s = -1.9 u = 21.21 a = -9.8 v = ?
v2 = u2 + 2as
v2 = 450 + 37.24
v = 22.07 ms-1 down
net v = 30.61
∆ke = final ke - initial ke
= 374.90 - 360 = 14.90
Post by: Gogo14 on February 02, 2017, 05:43:12 pm
1. For the travelator, there is no acceleration as everything is going at the same speed, therefore F = 0 therefore work = 0Thanks, for question 9, the answer says its 374J ?
9. You can use Work = Fd
F = ma, a = 9.8, F = .8*9.8 = 7.84
W = Fd = 7.84*1.9 = 14.90
Alternative Way:
Net work done = change in kinetic energy
You should resolve the speed into vertical and horizontal components.
v = 30 ms-1
Horizontal speed = 30cosx = 21.21 ms-1
Vertical speed = 30sinx = 21.21 ms-1
Horizontal speed is unchanging (no force acting on it)
s = -1.9 u = 21.21 a = -9.8 v = ?
v2 = u2 + 2as
v2 = 450 + 37.24
v = 22.07 ms-1 down
net v = 30.61
∆ke = final ke - initial ke
= 374.90 - 360 = 14.90
Post by: lzxnl on February 02, 2017, 11:37:26 pm
Thanks, for question 9, the answer says its 374J ?
I get the same answer as Shadowxo.
Post by: Shadowxo on February 03, 2017, 01:31:01 pm
Thanks, for question 9, the answer says its 374J ?
I think they were treating the d in W=Fd as distance not displacement (although I believe displacement, not distance, should be used)
Post by: lzxnl on February 03, 2017, 02:20:15 pm
What is work? Work is the ability of an object to exert a force over a displacement parallel to the force. Any object that is capable of doing work then has energy (this is my preferred definition of energy). Why is this directional requirement important? Defining it this way allows energy to be mathematically conserved, which makes it a physically useful definition.
So, if work is exerting a force in the direction of motion, forces that are constantly perpendicular to an object's motion should not do work. And indeed, uniform circular motion arises when the net force is ALWAYS perpendicular to the object's motion (and satisfies a magnitude requirement); the speed and thus the energy do not change in such a motion.
Now, because 2D motion can be separated into its two independent dimensions, we can consider the net force, the gravitational force, as acting on the horizontal and vertical components separately. The gravitational force is perpendicular to the horizontal velocity component, so no work is done there. It only affects the vertical velocity component. Therefore, as far as gravity is concerned, the object is only moving vertically. (it is possible to find a frame of reference such that the object is indeed only moving vertically; this frame of reference is best pictured by someone running straight under the object), and so the work it does on the object should only involve the vertical displacement.
Final question. Do we use displacements or distances? Well, suppose you throw a ball up. Gravity is clearly slowing it down when the motion is opposing the direction of gravity, and speeds it up when the motion is in the direction of gravity. Therefore, the direction of motion is important and you use displacements. It is possible to prove (I won't though) that gravity is a conservative force, which in this problem means that you only need to know the initial and final state to calculate the work; the path doesn't matter. In other words, you only need to know that the ball had a vertical displacement of 1.9 m. Any calculation involving the distance displays a fundamental misunderstanding of physics. Indeed, calculating the actual distance through the air travelled by the javelin along its parabolic trajectory isn't trivial; I invite any spesh students to do so as a preview of the coming calculus topics.
Post by: -273.15 on February 05, 2017, 11:52:50 am
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept
Thank you!
Post by: Syndicate on February 05, 2017, 11:57:44 am
Helloo,
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept
Thank you!
This pdf is quite helpful.
Post by: lzxnl on February 05, 2017, 11:59:30 am
Helloo,
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept
Thank you!
Hi there,
Suppose I want to lift a box. Suppose I want to lift it up fast. Then, the vertically upwards net force needs to be large. This means that the force I'm applying to the box has to be a lot larger than its weight.
Suppose I want to lower a box. Suppose I want it to fall fast. Then, the vertically downwards net force needs to be large. This means that the force I'm applying to the box has to be a lot smaller than its weight.
Now replace lifting/lowering the box with the lift applying a normal reaction force onto the box, which plays the same role. Does that help? The normal force is functionally equivalent to the force applied by a person onto said box.
Post by: -273.15 on February 05, 2017, 12:53:27 pm
Hi there,
Suppose I want to lift a box. Suppose I want to lift it up fast. Then, the vertically upwards net force needs to be large. This means that the force I'm applying to the box has to be a lot larger than its weight.
Suppose I want to lower a box. Suppose I want it to fall fast. Then, the vertically downwards net force needs to be large. This means that the force I'm applying to the box has to be a lot smaller than its weight.
Now replace lifting/lowering the box with the lift applying a normal reaction force onto the box, which plays the same role. Does that help? The normal force is functionally equivalent to the force applied by a person onto said box.
Thank you so much !
Post by: Guideme on February 09, 2017, 08:05:39 pm
2. Explain in terms of the kinetic particle model why you can put your hand safely in a 300C oven for a few seconds, while if you touch a metal tray in the same oven your hand will be burned
3. How does evapouration of water cause a reduction i the temperature of the surrounding air?
I am findinng Physics very difficult and my teacher just reads from the textbook, so it would be nice if anyone can provide an answer and explain it in detail so i can understand it thank you!
Post by: lzxnl on February 09, 2017, 09:14:09 pm
1. Explain the importance of keeping a lid on a simmering saucepan of water in terms of latent heat of vapoiurisation?
2. Explain in terms of the kinetic particle model why you can put your hand safely in a 300C oven for a few seconds, while if you touch a metal tray in the same oven your hand will be burned
3. How does evapouration of water cause a reduction i the temperature of the surrounding air?
I am findinng Physics very difficult and my teacher just reads from the textbook, so it would be nice if anyone can provide an answer and explain it in detail so i can understand it thank you!
1. Assuming you want to boil the water, if you keep the water exposed to the atmosphere, then the water's rate of heating will drop a lot due to heat exchange with the cooler atmosphere (heat flows from hot to cold, remember). The issue is, you need to input a certain amount of energy into the water (latent heat of vapourisation) for the water to evaporate, so your task is made doubly hard if you don't keep a lid on.
2. The kinetic particle model suggests that temperature is a measure of the average kinetic energy of particles. Now, for your hand to get hurt in that oven, the gas particles in the oven have to collide with your hand and impart enough energy to heat your hand. The problem is, gas particles aren't very concentrated, so the energy transfer won't be very fast. In contrast, particles are a lot more concentrated in a metal, so there will be a LOT more particle collisions if you touch a hot metal tray to heat your hand and the heat transfer will be much faster.
3. Evaporation of water requires absorbing energy from the surroundings. To conserve energy, this energy must come from the atmosphere, which will consequently decrease its temperature as its energy drops.
Post by: Ramones on February 27, 2017, 11:58:44 am
Post by: Syndicate on March 01, 2017, 07:57:48 pm
Just a question about special relativity: Why does time appear to be "slower" for the object in motion relative to the observer, which is technically (not exactly) stationary. I know that a longer distance is being travelled, by lets say a light beam, due to it's horizontal speed (which means it takes a longer amount of time). Shouldn't the time be passing faster, in order to reach it's destination in the same time interval as the clock (where one tick = the light beam travelling back and forth) held by the observer which we are assuming stationary in this case. I don't really know why the time get slower as we approach c. Can someone explain this in terms of words? I know how it works using the formula, but don't really understand the concept that well.
Thanks :)
Post by: -273.15 on March 04, 2017, 04:34:22 pm
Im stuck with questions where there are objects in a system
e.g. trailer and truck and have to find acceleration, tension, with and without friction
could someone please give me some tips / rules for working out these questions / any helpful info???
thanks heaps
Post by: Shadowxo on March 04, 2017, 04:49:44 pm
Hey guys,
Im stuck with questions where there are objects in a system
e.g. trailer and truck and have to find acceleration, tension, with and without friction
could someone please give me some tips / rules for working out these questions / any helpful info???
thanks heaps
Often the easiest way is to consider it as a whole system. eg if the truck is pulling the trailer with a force of 2000N and their combined weight is 1000kg, F=ma => a = 2ms-2 ignoring friction. If not ignoring friction, you have to use the net force. If the friction on each of the truck and trailer is 250N, F net= (2000-250-250)=ma
Then if you want to find the tension of the connection between the truck and trailer, use F=ma again, where a= 2ms-2 (from earlier) and the mass of the trailer only, eg 250kg. F=ma = 500N. If not ignoring friction, F net = 500 = F (from truck) - friction => F from truck = 750N
Often drawing diagrams can help a lot too.
Often questions are similar to that. If you get stuck on a specific question I can answer in a bit more detail :)
Hope this helps :)
Post by: -273.15 on March 04, 2017, 05:12:17 pm
Often the easiest way is to consider it as a whole system. eg if the truck is pulling the trailer with a force of 2000N and their combined weight is 1000kg, F=ma => a = 2ms-2 ignoring friction. If not ignoring friction, you have to use the net force. If the friction on each of the truck and trailer is 250N, F net= (2000-250-250)=ma
Then if you want to find the tension of the connection between the truck and trailer, use F=ma again, where a= 2ms-2 (from earlier) and the mass of the trailer only, eg 250kg. F=ma = 500N. If not ignoring friction, F net = 500 = F (from truck) - friction => F from truck = 750N
Often drawing diagrams can help a lot too.
Often questions are similar to that. If you get stuck on a specific question I can answer in a bit more detail :)
Hope this helps :)
Thank you so much this helped a lot :)
Post by: dylang99 on March 06, 2017, 09:16:23 pm
Any help would be much appreciated:
A high-energy physicist detects a particle in a particle accelerator that has a half-life of 20 s when travelling at 0.99c.
(a) Calculate the particle’s half-life in its rest frame.
Thank you in advance :)
Post by: Syndicate on March 06, 2017, 09:36:49 pm
Hi All,
Any help would be much appreciated:
A high-energy physicist detects a particle in a particle accelerator that has a half-life of 20 s when travelling at 0.99c.
(a) Calculate the particle’s half-life in its rest frame.
Thank you in advance :)
L= 20
L0 = ?
v= 0.99c
L= L0Y
20 = L0 x 7.089
L0 = 2.82 seconds
Post by: Mattjbr2 on March 08, 2017, 05:18:05 pm
Is this merely a rounding error in my calculations? The worked solutions are clearly wrong, so I can't rely on them.
I don't know what I'm missing in this question.
Note that the question says 20 degrees to the vertical, which means 70 degrees to the horizontal. So I used 70 in my calculations.
The book says 78.9kg.
Edit: The book used 10ms^-2 for g. So my method is correct. Never mind! ;D
Post by: sweetiepi on March 08, 2017, 05:26:03 pm
Hi guys,I got 80.92kg as my answer, without rounding, however, I tried again with g=10 and got 78.9.
Is this merely a rounding error in my calculations? The worked solutions are clearly wrong, so I can't rely on them.
I don't know what I'm missing in this question.
Note that the question says 20 degrees to the vertical, which means 70 degrees to the horizontal. So I used 70 in my calculations.
The book says 78.9kg.
Therefore you are correct, as the new study design dictates that g=9.8, not g=10 :)
Hope this helps! :)
Post by: Mattjbr2 on March 08, 2017, 05:27:52 pm
I got 80.92kg as my answer, without rounding, however, I tried again with g=10 and got 78.9.
Therefore you are correct, as the new study design dictates that g=9.8, not g=10 :)
Hope this helps! :)
Cheers :)
Post by: Gogo14 on March 19, 2017, 06:13:05 pm
Thanks
Post by: vcestressed on March 19, 2017, 07:49:00 pm
Post by: Syndicate on March 19, 2017, 07:57:44 pm
Is it only 'electricity' and 'motion' that is assumed knowledge from unit 1 & 2 for physics 3 & 4 for the new study design? Would appreciate if someone could clear this up for me. Thanks :D
Yep. Basically everything from 1D motion, which is then used to extend upon 2D physics (projectile motion, circular motion, special relativity (not sure if I should have included that here)). For electricity, I found that half of the content covered in units 1/2 is unnecessary (we go over different stuff, however, the formulas are still used).
Post by: vcestressed on March 19, 2017, 08:04:54 pm
Yep. Basically everything from 1D motion, which is then used to extend upon 2D physics (projectile motion, circular motion, special relativity (not sure if I should have included that here)). For electricity, I found that half of the content covered in units 1/2 is unnecessary (we go over different stuff, however, the formulas are still used).
Thank you!
Post by: Alexicology on March 19, 2017, 10:15:03 pm
How do you do q8,9
Thanks
Hi, I'm a current Year 12 student,
For question 8, use Coulomb's Law to find the force of repulsion. So F=kq1q2/r^2
Sub in k= 9x10^9, q1=1.602x10^-19, q2=1.602x10^-19 and r=2.5x10^-15
Therefore, force of repulsion= 37N
For Q9, use F=k1k2/r2 again to find q1
q1= 6.24x10^7C
Then find number of electrons transfered by q1/qe
Therefore number of electrons= 3.9x10^26 electrons
Post by: Mattjbr2 on March 20, 2017, 08:03:06 pm
I don't understand why the book says the answers for 3 and 4 aren't D and B, respectively.
Edit: The book claims C and C
Post by: Syndicate on March 20, 2017, 08:19:19 pm
Can someone please confirm for me the correct answers to these questions, as well as explain how they arrived at said answers?
I don't understand why the book says the answers for 3 and 4 aren't D and B, respectively.
Edit: The book claims C and C
You are correct.
Post by: Mattjbr2 on March 20, 2017, 08:44:38 pm
You are correct.
You sure?
Post by: Syndicate on March 20, 2017, 08:57:27 pm
You sure?
Yep.
Since Ffricition = 0N, the horizontal component of the normal force must be responsible for the net force. The horizontal component of the normal force is directly in front of the angle (which is vertically opposite to the angle beneath).
4) N = mg/cos15
mg = Ncos15
Post by: dylang99 on March 25, 2017, 05:40:00 pm
Any help would be greatly appreciated!
Post by: Syndicate on March 26, 2017, 02:09:26 pm
The Sun orbits the centre of our galaxy, the Milky Way, at a distance of 2.2 × 1020 m from the centre with a period of 2.5 × 108 years. e mass of all the stars inside the Sun’s orbit can be considered as being concentrated at the centre of the galaxy. e mass of the Sun is 2.0 × 1030 kg. If all the stars have the same mass as the Sun, how many stars are in the Milky Way?
Any help would be greatly appreciated!
Number of stars = total mass of stars/ mass of sun (as each star has an equal weight).
I am going to assume that you mean the number of stars in a sphere (where the sphere = the total area between the sun's orbit and the milky way).
Msun = \(2.0 \times 10^{30} \) kg
T = \( 7.884 \times 10^{15} \) seconds (assuming each year has 365 days)
r = \( 2.2 \times 10^{20} \) metres
M = total mass of stars
So firstly we need to calculate the mass of the stars in the area by equating Fg to Fc because all of the masses of the stars are concentrated at the centre of the galaxy.
Post by: Gogo14 on April 06, 2017, 10:59:41 am
Post by: cosecant on April 08, 2017, 08:59:24 pm
How do u do this q?
since the electrons emerge with no deflection and are subjected to both the magnetic field and electric field, the magnetic force must be balanced by the electric force, thus qE=qvB. Hence, E=vB
since E=V/d (we are trying to find the distance - d)
-> V/d=vB. d is the unknown, we have V and B so we need to find v first.
since work is being done on the electrons by the electric field
qV= (mv2)/2
q =1.6*10-19, m=9.1*10-31, V = 3*103
solving, v equals to 3.248*107 m/s
sub v= 3.248*107, V=3*103 and B= 1.6*10-3 into the equatio previously derived V/d=vB
rearranging, d=V/vB
d= (3*103)/(3.248*107*1.6*10-3)
d= 0.0577m
=5.8cm
Post by: TooLazy on April 12, 2017, 06:27:45 pm
and why it is the answer
Post by: Shadowxo on April 16, 2017, 06:36:18 pm
Could someone please explain how to get the answer in Q16,
and why it is the answer
You only posted 14, do you want to know 14 or 16? :)
Post by: TooLazy on April 17, 2017, 10:47:04 am
You only posted 14, do you want to know 14 or 16? :)
Sorry, I meant 14 :)
Post by: wyzard on April 17, 2017, 12:02:59 pm
Could someone please explain how to get the answer in Q14,
and why it is the answer
Simply equate the gravitational forces acting on X by each planet using Newton's law of Gravity, then you'll find the term G and R cancels out, leaving you only M and m. Rearranging the M and m, you can find M/m to be 16.
Post by: nicholas9027 on April 17, 2017, 04:19:33 pm
I need help on this Wave Question
1.Calculate the longest and shortest time for a radio signal travelling at the speed of light to go from the Earth to a space probe when the space probe is a) near Mars and (b) near Neptune.
Radius of Earth's orbit about the Sun= 1.49 x 10^11m
Radius of Mar's orbit about the Sun=2.28 x 10^11m
Radius of Neptune's orbit about the Sun= 4.50 x 10^12m
I've found the radius between Earth and Mars is 0.79 x 10^11m i think.
Thanks in advance :)
Post by: Shadowxo on April 19, 2017, 12:04:05 pm
Hey guys
I need help on this Wave Question
1.Calculate the longest and shortest time for a radio signal travelling at the speed of light to go from the Earth to a space probe when the space probe is a) near Mars and (b) near Neptune.
Radius of Earth's orbit about the Sun= 1.49 x 10^11m
Radius of Mar's orbit about the Sun=2.28 x 10^11m
Radius of Neptune's orbit about the Sun= 4.50 x 10^12m
I've found the radius between Earth and Mars is 0.79 x 10^11m i think.
Thanks in advance :)
I'm presuming this is the way they want you to do this.
a) So the shortest distance between Mars and Earth is when there is a straight line connecting the sun, earth, and mars, ie earth is partway between sun and mars, so they're on the same side of the sun and they're closest. So, the distance would be 2.28 x 1011 - 1.49 x 1011 = 7.9 x 1010 which is what you found out. Just use the speed of sound to figure out the time it would take to travel that distance.
The longest distance between Mars and Earth is when they're on opposite sides of the sun, directly opposite each other. So the distance between them would be 2.28 x 1011 + 1.49 x 1011 = 3.77 x 1011 (distance between Mars and Sun + distance between Earth and Sun). Then, do the same thing to determine the time it would take to travel that distance.
Repeat for b)
Post by: vcestressed on April 23, 2017, 07:39:52 pm
Post by: Syndicate on April 23, 2017, 07:49:16 pm
quick question. . . why is that good conductors of heat do NOT necessarily have high specific heat capacity?
They are both different things. Good conductors refers to how much heat can the substance absorb, whereas specific heat capacity refers to the amount of joules of energy required to raise the substance's temperature by 1C. ie. Copper is a good conductor of heat, but it's specific heat capacity is only 0.385 J C^-1 g^-1
Post by: Rusten on May 01, 2017, 07:31:35 pm
If the Moon were to be put into a new orbit of twice its current radius, what would its potential energy (increase, decrease, no change).
My teacher is saying increase, because E=mgh and double height means double energy... However, given that gravitational fields fall under the inverse square law, would it not be the case that doubling the radius would mean 1/4 of the original 'g' value (grav field strength). Hence, subbing it back into E=mgh, you'd end up with HALF of what the original energy was, not double. Am I missing something here? Thanks in advance.
Post by: RuiAce on May 01, 2017, 07:39:31 pm
We got a questions asking:Firstly, the inverse square law is related to Newton's Law of Universal Gravitation, i.e. \(F=\frac{Gm_1m_2}{d^2} \). F is an attractive "force". Not the work done.
If the Moon were to be put into a new orbit of twice its current radius, what would its potential energy (increase, decrease, no change).
My teacher is saying increase, because E=mgh and double height means double energy... However, given that gravitational fields fall under the inverse square law, would it not be the case that doubling the radius would mean 1/4 of the original 'g' value (grav field strength). Hence, subbing it back into E=mgh, you'd end up with HALF of what the original energy was, not double. Am I missing something here? Thanks in advance.
Post by: -273.15 on May 07, 2017, 04:59:03 pm
Could someone please show me how the formula is obtained??
Thanks :)
Post by: Shadowxo on May 07, 2017, 09:24:15 pm
I don't understand how the formula Ek=GMm/2R is derived from Ek=1/2mv^2, GMm/r^2 and mv^2/r
Could someone please show me how the formula is obtained??
Thanks :)
Post by: captkirk on May 08, 2017, 12:59:50 am
QQ. How can you improve a DC generator? (suggestions of improvement)
Post by: -273.15 on May 09, 2017, 07:22:43 pm
Also could someone explain why there is no current flow before it reverses direction?
Thank you :)
Post by: Gogo14 on May 10, 2017, 04:04:20 pm
Hi!Having stronger magnets, a longer coil, more coils, more planes of coils, more split rings, more efficient brushes(e.g. More smooth carbon brushes)
QQ. How can you improve a DC generator? (suggestions of improvement)
I was doing a vcaa question from the 2014 exam and you had to describe the direction of the current on a particular section of coil in a DC motor before during and after a quarter turn. I'm confused as i thought the split ring commutator reverses the direction of the current after each half turn, not every quarter turn?a full turn =360degrees, a quater turn =90 degrees nad hence there would not a magnetic force generated as the current is runnign parallel to the magnetic field. There is no current because at every 1/4 turn, the split ring commutator is split. This means that the wires are not in contact with teh commutation, but the split (gap). Hence there is no complete circuit and thus no current
Also could someone explain why there is no current flow before it reverses direction?
Thank you :)
Post by: captkirk on May 21, 2017, 12:12:53 am
Answer is 4.3N
How?!?!????
Post by: Logic on May 21, 2017, 09:06:34 pm
Post by: captkirk on May 21, 2017, 11:15:47 pm
Where's everyone up to in the course btw?Howdy, I'm starting special relativity tomorrow :D
Post by: mouldycarrots on May 22, 2017, 07:55:13 pm
Where's everyone up to in the course btw?We're doing transformers and transmission at the moment
Post by: princessofpersia on May 22, 2017, 08:15:47 pm
Where's everyone up to in the course btw?
we're currently doing circular motion on vertical planes
Post by: princessofpersia on May 22, 2017, 08:30:12 pm
A block on the table is accelerated by a falling weight. Calculate the tension in the cord if the block experiences a frictional force of 1.5N as it slides on the table (blockA=2.0kg, blockB=0.5kg)
Answer is 4.3N
How?!?!????
Fnet on system= (0.5 x 10) - 1.5 (friction)
=3.5N
Acceleration of 2.0kg............3.5N/(2kg+0.5kg)= 1.4m/s^2
Fnet on block= Tension + friction
mass x acceleration= tension -1.5
2.8= tension -1.5
tension = 4.3N
Post by: BlinkieBill on June 04, 2017, 12:53:14 pm
I am doing a projectile distance practical for the epi. We're shooting nerf gun bullets at different angles and heights above the ground.
I was wondering what a good research question/aim would be for it. My teacher said something about terminal velocity and comparing practical results to theoretical (using formulas) results, but I'm not too sure about the research question still.
Any help is appreciated. Thanks
Post by: Guideme on June 05, 2017, 09:38:24 pm
In general more heat is lost through the roof of a house then through the walls. Explain why this is so?
Often weather reports for skiing resorts will use the term 'wind chill factor'/ Explain the meaning of this term using appropriate thermodynamics principles and mechanisms
Post by: TooLazy on June 09, 2017, 01:33:26 pm
Post by: Alwin on June 10, 2017, 10:37:32 am
Please help me with these two questions Thank you a lot!
In general more heat is lost through the roof of a house then through the walls. Explain why this is so?
Often weather reports for skiing resorts will use the term 'wind chill factor'/ Explain the meaning of this term using appropriate thermodynamics principles and mechanisms
If you think about how heat would travel (does it rise or does it fall) I think you'll be able to see why more heat is lost at the roof.
In terms of the wind chill factor, it comes down to the idea that you have air flowing over your skin rather than being stationary around you. A loose analogy would be letting your coffee cool down by letting it rest, or stirring your coffee (making it cool faster). Hopefully you should be able to take it from here :)
I just want to confirm, when an object is decelerating the net force is in the direction opposite to its movement?
Yup! That would be correct :)
Post by: Mattjbr2 on June 12, 2017, 06:45:19 pm
Is the book's answer correct? Doesn't the flux increase when you increase the number of turns of wire? I'm getting conflicting information from multiple places: https://physics.stackexchange.com/questions/219881/is-the-number-of-turns-in-a-loop-part-of-the-magnetic-flux the dude says N (number of turns of wire) affects ϕ. Also: http://i.imgur.com/hzwGz0n.png the book even plainly states that. However, nowhere in the book is N included in the ϕ=BA equation.
What's going on!?
Post by: Mattjbr2 on June 13, 2017, 06:09:56 am
Post by: sweetiepi on June 13, 2017, 08:03:22 am
http://i.imgur.com/XtCj6OP.png
Is the book's answer correct? Doesn't the flux increase when you increase the number of turns of wire? I'm getting conflicting information from multiple places: https://physics.stackexchange.com/questions/219881/is-the-number-of-turns-in-a-loop-part-of-the-magnetic-flux the dude says N (number of turns of wire) affects ϕ. Also: http://i.imgur.com/hzwGz0n.png the book even plainly states that. However, nowhere in the book is N included in the ϕ=BA equation.
What's going on!?
Hey, sorry for my late reply,
I'd definitely go with the with equation ϕ=nBA because the number of coils does increase the flux generated.
As there is the new study design, books are churned out very rapidly and generally contain a few mistakes. :)
Post by: Syndicate on June 14, 2017, 09:23:56 pm
Hey, sorry for my late reply,The number of coils doesn't exactly effect the magnetic flux as the area remains the same (from a VCE point of a view, you might be correct from a first year perspective, but I am not really sure of that) . Magnetic flux is defined as the amount of magnetic fields going through a certain area, so the amount of coils doesn't affect it.
I'd definitely go with the with equation ϕ=nBA because the number of coils does increase the flux generated.
As there is the new study design, books are churned out very rapidly and generally contain a few mistakes. :)
(This is directly out of my book as well, so I am not really sure if I am correct either :P)
Post by: zsteve on June 15, 2017, 08:19:29 pm
- Flux is related to the total 'amount' of field passing through a given surface area (so
- The presence of multiple coils does affect the current/potential produced, but this is accounted for when dealing with Faraday's Law
Post by: Mattjbr2 on June 17, 2017, 05:43:21 pm
Post by: Syndicate on June 17, 2017, 05:53:25 pm
Therefore is my book's answer correct or incorrect?
Your book is right with the equation part, but, for question 2 (faraday's experiment question), I believe he used a coil with many turns as it increases the strength of static magnetic field around the coil, which thereby results in a greater EMF induced. They didn't have modern technology to measure the induced EMF, so if only a small EMF was induced, then they wouldn't really be able to calculate it (or know if something really happened or not).
Post by: Mattjbr2 on June 17, 2017, 06:22:55 pm
Your book is right with the equation part, but, for question 2 (faraday's experiment question), I believe he used a coil with many turns as it increases the strength of static magnetic field around the coil, which thereby results in a greater EMF induced. They didn't have modern technology to measure the induced EMF, so if only a small EMF was induced, then they wouldn't really be able to calculate it (or know if something really happened or not).What about question 3? The back of the book gives 3.0x10^-5 Wb but the worked solutions give 3.0x10^-3 Wb and neither took into consideration the 50 turns.
Post by: Syndicate on June 17, 2017, 09:03:34 pm
What about question 3? The back of the book gives 3.0x10^-5 Wb but the worked solutions give 3.0x10^-3 Wb and neither took into consideration the 50 turns.
It is most likely a misprint. The solutions with 3.0 x 10^-3 Wb is the correct one.
Post by: gamma032 on June 18, 2017, 06:28:42 pm
I've been trying to calculate the uncertainty in a couple of calculations, but I'll just present the hardest.
I'm finding the gravitational potential energy (m*g*h) of an object, and I measured:
- The measurement uncertainty in the mass to be ±0.005kg
- The uncertainty in the height to be ±0.0005m
I'm using g=9.8.
What would the final uncertainty of the GPE calculations be?
Post by: zsteve on June 18, 2017, 07:51:54 pm
Hi everyone!
I've been trying to calculate the uncertainty in a couple of calculations, but I'll just present the hardest.
I'm finding the gravitational potential energy (m*g*h) of an object, and I measured:
- The measurement uncertainty in the mass to be ±0.005kg
- The uncertainty in the height to be ±0.0005m
I'm using g=9.8.
What would the final uncertainty of the GPE calculations be?
From memory, what one must do when adding two quantities is to deal with fractional error. Consider c=a+b. Then the fractional error in c is given by:
Where a, b, c are the mean values of a, b, c; and the Delta values are the error margins.
This is also how we deal with errors for division :)
Post by: lzxnl on June 19, 2017, 07:50:23 pm
From memory, what one must do when adding two quantities is to deal with fractional error. Consider c=a+b. Then the fractional error in c is given by:
Where a, b, c are the mean values of a, b, c; and the Delta values are the error margins.
This is also how we deal with errors for division :)
Actually, you've quoted the rule for products. All of these rules originate from calculus.
If z = xy, then ln z = ln x + ln y
So differentiating gives
dz/z = dx/x + dy/y
If you associate dz as change in z, you get the rule for multiplication. Division works the same.
One way of thinking about it is, imagine I'm multiplying two numbers and one of them is 1% off while the second one is 2% off. The product is then roughly 3% off.
If z = x + y, dz = dx + dy. So you add the absolute errors; there's nothing more to it.
Post by: Gogo14 on July 01, 2017, 10:26:29 pm
Post by: KingKunta on July 09, 2017, 03:31:06 pm
just curious would you advise the use of edrolo in learning all Unit 3&4 physics content (simply based on the fact that my teacher cannot teachto save her own life, she just an overall useless person)
cheers.
ps: the guy who goes through it is guy named Andy Matthews.
Post by: Alevine on July 19, 2017, 05:07:24 pm
how can photons have momentum if it doesnt have massHi Gogo 14!
Fantastic question. It can seem bizarre that photons could even have momentum given that we're (probably) very accustomed to considering momentum in the 'classical' sense. That is, we're used to momentum being defined as the product of a given object's mass and velocity:
p = mv
However, this is not the only definition we have of momentum. We can also define the momentum as p = h / lambda
Whilst light does not have an associated mass, it certainly does however have a wavelength - and (using this formula) we can use that wavelength to calculate the photon's momentum.
In terms of WHY photons can still have momentum (despite being mass-less), a great experiment to consider is Compton Scattering. With Compton Scattering, we see in some situations that when photons of light are involved in a "collision" of one form or another they have a tendency to behave similarly to particles. In the experiment, photons of light collide with electrons in a graphite block. The interaction that follows is very similar to one billiard ball simply rolling into another. Once the photon hits the electron, the electron 'rolls' away (we get recoil electrons) and the photon loses some of its energy from the collision, subsequently increasing it's wavelength (since E = h / lambda, a decrease in the photon's energy means an increase in the photon's wavelength).
I hope some of that helps you get a sense of where all this comes from? Even though the idea is strange, there is more than one way that we can think about momentum.
Post by: Alevine on July 19, 2017, 05:20:28 pm
hey guys,Lmao hey KingKunta hahahah
just curious would you advise the use of edrolo in learning all Unit 3&4 physics content (simply based on the fact that my teacher cannot teachto save her own life, she just an overall useless person)
cheers.
ps: the guy who goes through it is guy named Andy Matthews.
Soooo Edrolo. Um, it's okay?
Personally I kinda....idk
I wasn't the biggest fan of Edrolo in year 12 but it's not a bad resource. That being said, (and again, this is just my personal opinion) I think Edrolo is much better for revision than it is for learning things for the first time.
Even though it's arduous, reading through a (good) year 12 physics textbook, as well as doing tOns of practice questions and regular revision is probably the best way to go in terms of learning and mastering things for yourself.
No self promo lol but maybe this article will help? https://atarnotes.com/vce-physics/
Hope everything works out!
Post by: Gogo14 on July 20, 2017, 11:44:09 am
Hi Gogo 14!Ok, now im a bit confused about what momentum is because I always thought that it was mass times velocity. So what really is momentum?
Fantastic question. It can seem bizarre that photons could even have momentum given that we're (probably) very accustomed to considering momentum in the 'classical' sense. That is, we're used to momentum being defined as the product of a given object's mass and velocity:
p = mv
However, this is not the only definition we have of momentum. We can also define the momentum as p = h / lambda
Whilst light does not have an associated mass, it certainly does however have a wavelength - and (using this formula) we can use that wavelength to calculate the photon's momentum.
In terms of WHY photons can still have momentum (despite being mass-less), a great experiment to consider is Compton Scattering. With Compton Scattering, we see in some situations that when photons of light are involved in a "collision" of one form or another they have a tendency to behave similarly to particles. In the experiment, photons of light collide with electrons in a graphite block. The interaction that follows is very similar to one billiard ball simply rolling into another. Once the photon hits the electron, the electron 'rolls' away (we get recoil electrons) and the photon loses some of its energy from the collision, subsequently increasing it's wavelength (since E = h / lambda, a decrease in the photon's energy means an increase in the photon's wavelength).
I hope some of that helps you get a sense of where all this comes from? Even though the idea is strange, there is more than one way that we can think about momentum.
Post by: Syndicate on July 20, 2017, 12:39:12 pm
Ok, now im a bit confused about what momentum is because I always thought that it was mass times velocity. So what really is momentum?
I don't know how correct I am, but I think because photons have energy, it can have momentum due to the mass-energy equivalence theory. So if the photons have x amount of energy, than it will have Y amount of mass (which is basically X divided by c, the speed of light, squared). Since they are already moving at the constant speed (speed of light), they will have momentum.
Post by: Gogo14 on July 26, 2017, 03:28:40 pm
1. I dont really undersstand the concept of natural frequency, i know that its the oscillation that an object will naturally take, but why? I just don't understand the concept it kinda seems wierd to me.
2. Question 3 attached, Why is u moving up?
3. Q19. Dont know how to do
Post by: gamma032 on July 27, 2017, 03:59:04 pm
I can understand why a constant DC supply would not induce an emf (no change in flux through the secondary coil) but wouldn't a variable DC supply, such as that generated by a generator with a slit ring, still be able to induce an emf?
Post by: Bri MT on July 30, 2017, 06:53:48 pm
Ok, now im a bit confused about what momentum is because I always thought that it was mass times velocity. So what really is momentum?
Not sure if this is a proper definition, but I think of momentum in relation to inertia so my definition is " A measure of a body's resistance to change in it's motion"
Thanks also I have a few qs about waves:2.
1. I dont really undersstand the concept of natural frequency, i know that its the oscillation that an object will naturally take, but why? I just don't understand the concept it kinda seems wierd to me.
2. Question 3 attached, Why is u moving up?
3. Q19. Dont know how to do
Q ans. The gradient is positive
understanding: The graph shows a snapshot of the wave at that current moment. From the graph we can see that the wave is transverse therefore the only possible answers are up, down, or not moving. As the particle is not at a peak, it hasn't reached the end of it range in either direction, so we know it isn't still. Based on the location of the particles surrounding it, we can deduce that it is moving up.
3. Given that the location of the antinode is the x-co-ordinate of the location of the peak, the distance between two consecutive antinodes is the wavelength. If you're still stuck let me know.
Hey everyone. In regards to transformers, my textbook and worked solutions for questions indicate that they only work with an AC input current.
I can understand why a constant DC supply would not induce an emf (no change in flux through the secondary coil) but wouldn't a variable DC supply, such as that generated by a generator with a slit ring, still be able to induce an emf?
Your textbook has generalised. You are right a DC power supply can induce emf through a transformer, when it is changing. This is something to be careful of in questions as when it is connected or disconnected there will be emf generated, and listing this can get you marks. In your situation with a slip ring, I'm not sure if that would still be referred to as DC???? but yes there would be emf generated when the current changes direction every half turn.
Post by: A TART on July 30, 2017, 10:21:32 pm
Quote
A brand-new Rolls Royce rolls off the back of a truck as it is being delivered to its owner. The truck is travelling along a straight road at a constant speed of 60 km h−1. The Rolls Royce slows down at a constant rate, coming to a stop over a distance of 240 m. It is a full minute before the truck driver realises that the precious load is missing. The driver brakes immediately, leaving a 25 m long skid mark on the road. The driver’s reaction time (time interval between noticing the problem and depressing the brake) is 0.5 s.
My Working:
(http://i.imgur.com/LdpDWty.jpg)
Solution given in the textbook:
Quote
It is important to remember that, at the instant that the Rolls Royce rolls off the truck, it is moving in the same direction as the truck.
After 1 minute, it has moved a distance of 1000 m (a constant speed of 60 km/h is equal to 1 km/min).
During the driver’s reaction time, the distance moved by the truck is 8.3 m (the distance moved in 0.5 s at a constant speed of 60/h 16.67 m/s 0.5 s).
The braking distance of the truck is 25 m.
The total distance moved by the truck is 1000 m+8.3 m+25 m+1033 m.
The distance moved by the Rolls Royce is 240 m (in the same direction as that of the truck). The Rolls Royce is, therefore, 793 m behind the stopped truck.
Wouldn't the car go in the opposite direction?
PS: What does everyone think of the Jacaranda textbook? Personally I hate it.
Post by: Shadowxo on July 31, 2017, 03:01:56 pm
(Just a little explanation but that's the basic idea)
Post by: -273.15 on August 12, 2017, 01:11:20 pm
Could someone please explain to me why the stopping voltage in the photoelectric effect is negative?
Thanks so much :)
Post by: sweetiepi on August 12, 2017, 01:19:31 pm
Hello :)I'm a tad rusty on the PE effect, however I believe that the stopping voltage is negative is due to the fact that a positively charged plate will attract the emitted electrons, a neutral plate will attract any electrons headed towards it, and an increasingly negatively charged plate will 'repel' the electrons, hence the current (flow of electrons) becomes zero. :)
Could someone please explain to me why the stopping voltage in the photoelectric effect is negative?
Thanks so much :)
Post by: -273.15 on August 13, 2017, 11:17:36 am
another question sorry
For a photoelectron to be emitted (in Photoelectron effect), the frequency of the electromagnet radiation and thus the energy of the photon must be greater than the threshold frequency and thus the work function.
However, can these be equal or does it have to be greater in order for photocurrent to be observed?
If they were equal, would it just mean that no excess kinetic energy is released??
Thank you!
Post by: Gogo14 on August 15, 2017, 05:05:32 pm
http://imgur.com/a/PFS6W
Dont really understand the answers thanks
Post by: Bri MT on August 15, 2017, 07:42:14 pm
Hey guys can you guys help me with q8-10
http://imgur.com/a/PFS6W
Dont really understand the answers thanks
You need to look at the graph and identify which letter on it best reflects the question. For example, in a compression the particles are close together, and in the graph the position where they are close together is at "c".
Post by: Shadowxo on August 15, 2017, 08:18:40 pm
Post by: chantelle.salisbury on August 17, 2017, 10:02:32 am
the time has come to do the extended investigation. i was wanting to do something regarding the young's experiment. however, i didnt know what or how to turn this into a good question. are there any suggestions?
i have thought about investigating the width of the bands produced and then having 2 independent variables to investigate. would this be sufficient for the entire prac considered it is constructed and investigated well?
thankyou :)
Post by: Syndicate on August 19, 2017, 02:26:57 pm
hiThat should be fine. I think a good question can be (I am not sure what your independent variables are): How does visible light of different wavelengths affects the width of the bands produced on the screen?
the time has come to do the extended investigation. i was wanting to do something regarding the young's experiment. however, i didnt know what or how to turn this into a good question. are there any suggestions?
i have thought about investigating the width of the bands produced and then having 2 independent variables to investigate. would this be sufficient for the entire prac considered it is constructed and investigated well?
thankyou :)
So you can investigate the affects of red and blue light for instance.
Post by: BlinkieBill on August 26, 2017, 03:56:16 pm
Why is the single slit (in the double slit experiment) required to make the light sources coherent?
Thanks
Post by: Syndicate on August 26, 2017, 04:39:12 pm
Hey,
Why is the single slit (in the double slit experiment) required to make the light sources coherent?
Thanks
Incoherent light waves will not have the same frequency (as basically the troughs and crests will be randomly distributed). When the incoherent light wave passes through a single slit, the majority of the wave focuses at the centre, hence making it coherent. How? Lets think about a bunch of marbles and a small opening. If you try to push all the marble through the opening at the same time, will they get through? Offcourse not. Whereas, if you put them in one by one, you will see that all the marbles get through (mainly aiming the centre) are the same distance apart (assuming you are releasing them at the same time). This is basically how a coherent light wave forms. A requirement to produce an interference pattern is for a wave to have the same frequency (and amplitude) when passing double slits in order to produce nodal and anti-nodal regions.
Post by: captkirk on September 13, 2017, 08:44:15 pm
So in my experiment I dropped a car down a inclined plane from 30°, 35° and 40° (Twice).The car was 800grams. I dropped the car with no weight so 800grams then I added a 1kg weight towards the middle of the car and dropped it from same angle. (Recorded the time it went down)
So my question is I want to calculate the acceleration right. I was going to use a=gsin(θ) but then I realised I need to use this formula 6 times. The problem was that for 30° for example I was going to use a=9.8sin(30) but then I looked back at my results and I saw the problem. What was I going to do with the 30° angle with 1kg extra weight on the car.? According to that calculation^ the acceleration would be the same as the car with weight but it's supposed to be different.
My partner suggested that we should use
v=u+at^2 then transpose to a=v/t^2. Then use a=f/m.
I feel like this is wrong. Can someone please help me out.
If you didn't understand my experiment I can explain again in further detail.
Post by: Shadowxo on September 13, 2017, 09:13:54 pm
Hey everyone, I need help with my physics epi.For a larger mass, the force is greater. So F/m is constant (on the same incline). It's like that experiment with the feather and the heavy ball, they both fall at 9.8 m/s2 (neglecting air resistance). So, you're right in that the acceleration just depends on the angle :) You could take into account friction but again that wouldn't be different for different weights (N=mgcos(theta) on an incline so F/m = gcos(theta)).
So in my experiment I dropped a car down a inclined plane from 30°, 35° and 40° (Twice).The car was 800grams. I dropped the car with no weight so 800grams then I added a 1kg weight towards the middle of the car and dropped it from same angle. (Recorded the time it went down)
So my question is I want to calculate the acceleration right. I was going to use a=gsin(θ) but then I realised I need to use this formula 6 times. The problem was that for 30° for example I was going to use a=9.8sin(30) but then I looked back at my results and I saw the problem. What was I going to do with the 30° angle with 1kg extra weight on the car.? According to that calculation^ the acceleration would be the same as the car with weight but it's supposed to be different.
My partner suggested that we should use
v=u+at^2 then transpose to a=v/t^2. Then use a=f/m.
I feel like this is wrong. Can someone please help me out.
If you didn't understand my experiment I can explain again in further detail.
Also, for your experiment you should compare your experimental values (using the data from the experiment) to theoretical (a=g sin(theta)). You could find reasons for why your experimental values different from theoretical - eg friction as mentioned above, or air resistance etc
Hope this helps :)
Post by: Willba99 on September 13, 2017, 09:14:39 pm
Hey everyone, I need help with my physics epi.
So in my experiment I dropped a car down a inclined plane from 30°, 35° and 40° (Twice).The car was 800grams. I dropped the car with no weight so 800grams then I added a 1kg weight towards the middle of the car and dropped it from same angle. (Recorded the time it went down)
So my question is I want to calculate the acceleration right. I was going to use a=gsin(θ) but then I realised I need to use this formula 6 times. The problem was that for 30° for example I was going to use a=9.8sin(30) but then I looked back at my results and I saw the problem. What was I going to do with the 30° angle with 1kg extra weight on the car.? According to that calculation^ the acceleration would be the same as the car with weight but it's supposed to be different.
My partner suggested that we should use
v=u+at^2 then transpose to a=v/t^2. Then use a=f/m.
I feel like this is wrong. Can someone please help me out.
If you didn't understand my experiment I can explain again in further detail.
I believe that the integral idea from your experiment should be that weight has no effect on acceleration. Have you ever seen the video where they drop a feather a bowling bowl in a vacuum, and they fall at the same rate? i would suggest that if your experiment gives you drastically different values when you add the weight then you're doing something wrong
Post by: Willba99 on September 13, 2017, 09:15:47 pm
For a larger mass, the force is greater. So F/m is constant (on the same incline). It's like that experiment with the feather and the heavy ball, they both fall at 9.8 m/s2 (neglecting air resistance). So, you're right in that the acceleration just depends on the angle :) You could take into account friction but again that wouldn't be different for different weights (N=mgcos(theta) on an incline so F/m = gcos(theta)).
Also, for your experiment you should compare your experimental values (using the data from the experiment) to theoretical (a=g sin(theta)). You could find reasons for why your experimental values different from theoretical - eg friction as mentioned above, or air resistance etc
Hope this helps :)
lol beat me to it
Post by: captkirk on September 13, 2017, 10:38:10 pm
I believe that the integral idea from your experiment should be that weight has no effect on acceleration.My research question is measuring energy released. So going to calculate the potential energy and kinetic energy on the incline.
Also my values are similar with the weight ontop of car. So I don't think I'm doing anything wrong just yet lol.
On a side note I'm going to calculate the components (perpendicular and parallel) acceleration (as said before) and the energys. Should be enough I think.
For a larger mass, the force is greater. So F/m is constant (on the same incline). It's like that experiment with the feather and the heavy ball, they both fall at 9.8 m/s2 (neglecting air resistance). So, you're right in that the acceleration just depends on the angle :) You could take into account friction but again that wouldn't be different for different weights (N=mgcos(theta) on an incline so F/m = gcos(theta)).
Also, for your experiment you should compare your experimental values (using the data from the experiment) to theoretical (a=g sin(theta)). You could find reasons for why your experimental values different from theoretical - eg friction as mentioned above, or air resistance etc
Hope this helps :)
Thanks for the detailed answer but I just got completely confused. So do I go and use a=gsin(theta) or my partners method (a=v/t^2)
So which method is correct for 30°
Method 1) a=gsin(theta) therefore a=9.8sin(30)=4.9
Therefore acceleration= 4.9ms^-2
Or
Method 2)v=s/t therefore 1.5/0.86=1.74 therefore a=v/t^2 therefore 1.74/0.86^2=2.35
Therefore acceleration= 2.35ms^-2
Which method is correct 1 or 2?
Edit those were some of my results so don't get confused by numbers. ;D
Post by: Shadowxo on September 13, 2017, 11:04:46 pm
Thanks for the reply's.You may need to look over / revisit some equations.
My research question is measuring energy released. So going to calculate the potential energy and kinetic energy on the incline.
Also my values are similar with the weight ontop of car. So I don't think I'm doing anything wrong just yet lol.
On a side note I'm going to calculate the components (perpendicular and parallel) acceleration (as said before) and the energys. Should be enough I think.
Thanks for the detailed answer but I just got completely confused. So do I go and use a=gsin(theta) or my partners method (a=v/t^2)
So which method is correct for 30°
Method 1) a=gsin(theta) therefore a=9.8sin(30)=4.9
Therefore acceleration= 4.9ms^-2
Or
Method 2)v=s/t therefore 1.5/0.86=1.74 therefore a=v/t^2 therefore 1.74/0.86^2=2.35
Therefore acceleration= 2.35ms^-2
Which method is correct 1 or 2?
Edit those were some of my results so don't get confused by numbers. ;D
v = u + at (not at2)
s=vt only works for a constant or average velocity, not when you want to calculate the final velocity
a=9.8sin(30) is the acceleration you expect - the theoretical value. It doesn't use your data, just is an expectation.
To find your experimental acceleration, think of the different physics formulas - which has all the variables you want? You know the distance travelled, initial speed, time taken and want to find acceleration. You'd use this formula for your results :)
If you want to find v instead for energy, you'd either use the acceleration from before or another physics formula
Post by: captkirk on September 13, 2017, 11:29:32 pm
You may need to look over / revisit some equations.
v = u + at (not at2)
s=vt only works for a constant or average velocity, not when you want to calculate the final velocity
a=9.8sin(30) is the acceleration you expect - the theoretical value. It doesn't use your data, just is an expectation.
To find your experimental acceleration, think of the different physics formulas - which has all the variables you want? You know the distance travelled, initial speed, time taken and want to find acceleration. You'd use this formula for your results :)
If you want to find v instead for energy, you'd either use the acceleration from before or another physics formula
Oops I just realised lol sorry I meant v=u+at.
Ok so I'm going to use one of the 5 formulas. Thanks!
Post by: waldo2000 on September 19, 2017, 12:30:22 pm
and we got the graphs that look like one period of a positive sine graph.
What can we do with this graph? use emf = -N d(flux)/dt to calculate what the change in flux was? is this the equation for average emf? and do we use the peak values for this?
is finding the area under the graph to calculate the strength of the magnets possible???
Post by: Bri MT on September 19, 2017, 07:33:46 pm
so we did a prac where we dropped magnets through solenoids the other day,
and we got the graphs that look like one period of a positive sine graph.
What can we do with this graph? use emf = -N d(flux)/dt to calculate what the change in flux was? is this the equation for average emf? and do we use the peak values for this?
is finding the area under the graph to calculate the strength of the magnets possible???
Yes, that is the equation for average emf.
It is possible to start square counting. If you know how to, you can find the area by integrating. If you don't, count squares.
Post by: simrat99 on September 20, 2017, 07:15:11 pm
Does an increase in intensity affect the Kinetic Energy of the photoelectrons in the Phototelectric effect?
Thanks :)
Post by: BlinkieBill on September 20, 2017, 07:32:50 pm
Does an increase in intensity affect the Kinetic Energy of the photoelectrons in the Phototelectric effect?An increase in intensity does not affect the kinetic energy, it increases the no. of photoelectrons (current)
however, an increase in the frequency of light does increase the kinetic energy of the photoelectrons
Post by: princessofpersia on September 30, 2017, 05:49:58 pm
did you learn about light dispersion in concave and convex lenses.
I'm asking b/c edrolo teaches it, but we never covered it in class.
Post by: Syndicate on September 30, 2017, 05:55:02 pm
Guys,
did you learn about light dispersion in concave and convex lenses.
I'm asking b/c edrolo teaches it, but we never covered it in class.
It is part of unit 2, I doubt you'll have to know it for the exams (best to confirm with your teacher though).
Post by: sweetiepi on September 30, 2017, 05:56:44 pm
Guys,Hey!
did you learn about light dispersion in concave and convex lenses.
I'm asking b/c edrolo teaches it, but we never covered it in class.
According to the VCAA study design this is what is assessable in regards to convex and concave lenses for unit 2:
"- describe image formation using pinhole cameras and convex and concave lenses
...
•distinguish between short-sightedness and long-sightedness, and explain their correction by concave and convex lenses, respectively"
I hope this helps :)
Post by: Bri MT on September 30, 2017, 08:06:13 pm
Guys,
did you learn about light dispersion in concave and convex lenses.
I'm asking b/c edrolo teaches it, but we never covered it in class.
From the study design (unit 4) "• investigate and explain theoretically and practically colour dispersion in prisms and lenses with reference to
refraction of the components of white light as they pass from one medium to another"
To be safe, I would understand what direction of bending occurs, and the impact of changing colour on the degree of bending/diffraction.
Post by: princessofpersia on October 01, 2017, 11:35:14 am
and just to be safe I think I'll learn the affect of bending and colour changing as miniturtle suggested.
Post by: TooLazy on October 01, 2017, 03:07:41 pm
Like calculating resistance across various components in series/parrallel.
I've noticed that much of what we did in unit 1/2 in that regard is absent in the unit 3/4 course. Is it safe to presume that we wont be asked any of those sorts of questions in the exam?
Post by: sweetiepi on October 01, 2017, 03:19:09 pm
Do we have to know circuits.Hey!
Like calculating resistance across various components in series/parrallel.
I've noticed that much of what we did in unit 1/2 in that regard is absent in the unit 3/4 course. Is it safe to presume that we wont be asked any of those sorts of questions in the exam?
I just skimmed through the VCAA study design, and I found no real mention of the circuitry outside of units 1 and 2. As I only skimmed through, I'd ask your teacher just to be 100% sure, however I believe it is highly unlikely to be assessed, given that I couldn't find much on it in the 3/4 design :)
Post by: kevinkyle28 on October 01, 2017, 05:58:00 pm
Post by: sweetiepi on October 01, 2017, 06:02:11 pm
For the pre written notes for the physics exam, are you allowed to bring in 1 double sided a4 (so basically 2 pages) or are you allowed to bring in 2 double sided a4 (so basically 4 pages)According to VCAA: "Pre-written notes (one folded A3 sheet or two A4 sheets bound together by tape)". I'm pretty sure this is the same specifications that we had last year, therefore definitely 2 double-sided A4 pages are allowed :)
Post by: Bri MT on October 01, 2017, 07:11:45 pm
Do we have to know circuits.
Like calculating resistance across various components in series/parrallel.
I've noticed that much of what we did in unit 1/2 in that regard is absent in the unit 3/4 course. Is it safe to presume that we wont be asked any of those sorts of questions in the exam?
You may need to add up the resistance of wires when looking at the transmission of electrical energy across power lines. We haven't had to do any calculations involving circuits in series. You do need to be able to use formulas such as ohms law
Post by: Rowan1999 on October 01, 2017, 10:48:56 pm
Quick question regarding one of the special relativity questions from the 2015 VCAA exam (Question 5) (Answer is D)
Spacecraft S66 is travelling at high speed away from Earth carrying a highly accurate atomic clock. Another spacecraft, T50, is travelling in the opposite direction to S66, as shown in Figure 1.
An observer, E, on Earth emits a short radio pulse to spacecraft S66, which reflects it directly back towards the
observer. The time elapsed for E between sending and receiving the pulse is 20.0 ms.
Which one of the following is true?
A. According to E, spacecraft S66 was more than 3000 km away when the pulse reached it.
B. According to E, the pulse took longer to reach spacecraft S66 than it did to return from spacecraft S66 to E.
C. The 20.0 ms interval measured by E is not a proper time because the radio pulse travelled away and back.
D. According to spacecraft S66, the time interval between the signal being sent and being received back by E is
greater than 20.0 ms.
I'm able to discount option A, but I'm having a hard time understanding why options B and C are false. I understand that at such high speeds, time dilation will occur, but how is that the 20.0 ms time observed by E is proper time? Is this because E is in the same frame of reference, space?
Also, given that this is a radio pulse, would not all observers see this travelling at the speed of light, hence making the Lorentz factor 0 and thereby time dilation undefined?
Cheers
Post by: Syndicate on October 02, 2017, 02:09:39 pm
Hey,
Quick question regarding one of the special relativity questions from the 2015 VCAA exam (Question 5) (Answer is D)
Spacecraft S66 is travelling at high speed away from Earth carrying a highly accurate atomic clock. Another spacecraft, T50, is travelling in the opposite direction to S66, as shown in Figure 1.
An observer, E, on Earth emits a short radio pulse to spacecraft S66, which reflects it directly back towards the
observer. The time elapsed for E between sending and receiving the pulse is 20.0 ms.
Which one of the following is true?
A. According to E, spacecraft S66 was more than 3000 km away when the pulse reached it.
B. According to E, the pulse took longer to reach spacecraft S66 than it did to return from spacecraft S66 to E.
C. The 20.0 ms interval measured by E is not a proper time because the radio pulse travelled away and back.
D. According to spacecraft S66, the time interval between the signal being sent and being received back by E is
greater than 20.0 ms.
I'm able to discount option A, but I'm having a hard time understanding why options B and C are false. I understand that at such high speeds, time dilation will occur, but how is that the 20.0 ms time observed by E is proper time? Is this because E is in the same frame of reference, space?
Also, given that this is a radio pulse, would not all observers see this travelling at the speed of light, hence making the Lorentz factor 0 and thereby time dilation undefined?
Cheers
- B is wrong, as the times to reach the Spacecraft, and travel back to E will be the same.
- C is wrong, as proper time is defined as the time measured between two events occurring in the same position in space. Therefore 20 ms is the proper time, as it occurred in E's frame (remember he is the one who emitted the pulse. He is part of the frame in which the radio pulse is in).
- according to Spacecraft S66, E is moving away from it, so the radio pulse will have to travel a greater distance in its perspective. Therefore, in this case the time has been dilated. So D is correct, as time measured will be grater for him.
As for the Lorenz factor, you are using it in a wrong sense. So we use the velocity of the spacecraft or E, depending on which one is defined at rest. The proper time is the time for the radio pulse to reach the spacecraft, and return to E in its frame of reference. However, we are testing for which observer sees the dilated time on their clock.
Post by: chantelle.salisbury on October 05, 2017, 10:15:52 am
i was just wondering what is an achievable goal for physics. i have an average of round about 80% and not too sure of the ranking but probably somewhere in the high middle. what score round abouts could i achieve if i dont loose too many marks on the exam?
thankyou
Post by: princessofpersia on October 05, 2017, 02:45:17 pm
I wanted to ask something in regards to the 2017 VCAA sample exam. in q14 they say that measurement were taken with a metre ruler, 'graduautesto 5 cm intervals and held by hand'
so would the uncertainity by 2.5cm? [half of the smallest unit on the ruler (5cm)]
Post by: Bri MT on October 05, 2017, 04:33:03 pm
Hello
i was just wondering what is an achievable goal for physics. i have an average of round about 80% and not too sure of the ranking but probably somewhere in the high middle. what score round abouts could i achieve if i dont loose too many marks on the exam?
thankyou
Are these company SACs? Which company?
Have you done a prac exam? What score?
hey ya all!!
I wanted to ask something in regards to the 2017 VCAA sample exam. in q14 they say that measurement were taken with a metre ruler, 'graduautesto 5 cm intervals and held by hand'
so would the uncertainity by 2.5cm? [half of the smallest unit on the ruler (5cm)]
I would think that it would be +/- 1.25cm seeing as it is analog but I could be wrong...
Post by: princessofpersia on October 05, 2017, 08:11:30 pm
I would think that it would be +/- 1.25cm seeing as it is analog but I could be wrong...
Why 1.25?
I thought analogue was 1/2 of smallest division, so 5/2, in other words
+/- 2.5cm?
I think I am missing something.
Post by: Bri MT on October 06, 2017, 10:06:21 am
Why 1.25?
I thought analogue was 1/2 of smallest division, so 5/2, in other words
+/- 2.5cm?
I think I am missing something.
No, I've checked my notes and you're right.
My confusion was coming from the fact that with an analog instrument you can state the value as being on a marking or halfway between two markings, however the uncertainty is still half the distance between two markings, not a quarter of the distance :)
Sorry for the unnecessary confusion
Post by: TooLazy on October 06, 2017, 03:22:10 pm
For example if I do VCAA 2012, exam 1, part 1 of section A ( so just motion)
how long should I take. Ive been generally averaging 35-40 mins, is this good or bad?
Post by: Syndicate on October 06, 2017, 05:04:27 pm
Recently I have been doing some of the past vcaa papers but because the sections are split up over two exams, and some sections arent relative anymore, I have been having trouble working out what a good time indicator is.
For example if I do VCAA 2012, exam 1, part 1 of section A ( so just motion)
how long should I take. Ive been generally averaging 35-40 mins, is this good or bad?
This year's exam is 130 marks. We only get about 150 minutes, so that's about 1 minute, 19 seconds per question. So, first of all, count how many marks there are in the "useful" section, and then multiply it with 1.15 (which is 69 seconds). That should indicate the amount of time you should be spending upon that section.
Post by: KiNSKi01 on October 10, 2017, 05:17:46 pm
Post by: Syndicate on October 10, 2017, 05:44:55 pm
How can you calculate acceleration when you aren't given mass. Or is there another way of answering question 71 without F=ma?
2005 exam says that the mass of the rocket is 0.2 kg. It is probably not listed here, as it is a follow up question.
Post by: KiNSKi01 on October 10, 2017, 05:49:01 pm
Post by: Sigma on October 10, 2017, 08:50:29 pm
Post by: chantelle.salisbury on October 10, 2017, 08:52:43 pm
just wondering whether there are any hints or tips on how to study for physics? i know to do heappps of questions but should i be relooking over notes etc before i do this? i dont know..... glad of any help!
Post by: Bri MT on October 11, 2017, 07:17:38 am
hello..Making a reallly good summary sheet can be a good form of revision
just wondering whether there are any hints or tips on how to study for physics? i know to do heappps of questions but should i be relooking over notes etc before i do this? i dont know..... glad of any help!
Post by: TooLazy on October 11, 2017, 12:35:17 pm
Post by: KiNSKi01 on October 11, 2017, 04:24:08 pm
Making a reallly good summary sheet can be a good form of revision
When you are required to make a cheat sheet anyway for the exam, is it still worthwhile making a separate summary sheet?
Post by: Bri MT on October 11, 2017, 05:12:39 pm
When you are required to make a cheat sheet anyway for the exam, is it still worthwhile making a separate summary sheet?
I'm not saying to make a separate one, I'm suggesting you improve your already existing one and make it better; and to then use that better version in your exam
Post by: peanut on October 17, 2017, 11:03:39 am
2 A4 double-sided pages OR
1 A3 double-sided page
Is this correct?
Post by: sweetiepi on October 17, 2017, 11:06:10 am
Just to confirm, the cheat can be either:Yup!
2 A4 double-sided pages OR
1 A3 double-sided page
Is this correct?
If you use the two A4 pages, they must be bound by tape :)
Post by: KiNSKi01 on October 22, 2017, 01:31:58 pm
A 10m shelf with as mass of 50kg has supports positioned at each end. A mass of 100kg on the shelf is 3 metres from the left-end, a mass of 150kg is 3 metres from the right-end and an additional 200kg mass is 1.5 metres from the right-end.
How do I calcualte the forces on the shelf due to the left and right supports?
Post by: Baylsskool on October 24, 2017, 09:50:41 pm
Use this link for the photo of why I'm asking https://m.imgur.com/a/QL5F2
Post by: TooLazy on October 28, 2017, 02:23:45 pm
Don't know how to calculate answer for this question:
A 10m shelf with as mass of 50kg has supports positioned at each end. A mass of 100kg on the shelf is 3 metres from the left-end, a mass of 150kg is 3 metres from the right-end and an additional 200kg mass is 1.5 metres from the right-end.
How do I calcualte the forces on the shelf due to the left and right supports?
I used g as 10m/s^2 to make things easier
Also not sure if its correct, haven't done these questions in ages
Post by: BlinkieBill on October 29, 2017, 02:17:58 pm
need help with answering this question.
Why is AC is preferred over DC for electricity transmission?
Thanks
Post by: Bri MT on October 29, 2017, 03:22:40 pm
Hey,Think about everything you've learnt about power loss in transmission.
need help with answering this question.
Why is AC is preferred over DC for electricity transmission?
Thanks
Transformers can be used to dramatically reduce this loss (explain how), and transformation only works for varied input (explain why and how this is relevant to AC/DC considerations)
Post by: Sigma on October 29, 2017, 03:31:14 pm
AC Circuit breakers are cheaper than DC Circuit breakers.
The repairing and maintenance of AC sub station is easy and inexpensive than DC Substation.
The Level of AC voltage may be increased or decreased(step up and Step down) transformers.
Post by: -273.15 on October 30, 2017, 04:31:38 pm
do you think its likely that they will ask us a question on inclined planes???
Post by: Bri MT on October 30, 2017, 05:11:39 pm
Hi guys
do you think its likely that they will ask us a question on inclined planes???
Probably. This might be in the form of banked tracks, connected bodies, etc.
Post by: TooLazy on October 31, 2017, 01:21:43 pm
Hey,
need help with answering this question.
Why is AC is preferred over DC for electricity transmission?
Thanks
3 main points:
-AC is preferred over DC because it can be transformed. Transformers work on the principle of electromagnetic induction.
More detail: Alternating current (AC) is constantly producing a changing magnetic flux, which induces a current in the secondary coil of the transormer
-The voltage can then be stepped up (using a transformer).
-This leads to lesser power loss in the transmission lines as, there is less current running through them.
(According to the equation Ploss=I^2 R)
Post by: KiNSKi01 on October 31, 2017, 05:11:49 pm
Post by: Bri MT on November 01, 2017, 03:21:58 pm
How do I do this question. I can't get the answer so I'm guessing I'm looking at the question the wrong way :-\. Im assuming you find 35% of 200 MW and then equate that to the work done over time taken. However when I get to this stage the only equation I know to calculate work done is Force x distance which doesn't make sense in this context. Help!!If the power station has an efficiency of 35% then 35% input energy = 200MW
Use that to find how much energy is input, and therefore how much fuel must be burnt
Post by: Cassidyhogi on November 02, 2017, 08:23:01 am
Describe how matter is converted to energy by nuclear fusion in the Sun.
thank you
Post by: chantelle.salisbury on November 02, 2017, 01:51:29 pm
Can someone please answer this question.umm... is this on the study design? if so i better be looking it up! :-\ :o
Describe how matter is converted to energy by nuclear fusion in the Sun.
thank you
i have a photoelectric q. that im stuck on
c. evaluate a value for W if the most energetic electrons have an energy of 1.9ev, and the incident light has a wavelength of 4.14*10^7m.
d. calculate the de broglie wavelength of 1.9eV electrons.
the ans.
c. 1.1eV - the incident photons have an energy of 3ev E=hc/wavelength hence, W=1.1
d. 8.9*10^10m - find momentum, then use wavelength =h/p
can anyone help here? thankyou
Post by: Syedali_ on November 02, 2017, 05:11:41 pm
Do you just approximate it or is there an actual method to draw it ?
Post by: chantelle.salisbury on November 02, 2017, 05:26:51 pm
I need some clarification in regards to some line of best fit questioni would wait till someone else replied to this... but what i have been told by my teacher and at colin hopkins lecture is to ensure that you go through two points that you plot (obviously the most suitable ones depending on the question)! also use a see-through / clear ruler so that you can see whereabouts the line of best fit would be placed most correctly..
Do you just approximate it or is there an actual method to draw it ?
Post by: jstrauss on November 02, 2017, 10:11:53 pm
umm... is this on the study design? if so i better be looking it up! :-\ :o
i have a photoelectric q. that im stuck on
c. evaluate a value for W if the most energetic electrons have an energy of 1.9ev, and the incident light has a wavelength of 4.14*10^7m.
d. calculate the de broglie wavelength of 1.9eV electrons.
the ans.
c. 1.1eV - the incident photons have an energy of 3ev E=hc/wavelength hence, W=1.1
d. 8.9*10^10m - find momentum, then use wavelength =h/p
can anyone help here? thankyou
So total energy equals planks constant * frequency of light and f=c/lambda so E=hc/lambda (wavelength). Therefore you have the total energy. E = Ek + W, where E total energy and Ek is kinetic energy. So, (4.14x10^-15)(3x10^8) / 4.14x10^7 = 3 eV. Therefore, 3 = 1.9 + W, W= 1.1
Wavelength = constant / momentum as the answer states. Using a derived fromula, p = v(2mEk) and coverting the 1.9 eV to J, p = 7. 44 x 10^-25. Therefore wavelength = 8.9 x 10^-10
Post by: chantelle.salisbury on November 03, 2017, 08:26:28 am
So total energy equals planks constant * frequency of light and f=c/lambda so E=hc/lambda (wavelength). Therefore you have the total energy. E = Ek + W, where E total energy and Ek is kinetic energy. So, (4.14x10^-15)(3x10^8) / 4.14x10^7 = 3 eV. Therefore, 3 = 1.9 + W, W= 1.1awesome thankyou alot... :)
Wavelength = constant / momentum as the answer states. Using a derived fromula, p = v(2mEk) and coverting the 1.9 eV to J, p = 7. 44 x 10^-25. Therefore wavelength = 8.9 x 10^-10
Post by: smiley123 on November 03, 2017, 04:38:17 pm
Post by: Bri MT on November 03, 2017, 05:12:03 pm
Hi everyone! I'm really confused about the right hand rule. Can someone please explain why the right hand rule is different for the green conductors when they're exactly the same? The only difference is the direction of the current of the conductor that they are parallel to, but how does this affect the direction of magnetic field for the green conductor?
Ok, so the difference is that the magnetic field next to them is opposite in direction. Therefore the force is opposite in direction.
Draw more circle around the neighbouring wire if you cant see the direction of the magnetic field through the green one
Post by: TooLazy on November 05, 2017, 07:33:15 am
Like drawing a magnetic field, or plotting a graph for the photoelectric effect with a line of best fit.
Post by: princessofpersia on November 05, 2017, 11:58:01 am
Does anyone know if we can use pencil for diagrams and graphs?
Like drawing a magnetic field, or plotting a graph for the photoelectric effect with a line of best fit.
Yes we can :)
Post by: BlinkieBill on November 06, 2017, 08:31:47 pm
just wanted to confirm that relativistic mass isnt in the course?
Post by: Bri MT on November 07, 2017, 07:22:58 am
Hey
just wanted to confirm that relativistic mass isnt in the course?
It's not, but the formula also isn't difficult you just use the Lorentz transformation so if it reassures you you can take a cm to write it down
Post by: chantelle.salisbury on November 07, 2017, 03:13:03 pm
Can someone please answer this question.
Describe how matter is converted to energy by nuclear fusion in the Sun.
thank you
Did anyone end up replying to this?
i would use the formula that E=mc^2, tho not entirely sure how to 'describe'.
Post by: Bri MT on November 07, 2017, 04:42:45 pm
Did anyone end up replying to this?
i would use the formula that E=mc^2, tho not entirely sure how to 'describe'.
If you are doing the fission/fusion extended study, you should refer to the 4 fusion formulae & list the energy released by each one. Also provide an overall energy release, and describe mass-energy equivalence
Post by: TooLazy on November 08, 2017, 04:04:43 pm
It's not, but the formula also isn't difficult you just use the Lorentz transformation so if it reassures you you can take a cm to write it down
Relativistic mass is on the course
Post by: Bri MT on November 08, 2017, 05:04:09 pm
Relativistic mass is on the course
Where?
"Einstein’s theory of special relativity
• describe Einstein’s two postulates for his theory of special relativity that:
– the laws of physics are the same in all inertial (non-accelerated) frames of reference
– the speed of light has a constant value for all observers regardless of their motion or the motion of the source
• compare Einstein’s theory of special relativity with the principles of classical physics
• describe proper time (t0 ) as the time interval between two events in a reference frame where the two events
occur at the same point in space
• describe proper length (L0 ) as the length that is measured in the frame of reference in which objects are at rest
• model mathematically time dilation and length contraction at speeds approaching c using the equations:
• explain why muons can reach Earth even though their half-lives would suggest that they should decay in the
outer atmosphere."
I can't find it in the study design?
The Edrolo person also said it isn't on the course this year as well
Post by: TooLazy on November 08, 2017, 06:17:50 pm
Where?
"Einstein’s theory of special relativity
• describe Einstein’s two postulates for his theory of special relativity that:
– the laws of physics are the same in all inertial (non-accelerated) frames of reference
– the speed of light has a constant value for all observers regardless of their motion or the motion of the source
• compare Einstein’s theory of special relativity with the principles of classical physics
• describe proper time (t0 ) as the time interval between two events in a reference frame where the two events
occur at the same point in space
• describe proper length (L0 ) as the length that is measured in the frame of reference in which objects are at rest
• model mathematically time dilation and length contraction at speeds approaching c using the equations:
• explain why muons can reach Earth even though their half-lives would suggest that they should decay in the
outer atmosphere."
I can't find it in the study design?
The Edrolo person also said it isn't on the course this year as well
My teacher said it was so not 100% sure.
Also Ive watched edrolo videos on it...? Like the guy solves questions involving relativistic mass etc
Post by: Bri MT on November 08, 2017, 07:53:13 pm
My teacher said it was so not 100% sure.
Also Ive watched edrolo videos on it...? Like the guy solves questions involving relativistic mass etc
I think that might be because it helps for understanding the Ek equation & because it was assessed in the past.
Might be one of things where it's the two different Edrolo people?
Mass dilation isn't on the formula sheet whereas length contraction and time dilation. It is, however, a bit strange to exclude mass (only) given how similar the calc is to time....
Post by: TooLazy on November 08, 2017, 09:42:46 pm
I think that might be because it helps for understanding the Ek equation & because it was assessed in the past.
Might be one of things where it's the two different Edrolo people?
Mass dilation isn't on the formula sheet whereas length contraction and time dilation. It is, however, a bit strange to exclude mass (only) given how similar the calc is to time....
Apparently the bio exam had stuff that wasnt explicitly stated on the study design.
Since both physics and bio are new designs this year, I definitely wouldnt take any risks.
and yeah also, vcaa are pretty dog so...
Post by: chantelle.salisbury on November 10, 2017, 12:24:35 pm
just wondering if someone could help me with uncertainty percentages and absolute errors? firstly how do i calculate these?
this is a sample q.
An atomizer sprays a fine mist of spherical oil droplets. the radius of one of the droplets is 1.37(plus minus) .02 micro m
calculate the volume (including uncertainty) V of the oil droplet, given V=4/3pie.r^2
Post by: Bri MT on November 10, 2017, 12:56:10 pm
Hey..The absolute uncertainty is 0.02 *10^-6m.
just wondering if someone could help me with uncertainty percentages and absolute errors? firstly how do i calculate these?
this is a sample q.
An atomizer sprays a fine mist of spherical oil droplets. the radius of one of the droplets is 1.37(plus minus) .02 micro m
calculate the volume (including uncertainty) V of the oil droplet, given V=4/3pie.r^2
The percentage uncertainty is (0.02/1.37)*100
Your percentage uncertainty on the sphere would be 2*((0.02/1.37)*100) (it is times 2 because r is squared)
More info:https://atarnotes.com/forum/index.php?topic=174013.0
Post by: -273.15 on November 11, 2017, 11:23:14 am
in terms of the uncertainty value, both itute and vicphysics solutions had different values
I also found a different value
Im not sure how to find the uncertanty correctly? What values have others calculated
Also do you think vcaa would be a bit lenient with uncertainty values?
Post by: manishmao on November 11, 2017, 06:02:11 pm
Jason and Kylie are sitting at the northern and southern ends respectively of a train carriage travelling north at
a high speed. Each holds a torch that they turn on and off. Harold is standing on a platform beside the train. As
the midpoint of the carriage passes Harold, he observes simultaneous light flashes from both Jason and Kylie.
Which one of the following statements is true?
A. To an observer inside the carriage, located at its midpoint, Jason and Kylie turned on their torches at the
same time.
B. To an observer inside the carriage, located at its midpoint, Jason turned on his torch before Kylie.
C. To an observer inside the carriage, located at the midpoint, Kylie turned on her torch before Jason.
D. It does not make any sense to ask in which order Jason and Kylie turned on their torches, because Einstein
showed that time is relative.
Answer is C
Post by: KiNSKi01 on November 11, 2017, 09:48:32 pm
Any help with this question?
Jason and Kylie are sitting at the northern and southern ends respectively of a train carriage travelling north at
a high speed. Each holds a torch that they turn on and off. Harold is standing on a platform beside the train. As
the midpoint of the carriage passes Harold, he observes simultaneous light flashes from both Jason and Kylie.
Which one of the following statements is true?
A. To an observer inside the carriage, located at its midpoint, Jason and Kylie turned on their torches at the
same time.
B. To an observer inside the carriage, located at its midpoint, Jason turned on his torch before Kylie.
C. To an observer inside the carriage, located at the midpoint, Kylie turned on her torch before Jason.
D. It does not make any sense to ask in which order Jason and Kylie turned on their torches, because Einstein
showed that time is relative.
Answer is C
Yo manishmao!
<reference to other forum removed>
-"The train is travelling north, and Kylie is at the south. The torch is turned on when the MIDPOINT of the train reaches Harold. At this point, Jason is now travelling AWAY from Harold, whereas Kylie is still travelling towards.
When the light ray comes out of their torches, it will bend towards the northerly direction (c is constant in all frames of references). If Jason's light ray is bent north, it's only getting further away from Harold. Whereas if KYLIE's light ray is bent north, it's actually doing the opposite and getting closer."
<reference to other forum removed>
Moderator action - insanipi: Removed references to rival forums.
Post by: Nicko912 on November 13, 2017, 10:18:02 am
Post by: TooLazy on November 13, 2017, 10:31:13 am
Just asking to be sure, but with physics are the answers accepted when their reasonable? I know with chemistry you have to simplify to the correct number of sigfigs, I've been doing sigfigs for physics too just for memory sake, do they accept this too?
Yeah I think there is just one question, where they test you on sig figs. Other than that i think if your answers are reasonable you shouldnt have an issue
Post by: sweetiepi on November 13, 2017, 10:37:14 am
Just asking to be sure, but with physics are the answers accepted when their reasonable? I know with chemistry you have to simplify to the correct number of sigfigs, I've been doing sigfigs for physics too just for memory sake, do they accept this too?Last year I believe it was perfectly fine to just be sensible with sigfigs, however best practice is to always be careful with sigfigs :)
Post by: Nicko912 on November 13, 2017, 10:44:23 am
Yeah I think there is just one question, where they test you on sig figs. Other than that i think if your answers are reasonable you shouldnt have an issue
Thanks for the responses!
Yeah mainly to what I was getting at, if it's safer to use sigfigs or to just use reasonable answers, the difference between a sigfig answer and a calculator answer can be quite different at times. I've seen in VicPhysics solutions they consistently use sigfigs, but they're not VCAA, so I guess it doesn't matter at all that much.
Post by: yewyewyewy on November 14, 2017, 01:05:32 pm
General questions about voltage;The negative symbol implies direction not magnitude so whilst 2V=-2V in terms of magnitude the understanding is that the latter flows in the opposite direction. 3 marks indicates that the examiner is looking for 3 distinct points, look to include relevant physics principles supported by formulas, so perhaps you could say:
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
- F=ma=mv/t
- Therefore impact force is inverse proportional to change in time
- Seatbelts and crumpling increase the collision time therefore decreasing the force of the collision
Post by: simrat99 on November 14, 2017, 04:08:54 pm
Post by: kalopsia on November 14, 2017, 05:59:49 pm
Hi, just to confirm, we are not required to know about the Michelson Morley experiment, right?
No, I don't think so. The study design states that we have to 'compare Einstein’s theory of special relativity with the principles of classical physics', so it would probably help if you know about it though.
Post by: Petra24 on November 14, 2017, 09:25:36 pm
Post by: sweetiepi on November 14, 2017, 09:30:16 pm
Hi, just to avoid any issues tomorrow, was the cheat sheet for physics supposed to be 2 A4 double sided sheets or single sided sheets?Double sided!
Make sure they are taped together! :)
Post by: Petra24 on November 14, 2017, 09:37:16 pm
Double sided!Okie dokie. Thank you very much! :D :D
Make sure they are taped together! :)
Post by: gamma032 on November 14, 2017, 10:09:32 pm
"12.5 Hz, rounded to 12 Hz, using the rule of rounding to the nearest even number."
Anyone heard of this before?
Post by: KiNSKi01 on November 17, 2017, 11:29:06 pm
For this question I couldn't find solutions so can someone please tell me if i am right: 15=A; 16=B; 17=D
MORE IMPORTANTLY, what is the best way to explain these answers or is it simply a matter of stating momentum is conserved, KE is converted to EPE etc. (I think it is something worth putting on my cheat sheet for 1/2 exam)
Post by: Shadowxo on November 18, 2017, 10:44:48 am
Yo everyone!What's the question? (The picture doesn't include the first part where it describes what's happening to the system)
For this question I couldn't find solutions so can someone please tell me if i am right: 15=A; 16=B; 17=D
MORE IMPORTANTLY, what is the best way to explain these answers or is it simply a matter of stating momentum is conserved, KE is converted to EPE etc. (I think it is something worth putting on my cheat sheet for 1/2 exam)
Just want to make sure my answers are for the right situation :)
Post by: KiNSKi01 on November 18, 2017, 11:07:04 am
Post by: occidit on November 18, 2017, 11:42:26 am
That was all of the question :(
Looks like A, B, B.
If it is elastic, KEfinal = KEinitial, however in the middle some of it is transferred to spring potential energy then back into KE which is the drop.
B is correct for the second and third as momentum is always conserved per the law of conservation of momentum, whether or not it's an elastic collision.
Post by: KiNSKi01 on November 18, 2017, 11:55:09 am
Thought it was only elastic collision where momentum is conserved
Post by: Shadowxo on November 18, 2017, 12:06:31 pm
Cheers occidit for the help!Momentum is always conserved within a closed system (so basically every collision, and not where a force is acting on an object).
Thought it was only elastic collision where momentum is conserved
Elastic just means kinetic energy is conserved :)
Post by: KiNSKi01 on November 18, 2017, 12:10:13 pm
Post by: Syndicate on November 18, 2017, 12:37:34 pm
I don't really get that because KE=1/2mv^2 and In an inelastic collision KE is lost- meaning velocity decreases. And since momentum is m(v-u) I don't understand how momentum is conserved :-\. But i'm happy just to accept it lol
If velocity of one object decreases, then the other's increases in a closed system.
Post by: KiNSKi01 on November 18, 2017, 12:40:54 pm
If velocity of one object decreases, then the other's increases in a closed system.
But inelastic collision are where KE is transferred into other forms of energy (e.g sound) so how could another object's velocity reach the same level as what the first object achieved if not all KE is transferred
Post by: Syndicate on November 18, 2017, 12:59:45 pm
But inelastic collision are where KE is transferred into other forms of energy (e.g sound) so how could another object's velocity reach the same level as what the first object achieved if not all KE is transferred
Momentum = mv
Kinetic Energy = 1/2mv^2
Whilst you can have negative momentum, it is the opposite for kinetic energy (because velocity will always be squared)
Example: Assume there are two toy cars. Car A (mass 1 kg) travelling towards Car B at 2 m/s to the right, whilst Car B (mass 3kg) is travelling towards it at 3 m/s to the left (Notice the differences in mass). After collision Car A travels 1 m/s to the left, and Car B travels 2 m/s to the left. (Assume right is positive)
Momentum.
Initial momentum of the system:
Car A: p = mv = 2 kg m/s
Car B: p = mv = -9 kg m/s
initial momentum = -7 kg m/s
final momentum
Car A: p = mv = -1 kg m/s
Car B: p = mv = -6 kg m/s
final momentum of system = -7 kg m/s
Therefore momentum is conserved.
Now as for kinetic energy within the system
initial kinetic energy = 0.5 x 1 x 2^2 + 0.5 x 2 x (-3)^2
= 11 J
final kinetic energy = 0.5 x 1 x (-1)^2 + 0.5 x 2 x (-2)^2
= 4.5 J
Ke is lost, therefore this is an inelastic collision (its being converted to heat, sound etc...).
When I said that if one object loses velocity, and the other gains it, it doesn't necessarily have to be the same amount (So in this case, Car B becomes slower by 1 m/s, but Car A did not "gain" that 1 m/s, instead it's change in velocity was -3 m/s). Mass plays a major role in determining its final velocity after collision.
Post by: Theodric_Ironfist on November 18, 2017, 01:07:12 pm
But inelastic collision are where KE is transferred into other forms of energy (e.g sound) so how could another object's velocity reach the same level as what the first object achieved if not all KE is transferred
This page might answer your question.
Post by: KiNSKi01 on November 18, 2017, 01:10:31 pm
We were never really taught why momentum can be conserved yet KE can be lost...sort of just had to accept it ;)
Never really thought I should just do some calculations to prove it to myself ahah
Post by: KiNSKi01 on November 20, 2017, 07:22:33 pm
11.11 I thought pressure was inversely proportional to volume and therefore I don't understand why the pressure increases. I get 0.514 metres cubed by doing PxV is a constant :-\
11.13 How are you meant to answer this question when all three of pressure volume and temperature change
EDIT: Whoops I just realised all you do is PV=nRT for 11.13
Post by: Shadowxo on November 20, 2017, 09:06:16 pm
Yo can someone help me with 11.11 and 11.13 eYou're right, pressure is inversely proportional to volume but you have to keep the other variables (temp and number of moles) constant. So as the pressure inside the tyre increases, the air that started off in the tyre decreases in volume (gets squeezed into a smaller space by the new air).
11.11 I thought pressure was inversely proportional to volume and therefore I don't understand why the pressure increases. I get 0.514 metres cubed by doing PxV is a constant :-\
11.13 How are you meant to answer this question when all three of pressure volume and temperature change
EDIT: Whoops I just realised all you do is PV=nRT for 11.13
For the air inside the tyre, we want to find the new volume it occupies (to then find how much volume the new air takes up). Since it's the same air, the number of moles is constant so we can use PV=constant
P1V1=P2V2
V2=P1V1/P2
=105*0.150/360
=0.04375 m3
Note that since it's P1/P2 you can use whichever units you choose for pressure but they must be the same.
Volume of new air inside the tyre (at pressure of 360kPa) is the remaining volume = 0.150 - 0.04375 = 0.10625 m3
Volume at atm pressure: use the same equation as number of moles & temp stays the same
V1=P2V2/P1
=360*0.10625/105
=0.364 m3
Also note that you need to take into account sig figs if applicable
Post by: KiNSKi01 on November 20, 2017, 09:17:04 pm
Didn't consider having to subtract the difference between the volumes
Post by: Unsplash on December 06, 2017, 12:48:33 pm
Post by: Syndicate on December 06, 2017, 03:20:10 pm
I'm having trouble with this question. So far I have got, Fnet = 360 + R * sin(20) where R is the reaction force. I have also got mg = R * cos (20) but cannot figure out what to do next. No matter what way I substitute it in, I always get two variables. I have a feeling it has something to do with "only just" taking the corner. Could someone point me in the right direction?
I will reply soon with the worked solution, however, your starting step would be to resolve for the horizontal forces of the system.
Note this is not a banked track, as the surface is horizontal, and "only just" implies that the system is in equilibrium.
Therefore, mv^2/r = mgsin(20) (the horizontal component of weight force) + 360
Post by: Vaike on December 06, 2017, 03:45:01 pm
Therefore, mv^2/r = mgsin(20) (the horizontal component of weight force) + 360
Unfortunately, the question doesn't seem to supply the velocity of the cyclist nor the radius of the bend, and I can't see any way to calculate these values, so I'm not sure this approach would work.
From what I can tell, there is no way to work this out properly. Is this a textbook question FelixHarvey? I won't be surprised if so, there are quite a few that leave out necessary information.
Post by: Unsplash on December 06, 2017, 04:32:21 pm
From what I can tell, there is no way to work this out properly. Is this a textbook question FelixHarvey? I won't be surprised if so, there are quite a few that leave out necessary information.
Yes, this is a question from the Jacaranda textbook. Thanks for your help!
Post by: KiNSKi01 on December 06, 2017, 08:41:51 pm
Show that the total energy of an electron has been accelerated to a speed of 0.98c is about 4x10^-13 J
Post by: Vaike on December 06, 2017, 08:57:52 pm
How do i approach this question?
Show that the total energy of an electron has been accelerated to a speed of 0.98c is about 4x10^-13 J
Hi!
Since the particle is traveling at very high speeds, we must take relativistic effects into account; as a particle approaches light speeds it's total energy is composed of both it's rest energy, E0 and it's kinetic energy, Ek. If you use the relevant formulae, adding them together, you get an equation for the total energy of the particle. Plug in the numbers and you should get the answer :)
Post by: Syndicate on December 06, 2017, 11:25:24 pm
I'm having trouble with this question. So far I have got, Fnet = 360 + R * sin(20) where R is the reaction force. I have also got mg = R * cos (20) but cannot figure out what to do next. No matter what way I substitute it in, I always get two variables. I have a feeling it has something to do with "only just" taking the corner. Could someone point me in the right direction?
The question you have posted is missing some information (there might be a diagram somewhere?). I have checked the worked solutions as well just to confirm, and they have seemed to use v = 9 m/s and radius of 10m to get 59.4 kg for the cyclist.
Post by: Unsplash on December 07, 2017, 12:10:52 pm
The question you have posted is missing some information (there might be a diagram somewhere?). I have checked the worked solutions as well just to confirm, and they have seemed to use v = 9 m/s and radius of 10m to get 59.4 kg for the cyclist.
No diagram that I can see. I've checked both the pdf and the online textbook through JacPlus. I've just skipped the question. Thanks for your help!
Post by: viviianle on December 18, 2017, 12:47:45 pm
--
One of the fastest objects ever made on Earth was
the Galileo Probe which, as a result of Jupiter’s huge
gravity, entered its atmosphere in 1995 at a speed of
nearly 50 000 m/S. Give an estimate of the Lorentz
factor for the probe to nine decimal places.
------
answer is 1.4*10^-8
Post by: sweetiepi on December 18, 2017, 01:01:15 pm
please helpThe equation we're using is: 1/sqrt(1-v^2/c^2)
--
One of the fastest objects ever made on Earth was
the Galileo Probe which, as a result of Jupiter’s huge
gravity, entered its atmosphere in 1995 at a speed of
nearly 50 000 m/S. Give an estimate of the Lorentz
factor for the probe to nine decimal places.
------
answer is 1.4*10^-8
c= speed of light, which is 3.0*10^8 (afaik, could be slightly off)
Therefore 1/sqrt(1-(50000)^2/(3.0x10^8)^2)
Lorentz's factor= 1.000000014 (9dp)
Disclaimer: I'm not 100% certain of my answer on this question
Post by: KiNSKi01 on December 19, 2017, 01:32:58 pm
The equation we're using is: 1/sqrt(1-v^2/c^2)
c= speed of light, which is 3.0*10^8 (afaik, could be slightly off)
Therefore 1/sqrt(1-(50000)^2/(3.0x10^8)^2)
Lorentz's factor= 1.000000014 (9dp)
Disclaimer: I'm not 100% certain of my answer on this question
Yep i got same answer (1.0000000138889 exactly)
Post by: skrt skrt on January 01, 2018, 12:21:02 am
Answer is 12.9N
Thanksss
Post by: kalopsia on January 01, 2018, 09:23:34 am
This is question 10 (7.1)in the hinemann year 12 book.A tennis ball of mass 57.5g is tested for compliance with tennis regulations by being dropped from a height of 251cm onto concrete. A bounce height of 146cm is deemed acceptable. Find the magnitude of the average force on the ball if it is in contact with the floor for 0.0550s.
Answer is 12.9N
Thanksss
Please find my working out attached.
The average force is f=mΔv/Δt. So you have to find the Δv, which is the change in velocity the ball experiences when it is in contact with the ground.
Post by: skrt skrt on January 02, 2018, 11:27:34 pm
For the cart to be towed at a constant speed, what must be the magnitude of the applied force?
Post by: chooby on January 02, 2018, 11:35:53 pm
Hi guys need help with this question. A skier wishes to have skiing equipment of mass 200kg transported tot he top of the downhill ski slope. The items are pulled on a cart by a rope with a force Fa and in doing so, a constant friction force of 100N is encountered.Here you would need to use Newton's second law (F=ma) and forces addition. Through newtons second law, we know the net force is 0 as there is a constant speed and there for 0 acceleration. Through force addition we know that net force = Fa - 100 (assuming that Fa is the positive direction). Then you equate so that 0 =Fa - 100 , then Fa = 100N. Direction not needed as only magnitude was asked for.
For the cart to be towed at a constant speed, what must be the magnitude of the applied force?
(https://uploads.tapatalk-cdn.com/20180102/343b8b5017a6ecb67c53ab2fb2173a85.jpg)
Post by: skrt skrt on January 02, 2018, 11:54:04 pm
Post by: chooby on January 03, 2018, 12:05:34 am
Fg = mg, so that is ( 20 )*( 9.8 ) which means Fg = 196N.
We can break Fg down into its vertical and horizontal component. However, we only need to find the horizontal compenent as it is parallel to the direction of motion. So using trig, the horizontal component is 67.04N in the same direction as the frictional force.
The rest is the same with net force = 0 due to newtons second law but now net force is also Fa - 100 - 67.04. So when you equate these, Fa = 167.04N(https://uploads.tapatalk-cdn.com/20180102/68725268e54439f9a18132460a73c637.jpg)
Post by: skrt skrt on January 14, 2018, 01:35:17 am
Thanks
Post by: Calebark on January 14, 2018, 02:09:21 am
Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks
Even though we use horizontal and vertical components, you need to remember that they're still measuring the same thing overall - the motion of a single object. A single object can't have two different time measurements. It might be easier to answer your question (as I may have missed the point) if I explain why we only use the vertical component of motion to measure time.
In VCE Physics, we assume that air resistance in projectile motion is negligible; it doesn't matter. This means that there is nothing slowing down the object in the horizontal component. However, overall, it DOES slow down, which shouldn't happen if we just consider the horizontal component (can you recall Newton's First Law of motion?), meaning the vertical component is acting upon the object to slow it down. What's a force that's always acting on an object in the vertical component? Gravity. Gravity is the only thing slowing an object down in the vertical component, meaning gravity is the only thing slowing down an object in projectile motion.
Post by: Bri MT on January 14, 2018, 09:19:23 am
Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks
Once the body hits the ground/other surface it stops moving
Edit: (We don't think about skidding, bouncing etc. in this course)
Post by: A TART on January 19, 2018, 11:39:30 pm
I tried finding the change in gravitational field strength (which will be equal to the kinetic energy). However, that graph only goes up to 10,000km, and in order to find enough information for the field strength, I need to go to about 14,000Km (I think the question assumes you know that the surface is around 6000Km from the core).
Is there another approach or am I missing something?
EDIT:I think I'm onto something
EDIT 2: I got it.
Post by: Yertle the Turtle on January 19, 2018, 11:47:01 pm
Hey people, just wondering why the time is the same for horizontal and vertical component for motion.It is because the entire flight of the projectile will take a given time, and the horizontal and vertical components are simply breaking that individual motion up into parts so that it can be analysed more easily.
Thanks
Post by: Bri MT on January 19, 2018, 11:50:32 pm
I'm not sure how to do this question: (attached)
I tried finding the change in gravitational field strength (which will be equal to the kinetic energy). However, that graph only goes up to 10,000km, and in order to find enough information for the field strength, I need to go to about 14,000Km (I think the question assumes you know that the surface is around 6000Km from the core).
Is there another approach or am I missing something?
EDIT: I think I'm on something
Reread the question. The maximum distance is 8000 km from the center (the starting position is 600km + radius of Earth away). Thus all of the needed information is provided.
Finding the area under the graph (between the two distances) effectively finds you g * h so then you just need to multiply that by m to get the change in gravitational potential energy and therefore the initial kinetic energy
Post by: A TART on January 20, 2018, 12:26:17 am
Reread the question. The maximum distance is 8000 km from the centre (the starting position is 600km + radius of Earth away). Thus all of the needed information is provided.
Finding the area under the graph (between the two distances) effectively finds you g * h so then you just need to multiply that by m to get the change in gravitational potential energy and therefore the initial kinetic energy
Yeah... I actually found the Mass of Earth then did the question. Then I realised the question stated "8000Km from the CENTRE of eath" not surface XD
Post by: Bri MT on January 20, 2018, 07:32:02 am
Yeah... I actually found the Mass of Earth then did the question. Then I realised the question stated "8000Km from the CENTRE of eath" not surface XD
Not sure if this was a typo or not, but you should be using the mass of the communications satellite (not earth) in your equation.
As you are finding the change in max GPE of the satellite all the variables you're using in the equation should be relevant to it.
The only impact Earth's mas here is that it influences g - but that's been given to you anyway
Post by: Abi21 on January 22, 2018, 07:55:25 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: KiNSKi01 on January 24, 2018, 04:33:03 pm
Hi, I'm struggling with two physics questions, hoping for a little bit of help please!
Thanks in advance!
Check my response in other thread :)
Post by: KiNSKi01 on January 24, 2018, 04:48:48 pm
A block weighing 2kg is accelerated by a falling weight (0.5kg). Calculate the tension in the cord if 2kg block experiences a fricitional force of 1.5N as it slides on the table.
First I considered the whole body assuming it experiences uniform acceleration. The only two forces acting on the whole body are the frictioon force and gravity. Therefore I calculated the resultant force to be 0.5 x 9.8 subtract 1.5. Meaning the system was accelerating at 1.36 m/s^2.
Then I looked at the forces acting on the 2kg block (drag and tension). and then Fresultant= 2 x 1.36 and calculated the tension to be 4.22 N.
The answer is 4.3N (yeah i know HUGE difference) but this answer couldn't be due to rounding ???
EDIT they used g as 10 instead of 9.8 -_____- :P
Post by: Yertle the Turtle on January 25, 2018, 06:26:59 am
EDIT they used g as 10 instead of 9.8 -_____- :PI HATE it when they do that. Checkpoints do that and it is the most annoying thing ever!
Post by: sweetiepi on January 25, 2018, 11:11:52 am
EDIT they used g as 10 instead of 9.8 -_____- :P
I HATE it when they do that. Checkpoints do that and it is the most annoying thing ever!Some of the older checkpoints questions could be from the old study design, where g was 10, instead of 9.8. Also note that is obviously the case in past VCAA exams before last year's one. :)
Otherwise,your working is right KiNSKi for using 9.8 instead of 10. :)
(Sorry that I didn't see this before aha)
Post by: Guideme on January 30, 2018, 11:44:54 am
I think the answer is 55m/s but it seems too simple to be true haha.
Help pls
Thx in advance
Post by: Yertle the Turtle on January 30, 2018, 11:51:19 am
In a tennis match at Kooyong a girl receives a serve which is traveling at 30m/s south just prior to hitting her racket. Immediately after leaving her racquet it is travelling at 25m/s north. Specify fully the change in velocity she gave the ballYou use the formula v=v2-v1 and specify either North or South as being positive. Then you simply do v=25-(-30), so v=55m/s North, since a velocity requires a direction as well as a magnitude. Hope this helps.
I think the answer is 55m/s but it seems too simple to be true haha.
Help pls
Thx in advance
Post by: Vaike on January 30, 2018, 12:05:05 pm
In a tennis match at Kooyong a girl receives a serve which is traveling at 30m/s south just prior to hitting her racket. Immediately after leaving her racquet it is travelling at 25m/s north. Specify fully the change in velocity she gave the ball
I think the answer is 55m/s but it seems too simple to be true haha.
Help pls
Thx in advance
Marvin's on the money. However I thought I'd still post this, as kind of a guide on how to best set out your working, It's pretty easy to mess this stuff up in a rush, so I found it really helpful to always clearly set out my working like this to avoid any potential errors.
Spoiler
(https://imgur.com/JxtCegN.png)
Remember, velocity is a vector, this means it requires both direction and magnitude! Drawing a diagram usually helps with this, just make sure to draw the arrows the correct relative sizes!
(http://www.abdsphysics.com/uploads/6/5/0/9/65090265/4802565.png?540)
Post by: skrt skrt on February 06, 2018, 06:07:03 pm
Post by: Calebark on February 06, 2018, 06:31:03 pm
Good resources for learning special relativity?
I didn't do the new study design, so I only know of the YouTube Channel ran up the new Edrolo physics guy, which is located here.
Post by: sweetiepi on February 06, 2018, 06:37:53 pm
I didn't do the new study design, so I only know of the YouTube Channel ran up the new Edrolo physics guy, which is located here.That guy is 100% a baller. Definitely rate this one too :)
Post by: skrt skrt on February 14, 2018, 05:31:08 pm
Initial horizontal velocity from both objects is 5m/s and 2m/s, and the height is 1.8m
Post by: Yertle the Turtle on February 14, 2018, 06:03:55 pm
Why is that if two projectiles are launched from the same height have the same time to fall to the ground?This is because the acceleration due to gravity is the same for all objects is 9.8m s^-2 and if you are disregarding air resistance they will all fall at the same time.
Initial horizontal velocity from both objects is 5m/s and 2m/s, and the height is 1.8m
Post by: Shadowxo on February 14, 2018, 08:40:25 pm
Why is that if two projectiles are launched from the same height have the same time to fall to the ground?I'd also just like to add that you can split the acceleration and speed etc into horizontal and vertical directions. Vertically, they both start off at the same height (1.8m), and the same vertical velocity (0 m/s) and having the same downwards acceleration from gravity (9.8 m/s^2) - gravity only acts vertically.
Initial horizontal velocity from both objects is 5m/s and 2m/s, and the height is 1.8m
Therefore their horizontal displacement / velocity / acceleration is irrelevant when calculating when they'll hit the ground (but it will determine how far the ball travels)
Post by: hemalw0914 on February 15, 2018, 11:16:58 pm
1)Give two reasons why you feel cooler when the wind is blowing than you would in still air at the same temperature.
2) In humid weather, evaporation of perspiration takes place as it does in dry weather. However, the cooling effect is greatly reduced. Why?
3) In hot weather, sweat evaporates from the skin. Where does the energy required to evaporate the sweat come from?
Post by: skrt skrt on February 16, 2018, 08:02:29 pm
Post by: blasonduo on February 16, 2018, 08:14:47 pm
Thanks for the responses but why doesnt mass affect this, like mg?Hey!
I assume you're asking why different masses hit the ground at the same time?
When I drop a feather and a bowling ball at the same height (without air resistance) they will both accelerate at 9.8 metres per second, so if their accelerations are constant they must fall at the same velocity!
In regards to F = ma, because acceleration is constant but mass if very different, it means the force is different. Remember, the force isn't how fast it goes, but how hard it hits the ground, so back to the ball and feather, they would land at the same time, but one would hurt MUCH more as its force is different :))
Post by: lzxnl on February 19, 2018, 01:03:21 pm
Thanks for the responses but why doesnt mass affect this, like mg?
This is actually a very interesting question that led Einstein to formulate the theory of general relativity. Here's some background.
You may know/will learn that the centripetal force required to move in a circle of radius r at speed v is
The force that you measure is proportional to your mass, like with gravity. It is what we call an inertial or fictitious force. No physical force, aside from gravity, is proportional to the mass. However, all inertial and fictitious forces are proportional to the mass. This led Einstein to believe that gravity isn't a force, but is just acceleration, like the centripetal acceleration above. In his model, free fall is motion with no acceleration. This may seem counterintuitive, but in a vacuum, if you're free falling and you let go of something, that object 'falls with you', because gravity accelerates it at the same rate. Therefore, ironically, when you're free falling, you can't detect the influence of gravity.
Tl;dr, gravity is special, it's the only force you'll study that has a constant acceleration for all masses.
Need help with some homework - Thermodynamics U1
1)Give two reasons why you feel cooler when the wind is blowing than you would in still air at the same temperature.
2) In humid weather, evaporation of perspiration takes place as it does in dry weather. However, the cooling effect is greatly reduced. Why?
3) In hot weather, sweat evaporates from the skin. Where does the energy required to evaporate the sweat come from?
1. Convection: the wind is cold, so more cold air hits your skin and draws heat away. Evaporation: the wind draws warm water vapour away from your skin.
2. In humid weather, you're surrounded by warm water vapour. The water from your body can't really escape because the vapour pressure is high.
3. The hot weather and your body both supply energy to evaporate the water on your skin.
Post by: KiNSKi01 on March 01, 2018, 08:33:19 pm
The question is "At the instant that m1 hits the ground the string breaks. Find the time taken for m2 to come to rest". (All relevant values are provided in the attachment)
I know you have to start by calculating the the acceleration of m2 however I can't even really visualise what would happen and consequently don't know what to do next
Thanks for the help!! ;D
Post by: Yertle the Turtle on March 01, 2018, 08:49:27 pm
Hey can I have some assistance with this question?I think that you would first calculate the net force on the system, which is coming from m1, due to gravity. That would be Fnet=2*9.8=19.6N
The question is "At the instant that m1 hits the ground the string breaks. Find the time taken for m2 to come to rest". (All relevant values are provided in the attachment)
I know you have to start by calculating the the acceleration of m2 however I can't even really visualise what would happen and consequently don't know what to do next
Thanks for the help!! ;D
Then you find the total acceleration of the system, as a=Fnet/m. Therefore a=19.6/5= 3.92m s^-2
I don't know where you would go from here, beyond working out the time until m1 hits the ground. Is that all the information given in the question?
Post by: KiNSKi01 on March 01, 2018, 09:19:41 pm
Post by: Bri MT on March 01, 2018, 09:30:48 pm
Hey can I have some assistance with this question?
The question is "At the instant that m1 hits the ground the string breaks. Find the time taken for m2 to come to rest". (All relevant values are provided in the attachment)
I know you have to start by calculating the the acceleration of m2 however I can't even really visualise what would happen and consequently don't know what to do next
Thanks for the help!! ;D
How I would approach this is by splitting it into 2 stages.
1. M1 is falling
2.M2 is falling
1.
- consider the force acting to make m1 fall and the force resisting this motion
- use F = ma
- find the time taken for M1 to hit the ground
- find the displacement of M2
2.
- find the acceleration of M2 in the relevant direction
- find the time taken for M2 to reach the end of the ramp
I haven't stepped this through, but I hope this helps :)
I'd draw a seperate diagram for each stage to assist
Post by: KiNSKi01 on March 01, 2018, 09:39:14 pm
Post by: skrt skrt on March 09, 2018, 08:38:35 pm
A stone is thrown from the top of a 15m high cliff above the sea at an angle of 30degrees.
Calculate the time taken for the stone to reach the sea. (answer is 3 seconds)
I first found the initial vertical velocity which was 10m/s and was hoping to find the find final velocity using v^2=u^2+2as then substitute my values into v=u+at. However I keep getting a negative number to find the final velocity which I cant square root.
I substituted u=10
a=-9.8
s=15
Please tell me what I'm doing wrong.
Thank you in advance
Post by: blasonduo on March 09, 2018, 08:49:55 pm
Hey people just need help with this VCAA problem.
A stone is thrown from the top of a 15m high cliff above the sea at an angle of 30degrees.
Calculate the time taken for the stone to reach the sea. (answer is 3 seconds)
I first found the initial vertical velocity which was 10m/s and was hoping to find the find final velocity using v^2=u^2+2as then substitute my values into v=u+at. However I keep getting a negative number to find the final velocity which I cant square root.
I substituted u=10
a=-9.8
s=15
Please tell me what I'm doing wrong.
Thank you in advance
Hey!
Remember that s is the change in y, and you have set the positive direction to be upwards, so your final displacement (in the y-direction) should be negative! (-15) This gives a positive number :)
EDIT How did you figure out the initial velocity was 10? From the values, it is impossible to figure this out with just acceleration and the height of the cliff only. (the angle would help, but there have been no given values of its velocity, it could take as long as it wants to hit the sea depending on its initial velocity.
Post by: skrt skrt on March 09, 2018, 10:18:12 pm
Thanks for the reply, makes things clear
Post by: Khadijah on March 13, 2018, 09:59:53 pm
Question 13
In the fusion process, a proton of rest mass 1.673 × 10–27 kg and a neutron of rest mass 1.675 × 10–27 kg combine
to form a deuterium nucleus of rest mass 3.344 × 10–27 kg, with a release of energy.
According to Einstein’s postulate of the equivalence of mass and energy, which one of the following is the best
estimate of the energy released in this interaction?
A. 1.2 × 10–21 J
B. 3.6 × 10–13J
C. 4.0 × 10–3 J
D. 3.6 × 1014 J
Post by: Bri MT on March 18, 2018, 09:48:13 am
Could someone help in this question-
Question 13
In the fusion process, a proton of rest mass 1.673 × 10–27 kg and a neutron of rest mass 1.675 × 10–27 kg combine
to form a deuterium nucleus of rest mass 3.344 × 10–27 kg, with a release of energy.
According to Einstein’s postulate of the equivalence of mass and energy, which one of the following is the best
estimate of the energy released in this interaction?
A. 1.2 × 10–21 J
B. 3.6 × 10–13J
C. 4.0 × 10–3 J
D. 3.6 × 1014 J
All you need to do here is find the change in mass and plug that into E=mc^2.
(Proton rest mass) + (neutron rest mass) - (deuterium rest mass) = m
c= 3*10^8
Let me know if you're still struggling after this :)
Post by: skrt skrt on March 18, 2018, 11:53:25 am
A ball of mass of 2.9kg swings on a string length of 2.2m and moves at a speed of 4.4m/s in a horizontal circular path.
Find the angle.
Thanks for the help.
(Btw you can find this question on http://www.physicseducation.com.au, really good website)
Post by: KiNSKi01 on March 18, 2018, 01:17:16 pm
All you have to do is substitute the values into this equation and rearrange for theta:
tan(theta)= v^2 / rg . In this question the mass is irrelevant ;D
You get an answer of 41.92 degrees! (please correct me if the answer is wrong because I couldn't find the solution in the website)
Post by: skrt skrt on March 18, 2018, 01:26:14 pm
Post by: KiNSKi01 on March 18, 2018, 02:33:04 pm
Post by: jazzycab on March 18, 2018, 11:04:28 pm
Hey guys, I seem to be struggling be with conical pendulum question.I actually think the problem is significantly more complicated than that, as we don't know the radius (the length of the pendulum is given).
A ball of mass of 2.9kg swings on a string length of 2.2m and moves at a speed of 4.4m/s in a horizontal circular path.
Find the angle.
Thanks for the help.
(Btw you can find this question on http://www.physicseducation.com.au, really good website)Sup skrt skrt!
All you have to do is substitute the values into this equation and rearrange for theta:
tan(theta)= v^2 / rg . In this question the mass is irrelevant ;D
You get an answer of 41.92 degrees! (please correct me if the answer is wrong because I couldn't find the solution in the website)
There are two forces acting on the ball: its weight and the tension up the string. Given that the path of the ball is horizontal circular motion, the ball is in equilibrium vertically. That is, the vertical component of the tension must be equal in magnitude to the weight force. In addition, the centripetal force is a result of the horizontal component of the tension force (see diagram below).
(https://i.imgur.com/U6kdp2W.png)
My solution contains some Specialist Maths ideas:
Because \(4g^2l^2>0\), we know that \(v^4+4g^2l^2>v^4\), which means that \(\sqrt{v^4}=v^2<\sqrt{v^4+4g^2l^2}\).
Additionally, because \(\theta\in\left[0,\frac{\pi}{2}\right]\), we also know that \(\cos{\left(\theta\right)}\in\left[0,1\right]\). That is, \(\cos{\left(\theta\right)}\) is positive.
This means that:
Finally, substituting in \(m=2.9, l=2.2, v=4.4\) and \(g=9.8\) gives \(\theta\approx 49.67^\circ\).
Post by: skrt skrt on March 19, 2018, 07:57:53 am
Post by: jazzycab on March 19, 2018, 09:06:42 am
umm, thanks for the response but how are we meant to do all that in exam conditions haha?
Given the level of maths required for it, I would suggest it wouldn't be assessable, especially given that there is no requirement that you complete any maths in VCE in order to do Physics (although I would personally recommend that you do Methods at least). You might get a similar, but simpler problem, where the radius is known instead, however (which you would solve by substituting straight into \(\tan{\left(\theta\right)}=\frac{v^2}{gr}\)).
Post by: Shadowxo on March 19, 2018, 05:26:27 pm
SnipIt's been a while since I've done physics and am very rusty with these formulas in particular, but wouldn't it be possible to do KiNSKi's way but substitute r=2.2sin(theta)?
Post by: jazzycab on March 19, 2018, 08:05:01 pm
It's been a while since I've done physics and am very rusty with these formulas in particular, but wouldn't it be possible to do KiNSKi's way but substitute r=2.2sin(theta)?
Yes, that's effectively what I did. The problem you run into though, is having sin and tan in the same equation. To solve that (without CAS) you need to use trig identities (which are certainly not in the Physics course)
Post by: Shadowxo on March 19, 2018, 09:07:20 pm
Yes, that's effectively what I did. The problem you run into though, is having sin and tan in the same equation. To solve that (without CAS) you need to use trig identities (which are certainly not in the Physics course)Ah, I see. I missed it when I was reading through.
skrt, what I would do to make it a bit easier is substitute in the values earlier as it makes it a bit easier. I'd substitute in around
Then you can continue by completing the square (as jazzycab has done) or by using the quadratic formula.
Sometimes you'll want to substitute in later as things can cancel but in this case it's a lot easier to just substitute in as you're left with a quadratic equation.
Usually you won't be given a question like this though.
Post by: skrt skrt on March 20, 2018, 05:21:08 pm
Thanks
Post by: Shadowxo on March 20, 2018, 10:05:51 pm
Hey people :), just for question 2b for the 2014 physics exam I'm not understanding the numbers that are being inputted into Eg=Es. If someone could explain where their getting the numbers from that would be great.There are four 50g masses on it = 200g = 0.2kg
Thanks
They used g=10 m/s^2
Change in height = max extension = h
k was found to be 5.0 N/m
Hope this helps, let me know if you're still confused :)
Post by: skrt skrt on March 21, 2018, 06:08:20 pm
Thanks :D
Post by: skrt skrt on March 22, 2018, 05:16:55 pm
Post by: KiNSKi01 on March 22, 2018, 05:55:58 pm
Post by: melb1234 on April 01, 2018, 10:19:48 am
a. Calculate time for sound to travel from A to B.
b. Determine the time for sound to travel from A to C and back to A.
c. Calculate the time difference, Δt, for the two returning sounds.
i'm not sure how to do calculations involving wind, how does the wind affect person B in relation to person in in part a)
thanks in advance!
Post by: skrt skrt on April 11, 2018, 11:11:21 pm
I've kinda modified the question myself. (Q3 b 2014)
So if they gave you:
h=3.7m
vx=17.32
uy=10
angle=30 deg
(Gotta look at the question itself or its pretty confusing :P)
How would you find the total time?
(I used s=ut+0.5at^2, then the quadratic formula but i keep getting the wrong number)
Post by: VinnyD on April 11, 2018, 11:39:02 pm
Need help with this projectile motion question.
I've kinda modified the question myself. (Q3 b 2014)
So if they gave you:
h=3.7m
vx=17.32
uy=10
angle=30 deg
(Gotta look at the question itself or its pretty confusing :P)
How would you find the total time?
(I used s=ut+0.5at^2, then the quadratic formula but i keep getting the wrong number)
You would have to use the initial vertical speed as U and v=0 at the top (talking about the motion vertically). Then find the time to get to v=0 and double it.
You can find the initial vertical speed by using trig
Post by: skrt skrt on April 12, 2018, 09:20:44 am
So you wouldn't double the time.
Post by: KiNSKi01 on April 12, 2018, 06:20:09 pm
"An electron moves due North in a horizontal plane with uniform speed. It enters a uniform field directed due South in the same plane. Which one of the following statements concerning the motion of the electron in the magnetic field is correct?'
Answer= It continues to move North with its original speed
Help please :-\
Post by: Bri MT on April 13, 2018, 05:41:42 pm
I have no clue what this question is on about or how to visualise it:
"An electron moves due North in a horizontal plane with uniform speed. It enters a uniform field directed due South in the same plane. Which one of the following statements concerning the motion of the electron in the magnetic field is correct?'
Answer= It continues to move North with its original speed
Help please :-\
Start out by drawing a diagram. Mark "up" on your page as north and draw the electron with an arrow pointing upwards. Then draw the field lines running south.
Notice that the electron is moving parallel to the field lines; therefore the field lines will have no impact on the electrons motion
Post by: KiNSKi01 on April 13, 2018, 08:28:54 pm
got confused by the wording of the question
Post by: DrDin on April 23, 2018, 11:10:01 pm
Post by: Yertle the Turtle on April 24, 2018, 07:52:14 am
Hello everyone! I have a special relativity question that I can't wrap my head around. Particularly part d). An answer with an explanation will be much appreciated. Thanks!Hey there, welcome to ATAR Notes! :)
The speed of sound varies depending on your speed relative to it. Therefore in these questions you would find your speed relative to the source of the sound and then add/subtract from the speed of sound.
Hope this helps :)
Post by: DrDin on April 24, 2018, 05:10:14 pm
Hey there, welcome to ATAR Notes! :)Thanks for the welcome!
The speed of sound varies depending on your speed relative to it. Therefore in these questions you would find your speed relative to the source of the sound and then add/subtract from the speed of sound.
Hope this helps :)
Yep that makes sense. But the for the last one I don't get the correct answer (I may be misinterpreting the question). The answer is apparently 340ms^-1. I don't understand why the speed relative to sound in this case is 0.
Post by: Shadowxo on April 24, 2018, 08:03:37 pm
Thanks for the welcome!I think it's because you're about to overtake the fire truck, meaning you're right beside it. The sound waves you hear will be travelling perpendicular to your motion rather than from in front / behind you. This means for this question your speeds don't matter because your left/right velocity is 0.
Yep that makes sense. But the for the last one I don't get the correct answer (I may be misinterpreting the question). The answer is apparently 340ms^-1. I don't understand why the speed relative to sound in this case is 0.
Hope that makes sense haha
Post by: Richard Feynman 101 on April 24, 2018, 08:51:17 pm
Answers are incorrect.
Post by: DrDin on April 25, 2018, 08:55:58 pm
Know the authors of the book.Okay? Do you know the correct answer?
Answers are incorrect.
I think it's because you're about to overtake the fire truck, meaning you're right beside it. The sound waves you hear will be travelling perpendicular to your motion rather than from in front / behind you. This means for this question your speeds don't matter because your left/right velocity is 0.Your explanation makes a lot of sense. But now I'm confused as the Richard Feynman 101 is saying the answers are incorrect D:
Hope that makes sense haha
Post by: Seno72 on April 26, 2018, 11:40:29 pm
Post by: skrt skrt on May 06, 2018, 04:44:51 pm
thanks
Post by: Seno72 on May 07, 2018, 07:15:25 am
Hey people, could someone please explain why Fg=Fc. This reminds of Es=Ek for spring questions(kinetic energy converts to spring potential and vice versa). However I don't see the connection between Fg=Fc, isn't Fc the sum of all force hence its not a real force?
thanks
I think you mean Fnet=Fc (Centripetal force)? Fnet and Fc are the vector some of multiple forces, sometimes including the Fg.
Post by: Shadowxo on May 09, 2018, 02:57:10 pm
Hey people, could someone please explain why Fg=Fc. This reminds of Es=Ek for spring questions(kinetic energy converts to spring potential and vice versa). However I don't see the connection between Fg=Fc, isn't Fc the sum of all force hence its not a real force?In general this isn't true, but it can be the case for things orbiting the earth, or if something's swinging vertically and is at the top of its swing. What situation is it referring to?
thanks
Post by: lzxnl on May 09, 2018, 07:06:10 pm
Post by: skrt skrt on May 09, 2018, 07:30:11 pm
I was just wondering why the two equate each other, sorry for the late reply :)
Post by: lzxnl on May 09, 2018, 11:10:47 pm
(GMm/r^2)=(mv^2/r), this is the equation used to find Kepler's laws for natural or artificial satellites,
I was just wondering why the two equate each other, sorry for the late reply :)
One side is the magnitude of the gravitational force, which is directed inwards. The other side is the centripetal force requirement for circular motion. As we're assuming circular orbits, the condition for moving in such an orbit is Fnet = mv^2/r, and this force must be directed inwards. The gravitational force must then meet this condition in direction and magnitude. Otherwise you don't move in that orbit at that speed or orbital radius.
Post by: skrt skrt on May 10, 2018, 07:55:45 am
Post by: Seno72 on May 15, 2018, 08:12:13 am
Post by: sweetiepi on May 15, 2018, 08:20:29 am
Hey guys, what's the purpose of slip rings in an alternator. And also, for every half turn, do the slip rings reverse their polarity?The purpose of slip rings is to maintain a continuous connection between the wire and armature, in comparison to a split-ring commutator, which does not.
Slip rings do in fact reverse the direction of the current halfway.
Post by: Seno72 on May 15, 2018, 09:29:41 am
The purpose of slip rings is to maintain a continuous connection between the wire and armature, in comparison to a split-ring commutator, which does not.
Slip rings do in fact reverse the direction of the current halfway.
Thanks
Post by: Seno72 on May 17, 2018, 01:18:54 pm
Post by: KiNSKi01 on May 17, 2018, 06:36:52 pm
Hey guys,
I'm not sure if this is the right place to ask. I'm thinking of picking up Physics Unit 2 without Unit 1. Is this manageable?
In year 10, my school taught general science which included topics of motion and electricity. Is there anything new taught in Unit 1 Physics that I would need to catch up on? Any tips?
Thanks! :D
Eh, you would have to do some catching up if your school is going to do exams on the whole 1/2 course content; however, in saying that considering you already have knowledge in motion and electricity (two topics relevant to 3/4) what you've missed out in unit one isn't that important. So i would say if you want to do physics and you think you can do well definitely pick it up now rather than later
Post by: Guideme on May 19, 2018, 03:28:49 pm
Can anyone help me with 39 a and b thank you in advance
Post by: IonianDeo on May 22, 2018, 04:45:47 pm
Post by: Bri MT on May 22, 2018, 05:35:34 pm
How would you go about approaching this question?
Ignore the piece of vertical wire and just use the right hand rule for the solenoid
Post by: IonianDeo on May 22, 2018, 05:48:32 pm
Post by: Floatzel98 on May 22, 2018, 07:13:06 pm
Why is it that at the end of parallel oppositely charged plates that the electric field lines begin to bulge outwards? What causes the bulging?Think about all the individual point charges that make up the plates. In the center of the plates, far away from the edges, the electric field looks uniform. If we think of the electric field of a two point charges together, they cancel (or deflect) in the direction connecting them (see image). If you put heaps of these next to each other, the next result is an electric field perpendicular to the line of charges (perpendicular to the plate). So in the center of the plate, we see a uniform electric field. Now consider the very last charge, at the very edge of the plate. There is no charge adjacent to this one and so there is only cancellation of the electric field on one side. The resultant electric field is a sum of a vector perpendicular to the plate and one 'diagonally' away (i.e the electric field of a point charge). Combined with the same effect on the oppositely charge plate, this gives the bulge effect in the electric field at the edges of the plate.
Hopefully that helps :)
Post by: IonianDeo on May 22, 2018, 07:26:20 pm
Post by: sweetiepi on May 22, 2018, 07:49:31 pm
In gravitational, electric and magnetic fields, what is meant by a uniform field and a static field? What are the differences?Not sure about static fields but here's what I remember about uniform fields:
A uniform gravitational field means that there is no gravitational effect at all when in free fall.
A uniform magnetic field is when magnetic induction has the same magnitude and the same direction at all the points in the region.
A uniform electric field is when there are two straight parallel electrodes or plates that have a potential difference between them.
Post by: Floatzel98 on May 22, 2018, 08:00:44 pm
In gravitational, electric and magnetic fields, what is meant by a uniform field and a static field? What are the differences?An example of a uniform and static magnetic field would be,
This field doesn't change in direction or intensity at any point, and it doesn't change in time. It is just a constant value in one direction for all times. The electric field of a charged parallel plate capacitor is uniform and static. But while it is being charged, it is no longer static, as the magnitude of the field (current) is changing in time.
Post by: IonianDeo on May 22, 2018, 08:59:47 pm
How do we know what part of a field line passes through the points P and Q from the solenoid?
Post by: Bri MT on May 23, 2018, 08:02:31 am
I am confused about 'Earth has a horizontal magnetic field'. What does it mean?
How do we know what part of a field line passes through the points P and Q from the solenoid?
1. The solenoid is positioned relative to the Earth's surface in a manner such that the Earth's magnetic field goes left to right or right to left across the page
2. Use the right hand rule to draw the magnetic field produced by the solenoid.
You'll know which way the current is travelling through the solenoid because of the battery
Post by: Seno72 on June 04, 2018, 06:02:07 pm
Post by: skrt skrt on June 06, 2018, 04:49:11 pm
So if we have an AC power supply connected to a step up transformer which the voltage is transferred through transmission lines and then to a step down transformer, which values does the formula Vdelivered=Vsupply-Vdrop equate to. Using the notion that V1 and V2 are used for the first transformer and V3 and V4 is for the second transformer.
(2015 Q16 figure 2 for a visual)
What is the function of slip rings in the AC generator?
(looked at definitions didn't really get it)
Thanks
Post by: S200 on June 06, 2018, 06:20:13 pm
Sorry, can't really help ya on the first part...
What is the function of slip rings in the AC generator?
(looked at definitions didn't really get it)
But slip rings just ensure that the current out stays as AC.
If you use split ring rather than slip rings, you would end up with DC.
Post by: k.chen on June 06, 2018, 06:24:59 pm
Post by: S200 on June 06, 2018, 06:28:20 pm
Just a quick question how do you work out maximum height, final vertical and horizontal velocity at landing of a asymmetric projectile motion when only provided with time and distanceCould you post the question K.Chen?
I would assume using one of Newtons formulae... Something with S (displacement) and T (Time) in it...
Post by: skrt skrt on June 06, 2018, 06:29:18 pm
Post by: S200 on June 06, 2018, 06:30:34 pm
pm me question and i could help :)Just a bit of general knowledge...
You can't attach files to PM's.
Post by: skrt skrt on June 06, 2018, 06:34:10 pm
Post by: banksduyvu on June 09, 2018, 06:17:48 pm
I don't quite understand why violet light refracts more than red light. Can someone please explain this phenomenon to me? Greatly appreciated !
Post by: Floatzel98 on June 09, 2018, 07:00:07 pm
UNIT 4 - LIGHTFor these cases, the refractive index (and phase velocity) is dependent on the wavelength/frequency of light. So each frequency will have a different refractive index for a given material. From Snell's law we can then see that different wavelengths will refract by different angles
I don't quite understand why violet light refracts more than red light. Can someone please explain this phenomenon to me? Greatly appreciated !
Post by: skrt skrt on June 19, 2018, 07:16:17 pm
Thanks
Post by: S200 on June 19, 2018, 08:12:27 pm
Quick question, does anyone know the best way to wrap cooper wire around a tube?? (for my practical)I don't know the best way, but you could prolly jerry-rig the tube to a drill somehow?
Thanks
Then, you can just hook one end of the wire to the tube, and then just hold the remainder firmly enough to give it some tension and wrap-direction...
Just an idea...
Post by: layalasmith on June 24, 2018, 01:22:22 pm
Post by: G-Fr3sh on June 24, 2018, 02:46:07 pm
What is the reaction force to weight force acting on object in free-fall? What would the the two reaction-action pairs be in relation to Newton's third law?
I don't think that there is reaction force for an object in free fall due to no object pushing the object in the reverse direction **aside air resistance which is ignored in vce physics**. I could be wrong, try to email your teacher
Post by: sprout on June 24, 2018, 03:11:45 pm
I don't think that there is reaction force for an object in free fall due to no object pushing the object in the reverse direction **aside air resistance which is ignored in vce physics**. I could be wrong, try to email your teacher
^ There's no normal force for objects that are free falling (if there was, the object wouldn't be accelerating at 9.8m/s^2!)
Post by: layalasmith on June 24, 2018, 06:06:50 pm
Post by: S200 on June 24, 2018, 06:11:40 pm
Post by: layalasmith on June 24, 2018, 07:15:04 pm
Unless you are referring to when the ball hits the earth, there is no reaction force
So if I understand this correctly, when the ball is in air, there is no reaction force?
Post by: sprout on June 24, 2018, 07:28:53 pm
So if I understand this correctly, when the ball is in air, there is no reaction force?There's no normal force.
I think you might be getting confused by newton's law (forgot which one) that states for every force there's an equal and opposite one. That's the force on the earth by the ball. In questions where the object is against a table etc, the normal force is this equal and opposite force, so that might be confusing you.
Post by: S200 on June 24, 2018, 07:53:13 pm
So if I understand this correctly, when the ball is in air, there is no reaction force?Indeed. The only possible opposing force is air resistance, which IS ignored in VCE Physics.
Post by: skrt skrt on June 24, 2018, 09:21:19 pm
Hope that answers your question
Post by: layalasmith on June 25, 2018, 06:24:17 pm
Post by: IonianDeo on June 30, 2018, 05:26:07 pm
Post by: Bri MT on June 30, 2018, 05:53:24 pm
Why is option A correct instead of B?
Force = current * magnetic field strength * length of the wire
As the armature rotates T/4 none of these quantities change until at T/4 when the wires are in the gap of the split ring communtator
When the armature rotates from T/4 to 3T/4, the current is running through the section of wire AB in the opposite direction
(- Force = - current * magnetic field strength * length of wire)
Hope this helps :)
Post by: IonianDeo on June 30, 2018, 07:14:34 pm
Force = current * magnetic field strength * length of the wire
As the armature rotates T/4 none of these quantities change until at T/4 when the wires are in the gap of the split ring communtator
When the armature rotates from T/4 to 3T/4, the current is running through the section of wire AB in the opposite direction
(- Force = - current * magnetic field strength * length of wire)
Hope this helps :)
Thanks, it helps a lot.
So if the y-axis value was recording torque on the coil instead, would Graph B be the most appropriate answer option or is it completely different?
Post by: Bri MT on June 30, 2018, 08:25:26 pm
Thanks, it helps a lot.
So if the y-axis value was recording torque on the coil instead, would Graph B be the most appropriate answer option or is it completely different?
torque = F perpendicular r
We've just discussed how F is behaving (& we know it's always vertical in this case), so let's look at r
r isn't AB - the armature is rotating around the dotted line and r would be .5*BC or equivalently .5*AD
the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal
So lets define theta as the angle between the armature and the horizontal plane
(if this explanation didn't make sense let me know)
In this case, torque = Force * .5*BC*cos(theta)
As you can see, this is best represented by the graph B
Post by: IonianDeo on July 01, 2018, 07:46:01 pm
torque = F perpendicular r
We've just discussed how F is behaving (& we know it's always vertical in this case), so let's look at r
r isn't AB - the armature is rotating around the dotted line and r would be .5*BC or equivalently .5*AD
the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal
So lets define theta as the angle between the armature and the horizontal plane
(if this explanation didn't make sense let me know)
In this case, torque = Force * .5*BC*cos(theta)
As you can see, this is best represented by the graph B
I didn't really understand what you meant by this:
"the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal"
Post by: Bri MT on July 02, 2018, 04:48:11 pm
I didn't really understand what you meant by this:
"the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal"
By the right hand rule (slap version), we know that the force acting on the wire is acting in a vertical direction.
When calculating torque, we are interested in the component of r perpendicular to F (or equivalently the component of F perpendicular to r)
Perpendicular refers to being 90 degrees away, and anything is the horizontal direction is perpendicular to anything in the vertical direction (& vice versa)
Post by: Seno72 on July 23, 2018, 06:45:29 pm
Indeed. The only possible opposing force is air resistance, which IS ignored in VCE Physics.
I think you need to know the effects of air resistance on a projectile, but you are right in that we don't need to know about air resistance in calculations.
Post by: Yertle the Turtle on July 23, 2018, 08:08:32 pm
I think you need to know the effects of air resistance on a projectile, but you are right in that we don't need to know about air resistance in calculations.Unless you are doing the Sport Module in U2 Physics, which covers drag coefficient and air resistance etc. Otherwise, no, it is just a variable that occurs in real life but is discounted in VCE.
Post by: geek123456 on July 31, 2018, 07:10:17 pm
I was wondering if anyone could help me solve the following two questions attached.
Cheers
Post by: Bri MT on July 31, 2018, 07:58:33 pm
Hi,
I was wondering if anyone could help me solve the following two questions attached.
Cheers
For the second one, split it into 3 segments:
1) acceleration from rest to 12 (find distance and time)
2) acceleration from 12 to rest (find distance and time)
3) left over distance and time associated with travelling at 12 m/s (use distance to find time)
Does that help?
Post by: geek123456 on July 31, 2018, 08:23:06 pm
For the second one, split it into 3 segments:thank you!
1) acceleration from rest to 12 (find distance and time)
2) acceleration from 12 to rest (find distance and time)
3) left over distance and time associated with travelling at 12 m/s (use distance to find time)
Does that help?
Post by: banksduyvu on August 07, 2018, 10:31:17 pm
Suppose Anna is moving towards a space station (that is not moving relative to an observer on Earth). When she is half way between Earth and the space station, she sends an electromagnetic pulse out. I don't understand how the signal reaches the space station first, when at the instant the light signal is sent out, she is exactly half way between Earth and the spacelab.
Special relativity kills me ~ . ~ Thank yoou !
Post by: Bri MT on August 08, 2018, 08:38:37 am
I need help with special relativity - simultaneity
Suppose Anna is moving towards a space station (that is not moving relative to an observer on Earth). When she is half way between Earth and the space station, she sends an electromagnetic pulse out. I don't understand how the signal reaches the space station first, when at the instant the light signal is sent out, she is exactly half way between Earth and the spacelab.
Special relativity kills me ~ . ~ Thank yoou !
The pulse moves at a speed relative to her. Let's say it moves at a speed of P, and Anna is moving at a speed of A.
Relative to Earth, the pulse is moving forwards at P + A
Relative to Earth, the pulse is m moving backwards at P - A.
It's like if you were on a train and threw a ball upwards you wouldn't need to run backwards as fast as the train is going to catch it, because the ball also has the train's velocity, just like you.
Post by: S200 on August 21, 2018, 05:48:00 am
Can anyone help with part B of this checkpoint question? (Question 12 of Chapter 6)
The answers say 0.31N, but I consistently get -0.09N.
I am using the formula \(N+mg=\frac{mv^2} {r}\), which is what the answer's say to use, but I still can't get the 0.31
Any help would be appreciated...
(and sorry for terrible pic quality)
Post by: Seno72 on August 21, 2018, 08:04:21 am
Hey,
Can anyone help with part B of this checkpoint question? (Question 12 of Chapter 6)
The answers say 0.31N, but I consistently get -0.09N.
I am using the formula \(N+mg=\frac{mv^2} {r}\), which is what the answer's say to use, but I still can't get the 0.31
Any help would be appreciated...
(and sorry for terrible pic quality)
Fnet= mv^2/r
Fg= mg
R = ? (reaction force)
We know that both the reaction and gravity force is facing downwards so for vector calculations, we will take downwards as positive, so:
Fnet= mg + R (take R to the subject side)
R = Fnet - mg
R = mv^2/r - mg
R = (0.05 x 2^2)/(0.25) - (9.8 x 0.05)
R = 0.31 N downwards.
This is what I got. You had the right equation, now you just need to transpose so to make R the subject. Then put in the correct values inside. (Make sure it's 0.25 metres not 0.50 metres as it's radius and make sure that 50/1000 equals 0.05) Maybe you made a mistake up there. Hope it helps:
Post by: S200 on August 21, 2018, 08:07:45 am
Thanks heaps.
Post by: Seno72 on August 26, 2018, 08:17:31 pm
Post by: S200 on August 26, 2018, 08:18:43 pm
Post by: skrt skrt on September 13, 2018, 08:11:55 am
Thanks
Post by: Bri MT on September 14, 2018, 10:38:19 am
Hey AN, really struggling to understand how electron's actually form standing waves, do they move around the nucleus like a standing wave or is it just one of those concepts where you can't imagine it abstractly.
Thanks
I wouldn't try to visualise the electron actually moving around in a wave shape (an definitely not in a neat circular orbit) - instead I think about it as the wave telling you about the probability of the electron being there.
Post by: VinnyD on September 26, 2018, 09:50:46 pm
what does the study design point mean "compare alternating voltage expressed as the root-mean-square (rms) to a constant DC voltage developingI'm pretty sure its just using the root mean square value (root 2 formula) to just convert AC into a constant output. Kind of making AC into DC. (ac has peaks higher than rms value, so the rms value gives us the average value for the ac current that's lower than the peak.) Hope this helps
the same power in a resistive component"? i cant find much information regarding this point in the book, can anyone help?
Post by: Yertle the Turtle on September 26, 2018, 10:14:47 pm
what does the study design point mean "compare alternating voltage expressed as the root-mean-square (rms) to a constant DC voltage developingJust to add to what VinnyD said, the RMS voltage is the DC equivalent of the AC voltage, being the average value of the sine curve. If you think about the average value of a sine wave it is at pi/4 radians, which is 1/sqrt(2), so to find RMS voltage from peak to peak voltage you multiply by 1/sqrt(2). Hope this helps. :D
the same power in a resistive component"? i cant find much information regarding this point in the book, can anyone help?
Post by: Ghost_ on October 03, 2018, 08:10:50 pm
In VCAA's answers for the 2017 VCE Physics exam (Q9c), they showed a graph that looked like a connect the dots instead of a smooth parabolic curve. Why didn't they sketch a smooth curve of best fit instead of just connecting the dots?
-Thanks
Post by: minhalgill on October 04, 2018, 10:30:24 pm
Post by: S200 on October 05, 2018, 02:15:52 am
Ans also always 3 sig fig, plus scientific notation if needed.
Post by: Yertle the Turtle on October 05, 2018, 02:28:43 pm
do we take the value of g to be 9.8 or 10??? and how many significant figures should our answers be to?As S200 has said, g is always 9.8. However, for the s.f question, you always use the part of the question with the fewest significant figures as your base. If the part of the question with the least precision has 3 s.f. then your answer should be to 3 s.f. You just always use the minimum level of accuracy that the question uses.
Post by: S200 on October 05, 2018, 03:23:06 pm
However, for the s.f question, you always use the part of the question with the fewest significant figures as your base. If the part of the question with the least precision has 3 s.f. then your answer should be to 3 s.f. You just always use the minimum level of accuracy that the question uses.What do you do when the question has min 4 sig fig?
Do you go four, or three?
Post by: Yertle the Turtle on October 05, 2018, 03:37:31 pm
What do you do when the question has min 4 sig fig?In that case you would go 4 sig figs. You always have to follow the question in what answers you give.
Do you go four, or three?
Post by: S200 on October 05, 2018, 03:50:12 pm
In that case you would go 4 sig figs. You always have to follow the question in what answers you give.Hmmm... :-\
Dat means all my answers are wrong then... :'(
I always go to 2 sig figs after the decimal (so three overall), and only at the very very end of all the working out...
Like, if I have \(12.3476 \times 15.9347\), which equals \(196.755302\), I would answer that as \(196.76\)
For larger numbers, like \(10456.789 \times 164.9876\), which equals \(1725240.52082\), I would answer \(1.73\times 10^6\).
How would you answer those two?
Post by: Bri MT on October 05, 2018, 07:42:29 pm
Hmmm... :-\
Dat means all my answers are wrong then... :'(
I always go to 2 sig figs after the decimal (so three overall), and only at the very very end of all the working out...
Like, if I have \(12.3476 \times 15.9347\), which equals \(196.755302\), I would answer that as \(196.76\)
For larger numbers, like \(10456.789 \times 164.9876\), which equals \(1725240.52082\), I would answer \(1.73\times 10^6\).
How would you answer those two?
I would answer as 196.755 and 1.725241 * 10^6 respectively.
Especially if this was a data analysis type question, I would not consider 3 sig figs as a standard "rule"
Post by: S200 on October 05, 2018, 07:52:23 pm
I would answer as 196.755 and 1.725241 * 10^6 respectively.Yeah ok.
Especially if this was a data analysis type question, I would not consider 3 sig figs as a standard "rule"
But doing 1.725241 * 10^6 seems to kinda defeat the purpose of scientific notation. You may as well just write it out in full. :-\
Post by: sweetiepi on October 05, 2018, 08:02:50 pm
Yeah ok.Scientific notation will always be the preferred format for something like that, I got stung a quite a few times for not writing out big/small numbers in scientific notation, even though it feels counter-intuitive :) (I did that recently on a lab report and lost marks I shouldn't have otherwise. #salty)
But doing 1.725241 * 10^6 seems to kinda defeat the purpose of scientific notation. You may as well just write it out in full. :-\
I would also not use the 3-sig figs rule as a rule, answers should typically depend on the sig figs on the values given in the question stem. :)
Post by: Ghost_ on October 06, 2018, 11:16:40 pm
-Thanks
Post by: KiNSKi01 on October 11, 2018, 09:05:17 pm
A steel bar is moved at a constant speed of v m/s perpendicular to magnetic field (directed into page). The steel bar is 0.8 m long. The magnetic field strength is 8 x 10^-3 T.
If the emf between the ends of the steel bar is 30 mV what is the constant speed?
I'm not sure how I can use the extra info regarding voltage to help calculate answer
Post by: VinnyD on October 11, 2018, 09:28:32 pm
YOO. What am I missing here? :/In this case i think its the special formula for the emf (E=LVB) where L= length, v=velocity and B= mag field. Very specific question though. Just rearrange for V as we are given everything else. Let me know if it's something different
A steel bar is moved at a constant speed of v m/s perpendicular to magnetic field (directed into page). The steel bar is 0.8 m long. The magnetic field strength is 8 x 10^-3 T.
If the emf between the ends of the steel bar is 30 mV what is the constant speed?
I'm not sure how I can use the extra info regarding voltage to help calculate answer
Post by: KiNSKi01 on October 13, 2018, 01:20:51 pm
Didn't know how to derive an equation from equations on formula sheet which would actually help :P
Post by: minhalgill on October 13, 2018, 02:57:33 pm
Post by: minhalgill on October 13, 2018, 03:07:28 pm
Post by: KiNSKi01 on October 13, 2018, 03:13:21 pm
why cant we use momentum to show whether the collision was inelastic or elastic? why do we have to use KE?
Yo!
Whether a collision is inelastic or elastic depends on whether energy is conserved. Using momentum will not show whether a collision was inelastic or elastic. Although within a system momentum may be conserved (basically always in the case of VCE questions), this doesn't necessarily mean energy is conserved ;D (in most questions the collision will be inelastic) as energy is dissipated/loss by thermal, sound etc.
Hope that helped!
Post by: sweetiepi on October 13, 2018, 03:14:29 pm
how many significant figuires should we use in phyics?Typically the number of significant figures that is used in the question.
Pretty sure there was a discussion on this recently.
Post by: minhalgill on October 13, 2018, 05:29:37 pm
Yo!
Whether a collision is inelastic or elastic depends on whether energy is conserved. Using momentum will not show whether a collision was inelastic or elastic. Although within a system momentum may be conserved (basically always in the case of VCE questions), this doesn't necessarily mean energy is conserved ;D (in most questions the collision will be inelastic) as energy is dissipated/loss by thermal, sound etc.
Hope that helped!
this helped alot, but im just confused about how the momentum can be conserved yet energy is not?
are clipping voltages in the study design?
mod edit: merged posts.
Post by: Bri MT on October 14, 2018, 09:42:45 am
this helped alot, but im just confused about how the momentum can be conserved yet energy is not?
Momentum is m*v whereas kinetic energy is .5*m*v^2
If you double the mass and halve the speed you'll get 2m*.5v = mv (momentum is conserved) and .5*2m*(.5v)^2 = .25*m*v^2 (kinetic energy is not conserved)
An example of this as a question is: you have a trolley being pushed along the ground with mass m and speed v. An object of mass m is dropped into the trolley from directly above it. Find the new speed of the trolley if momentum is conserved (2 marks). Is the collision elastic? (3 marks)
Post by: minhalgill on October 14, 2018, 01:20:37 pm
Post by: KiNSKi01 on October 14, 2018, 03:59:59 pm
are inverting amplifiers in the study design?
not exactly sure what you are referring to :P
https://www.vcaa.vic.edu.au/Documents/vce/physics/PhysicsSD-2016.pdf
Have a read of this and it might answer a few of your questions ;)
Post by: minhalgill on October 14, 2018, 06:35:04 pm
Post by: Bri MT on October 14, 2018, 09:04:10 pm
can someone please help with this question? i get why the emf graph is constant, i just dont get why its negative?
Lenz's law / the fact that there's a negative in the emf equation (-N * (delta phi)/(delta t)
Post by: minhalgill on October 14, 2018, 11:09:16 pm
do we always draw this reflected ray of light as well as the refracted one?
Post by: S200 on October 14, 2018, 11:14:20 pm
It's something to do with being less than the critical angle...
#HighlyTechnicalAnswer
Post by: KiNSKi01 on October 18, 2018, 08:48:46 pm
Mostly I know how to answer Lenz's Law questions BUT i am confused by q12c on the 2017 NTH exam (photo of question is attached)
I understand the change in flux and that the the magnetic field produced by the current must oppose this change but I am confused about the direction the current flows. When using the RH grip rule must the field inside or outside the coil oppose the change in flux or am I just totally off the mark with this question!??
Help is much appreciated ;D
Post by: Yertle the Turtle on October 18, 2018, 09:24:35 pm
HellooooThe change in flux relates to the change in flux within the loop. Therefore you would use the right hand grip rule to show you that the induced emf will produce a current that will cause a magnetic field coming out of the page in 23a. Therefore the current will travel from X to Y.
Mostly I know how to answer Lenz's Law questions BUT i am confused by q12c on the 2017 NTH exam (photo of question is attached)
I understand the change in flux and that the the magnetic field produced by the current must oppose this change but I am confused about the direction the current flows. When using the RH grip rule must the field inside or outside the coil oppose the change in flux or am I just totally off the mark with this question!??
Help is much appreciated ;D
Post by: KiNSKi01 on October 18, 2018, 10:25:04 pm
Just found out the reason I had trouble was because the solution Vic Physics provided was highly confusing (as per usual)
Post by: minhalgill on October 20, 2018, 10:48:44 pm
just wondering when we use which version of plancks constant?
thanks
Post by: KiNSKi01 on October 20, 2018, 11:17:20 pm
Personally, I always use the 6.63 x 10^34 J and convert if necessary to eV when I arrive to my final answer. (just divide by charge of electron). In some cases this wastes a very tiny bit of time but at least I am more confident that I won't make a stupid mistake as I always consider which units the answer is required to be in
Post by: Bri MT on October 21, 2018, 08:55:29 am
hi,
just wondering when we use which version of plancks constant?
thanks
Just match up the units to what you're doing. Eg. If you're trying to find the answer in Joules use the joules one.
Post by: minhalgill on October 21, 2018, 10:00:33 am
Post by: sweetiepi on October 21, 2018, 10:02:47 am
are 'further electronics' and 'materials and their use in structure' in the study design?Materials and their use in their structure was part of the 2002-2016 study design.
Pretty sure the same goes for further electronics.
So no, not in current SD as far as I am aware. :)
Post by: minhalgill on October 21, 2018, 07:25:55 pm
in question 1 (b) (one with the wagon and the engine) we take into account the frictional backwards force, yet in the question 1 (b) with the locomotive and the two trucks) we dont take into account the frictional force. why is that? it doesnt state anywhere in the question that theres no frictional force
thanks
Post by: KiNSKi01 on October 21, 2018, 07:36:27 pm
All you have to do for this question is consider the whole system and do F=ma
F= 1 x 9.8 N (gravitational force) *tension is effectively zero if u consider the whole system
m= 1+ 4 kg
a= ?
Then you just solve for a ;)
It is important you clearly show in your working whether you are considering the whole system, m1 or m2
Post by: minhalgill on October 21, 2018, 07:55:35 pm
In another question in the marking scheme, theyve said there is tension and the gravitational weight force acting on the ball, yet they havent drawn tension in, so are we supposed to draw it in or not?
Post by: KiNSKi01 on October 21, 2018, 08:41:03 pm
Post by: minhalgill on October 22, 2018, 06:55:03 pm
thanks
Post by: Bri MT on October 23, 2018, 08:40:20 am
when finding the tension in T2 in part (b) why do we take into account the forces acting only on the first and second logs, why dont we take into account the driving force of the truck?
In another question in the marking scheme, theyve said there is tension and the gravitational weight force acting on the ball, yet they havent drawn tension in, so are we supposed to draw it in or not?
If I take an elastic band (nothing attached) and drag it through the air it will have no tension. If I then attach a weight to one end and hold that end to drag it through the air there will still be no tension. If I drag it through the air holding the side NOT attached to the weight there will tension and my band might snap.
The point isn't how fast the truck is going, the point is that the bit of rope attached to the log wants to slow down and the side of that rope that's attached to the truck doesn't want to do that. Btw, there is a net force of 0 on the truck. We know this because of "constant speed"
Could you show the second question? Whether you draw it in will likely depend on what the question is asking.
Post by: minhalgill on October 23, 2018, 12:47:51 pm
also, in question 7 wont the mass of the electron travelling be different from the mass of an electron at rest?
Post by: minhalgill on October 24, 2018, 02:13:12 pm
in this question i dont get why theyve drawn a reflected ray inside the glucose, as the angle of incidence is less than the critical angle? do we always draw the reflected ray?
Post by: S200 on October 24, 2018, 02:20:25 pm
hey,You literally asked the same thing 10 days ago.
in this question i dont get why theyve drawn a reflected ray inside the glucose, as the angle of incidence is less than the critical angle? do we always draw the reflected ray?
If you have ever done a practical with a light box, you'll see that some portion of the light can be reflected, as well as refracted.
Similar to seeing a double image when you are under water in a pool...
EDIT: Alternatively, the paper is suffering diplopia... :-\
Post by: Theodric_Ironfist on October 24, 2018, 03:05:51 pm
this helped alot, but im just confused about how the momentum can be conserved yet energy is not?
are clipping voltages in the study design?
mod edit: merged posts.
Momentum is conserved because its fundamentally a different quantity than kinetic energy. In an isolated system, which most of the questions are framed in, momentum is always conserved because the net force on all objects is zero (that's the definition of an isolated system). Newton's 2nd law in its original form states that
Thus since the net force is zero, in a collision the change in momentum is zero, so it's conserved (the change in momentum in this equation refers to the total momentum of the system, not the momentum of an object).
Kinetic energy isn't conserved because there are many forms of energy. Some of the original kinetic energy in your system will be present after a collision, however some of it is lost as sound, heat, etc. The TOTAL amount of energy in an isolated system is conserved, but the kinetic energy present at the beginning could transform into another form.
Post by: minhalgill on November 02, 2018, 01:04:56 pm
just wondering about 'uncertainties' and 'uncetainty bars' ive checked a couple books and im still confused. does anyone know a good resource where i can learn from them or can someone please explain them?
thanks
Post by: Bri MT on November 02, 2018, 01:06:09 pm
hi
just wondering about 'uncertainties' and 'uncetainty bars' ive checked a couple books and im still confused. does anyone know a good resource where i can learn from them or can someone please explain them?
thanks
Hey,
Can you give me a quick description of what ideas you anyway have & I'll correct any misunderstandings/provide more info?
Post by: minhalgill on November 02, 2018, 01:33:08 pm
Hey,
Can you give me a quick description of what ideas you anyway have & I'll correct any misunderstandings/provide more info?
my knowledge on uncetanties is very limitied. did not do this even once this year and my teacher had no idea what they are either, only discovered them now that im doing the 2017 vcaa exam and i see uncertainty bars. i understand theyre a +- a certain value but thats pretty much it
also in question 6 c of the vcaa 2013, why cant the spring potential energy be 0 at the top?
Post by: Bri MT on November 02, 2018, 01:46:42 pm
my knowledge on uncetanties is very limitied. did not do this even once this year and my teacher had no idea what they are either, only discovered them now that im doing the 2017 vcaa exam and i see uncertainty bars. i understand theyre a +- a certain value but thats pretty much it
also in question 6 c of the vcaa 2013, why cant the spring potential energy be 0 at the top?
I haven't brought up the question fully since I'm on my phone atm, but if you have a spring it has 0 elastic potential energy when its unstretched. However, if the spring changes from it's unstretched length (whether that's up or down) the elastic potential energy would increase.
Let me know if you don't understand that explanation or that doesn't answer your question & if so I'll draw up a diagram for you to help.
Post by: minhalgill on November 02, 2018, 01:54:38 pm
I haven't brought up the question fully since I'm on my phone atm, but if you have a spring it has 0 elastic potential energy when its unstretched.
can you not decide where to take the 0 of the spring potential energy? like for gravitational you can take the 0 for gravitational at the top or bottom, dont you do that for spring potential energy as well? because in the question it says 'they take the 0 of the elastic potential energy to be q'?
Post by: Bri MT on November 02, 2018, 02:24:01 pm
can you not decide where to take the 0 of the spring potential energy? like for gravitational you can take the 0 for gravitational at the top or bottom, dont you do that for spring potential energy as well? because in the question it says 'they take the 0 of the elastic potential energy to be q'?
You can't say that the elastic potential energy is zero there (the students incorrectly did this), we know that the elastic potential energy is zero when the spring is unstretched and that's at X
Post by: minhalgill on November 02, 2018, 02:42:54 pm
You can't say that the elastic potential energy is zero there (the students incorrectly did this), we know that the elastic potential energy is zero when the spring is unstretched and that's at X
i get it now! thanks!
also,
does anyone know how to do question 10? dont get why we wont double the answer we received in 9 to get 1.0?
Post by: Bri MT on November 02, 2018, 03:00:41 pm
i get it now! thanks!
also,
does anyone know how to do question 10? dont get why we wont double the answer we received in 9 to get 1.0?
There's an equation, f= nv/4l
170 = ((1)(340))/4l
340/170 = 4l
2= 4l
l = .5 m
Then we say "have n=2 (instead of 1) and see what the length is"
170= (2*340)/4l
(2*340)/170 = 4l
4=4l
l=1
Post by: minhalgill on November 02, 2018, 04:14:13 pm
There's an equation, f= nv/4l
170 = ((1)(340))/4l
340/170 = 4l
2= 4l
l = .5 m
Then we say "have n=2 (instead of 1) and see what the length is"
170= (2*340)/4l
(2*340)/170 = 4l
4=4l
l=1
yes, i got 1 as well, but the marking scheme says its 1.5?
Post by: Bri MT on November 02, 2018, 04:24:32 pm
yes, i got 1 as well, but the marking scheme says its 1.5?
The marking scheme says the answer is B, which is 1.0 m
Post by: minhalgill on November 02, 2018, 07:43:00 pm
Post by: Bri MT on November 02, 2018, 08:00:43 pm
for question 10, doesnt only the mass increase substantially? how does the speed increase slightly?
If it's being accelerated to close to the speed of light its speed is increasing
Post by: LOL_ABC on November 02, 2018, 08:28:53 pm
Post by: minhalgill on November 02, 2018, 08:44:42 pm
Post by: Bri MT on November 02, 2018, 08:49:19 pm
No, speed of light is always constant remember that fact if you dont want to bomb the exam :)
I'm not sure if this is a joke... ?
- the speed of light is constant (c) under set conditions. All the ray diagram stuff you do with light bending is dependent on light slowing down as it enters a different medium
- when the electron is being accelerated and approaching c, it is not yet at c (think of c as an asymptote) and its speed is increasing
- given that I duxed my school for physics I like to think that I didn't bomb the exam :P
Edit: this post is not meant as a criticism in any way shape or form :)
For question 4, wont it be A because the observer has to be in the same frame of reference?
To be in the same frame of reference they need to be travelling at the same velocity, so the observer should be stationary relative to the object
Post by: KiNSKi01 on November 02, 2018, 09:04:21 pm
For question 4, wont it be A because the observer has to be in the same frame of reference?
'location' in A doesn't mean in the same frame of reference (bit misleading)
B is the answer because if the observer is at rest relative to the object this means they are travelling at the same velocity (therefore they are in the same frame of reference)
Post by: hailstormb on November 02, 2018, 09:47:51 pm
Post by: Bri MT on November 02, 2018, 09:53:05 pm
How is working out shown on Section B of the exam? Do I have to show all lines of working out AND the final answer in the lines provided and then state the final answer in the box, or can I just show working out in the lines, and then the final answer goes in the box without it being stated in the provided lines?
I always showed both in the lines so that it was easy to see where the answer came from but I don't know if they'd actually penalise you for not doing that
Post by: minhalgill on November 02, 2018, 10:32:18 pm
is all the voltage in a circuit always used up or lost?
Post by: Bri MT on November 03, 2018, 09:46:39 am
does anyone know how to do 2?2. You need to consider whether length will be contracted or dilated & whether or not this occurs only in the direction of motion
is all the voltage in a circuit always used up or lost?
Voltage is a measure of difference in potential, which may help you conceptualise this. (The answer is yes)
Post by: KiNSKi01 on November 03, 2018, 09:23:19 pm
As in which experiments and their observations which support the wave model
Post by: sweetiepi on November 03, 2018, 11:24:17 pm
Yo what evidence do we need to know which supports the wave model of light?Experiment: Young's Double Slit experiment
As in which experiments and their observations which support the wave model
A couple of notable observations:
Diffraction- in Young's Double Slit, obstacles are placed in the path of a wave will not directly block the wave, rather the wave can bend around the barrier. Particles don't do this.
Superposition- Peaks add together in two ways:
- Constructive interference (two crests + two troughs that add up to make a big peak- an antinode)
- Destructive interference (a crest and a trough cancel out- node)
(This took me an embarrassing while to remember from what I've actually relearnt for one of my units this semester- I'm not 100% certain on how much of this is still in the SD ^-^ )
Post by: KiNSKi01 on November 04, 2018, 12:27:32 am
Correct me if I'm wrong but other than photoelectric effect, single-slit experiment can also support the particle model of light?
Just trying to make sure I don't get confused between double-slit and single-slit when discussing wave/particle models of light
Post by: sweetiepi on November 04, 2018, 09:22:02 am
Ok thankyouuuI don't know about the single-slit experiment (and I was finding conflicting info online), however for the particle nature of light, I'd hands-down say photoelectric effect is the main experiment/evidence ^-^
Correct me if I'm wrong but other than photoelectric effect, single-slit experiment can also support the particle model of light?
Just trying to make sure I don't get confused between double-slit and single-slit when discussing wave/particle models of light
Post by: Freddie Hg on November 04, 2018, 09:24:21 am
Ok thankyouuuThe single slit proves the dual nature of light as both waves and matter
Correct me if I'm wrong but other than photoelectric effect, single-slit experiment can also support the particle model of light?
Just trying to make sure I don't get confused between double-slit and single-slit when discussing wave/particle models of light
does anyone know how to do 2?for 2 you have to consider in what situation of the lorentz factor will have no change on mass length and time, just take a look at the equations
is all the voltage in a circuit always used up or lost?
mod edit (insanipi): merged posts :)
Post by: Bri MT on November 04, 2018, 09:45:01 am
The single slit proves the dual nature of light as both waves and matter
Absolutely - but I'm always cautious about words like "proves" in science subjects
In regards to these experiments make sure you know why They support a particular model & what we would expect, for example, from the double slit experiment if light didn't have wave properties
Post by: Freddie Hg on November 04, 2018, 10:44:29 am
Absolutely - but I'm always cautious about words like "proves" in science subjectsYeah i have to agree with you. rookie mistake, for minhaigill's questions what are the answers? i got b and c respectively for 1 and 2
In regards to these experiments make sure you know why They support a particular model & what we would expect, for example, from the double slit experiment if light didn't have wave properties
Post by: KiNSKi01 on November 04, 2018, 12:34:03 pm
The single slit proves the dual nature of light as both waves and matter
Ok sickkkk
Post by: AnonymooseUser on November 04, 2018, 12:50:14 pm
Say for example I had a final value for a problem of 10.16. How do I figure out the number of significant figures required? That is, how do I know whether my answer should be 10, 10.2 or 10.16 etc?
Also, can you be penalised in the exam for incorrect significant figures?
Thanks!
Post by: KiNSKi01 on November 04, 2018, 01:22:42 pm
Post by: Freddie Hg on November 04, 2018, 01:24:53 pm
I'm still a little bit confused on significant figures for the exam (I looked at the other thread on the front page), so was wondering if anyone could help.Miniturtle addressed this issue somewhere in the forum, i searched for it but couldnt find it. i think just round to the lowest number of sig figs
Say for example I had a final value for a problem of 10.16. How do I figure out the number of significant figures required? That is, how do I know whether my answer should be 10, 10.2 or 10.16 etc?
Also, can you be penalised in the exam for incorrect significant figures?
Thanks!
Post by: sweetiepi on November 04, 2018, 02:46:07 pm
I'm still a little bit confused on significant figures for the exam (I looked at the other thread on the front page), so was wondering if anyone could help.My general rule of thumb for sigfigs is to go with the lowest amount quoted in the question stimulus :)
Say for example I had a final value for a problem of 10.16. How do I figure out the number of significant figures required? That is, how do I know whether my answer should be 10, 10.2 or 10.16 etc?
Also, can you be penalised in the exam for incorrect significant figures?
Thanks!
Physics is not as strict as chemistry with sigfigs. You just have to be reasonable- i.e. don't quote 9sigfigs when only 2 or 3 are needed ^-^
Post by: Bri MT on November 04, 2018, 04:35:38 pm
Yeah i have to agree with you. rookie mistake, for minhaigill's questions what are the answers? i got b and c respectively for 1 and 2
I got the same :)
Post by: AnonymooseUser on November 04, 2018, 04:36:59 pm
My general rule of thumb for sigfigs is to go with the lowest amount quoted in the question stimulus :)
Physics is not as strict as chemistry with sigfigs. You just have to be reasonable- i.e. don't quote 9sigfigs when only 2 or 3 are needed ^-^
Miniturtle addressed this issue somewhere in the forum, i searched for it but couldnt find it. i think just round to the lowest number of sig figs
I'm fairly sure that in physics, unless specified you don't need to worry about sig figs
So as long as your answers are reasonable in terms of sig figs, vcaa will accept it?
Thank you for the responses by the way! :)
Post by: KiNSKi01 on November 05, 2018, 06:52:43 pm
I'm stuck on how frequency (a wave property) can explain the particle nature of light
Cheers :P
Post by: Bri MT on November 06, 2018, 07:59:32 pm
Yo can someone help me understand why E=hf is an equation relevant to the particle model of lightIf you think about frequency it just specifies how often something occurs in a period of time - it doesn't necessarily have to be related to wavelength (when we're talking about waves it is related to wavelength)
I'm stuck on how frequency (a wave property) can explain the particle nature of light
Cheers :P
Think this might be related to vibrations, but unfortunately I can't confidently answer you.
Post by: VinnyD on November 06, 2018, 08:17:57 pm
Yo can someone help me understand why E=hf is an equation relevant to the particle model of light
I'm stuck on how frequency (a wave property) can explain the particle nature of light
Cheers :P
Pretty sure its the quantised state of energy (Hf) that supports the particle model, whereas i don't think we know anything regarding the energy for waves in VCE. This is linked closely with the emission and absorption spectra as well as the atomic orbiting (all discrete amounts of energy that relate to hf)
Post by: KiNSKi01 on November 06, 2018, 09:15:38 pm
That's what I was thinking but not confident about
Think this might be related to vibrations, but unfortunately I can't confidently answer you.
Pretty sure its the quantised state of energy (Hf) that supports the particle model,
Quantised states of energy also explain why electrons can form standing waves around nucleus which are of a whole number of wavelengths- therefore supporting the wave model :o
Now I am confused :P which model does that support
Post by: VinnyD on November 06, 2018, 10:24:46 pm
That's what I was thinking but not confident about
Quantised states of energy also explain why electrons can form standing waves around nucleus which are of a whole number of wavelengths- therefore supporting the wave model :o
Now I am confused :P which model does that support
there's a lot of duality as far as i know. You can argue different things for experiments, supporting duality.
Post by: KiNSKi01 on November 06, 2018, 10:52:59 pm
cheers
Post by: lzxnl on November 07, 2018, 10:35:03 pm
Yo can someone help me understand why E=hf is an equation relevant to the particle model of light
I'm stuck on how frequency (a wave property) can explain the particle nature of light
Cheers :P
E = hf is more than defining the energy of a photon. It is an assertion that light exists in discrete packets of energy, and each packet of energy has size hf. It then links the particle and wave models of light by further asserting a direct relationship between a particle quantity, the photon energy, and a wave quantity, the wave frequency. That is the significance of this equation.
Post by: KiNSKi01 on November 07, 2018, 11:56:42 pm
makes more sense now
Post by: KiNSKi01 on November 08, 2018, 11:24:27 pm
For graphing, what are the main graph transformations we need to know to linearize results AND what areas of study do these likely relate to?
e.g a graph of Kinetic energy against velocity (m/s) would require an x^2 transformation
Post by: S200 on November 08, 2018, 11:59:55 pm
supWhy do you need to linerise results?
For graphing, what are the main graph transformations we need to know to linearize results AND what areas of study do these likely relate to?
e.g a graph of Kinetic energy against velocity (m/s) would require an x^2 transformation
Post by: KiNSKi01 on November 09, 2018, 12:05:06 am
I just want to know some of the transformations and which AoS they r likely linked to so I can chuck it on my cheat sheet
Post by: KiNSKi01 on November 10, 2018, 06:45:10 pm
solid line= current after intensity of light source is DOUBLED and frequency is KEPT THE SAME
I understand why increasing the intensity of light source will increase photo current but not max KE of photoelectrons and that the stopping voltage would remain the same
However, this graph (taken from 2017 NHT EXAM) shows that the voltage at which the saturation point occurs also increases (appears to increase from roughly 0.5 V to 1.0 V) Am I just getting confused by what the x axis means? Someone please explain ;)
Post by: Freddie Hg on November 10, 2018, 07:05:50 pm
Dashed line= original currentmaybe its a user error on your behald but i dont see a change between the two graphs x intercept
solid line= current after intensity of light source is DOUBLED and frequency is KEPT THE SAME
I understand why increasing the intensity of light source will increase photo current but not max KE of photoelectrons and that the stopping voltage would remain the same
However, this graph (taken from 2017 NHT EXAM) shows that the voltage at which the saturation point occurs also increases (appears to increase from roughly 0.5 V to 1.0 V) Am I just getting confused by what the x axis means? Someone please explain ;)
Post by: KiNSKi01 on November 10, 2018, 07:13:40 pm
Post by: Freddie Hg on November 10, 2018, 07:16:55 pm
no not the x axis intercepts but when the gradient of the graphs reach zerohuh. idk maybe they made a mistake or maybe because gradient equals y/x increasing y drecreases x to maintain the same rate of change therefore x decreases. idk man everytime i was taught this i leveled the graph off at the same point.
i guess i dont know.
so Bump!!
Post by: Bri MT on November 10, 2018, 07:44:07 pm
Maybe it's related to that they showed one as a straight line segments and the other one as gradual change & that just showing up differently?
Post by: KiNSKi01 on November 10, 2018, 07:49:40 pm
Maybe it's related to that they showed one as a straight line segments and the other one as gradual change & that just showing up differently?
Lol was going to ask why they did that too :P
So basically it's probably something I shouldn't be concerned about and I should be fine to assume the two graphs would level off at same point in similar questions?
Post by: Bri MT on November 10, 2018, 08:00:48 pm
Lol was going to ask why they did that too :P
So basically it's probably something I shouldn't be concerned about and I should be fine to assume the two graphs would level off at same point in similar questions?
Yes :)
Post by: KiNSKi01 on November 10, 2018, 08:04:59 pm
thankyouuuu
Post by: KiNSKi01 on November 12, 2018, 05:04:59 pm
Light emitted during electron transitions in an emission spectrum experiment and light absorbed by valence electrons during an absorption spectrum experiments are both experiments which demonstrate the particle nature of light.
Lol isn't this a bit of stretch? Or am I wrong?
Post by: Freddie Hg on November 12, 2018, 05:08:55 pm
Ok so according to an insight exam paperno this is correct.
Light emitted during electron transitions in an emission spectrum experiment and light absorbed by valence electrons during an absorption spectrum experiments are both experiments which demonstrate the particle nature of light.
Lol isn't this a bit of stretch? Or am I wrong?
to explain the quantised states which cause the absorption and emission bands we rely on the dual nature of light as both a particle and a wave. so these experiments do give an indication that light may behave like a particle
Post by: KiNSKi01 on November 12, 2018, 05:14:29 pm
Post by: DinWell on November 13, 2018, 04:55:00 pm
Post by: Freddie Hg on November 13, 2018, 05:01:47 pm
This is from NHT 2017 exam. Why is the answer X to Y? Doesn't that create a downwards flux?it creates an upwards magnetic field? maybe you should check your right hand grip rule
Post by: DinWell on November 13, 2018, 05:05:49 pm
it creates an upwards magnetic field? maybe you should check your right hand grip ruleI was doing the attached rule. Idk if I'm going crazy but I get a downwards flux.
Post by: Bri MT on November 13, 2018, 05:10:21 pm
I was doing the attached rule. Idk if I'm going crazy but I get a downwards flux.
Remember that the induced current will oppose the change.
Post by: DinWell on November 13, 2018, 05:14:08 pm
Okay, so from the diagram I showed a current going counter clockwise will create field lines going up. Doesn't this oppose the change which is that a downwards flux goes through the coil?
Remember that the induced current will oppose the change.
BTW I realise I'm wrong, I'm just trying to understand it.
Post by: DinWell on November 13, 2018, 05:37:17 pm
Okay, so from the diagram I showed a current going counter clockwise will create field lines going up. Doesn't this oppose the change which is that a downwards flux goes through the coil?Thanks guys. I think I got it. Even though current does flow from Y to X around the coil, it's from X to Y THROUGH THE METER. I didn't pay attention to those three words. I'm very sorry, please excuse my stupidity.
BTW I realise I'm wrong, I'm just trying to understand it.
Post by: davie18 on November 13, 2018, 05:41:18 pm
Post by: Bri MT on November 13, 2018, 05:48:05 pm
Thanks guys. I think I got it. Even though current does flow from Y to X around the coil, it's from X to Y THROUGH THE METER. I didn't pay attention to those three words. I'm very sorry, please excuse my stupidity.
Absolutely nothing to apologise for
Glad you figured it out in the end (and thank you for letting us know once you did); maybe this'll help people avoid making the same error on the exam :)
Post by: DinWell on November 13, 2018, 06:34:53 pm
Attached are the solutions to the 2018 NHT exam. For 6a the exact answer is 2.0203... which is rounded to 2.0 for the answer. Then for 6b, they use 2.0 for the calculations and get 50m as a result. But if you used the exact answer,, you'd get 50.5m, which will round to 51m. Am I supposed to use rounded figures? Will They accept both answers?
Post by: Bri MT on November 13, 2018, 06:40:02 pm
I have another question:
Attached are the solutions to the 2018 NHT exam. For 6a the exact answer is 2.0203... which is rounded to 2.0 for the answer. Then for 6b, they use 2.0 for the calculations and get 50m as a result. But if you used the exact answer,, you'd get 50.5m, which will round to 51m. Am I supposed to use rounded figures? Will They accept both answers?
Imo they should accept the answer with the more exact value.
Make sure you clearly show the values you have subbed into the formula and you should be fine
Post by: harold17 on November 21, 2018, 07:54:20 pm
Post by: Bri MT on November 21, 2018, 08:22:39 pm
Hey guys, I just finished 1&2 physics and I'm doing 3&4 physics next year. I had my orientation or commencement, whatever you like to call it. (2 periods to get an introduction to year 12 physics and get holiday homework) and my teacher said some things im not too sure about. He said the textbook (Heinemann) has a lot of unnecessary information. Just wondering how accurate this is? also we started some course work and he skipped through a lot of theory, only focusing on the formulas, making the review questions from the textbook pretty confusing (although i ended up reading all the theory). So, basically what im asking is, what do you recommend for learning content? Should i read and get notes for all parts of the textbook as well as math questions? Would be great help if someone could help me out :)
As a general VCE tip, I doubt there are any textbooks where all of the information is necessary.
I wouldn't recommend making notes for all of the textbook, for physics I made my notes in class, but if this isn't giving you the level of understanding you need I recommend using the study design dot points as your guide.
You should also practise questions that use the formulas and ensure that you understand the formulas. If, for example, you have the formula F= (mv^2)/r, it's very important that you understand what directions F and v are in.
Disclaimer: there are many different valid methods of studying and I can't guarantee that what worked for me will work for you
Post by: S200 on November 21, 2018, 09:18:34 pm
Hey guys, I just finished 1&2 physics and I'm doing 3&4 physics next year. I had my orientation or commencement, whatever you like to call it. (2 periods to get an introduction to year 12 physics and get holiday homework) and my teacher said some things im not too sure about. He said the textbook (Heinemann) has a lot of unnecessary information. Just wondering how accurate this is? also we started some course work and he skipped through a lot of theory, only focusing on the formulas, making the review questions from the textbook pretty confusing (although i ended up reading all the theory). So, basically what im asking is, what do you recommend for learning content? Should i read and get notes for all parts of the textbook as well as math questions? Would be great help if someone could help me out :)I was given the JacPlus book at the start of the year, and I probably opened it all of 5 times.
Almost everything I was taught was from the notes that the teacher had made themselves, and from YouTube videos that we were shown in class.
There is definitely a fair bit of over-information throughout the textbook, and I personally think that the explanations provided were either too technical or lacked clarity, so the more layman notes really worked well.
If you are worried that your teacher is not explaining clearly or you are missing something, you can watch YouTube videos on the subjects or as said above, make your own notes based off the study design...
Post by: captainbobted on December 02, 2018, 03:29:47 pm
Thank youuuuu! :D
Post by: S200 on December 02, 2018, 07:33:54 pm
I read this thing where it said that "electric field lines start and end at 90degrees to the surface, with no gap between the lines and the surface. What does mean? Whats the surface? Why is 90 degrees?imagine a positively charged plate parallel to a negatively charged plate.
Thank youuuuu! :D
How would you draw the conventional field lines in this instance?
Post by: dream chaser on December 03, 2018, 06:17:59 am
Can someone please explain to me why we need to calculate the centripetal force in this question. Woudn't the centripetal force always equal the tension force anyway. Or is that only when there is no angle involved? Also, what do they mean "with the horizontal"? And is the centripetal force always equal to the net force acting on the object?
Question is in the attachment. The question is related to Uniform circular motion.
Thanks
Post by: fun_jirachi on December 03, 2018, 06:44:45 am
ie look at this ∠, the angle to the horizontal is roughly 50 degrees (hope this explanation helps)
IN this case, the centripetal force isn't actually equal to the tension force! Note that with an angle you actually get a force triangle :)
Since you don't know the mass or velocity of the ball, you need to find some other way to get the centripetal force. It is in uniform circular motion, so it is a perfectly balanced force triangle. Drawn properly you should get that centripetal force is equal to 12cos10.
Hope this helps (hope i got it right too) :)
Post by: dream chaser on December 03, 2018, 06:58:46 am
Post by: S200 on December 03, 2018, 07:23:35 am
Okay, thanks for letting me know. Also, how did you get 50 degrees?I believe jirachi just approximated that number by looking at the symbol he used in his post. There is a limited number of angle symbols available... :)
The 50 is an example and is not related to the final answer....
Post by: dream chaser on December 03, 2018, 08:02:40 am
Post by: Bri MT on December 03, 2018, 08:54:28 am
Hi Guys,
Can someone please explain to me why we need to calculate the centripetal force in this question. Woudn't the centripetal force always equal the tension force anyway. Or is that only when there is no angle involved? Also, what do they mean "with the horizontal"? And is the centripetal force always equal to the net force acting on the object?
Question is in the attachment. The question is related to Uniform circular motion.
Thanks
To build on Jirachi's explanation, the centripetal force is the net force which points directly towards the center of the circle. However, because the string is at an angle it has a vertical component and a horizontal component (we only want the horizontal one since only the horizontal component points to the center of the circle).
To find this horizontal component (the centripetal force) we draw up a triangle. The hypotenuse of the triangle is the tension (12 N) and we know that the angle between the tension and the horizontal is 10 degrees.
cos(angle) = adjacent/hypotenuse
cos(10) = centripetal force/12
12cos(10) = centripetal force
Post by: studyingg on December 17, 2018, 06:20:12 pm
Post by: Seno72 on December 17, 2018, 06:38:25 pm
is the concept of true'weightlessness' still aplicable to gravity in the new SD?
Yep.
Post by: dream chaser on December 20, 2018, 02:05:45 pm
Gavin is at the golf driving range. He selects a club that will launch the ball with an initial vertical velocity of 43.3m/s to a maximum height of 93.75m. The effects of air resistance can be ignored. Find the time it takes to reach the maximum height.
When I answered it, I knew in terms of vertical motion(given that I take the initial direction of motion i.e when the ball is going upwards as the positive direction):
a=-9.8m/s^2
x=93.75m
u=43.3m/s
v=0m/s
t=?
So technically, you can use either x=ut+1/2at^2 or v=u+at to find the 't' value, as we have 4 pieces of information. However, when I tried both approaches, I got separate answers. What am I doing wrong?
Post by: studyingg on December 20, 2018, 02:40:20 pm
Need help with this question regarding projectile motion. All help will be appreciated. :)
Gavin is at the golf driving range. He selects a club that will launch the ball with an initial vertical velocity of 43.3m/s to a maximum height of 93.75m. The effects of air resistance can be ignored. Find the time it takes to reach the maximum height.
When I answered it, I knew in terms of vertical motion(given that I take the initial direction of motion i.e when the ball is going upwards as the positive direction):
a=-9.8m/s^2
x=93.75m
u=43.3m/s
v=0m/s
t=?
So technically, you can use either x=ut+1/2at^2 or v=u+at to find the 't' value, as we have 4 pieces of information. However, when I tried both approaches, I got separate answers. What am I doing wrong?
I also got two seperate answers using your approach. However, I think it's because you are not using the vertical component of the initial velocity. Was there a diagram in this question which contained an angle? I belive u should actually be u=sin(angle)*43.3
edit: or, if there isn't an angle you should actually first use v^2= u^2 +2as to find u and then proceed to find t
Post by: dream chaser on December 20, 2018, 02:51:47 pm
I also got two seperate answers using your approach. However, I think it's because you are not using the vertical component of the initial velocity. Was there a diagram in this question which contained an angle? I belive u should actually be u=sin(angle)*43.3
edit: or, if there isn't an angle you should actually first use v^2= u^2 +2as to find u and then proceed to find t
I am using the vertical component of the initial velocity. it is 43.3m/s(given as part of the question). Plus, no angle was given. No diagram was given either.
What 2 answers did you get anyway.
When I did it, I got 10.30s using x=ut+1/2at^2 and 4.42s using v=u+at. However, for the first approach, I got 10.30 with a negative infront. Did you have the same problem?
Post by: studyingg on December 20, 2018, 03:04:11 pm
I am using the vertical component of the initial velocity. it is 43.3m/s(given as part of the question). Plus, no angle was given. No diagram was given either.
What 2 answers did you get anyway.
When I did it, I got 10.30s using x=ut+1/2at^2 and 4.42s using v=u+at. However, for the first approach, I got 10.30 with a negative infront. Did you have the same problem?
i got 4.42 s using v=u+at , which I think is correct. I think you got -10.30 because you might have accidently used 9.8m/s^2 and not -9.8 m/s^2. (That's what I got with +9.8m/s^2). But even using the negative sign I got 5.04 s. It's a weird question tbh...
Are you sure 43.3 is the initial vertical component? Because I tried the way I described in my edit and I got 4.37s using both methods. If 43.3 is the initial vertical component then I think that there is an error in the question.
Post by: dream chaser on December 20, 2018, 03:19:03 pm
i got 4.42 s using v=u+at , which I think is correct. I think you got -10.30 because you might have accidently used 9.8m/s^2 and not -9.8 m/s^2. (That's what I got with +9.8m/s^2). But even using the negative sign I got 5.04 s. It's a weird question tbh...
Are you sure 43.3 is the initial vertical component? Because I tried the way I described in my edit and I got 4.37s using both methods. If 43.3 is the initial vertical component then I think that there is an error in the question.
There is probably an error then. It says in the question the initial vertical velocity is 43.3m/s.
In regards to the negative answer, I made a mistake in calculations.
Also, can you show me your working out if you use x=ut+1/2at^2 and you take a=-9.8m/s^2, how you get your answer?
Post by: studyingg on December 20, 2018, 03:39:45 pm
There is probably an error then. It says in the question the initial vertical velocity is 43.3m/s.
In regards to the negative answer, I made a mistake in calculations.
Also, can you show me your working out if you use x=ut+1/2at^2 and you take a=-9.8m/s^2, how you get your answer?
Well I rearranged the equation to form a quadratic, factorised out 4.9, and completed the square to get 5.04 and 3.79, I could take a photo if you want. I hate to be annoying but I really think that 43.3 is just the initial velocity and not the vertical component though.
Post by: dream chaser on December 20, 2018, 04:01:40 pm
Well I rearranged the equation to form a quadratic, factorised out 4.9, and completed the square to get 5.04 and 3.79, I could take a photo if you want. I hate to be annoying but I really think that 43.3 is just the initial velocity and not the vertical component though.
Thanks for showing me your working out. No need for the photo. I am also feeling the same way as to the vertical initial velocity but that is what it says on the question unless it is a typo error or something, Thanks for the help by the way :D
Post by: studyingg on December 20, 2018, 04:08:23 pm
Thanks for showing me your working out. No need for the photo. I am also feeling the same way as to the vertical initial velocity but that is what it says on the question unless it is a typo error or something, Thanks for the help by the way :D
All good!
Post by: dream chaser on December 20, 2018, 04:25:35 pm
All good!
Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?
Post by: studyingg on December 20, 2018, 04:31:52 pm
Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?honestly don't know, I've never encountered something like this in physics, probs cause the question is flawed. I'd go with the one that matched my v=u+at formula.
Post by: dream chaser on December 20, 2018, 04:34:55 pm
honestly don't know, I've never encountered something like this in physics, probs cause the question is flawed. I'd go with the one that matched my v=u+at formula.
Okay. Cheers for the help
Post by: lzxnl on December 20, 2018, 07:13:18 pm
Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?If you get two answers, NORMALLY that's because the first time is on the way up and the second time is on the way down. However, the question is about maximum height so you should only have one time. So, there is another possibility that the OP has hinted at here:
I am using the vertical component of the initial velocity. it is 43.3m/s(given as part of the question). Plus, no angle was given. No diagram was given either.
What 2 answers did you get anyway.
When I did it, I got 10.30s using x=ut+1/2at^2 and 4.42s using v=u+at. However, for the first approach, I got 10.30 with a negative infront. Did you have the same problem?
In this question, you're given FOUR pieces of information:
y = +93.75 m
u = 43.3 m/s
v = 0 m/s
a = -9.8 m/s2
t = ?
Given that the question says 'initial vertical velocity' and makes no mention of the horizontal velocity, this is, in fact, a one-dimensional problem.
The issue is, a 1D constant (but nonzero) acceleration problem is uniquely defined by any three of the five quantities. You're given four, so there's a decent chance you have too much information. And indeed, we see that the information you have does not satisfy the equation v2 = u2 + 2ay. Therefore, your question itself is physically impossible, explaining why you can't solve the problem.
Next time you encounter a constant acceleration problem with four pieces of information, check, using the equation that involves all four, that the four pieces of information make sense. Not all questions have answers. A simple one is trying to solve x = x + 1 for x. Obviously, this won't work.
Post by: dream chaser on January 03, 2019, 08:47:39 am
Just a quick question. If you include air resistance, would the horizontal velocity of an object in projectile motion, eg. a ball have constant velocity still?
Also, why does the vertical velocity of an object change, ignoring air resistance, throughout projectile motion?
Post by: Unsplash on January 03, 2019, 09:12:07 am
Just a quick question. If you include air resistance, would the horizontal velocity of an object in projectile motion, eg. a ball have constant velocity still?
No it would not have a constant velocity. The net force in the horizontal direction would be non-zero and thus deceleration would occur, and therefore the object changes velocity.
Also, why does the vertical velocity of an object change, ignoring air resistance, throughout projectile motion?The net force in the vertical direction is mg (assuming no air resistance), thus the forces are unbalanced and a change in acceleration will occur, therefore, the object changes velocity.
Post by: dream chaser on January 03, 2019, 09:27:45 am
No it would not have a constant velocity. The net force in the horizontal direction would be non-zero and thus deceleration would occur, and therefore the object changes velocity.
The net force in the vertical direction is mg (assuming no air resistance), thus the forces are unbalanced and a change in acceleration will occur, therefore, the object changes velocity.
Okay. Thank you for the help. Much appreciated.
Post by: dream chaser on January 03, 2019, 07:50:33 pm
Are my reasoning's for these questions okay? The question is related to projectile motion.
Question: A parcel is dropped from a height of 500m from a helicopter travelling at a speed of 20m/s.
(a) Describe the effects of air resistance on:
(i) the horizontal component of the motion of the parcel.
(ii) the vertical component of motion of the parcel.
(i)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- As the object descends more and more, the air resistance would be applied to a larger effect on the horizontal component of motion.
- The horizontal component of motion(its velocity) would decrease in value as it decelerates due to the horizontal net force becoming more and more non zero.
(ii)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- Unlike horizontal motion, the velocity in the vertical component would increase in value as initially, the velocity in the vertical component would be 0m/s due to it being dropped from rest.
- As the objects descends closer to the ground, the velocity in the vertical component of the object gets bigger due to the non zero net force. However, even though the velocity is getting greater, air resistance slows down the acceleration of the object meaning the vertical component of the velocity as it hits the ground won't be as high of a value as it would if air resistance was not involved at all(was negligible).
(b) Which of the horizontal or vertical components of the motion of the parcel is likely to experience the greater air resistance during:
(i) the first 2 seconds of its fall
(ii) the final 2 seconds of its fall?
(i)
My Answer: Horizontal Component.
- This is due to air resistance in the horizontal direction causing the horizontal velocity to decrease.
- The velocity in the vertical direction would still increase, even with the presence of air resistance involved.
(ii) My Answer: Vertical Component.
- This is due to air resistance causing the vertical velocity to accelerate slowly(not as much as it would if air resistance was not involved).
- Since in the final 2 seconds where the horizontal velocity would already be decelerating closer and closer to 0m/s, the effect in which air resistance would be applied to the horizontal component decreases as Air Resistance ∝ Velocity.
All replies would be much appreciated. Thanks :) :D
Post by: dream chaser on January 04, 2019, 11:44:07 pm
Really need help to do this question. The question is in regards to uniform horizontal circular motion.
When travelling around a roundabout, John notices that the fluffy dice suspended from his rear-vision mirror swing out. If John is travelling at 8.0 m/s and the roundabout has a radius of 5.0 m, what angle will the string connected to the fluffy dice (mass 100 g) make with the vertical?
One of the things I don't get is why the radius of the car in circular motion(i.e the roundabout) would be the same as the radius of the circular motion in which the dice makes. Also, will the dice be experiencing circular motion as the car?
Also, could someone please explain to me what happens to the smaller object when an smaller object within a larger object(eg. a dice in a car) goes around a roundabout if the larger object is going at a constant speed(uniform circular motion)
All help will be greatly appreciated. Thanks :)
Post by: dream chaser on January 05, 2019, 10:20:33 am
Post by: studyingg on January 05, 2019, 11:27:50 am
Would someone be able to help me with the question in my previous post for physics? Thanks
I'm not sure if I fully grasped this concept, but I thinkkkk the reason the radius is the same can be explained by the second law of motion
I'm sure you know that when two bodies are connected, they experience the same acceleration. You probably also know that in the cases of centripetal acceleration, the object (in this case -- the connected body) is deviating towards the centre C. The distance from the centre to the object is the radius, and because both the dice and car are experiencing the same centripetal acceleration, the radius is 5m for both. With response to your other question, (again I’m not too sure), but I would say, yes, the two objects are experiencing the same circular motion. The forces acting on the car are gravity and (perhaps) friction, while the forces acting on the dice are tension and gravity. Which means that the dice is a conical pendulum, but this is still a form of circular motion. Which leads me to say that, the smaller object and the larger object, when connected, will experience the same motion (and act as one system I guess). This is because acceleration is the same. I'm not 100% sure about this; however.
Post by: lzxnl on January 05, 2019, 11:25:32 pm
Hi Guys,
Are my reasoning's for these questions okay? The question is related to projectile motion.
Question: A parcel is dropped from a height of 500m from a helicopter travelling at a speed of 20m/s.
(a) Describe the effects of air resistance on:
(i) the horizontal component of the motion of the parcel.
(ii) the vertical component of motion of the parcel.
(i)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity). this relationship requires a few complicated assumptions but I'll let it slide
- As the object descends more and more, the air resistance would be applied to a larger effect on the horizontal component of motion. just say 'air resistance increases'
- The horizontal component of motion(its velocity) would decrease in value as it decelerates due to the horizontal net force becoming more and more non zero. the horizontal component would decay to zero; becoming more and more non-zero makes no sense here
(ii)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- Unlike horizontal motion, the velocity in the vertical component would increase in value as initially, the velocity in the vertical component would be 0m/s due to it being dropped from rest.
- As the objects descends closer to the ground, the velocity in the vertical component of the object gets bigger due to the non zero net force. However, even though the velocity is getting greater, air resistance slows down the acceleration of the object meaning the vertical component of the velocity as it hits the ground won't be as high of a value as it would if air resistance was not involved at all(was negligible).
Assuming your air resistance is proportional to the velocity, you can write a = g - kv. Initially, when v=0, the acceleration is positive and the object speeds up. Eventually, as v increases, the acceleration decreases and asymptotes towards zero; this limiting velocity is called 'terminal velocity'
(b) Which of the horizontal or vertical components of the motion of the parcel is likely to experience the greater air resistance during:
(i) the first 2 seconds of its fall
(ii) the final 2 seconds of its fall?
(i)
My Answer: Horizontal Component.
- This is due to air resistance in the horizontal direction causing the horizontal velocity to decrease.
- The velocity in the vertical direction would still increase, even with the presence of air resistance involved.
Answer is correct, reasoning isn't great. Rather, mention that the initial horizontal velocity is 20 m/s and in 2 seconds, the vertical velocity cannot accelerate to 20 m/s, so the horizontal air resistance will always be greater
(ii) My Answer: Vertical Component.
- This is due to air resistance causing the vertical velocity to accelerate slowly(not as much as it would if air resistance was not involved).
- Since in the final 2 seconds where the horizontal velocity would already be decelerating closer and closer to 0m/s, the effect in which air resistance would be applied to the horizontal component decreases as Air Resistance ∝ Velocity.
Your second point is good, but the point is really that the horizontal velocity will be small due to air resistance and no horizontal driving force, while the vertical velocity will be much larger as gravity will then approximately balance air resistance
All replies would be much appreciated. Thanks :) :D
Hi Guys,
Really need help to do this question. The question is in regards to uniform horizontal circular motion.
When travelling around a roundabout, John notices that the fluffy dice suspended from his rear-vision mirror swing out. If John is travelling at 8.0 m/s and the roundabout has a radius of 5.0 m, what angle will the string connected to the fluffy dice (mass 100 g) make with the vertical?
One of the things I don't get is why the radius of the car in circular motion(i.e the roundabout) would be the same as the radius of the circular motion in which the dice makes. Also, will the dice be experiencing circular motion as the car?
Also, could someone please explain to me what happens to the smaller object when an smaller object within a larger object(eg. a dice in a car) goes around a roundabout if the larger object is going at a constant speed(uniform circular motion)
All help will be greatly appreciated. Thanks :)
The dice is in the car. It essentially moves with the car, hence the radii and speeds are the same. It's like saying, if you're holding onto your phone as you run, your phone moves as fast as you do, because it's moving with you.
For your last question, it depends on the sizes of the objects involved. In this particular case, the dice is so small compared to the car that you can consider it part of the car.
Post by: dream chaser on January 07, 2019, 11:43:39 am
Post by: Ansaki on January 26, 2019, 01:02:49 pm
(a) moving with constant speed of 10 m/s
(b) accelerating at 1.5 m/s^2?
Post by: blasonduo on January 26, 2019, 01:16:32 pm
Ishtar is riding a motorised scooter along a level bike path. The combined mass of Ishtar and her scooter is 80 kg. The friction and drag forces that are acting total to 45 N. What is the magnitude of the driving force being provided by the motor if she is:
(a) moving with constant speed of 10 m/s
(b) accelerating at 1.5 m/s^2?
Hey!
a) We want the scooter to NOT accelerate, ie a = 0. If we want the the scooter to not accelerate, the forces on the scooter should cancel out, so the motor should provide a force of 45N
b) F = ma -> If we want the scooter to accelerate 1.5m/s^2 we need a net force in the direction of motion to be 120N meaning including the frictional forces, we need 120+45 N which is 165N
Hope this helps!
Post by: HolHen on January 28, 2019, 04:57:24 pm
I am going into year 12 this year and I'm not really sure how to approach the SACs that are going to be practical experiment based.
Any advise??
Thanks :)
Post by: nerdynajla0605 on February 02, 2019, 12:50:28 pm
Post by: HolHen on February 09, 2019, 05:49:48 pm
What are some topics to watch out for in Physics 1/2?
TBH I found most of the topics in 1/2 okay once I took some time to understand them.
Some of the ones that were a little tricky were specific and latent heat in thermodynamics (especially the questions that ask you to find the final temperature when adding ice etc. to a solution); potential spring energy; some of the rollercoater energy conversion questions; and radioisotopes.
Hope that helps good luck ;)
Post by: mcl0028 on February 11, 2019, 10:08:31 am
Post by: lzxnl on February 11, 2019, 09:25:41 pm
I dont get the difference between electric and magnetic feilds or electricity and magnetism in terms of fundamental principles and why there is another phenomena called magnetism and not just electricity. I understand it terms of special relativity and that helped make distinuishes for moving particles but how do dipoles exist etc. noone has been able to describe the magnetic phenomena to me and my teacjer just said "i cant answer that" pls help ded. i understand elctricity as the effect of charged particlesWell...you can't actually distinguish between electricity and magnetism because they're related, as you've mentioned, by special relativity. Indeed, they're part of a more unified object called the electromagnetic field tensor which I won't go into here.
Maxwell's equations for electromagnetism tell us that electric fields are generated by charges and magnetic fields are generated by currents. However, there is one other source of magnetism too that indirectly enter into Maxwell's equations:
Elementary particles have an inherent magnetic dipole moment, which is best thought of as being spinning magnets. These dipole moments have directions, so if you line up all of these elementary particle dipole moments, you get a magnet. This is how permanent magnets work, and probably what you're thinking of when you ask for how dipoles exist.
Post by: dream chaser on February 16, 2019, 10:44:58 am
Got a few questions I want answered. Look at the diagrams in the attachments when answering them. I know it is not part of the VCAA study design(I think) but will help me understand the topic better. By the way, for the vertical loop to loop diagram, assume that the object is at the top and bottom of the loop.
Q1. Are there any horizontal forces acting on the object in any of these diagrams. If so, what and in which diagrams.
Q2. In every position of each diagram, is the object travelling at a constant speed. If so, which one.
Q3. Do we need to either qualitatively and/or quantitatively about the motion of objects in the diagrams given when they are not at the top or at the bottom?
All help and replies will be very much appreciated. Thanks.
Post by: dream chaser on February 18, 2019, 09:39:37 am
Need help with these couple of questions.
Q1. An empty railway cart of mass 500 kg is moving along a horizontal low-friction track at a velocity of 3.0 m s−1 due south when a 250 kg load of coal is dropped into it from a stationary container directly above it.
(a) Calculate the velocity of the railway cart immediately after the load has been emptied into it.
(b) What happens to the vertical momentum of the falling coal as it lands in the railway cart?
(c) If the fully loaded railway cart is travelling along the track at the velocity calculated in (a) and the entire load of coal falls out through a large hole in its floor, what is the final velocity of the cart?
Q2. Two iceskaters, Melita and Dean, are performing an ice dancing routine in which Dean (with a mass of 70 kg) glides smoothly at a velocity of 2.0 m s−1 due east towards a stationary Melita (with a mass of 50 kg), holds her around the waist and they both move off together. During the whole move, no significant frictional force is applied by the ice.
(a) What is Dean’s momentum before making contact with Melita?
(b) Where is the centre of mass of the system comprising Dean and Melita 3.0 s before impact?
(c) What is the velocity of the centre of mass of the system before impact?
(d) Calculate the common velocity of Melita and Dean immediately after impact.
(e) What impulse is applied to Melita during the collision?
For question 1, how are we meant to solve (a)? I used m1u1 + m2u2 = The sum of m x v, however don't know what u2 given that 1 is for the railway cart and 2 is for the coal. We need u2 to solve for v. How do I do it?
For question 2, I don't get what question (b) is asking for. What do they mean 'where is the centre of mass comprising Dean and Melita'? And for (c), are they just asking for the combined velocity before impact?
I don't want anyone to answer all the questions on this post for me, just the parts from the last 2 paragraphs above as I am unsure as to how to do it. Once I know how to do it, I should be able to do the rest by myself.
Also, it would be great if someone could answer the questions from my previous post on the physics QandA thread. It will really help me out.
All help, replies and feedback will be very much appreciated. Thanks.
Post by: dream chaser on February 21, 2019, 03:30:08 pm
Q1. Say for instance there is a block that is held at the end of a spring vertically. At any position(including when it is/is not at equilibrium), will the Tension of the block be equal to the weight.
Q2. Say for instance a car collides with a spring making it compress. Even if it does not compress to its maximum compression, do we suggest that all the kinetic energy gets transferred into elastic potential energy?
All help will be much appreciated. Thanks.
Post by: Bri MT on February 22, 2019, 03:57:01 pm
Set 1
Q1 The two forces we would be considering here (ignoring resistive forces) are weight and the normal force. Remember that the normal force acts perpendicular to the surface the cart/trolley/whatever is on.
Q2 Think of your experiences in regards to speed - then consider how physics could explain this
Q3 is there a verb missing here?
Set 2
Q1.
you're right that you use conservation of momentum :) . Ignore the vertical velocity of the coal (it'll have 0 once it finishes falling anyway), and use the fact that the coal has 0 horizontal velocity
Q2.
I'm pretty sure it just wants to know their displacement using the velocity and time elapsed
and yeah it just wants the velocity of the Dean and Malta system
Set 3
Q1 Think about when the net force will equal 0
Q2 yeah for any calculation questions you would. If you need to describe things qualitatively you would be expected to know that energy transfers aren't 100% efficient.
Hope this helps! :)
Post by: dream chaser on February 23, 2019, 03:02:17 pm
Hey, I'm giving you hints to help you get started since I think you'll probably be okay with a few of them after that. If you still don't really understand feel free to let me know
Set 1
Q1 The two forces we would be considering here (ignoring resistive forces) are weight and the normal force. Remember that the normal force acts perpendicular to the surface the cart/trolley/whatever is on.
Q2 Think of your experiences in regards to speed - then consider how physics could explain this
Q3 is there a verb missing here?
Set 2
Q1.
you're right that you use conservation of momentum :) . Ignore the vertical velocity of the coal (it'll have 0 once it finishes falling anyway), and use the fact that the coal has 0 horizontal velocity
Q2.
I'm pretty sure it just wants to know their displacement using the velocity and time elapsed
and yeah it just wants the velocity of the Dean and Malta system
Set 3
Q1 Think about when the net force will equal 0
Q2 yeah for any calculation questions you would. If you need to describe things qualitatively you would be expected to know that energy transfers aren't 100% efficient.
Hope this helps! :)
Cheers miniturtle for the reply. Much appreciated.
For Set 1 Q3, the verb missing was know.
And for Set 3 Q1, would that mean that only at equilibrium where the fnet would be 0N?
Post by: Bri MT on February 24, 2019, 10:43:58 am
Cheers miniturtle for the reply. Much appreciated.
For Set 1 Q3, the verb missing was know.
And for Set 3 Q1, would that mean that only at equilibrium where the fnet would be 0N?
Set 1 Q3 We don't need to know about anything other than the top and the bottom (see page 39 of the study design) - but you might find a qualitative understanding useful for "getting" the topic.
Set 3 Q1 We know the upwards force being applied to the block is from the tension in the spring, and the downwards force is the weight of the block. When they equal, they'll cancel eachother out (Fnet = 0) and the acceleration will be zero. Yes, this is the case when the block is at the equilibrium position. Let's look at a few ways of thinking about this.
- At the equilibrium position, the kinetic energy is at a maximum because speed is at a maximum too. When the velocity is at a stationary point (eg a maximum or minimum), the acceleration is 0
- At the equilibrium position, the displacement from that position is zero: F=-kx, F=-k(0), F=0
If the spring being vertical is confusing you, this is a video I found useful when I was in VCE
If you think about Hooke's law, it wouldn't make sense for Fnet to be zero at any other time given that weight is constant
Post by: dream chaser on February 24, 2019, 02:50:59 pm
Set 1 Q3 We don't need to know about anything other than the top and the bottom (see page 39 of the study design) - but you might find a qualitative understanding useful for "getting" the topic.
Set 3 Q1 We know the upwards force being applied to the block is from the tension in the spring, and the downwards force is the weight of the block. When they equal, they'll cancel eachother out (Fnet = 0) and the acceleration will be zero. Yes, this is the case when the block is at the equilibrium position. Let's look at a few ways of thinking about this.
- At the equilibrium position, the kinetic energy is at a maximum because speed is at a maximum too. When the velocity is at a stationary point (eg a maximum or minimum), the acceleration is 0
- At the equilibrium position, the displacement from that position is zero: F=-kx, F=-k(0), F=0
If the spring being vertical is confusing you, this is a video I found useful when I was in VCE
If you think about Hooke's law, it wouldn't make sense for Fnet to be zero at any other time given that weight is constant
Thanks Miniturtle. Really appreciate the help. :)
Post by: huangggygy on February 26, 2019, 09:09:35 pm
temperature of the mixture?
If 100 g of water at 95 C is poured into a 500 g glass cup (with an initial temperature of 25
C), what is the final temperature of the water and the cup? (specific heat capacity of
glass = 8.42 × 10 2 J kg -1 K -1 ).
thanks ^^ need help on these two
Post by: lzxnl on February 26, 2019, 10:02:00 pm
If 0.20 kg of water at 90 C is mixed with 0.50 kg of water at 16 C, what is the finalFor the first question, let the final temperature be x degrees C, where 16 < x < 90. Then, the 0.2 kg of water transfers a certain amount of heat to the 0.5 kg water, given by the specific heat capacity * mass * temperature change (90 - x). This heat transferred is the same as the heat used to heat the 0.5 kg water, which is specific heat capacity * mass * temperature change (x - 16). Basically, the temperature change is inversely proportional to the mass. See what you get.
temperature of the mixture?
If 100 g of water at 95 C is poured into a 500 g glass cup (with an initial temperature of 25
C), what is the final temperature of the water and the cup? (specific heat capacity of
glass = 8.42 × 10 2 J kg -1 K -1 ).
thanks ^^ need help on these two
Do the same for the second problem, except now the specific heat capacities are different.
Post by: schoolstudent115 on March 02, 2019, 02:11:02 pm
I have tried working out that given the voltage difference is equal across the circuit, resistance is equal to: 2/0.06, as the total current must be 3 times 20 milliamps - This was apparently wrong.
In the same question , there was the same circuit, except the led's were in series, and I got the correct answer.
Post by: lzxnl on March 02, 2019, 09:45:37 pm
If each LED is identical, and each have a switch-on voltage ofYou mean switch on voltage of 20 mV? mA is not a unit of voltage.mA, then what resistance
is required for the LED's to have optimal light production?
I have tried working out that given the voltage difference is equal across the circuit, resistance is equal to: 2/0.06, as the total current must be 3 times 20 milliamps - This was apparently wrong.
In the same question , there was the same circuit, except the led's were in series, and I got the correct answer.
Post by: Srivathsa on March 10, 2019, 01:42:45 pm
Why is the electric field between two parallel charged plates uniform? I mean I get that their field lines are parallel, but why are they even parallel?
Also why does the field line curve outwards at the ends of the plates?
Post by: lzxnl on March 10, 2019, 08:00:29 pm
Hey goiz, I got a quick question,Hey there. It's actually not. The electric field between two INFINITE parallel plates (in both the x and y dirextions) is uniform. However, real plates have finite area, so you'll find the field isn't constant (near the middle is certainly different to being near the edge of the plates).
Why is the electric field between two parallel charged plates uniform? I mean I get that their field lines are parallel, but why are they even parallel?
Also why does the field line curve outwards at the ends of the plates?
In the infinite case, there are a few reasons why the field is constant.
1. There is no sense of distance. Your plates are infinitely sized. Getting closer to a plate is like zooming in on the plate...which, being infinitely sized, looks the same (try zooming in on an infinitely sized square. It will look the same). Hence, the field strength can't change.
2. If the plates are infinite and oriented horizontally, then if you move across horizontally, you find that the plates look the same. This means the field lines have to look the same as you move between the plates.
3. If you rotate the infinite plates about an axis perpendicular to both plates, you find the plates still look the same. Thus, the shape of the field lines can't be affected by such a rotation, and they must ALL be perpendicular from one plate to another (think about it)
This is an introduction to how symmetry can greatly simplify a physics problem. Ask if you're still stuck! This explanation isn't needed for VCE, don't worry.
Note: none of the above holds for finite-sized parallel plates. Infinite plates are an approximation and a model only.
Post by: juntyhee on March 11, 2019, 08:40:04 pm
Post by: dream chaser on March 18, 2019, 09:48:47 pm
What forces are acting on you when you are feeling zero gravity. Is it just the weight force meaning there is no reaction force. And is experiencing true weightlesness and zero gravity the same thing?
Thanks. All help will be much appreciated.
Post by: lzxnl on March 19, 2019, 10:13:08 pm
Hi Guys,
What forces are acting on you when you are feeling zero gravity. Is it just the weight force meaning there is no reaction force. And is experiencing true weightlesness and zero gravity the same thing?
Thanks. All help will be much appreciated.
From a general relativistic perspective, gravity isn't a force, so when you're free falling, there are no forces acting on you, which makes sense because if you let go of something, that object moves with you.
From a classical/VCE perspective, when you're free-falling, the net force acting on you is mg, and that's it. Zero gravity is, ironically, when you move under the full force of gravity, with NO reaction force. The 'no reaction force' bit leads to apparent weightlessness.
To experience true weightlessness, you need to have zero net gravitational force on you. This is achieved by either moving really, really, really far away from any matter, or by being in a location between the Earth, Moon and the Sun such that the gravitational attractions from all three cancel (for instance).
Post by: dream chaser on March 20, 2019, 05:35:33 am
From a general relativistic perspective, gravity isn't a force, so when you're free falling, there are no forces acting on you, which makes sense because if you let go of something, that object moves with you.
From a classical/VCE perspective, when you're free-falling, the net force acting on you is mg, and that's it. Zero gravity is, ironically, when you move under the full force of gravity, with NO reaction force. The 'no reaction force' bit leads to apparent weightlessness.
To experience true weightlessness, you need to have zero net gravitational force on you. This is achieved by either moving really, really, really far away from any matter, or by being in a location between the Earth, Moon and the Sun such that the gravitational attractions from all three cancel (for instance).
Thanks lzxnl. Really appreciate it. :)
One question. So like apparent weightlessness, does zero gravity mean when the only force acting on you is the weight force?
Post by: captainbobted on March 25, 2019, 08:47:10 pm
So for DC voltages, I understand that as you had more armatures the voltage smoothens out even further. Is this the reason why there are DC-emf graphs that have a 'consistent bumpy' emf AND also there are DC-emf graphs of a linear horizontal line (y=...say 8V), where the 'linear horizontal line' DC-emf graphs are DC generators with so many armatures that it basically smooths out perfectly?
Thank you :) :) :)
Post by: julia_atarnotes on April 15, 2019, 04:47:02 pm
Hello!
So for DC voltages, I understand that as you had more armatures the voltage smoothens out even further. Is this the reason why there are DC-emf graphs that have a 'consistent bumpy' emf AND also there are DC-emf graphs of a linear horizontal line (y=...say 8V), where the 'linear horizontal line' DC-emf graphs are DC generators with so many armatures that it basically smooths out perfectly?
Thank you :) :) :)
Yes! That sounds about right! The emf graphs from DC generators are consistently bumpy as they are the gradient of the flux graph - as dictated by Faraday's Law. If there is a generator creating a straight line for the emf graph though, it would be because there are many armatures because all the voltages from each armature would add up so that there would be a straight line. However, this scenario is not generally encountered in VCE Physics so I would not worry about it too much!
Post by: tayzerface on May 01, 2019, 10:24:12 am
Question from Unit 1 Physics, does anyone know find the effective resistance of this circuit?
TIA!
Post by: AlphaZero on May 01, 2019, 12:50:35 pm
Hey guys, first post on here.
Question from Unit 1 Physics, does anyone know find the effective resistance of this circuit?
TIA!
Hey there and welcome to the forums!
Just for clarity, I'm going to label the resistors from left to right as \(R_1,\ R_2,\ R_3\ \)and \(R_4\). The effective resistance of the circuit is given by \[R_T=R_1+\left(\frac{1}{R_2}+\frac{1}{R_3+R_4}\right)^{-1}\] since \(R_3\ \)and \(R_4\) are wired in series, which is wired in parallel to \(R_2\). The result can then be added to \(R_1\). Plugging in values, we obtain \[R_T=10+\left(\frac{1}{10}+\frac{1}{5+5}\right)^{-1}=15\ \Omega\]
Post by: tayzerface on May 01, 2019, 01:41:29 pm
Post by: randomnobody69420 on June 01, 2019, 06:39:54 pm
Post by: Bri MT on June 03, 2019, 11:51:28 am
any good physics EPI ideas? kinda desperate
There are some ideas from vicphysics
My main advice would be to go for something relatively simple and think about what topic you want to revisit or study in more depth :)
Post by: Jackson.Sprigg on June 03, 2019, 05:20:49 pm
And then for Q 8 why can it be said that Hubble's orbital speed is faster?? Wouldn't the one further out move faster? In the answers it says velocity is proportional to 1/sqrt(radius) but if I rearrange v^2/r = 4pi^2r/T^2 I get v is proportional to r and a different proportionality for a = v^2/r.
i just realised as writing this that I can rearrange Newton's law of universal gravitation to arrive at their answer by replacing F with mv^2/r but why do I have to do this? Is it because g isn't constant?
Thanks for any help it's greatly appreciated!
Sorry for all the questions but I'm quite confuddled and not sure if I'm even doing proportionalitys right.
Post by: S200 on June 03, 2019, 10:18:22 pm
And then for Q 8 why can it be said that Hubble's orbital speed is faster?? Wouldn't the one further out move faster? In the answers it says velocity is proportional to 1/sqrt(radius) but if I rearrange v^2/r = 4pi^2r/T^2 I get v is proportional to r and a different proportionality for a = v^2/r.I'm really terrible on the RMS side of things, but for Gravity you sorta have it.
i just realised as writing this that I can rearrange Newton's law of universal gravitation to arrive at their answer by replacing F with mv^2/r but why do I have to do this? Is it because g isn't constant?
Thanks for any help it's greatly appreciated!
Sorry for all the questions but I'm quite confuddled and not sure if I'm even doing proportionalitys right.
Obviously the force of gravity is inversely proportional to radius, and therefore radius and velocity cannot be proportional, because as radius increases, the the gravitational force relative to the earth decreases. This means that the closer something is to a large body, the more gravitational force it is experiencing (relative to mass of course). So the closer object (the Hubble) must travel be traveling faster to stay in orbit (velocity being tangential), whereas the other satellite is experiencing less gravitational "pull" and hence if it was traveling as fast as (or faster than) the hubble telescope it would break orbit and hammer off into space.
I believe your first error was that you were focusing on acceleration, and as gravity is the Force created by the mass undergoing the centripetal acceleration (not the dictionary definition, I know... :) ), you needed to equate forces rather than accelerations. It sound silly, 'cause the masses are the same so only the accelerations change, but see where i'm coming from?
Post by: JeKnYan on June 10, 2019, 08:18:43 pm
Post by: Jackson.Sprigg on June 11, 2019, 06:21:24 pm
i.e. If the question gives you 3 values with 2, 3 and 4 sig figs respectively and you use all these values in your working. Then you can only say for certain that your answer is as accurate as the least accurate value you used. In this case it would be the 2 sig figs and so that is how many you would use in your final answer.
Post by: Zealous on June 14, 2019, 10:08:40 pm
I believe you should base your significant figures on the most inaccurate measurement in the question.
i.e. If the question gives you 3 values with 2, 3 and 4 sig figs respectively and you use all these values in your working. Then you can only say for certain that your answer is as accurate as the least accurate value you used. In this case it would be the 2 sig figs and so that is how many you would use in your final answer.
Yep, this approach works fine. If you are unsure, usually 3-4 sig figs is more than enough - I'm quite sure the exam markers accept answers within a certain tolerance.
Post by: Terrapin on June 18, 2019, 10:45:24 am
into a rigid barrier at an initial speed of 5.0 m s−1.
The dodgem rebounds at a speed of 2.0 m s−1. It is
in contact with the barrier for 0.20 s. Calculate:
(a) the average acceleration of the car during its
interaction with the barrier
(b) the average net force applied to the car during
its interaction with the barrier.
Post by: Erutepa on June 18, 2019, 08:15:52 pm
A dodgem car of mass 200 kg is driven due southBefore others weigh in on these questions, it is important that you give them a try first and write down what your current thought processes you have in regards to them. It is important to identify specifically what things you do know and what things you don't know so that you can better learn and people can better explain the questions to fit your own confusions. ;D
into a rigid barrier at an initial speed of 5.0 m s−1.
The dodgem rebounds at a speed of 2.0 m s−1. It is
in contact with the barrier for 0.20 s. Calculate:
(a) the average acceleration of the car during its
interaction with the barrier
(b) the average net force applied to the car during
its interaction with the barrier.
And don't worry about getting things wrong - you won't be judged harshly at all.
Post by: julia_atarnotes on June 29, 2019, 11:56:28 am
What's the recommendation on decimal places/sig figs in the exam? I've always been taught it's two, just asking for your two cents
Hey! My past physics teacher was an examiner and he said that 2 decimal places is sufficient. Marks will only be taken off if the question asks you specifically to answer to the correct number of significant figures OR if you use too few significant figures eg. rounding 543.67 to 540. Using the "correct" number of significant figures can be tricky because a lot of physics constants that we are given only use 2-3 significant figures (eg. gravity, speed of light) which may not be enough to satisfy your examiner.
Post by: S200 on July 03, 2019, 11:27:54 pm
A dodgem car of mass 200 kg is driven due south
into a rigid barrier at an initial speed of 5.0 m s−1.
The dodgem rebounds at a speed of 2.0 m s−1. It is
in contact with the barrier for 0.20 s. Calculate:
(a) the average acceleration of the car during its
interaction with the barrier
(b) the average net force applied to the car during
its interaction with the barrier.
A.)
Acceleration is defined as the rate of change of velocity, or \(\frac {V_I - V_F}{Time}\).
We're given the initial and the final velocities, and also the full time of the collision.
Hence, substitute the values and calculate...
We're given the initial and the final velocities, and also the full time of the collision.
Hence, substitute the values and calculate...
Answer
\(\frac{5-(-2)}{0.20} = 35ms^{-1}\) in a negative direction, or away from the barrier
B.)
Force is merely \(Mass \times Acceleration\). The word 'average' is used to indicate that this is not an 'instantaneous' or precisely measured velocity. Thus, average Force is equal to the mass of the body multiplied by the average velocity over the defined time. \(F = m \frac{(v_f - v_i)}{t}.\)
Post by: eo to on July 22, 2019, 09:37:09 pm
It's from the Heinemann physics 12 4th edition textbook (chapter 8 review)
Post by: BAH0003 on July 23, 2019, 07:06:23 pm
Quick question, if anyone cna help that would be great:
"A Subaru travels with a uniform acceleration on a racetrack. It starts from rest and covers 400m in 16s."
What is the cars final speed in km h^-1?
Thanks
Post by: BAH0003 on July 23, 2019, 07:40:27 pm
Another question if anyone can answer.
"During its launch phase, a space rocket accelerates
uniformly from rest to 160 m s–1 upwards in 4.0 s,
then travels with a constant speed of 160 m s–1 for the
next 5.0 s"
How far (in km) does the rocket travel in this 9.0 s
period?
Thanks
Post by: ^^^111^^^ on July 23, 2019, 07:40:51 pm
Hi,
Quick question, if anyone cna help that would be great:
"A Subaru travels with a uniform acceleration on a racetrack. It starts from rest and covers 400m in 16s."
What is the cars final speed in km h^-1?
Thanks
Use the kinematic formula v= u+at
v is final speed
u is initial speed
a is acceleration
t is time
initial speed is 0 (at rest) so v=at. The acceleration is 25 m/s. This multiplied by 3600(number of seconds in an hour) = 90,000 meters (ideally it should be written as 90 km. In case you may need it other kinematic formulas are:
v^2 = u^2 + 2as
s = ut + 1/2*[a(t^2)]
s=1/2*(v+u)t
s=vt - 1/2*[a(t^2)]
v is final speed
u is initial speed
s is displacement
a is acceleration
t is time
Post by: ^^^111^^^ on July 23, 2019, 07:48:14 pm
Hi again,
Another question if anyone can answer.
"During its launch phase, a space rocket accelerates
uniformly from rest to 160 m s–1 upwards in 4.0 s,
then travels with a constant speed of 160 m s–1 for the
next 5.0 s"
How far (in km) does the rocket travel in this 9.0 s
period?
Thanks
This time use the kinematic formula s=ut + 1/2[a(t^2)]
Initial speed is zero. acceleration is 160 m/s. Time is 4 seconds. [This is for the first part. I will explain next shortly]. 4^2 = 16. 16 seconds *160 meters/seconds = 2.56 km
Do the exact same method as before and add the values together to find the displacement of the rocket.
Post by: milanander on July 28, 2019, 09:40:55 pm
Post by: ^^^111^^^ on July 28, 2019, 10:24:37 pm
Hi, current year 11 here doing physics 1/2. I'm hoping to get a head start for 3/4 next year and was wondering what are some key areas which I should start looking over now? Also what company would everyone recommend for practice exams? Thanks so much.
There is not much I can say about this but I am using the Heinemann Physics 3/4 textbook for Physics , so here I go.
- Heisenberg Uncertainity Principle
- Quantum Mechanics
- Photoelectric Effect
- Circular Motion
- Special Relativity
- Electromagnetic Induction
- Properties of mechanical waves
These are the topics that are supposed to be covered in Units 3/4. Sorry I don't think I really answered your question completely. It would be great if someone else can have a look at this.
Post by: redpanda83 on July 28, 2019, 11:10:20 pm
Hi, current year 11 here doing physics 1/2. I'm hoping to get a head start for 3/4 next year and was wondering what are some key areas which I should start looking over now? Also what company would everyone recommend for practice exams? Thanks so much.it would be really good to get head start on newtonian mechanics(classical physics for motion, refering to point 1 mentioned below), which you should be doing as your unit 2 atm or starting.
1. Kinematics, vectors, Forces(newtons laws), Energy(conservation of energy, kinetic energy, GPE, Spring potential energy stuff) momentum and impulse, projectile motion, circular motion.
2. Special Relativity
3. Fields and their patterns, Gravitational Fields, Electric Fields, Magnetic Fields.
4. DC motors, Power generation principles, Transformers and transmission. (this part relates a bit to your electrical circuit component)
5. Light (wave-particle duality) -
(i) light as a wave - wave principles(what is a wave?), wave phenomenon(interference, diffraction, dispersion and
refraction), standing waves (harmonics, look into rubens tube for this its fun!) - light is not dicrete but continuos
(ii) light as particle - photo electric effect (how solar panel works), light is emitted in quantised(discrete) packets of
energy and the amount of energy is dependent of frequency of the photon, how matter can also behave like wave
De Broglie’s wave–particle theory, emission spectra and energy levels of an atom.
I dont really find textbook helpful at all (I have both Jacaranda and Heinmann textbooks), but if you want you can use it as a guide. Khan academy, Lectures by Professor Walter lewins and some other resources are much better. Building things throughout the course will help you further you understanding i think, if you really wanna do it, even little simulations and thought experiments are good.
Hope it helps
Post by: Bri MT on July 29, 2019, 06:51:07 am
Hi, current year 11 here doing physics 1/2. I'm hoping to get a head start for 3/4 next year and was wondering what are some key areas which I should start looking over now? Also what company would everyone recommend for practice exams? Thanks so much.
Fields and projectile motion are probably some of the easier 3&4 topics to wrap your head around on your own. Circular motion could also fall into this category.
I'd also consider revising year 11 concepts that are applicable in year 12 such as conservation of momentum, elastic & inelastic collisions, and converting between different forms of energy (kinetic, gravitational potential, elastic potential).
Lots of topics have been listed for you to potentially try out - please don't feel like you have to get through all of them! If you're unsure what you could cover for a particular topic consider looking at the study design. There are also some past exam questions you can do from just a units 1&2 knowledge base.
Good luck :)
Post by: ^^^111^^^ on July 29, 2019, 10:50:41 am
Fields and projectile motion are probably some of the easier 3&4 topics to wrap your head around on your own. Circular motion could also fall into this category.
I'd also consider revising year 11 concepts that are applicable in year 12 such as conservation of momentum, elastic & inelastic collisions, and converting between different forms of energy (kinetic, gravitational potential, elastic potential).
Lots of topics have been listed for you to potentially try out - please don't feel like you have to get through all of them! If you're unsure what you could cover for a particular topic consider looking at the study design. There are also some past exam questions you can do from just a units 1&2 knowledge base.
Good luck :)
Yes this SO true. I merely just stated all the topics that could be potentially be studied for Units 3/4, not what is needed to study beforehand. And as redpanda said, there many resources out there so try to exhaust any that is necessary. All the best!
Post by: milanander on August 02, 2019, 10:09:07 pm
Post by: DrDusk on August 04, 2019, 11:47:47 pm
Hi, current year 11 here doing physics 1/2. I'm hoping to get a head start for 3/4 next year and was wondering what are some key areas which I should start looking over now? Also what company would everyone recommend for practice exams? Thanks so much.The most important thing is always be curious about what you're learning, most especially with Physics.
If you're sticking to just the syllabus your doing it wrong and it won't help you in developing an understanding of the subject.
Post by: milanander on August 05, 2019, 08:05:30 am
The most important thing is always be curious about what you're learning, most especially with Physics.
If you're sticking to just the syllabus your doing it wrong and it won't help you in developing an understanding of the subject.
If I wasn’t curious about physics I wouldn’t be doing it mate. 😂
Post by: DrDusk on August 05, 2019, 10:51:53 am
Post by: blyatman on August 05, 2019, 11:43:30 am
Most people just sit in the clsssroom just mindlessly taking in what the teacher is saying, without properly thinking about what's been said, and I've noticed this far too many timesHaha agreed. Not to mention, there's a few things that they teach (at least in the HSC counterpart of physics) which are incorrect - some are accidental, whereas some are intentional since the real reason might be too complicated. Though I wouldn't be too surprised if the same misconceptions were also carried over to the VCE.
Post by: milanander on August 09, 2019, 11:59:35 pm
Most people just sit in the clsssroom just mindlessly taking in what the teacher is saying, without properly thinking about what's been said, and I've noticed this far too many timesI mean, I definitely understand what you're saying (see this all the time at my school - students who care more about marks than content) and I'm sure you'd know more considering you have years more experience studying / teaching physics, but I don't think following the study design / syllabus is necessarily a bad thing? It was obviously written by a team of people who are very knowledgeable on physics so they'd obviously know what are the best areas to learn first.
If you were just grabbing bits and pieces of disjointed knowledge here and there out of sheer curiosity that's pretty pointless too. But that's just my very uninformed and unprofessional opinion ¯\_(ツ)_/¯
Post by: DrDusk on August 10, 2019, 03:38:08 am
I mean, I definitely understand what you're saying (see this all the time at my school - students who care more about marks than content) and I'm sure you'd know more considering you have years more experience studying / teaching physics, but I don't think following the study design / syllabus is necessarily a bad thing? It was obviously written by a team of people who are very knowledgeable on physics so they'd obviously know what are the best areas to learn first.
If you were just grabbing bits and pieces of disjointed knowledge here and there out of sheer curiosity that's pretty pointless too. But that's just my very uninformed and unprofessional opinion ¯\_(ツ)_/¯
Oh no I definitely agree with you. You should definitely follow the syllabus. What I meant is don't only follow the syllabus. I remember for Math's, my extension 2 math's tutor who *state ranked* had learnt literally half the HSC syllabus in year 11 and he extended himself onto a lot of uni maths which ultimately led to him getting a raw mark of 114/120. Same thing can be applied to Physics(although I'm not that familiar with the VCE Physics syllabus), but I'm inclined to say it's similar to the new HSC Physics syllabus, in which case it's true.
Also I'm only a first year uni student lol, you make me seem old ;D
Post by: randomnobody69420 on August 14, 2019, 09:01:49 pm
Explain, giving clear reasons, how the movement of the wheels relative to the road enables the car to accelerate
forward.
Explain, with the aid of a clear force diagram, how the rotation of the wheels results in the cyclist accelerating
forwards.
Post by: DrDusk on August 14, 2019, 10:15:27 pm
Found these questions in some old VCAA exams. Is this stuff still on the study design?There's nothing really 'special' about it. It's just a part of Classical Mechanics and so I would say it should definitely be a part of the syllabus. I'm just saying this off experience but I'm from NSW so I can't say it with 100% accuracy.
Explain, giving clear reasons, how the movement of the wheels relative to the road enables the car to accelerate
forward.
Explain, with the aid of a clear force diagram, how the rotation of the wheels results in the cyclist accelerating
forwards.
Post by: blyatman on August 14, 2019, 11:12:24 pm
It was obviously written by a team of people who are very knowledgeable on physics so they'd obviously know what are the best areas to learn first.You'd think so eh haha
Post by: DrDusk on August 15, 2019, 12:16:16 am
You'd think so eh haha
Exactly haha. You'd think if that was true they wouldn't make a syllabus like the old HSC one...
Post by: Just another student on August 15, 2019, 05:51:52 pm
Also are company exams worth doing or a waste of time?
To any 45+ study score achievers, how many practice exams did u do and which ones? (I know u should go quality over quantity, but just want to know how many I should do)
Thanks
Post by: Tau on August 15, 2019, 06:02:09 pm
Hi there does anyone have a list of questions to avoid for VCAA 2002- 2013 physics exams? Is it even worth doing the 2002-2005 exams or should I just start on more recent ones?
Also are company exams worth doing or a waste of time?
To any 45+ study score achievers, how many practice exams did u do and which ones? (I know u should go quality over quantity, but just want to know how many I should do)
Thanks
I would recommend doing past VCAA ones, starting from, say, roughly 2013 and working your way up. I personally left the more recent ones, on the new study design, until later. (Note that there are also a few years worth of NHT exams for Physics).
If your not sure what topics are on the new study design, then print out a copy and look - it’s also a great way to become familiar with what’s actually assessed.
Regarding other companies’ papers: I actually found most of them to be of a really poor standard. They either had mistakes, were poorly phrased, or just terrible question construction. Your priority should really be on VCAA past papers - even older ones will have relevant questions from areas like Special Relativity.
Good luck!
Post by: blyatman on August 16, 2019, 10:40:09 am
Exactly haha. You'd think if that was true they wouldn't make a syllabus like the old HSC one...
Should start a thread that states and corrects all the wrong things they teach in HSC physics lol.
Post by: harold17 on August 28, 2019, 12:06:07 pm
Post by: ImproviseAdaptOvercome on August 28, 2019, 04:48:47 pm
Post by: S200 on August 28, 2019, 06:15:23 pm
I'm in Yr 11 doing 3/4 physics so I have some time in my hands. Can anyone recommend good resources to prepare for my first VCE exam? I am really nervous...best resource I can give is the VCAA exams from the past few years and the VCAA NHT exams from the same period.
Post by: milanander on August 29, 2019, 11:11:45 pm
best resource I can give is the VCAA exams from the past few years and the VCAA NHT exams from the same period.
I know commercial resources get a bad rap (and some for good reason), but do you think there are some companies that actually produce resources on par or at least close to VCAA's standards? Cheers.
Post by: Sine on August 29, 2019, 11:44:16 pm
I know commercial resources get a bad rap (and some for good reason), but do you think there are some companies that actually produce resources on par or at least close to VCAA's standards? Cheers.I didn't do physics but for most other subjects imo rarely do company's produce exams at VCAA standard - however at both ends of the spectrum, either too easy or too hard. Some companies that do get close to VCAA lack the ability to produce good separator questions and usually just copy last years VCAA seperator question (or something from previous years). So usually teachers will focus on past exam concepts and what past students found difficult on the older exams so once it is time to complete these company exams those seperator questions are not too bad. Same goes if the student completes the past VCAA exams before the certain company exams - they have learnt about the "trick" required to solve the separator question before seeing it in the company exam. So really the biggest problem is the lack of creativity for those separator questions. The easy-moderate difficult questions on VCAA exams are easier to replicate because most peopel should be getting those right.
Post by: Tau on August 30, 2019, 12:17:21 am
I know commercial resources get a bad rap (and some for good reason), but do you think there are some companies that actually produce resources on par or at least close to VCAA's standards? Cheers.
From my experience doing trial papers last year, I have to say that the average quality of the papers was appalling. Riddled with mistakes, ambiguous question, misleading statements etc.
My recommendation would be to prioritise the VCAA exams instead. (Even the Checkpoints Questions that weren’t written by VCAA were of a much poorer standard, and often had mistakes in them)
(In Physics they don’t really tend to ask ‘separator’ questions as much as in maths subjects)
Post by: Just another student on August 30, 2019, 08:07:36 pm
From my experience doing trial papers last year, I have to say that the average quality of the papers was appalling. Riddled with mistakes, ambiguous question, misleading statements etc.
Great job on your physics score. Did you only do VCAA exams to revise or company ones too? If i many ask how many exams did u do?
Post by: Tau on August 30, 2019, 08:27:41 pm
Great job on your physics score. Did you only do VCAA exams to revise or company ones too? If i many ask how many exams did u do?
Cheers, I started off with company exams, and then dropped that when I realised it wasn't particularly beneficial. I probably sat about 6 full VCAA exams, but I did Checkpoints and went through a bunch of relevant questions from past study designs.
Post by: Just another student on September 28, 2019, 07:51:14 am
I have printed all the VCAA exams from 2002. However it looks like there is barely any relevant questions. From which year would you recommend starting VCAA exams?
Thanks
Post by: Ninjamagics on October 10, 2019, 05:27:59 pm
I feel like although i seem to get them right, I lack a structure or approach to finding what the area of the graph actually means
Post by: Erutepa on October 10, 2019, 06:07:37 pm
Hi everyone,For myself, the real point of doing practice exams is to simulate the real exam. As such, since all the exams before 2017 used a completely different structure, I haven't been doing any of them. I have only been doing exams from 2017 onwards and have only done company exams so far as I am saving the VCAA ones till last. You should be able to get yourself some relevant company exams from your teacher.
I have printed all the VCAA exams from 2002. However it looks like there is barely any relevant questions. From which year would you recommend starting VCAA exams?
Thanks
In Terms of the earlier exams, they are still useful, however moreso as a source of practice for content knowledge and lesser for doing as a timed exam and I have been relevant questions from these exams mainly through my checkpoint book throughout the year.
So I would recommend sticking to the 2017 and onwards exams for timed exams, and using those older exams just for practicing content knowledge if you need it. However, that is just my opinion and other may beg to differ. :)
A am struggling with an approach towards area under the graph questions.Here is a list of all the relevant area under the graphs that I can think of
I feel like although i seem to get them right, I lack a structure or approach to finding what the area of the graph actually means
- Area under a velocity time graph = displacement/distance (note that if the velocity time graph goes in the positive and the negative, the displacement is given as the magnitude of the area made by the positive velocity - the magnitude of the area made by the negative velocity, and the distance is given by the magnitude of the area made by the positive velocity + the magnitude of the area made by the negative velocity)
- Area under acceleration-time graph = velocity
- Area under force-time graph is the impulse
- Area under a force extension graph (for a spring) = change in spring potential energy
- Area under a gravitational force-radius graph = change in gravitational potential energy
Hopefully this makes sense, and if I have missed any graphs you should know, hopefully others can add them.
Post by: blyatman on October 10, 2019, 10:06:59 pm
A am struggling with an approach towards area under the graph questions.It sucks that they don't teach calculus in physics, because it would make stuff like this more intuitive.
I feel like although i seem to get them right, I lack a structure or approach to finding what the area of the graph actually means
\(v=\frac{dx}{dt}\), so \(dx = v \,dt\), and hence \(x=\int v\,dt\)
So, the area under a velocity curve is equal to the change in displacement.
Likewise, for the area under the gravitational force curve as a function of radius: the force is equal to negative of the rate of change of the potential, i.e. \(F=-\frac{d\Phi}{dr}\), so \(F\,dr = -d\Phi\), i.e. \(\int F\,dr = -\Phi\). So the area under the gravtiational force curve is equal to the negative of the change in potential energy.
You can obtain similar expressions for the other variables to figure out what the area under the curve represents.
Post by: qldpc on October 15, 2019, 03:43:20 pm
On a golf course, a green is 60m up a 10 degrees slope.
The golfer chose to use a 3 iron to make the approach. With a 3 iron, the ball leaves the ground at an angle of 20 degrees to the ground. With what speed must the golf ball leave the club such that the ball lands on the edge of the green.
Post by: Erutepa on October 15, 2019, 04:02:47 pm
DESPERATE HELP PLEASEWhile this thread is here to answer queries you have about physics, it's important that you do have a go at the question first and let us know what specifically you are finding difficult with the question such that we can give specific advice to better help you.
On a golf course, a green is 60m up a 10 degrees slope.
The golfer chose to use a 3 iron to make the approach. With a 3 iron, the ball leaves the ground at an angle of 20 degrees to the ground. With what speed must the golf ball leave the club such that the ball lands on the edge of the green.
Also note that, when asking questions, you shouldn't be demanding, and while perhaps not your intention, you should keep in mind noone is obliged you answer you're question. You're formatting of that post is also a bit unnecessary.
That being said, I would recommend turning this information into a diagram. You will also want to break up the problem into vertical and horizontal components.
Post by: DrDusk on October 15, 2019, 04:05:28 pm
DESPERATE HELP PLEASECould you please give us an image of your working out? That way we can instantly see what you know and don't know which makes it much easier for us to help.
On a golf course, a green is 60m up a 10 degrees slope.
The golfer chose to use a 3 iron to make the approach. With a 3 iron, the ball leaves the ground at an angle of 20 degrees to the ground. With what speed must the golf ball leave the club such that the ball lands on the edge of the green.
Post by: qldpc on October 15, 2019, 11:29:21 pm
While this thread is here to answer queries you have about physics, it's important that you do have a go at the question first and let us know what specifically you are finding difficult with the question such that we can give specific advice to better help you.
Also note that, when asking questions, you shouldn't be demanding, and while perhaps not your intention, you should keep in mind noone is obliged you answer you're question. You're formatting of that post is also a bit unnecessary.
That being said, I would recommend turning this information into a diagram. You will also want to break up the problem into vertical and horizontal components.
Hey Erutepa,
Yeah, I completely understand where you're coming from. I've been struggling with this question for the past two days and it was stressing me out like crazy. Like an idiot, I jumped on here and typed in the question. 100% I should have explained how I was tackling the question and what was causing me issues. Also, I agree with the fact that the post was aggressive and demanding, my apologies for not giving any thought to my post. Obviously no one is obliged to answer the question and looking back, the post was so overkill.
Cheers for pointing all of this out. I'll definitely take these comments on board if I ever post in a question thread again.
With regard to the question, I did end up solving it :) Like you said, a diagram is key to this question and allows the isolation of the components.
Thanks again
Could you please give us an image of your working out? That way we can instantly see what you know and don't know which makes it much easier for us to help.
I did end up solving it :) Thanks for offering to help though.
Mod Edit: Merged posts - try to quote multiple messages in the same post rather than make two seperate posts
Post by: stewartw20 on October 17, 2019, 03:55:07 pm
Guys please help I need this answered ASAP!!!
Post by: Poet on October 17, 2019, 05:09:42 pm
URGENT HELP REQUIRED, PLEASE SOLVE THIS QUESTION NOW!
Guys please help I need this answered ASAP!!!
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, 10:31:00 am
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.
I'm so sorry :'(
Post by: Just another student on October 21, 2019, 03:52:53 pm
Is there any sig fig rules for physics like there is for chem? like do we need to round to lowest number of sig figs? thanks
Post by: Tau on October 21, 2019, 04:05:53 pm
Hi all,
Is there any sig fig rules for physics like there is for chem? like do we need to round to lowest number of sig figs? thanks
There was a video overview of the 2017 exam by the Physics Chief Examiner in 2017. From what he said they’ll no longer really take off marks for significant figures, unless the question stipulates it (which one did in 2018). Just ensure you work to a reasonable level of accuracy and quote your answer to an appropriate level. They shouldn’t take off marks unless asked for in a specific form.
Post by: studyingg on November 09, 2019, 10:50:02 am
Post by: Ninjamagics on November 09, 2019, 04:31:43 pm
Q1) when can you use power loss to find power
Q2) referring to the vcaa 2017 Q4 b attached question, why is Rt not Rp+Rc
Post by: studyingg on November 09, 2019, 05:17:57 pm
hopefully 1 of u amazing individuals gets back to me in time,Q1/ Not sure how to answer this without a specific question...sorry :)
Q1) when can you use power loss to find power
Q2) referring to the vcaa 2017 Q4 b attached question, why is Rt not Rp+Rc
Q2/ Because the radius of orbit accounts for the distance between Pluto and Charon; if you were to add the radius of Pluto with that of Charon you would be falsely assuming that there is no distance between these planets. Drawing a diagram helps.
Post by: 3086 on November 10, 2019, 05:12:41 pm
hopefully 1 of u amazing individuals gets back to me in time,
Q1) when can you use power loss to find power
Q2) referring to the vcaa 2017 Q4 b attached question, why is Rt not Rp+Rc
You cannot use power loss to directly find power however power loss can be used to determine how much power has reached our desired destination (for example to light a globe at our house). This can be done by subtracting the power loss from the initial power supplied. Another way in which power loss can indirectly help us calculate power is if we add it to the power at our destination. This will help us determine the power we started with ie the supply power.
Post by: 3086 on November 10, 2019, 08:02:25 pm
-The current throughout the system remains constant (except when transformers are in use)
-For transformers a step up transformer requires simple multiplication of the most simplified ratio (ie if it is 12:240 the most simplified would be 1:20, so you would multiply by 20) and for step down you divide by the most simplified ratio (so divide by 20)
-To calculate the voltage at the desired location when transformers are involved, simply use the following equation:
Vdesired=(VsupplyxT1-IlinexRline/T2
For example: Vsupply=400 V
T1 and T2 are 1:5 and 5:1 respectively
I is 2 Amps and R is 5 Ohms
Then: 400x5-2x4/5 = 398 V as what we recieve (V desired)
Post by: Bri MT on November 10, 2019, 08:33:30 pm
Could someone please explain why the flight time is the same? Thanks :)
In this case the flight path is symmetrical and you can find the time of flight for the second half using initial vertical velocity of 0, the vertical displacement & acceleration due to gravity. These 3 factors are the same both times the paintball is fired
Post by: Matthew_Whelan on November 11, 2019, 02:01:48 pm
(I know power transmission is necessary to know)
Post by: EricAyl on December 23, 2019, 04:26:48 pm
Due to a lack of people enrolling in physics 3/4 at our school, I'm doing physics via virtual schooling (similar to distance ed, where you send assignments, sacs and homework online with a virtual schooling teacher) and I've been told that it is very difficult. I have a friend doing it with me so I hope we get through it, is there anything that I should be aware of regarding my situation? Any tips to survive or any resources that anyone has? Anything that would help me because I'm quite worried D: Thanks guys :)
Post by: milanander on December 24, 2019, 10:13:45 am
Hey all! :) I'm new to this whole forum thing so I'm not sure if I'm doing this right :P
Due to a lack of people enrolling in physics 3/4 at our school, I'm doing physics via virtual schooling (similar to distance ed, where you send assignments, sacs and homework online with a virtual schooling teacher) and I've been told that it is very difficult. I have a friend doing it with me so I hope we get through it, is there anything that I should be aware of regarding my situation? Any tips to survive or any resources that anyone has? Anything that would help me because I'm quite worried D: Thanks guys :)
I will be doing physics 3/4 through distance education as well — this year I did 1/2 through it and long story short, it definitely wasn’t the best experience and I fully considered either doing it through a nearby school that offered physics or dropping the subject entirely. I am also doing specialist maths through distance (though I thought the subject is slightly better organised) and my experience is that teachers are INCREDIBLY slow at responding to your queries, sometimes they don’t respond at all. Also the resources probably aren’t to the standard you would hope to receive.
I am yet to complete a 3/4 subject through distance education but from my experience with unit 1/2s, I think it is crucial to get some external resources yourself, maybe from a past student. I’ve also talked to my maths and science teachers at school and they have agreed to pass me any resources they have. I don’t know whether tutoring is necessary - I’m yet to get tutoring but maybe it’s something to consider. As for tips, just be really on top of everything (eg SAC’s, assignments, revision etc) yourself because you won’t have a teacher chasing you down like you would in a normal school.
Post by: Erutepa on December 24, 2019, 12:57:18 pm
Hey all! :) I'm new to this whole forum thing so I'm not sure if I'm doing this right :PWelcome to the forums EricAyl!
Due to a lack of people enrolling in physics 3/4 at our school, I'm doing physics via virtual schooling (similar to distance ed, where you send assignments, sacs and homework online with a virtual schooling teacher) and I've been told that it is very difficult. I have a friend doing it with me so I hope we get through it, is there anything that I should be aware of regarding my situation? Any tips to survive or any resources that anyone has? Anything that would help me because I'm quite worried D: Thanks guys :)
I haven't had any experience with virtual schooling or distance ed but I have had a friend do it. From his experience, the main challenge to overcome was a lack of motivation to study. I think normally one gets a lot of will to study from being in a class environment as having others striving to succeed around you pushes you to do the same. Doing distance ed, my friend found this lack of an actual class and teacher mean he struggled to consistently find the will to push himself to do more than the minimum required work and this was reflected in rather poor scores. Without having a teacher you see in person regularly or peers that can both inspire you to work harder and help you through difficult topics, the process of learning can become somewhat hard to do mostly independently. That being said, this is probably something that may not be that much of a problem for more self-motivated student.
To remedy this, I would recommend making sure your friend and yourself stick together and help each other through tough bits of content and keep each other excited about doing physics. Maybe schedule a study session with them a couple times a week after school at a library and get into the habit of going through things you don't understand and marking each others questions. Apart from motivating you, teaching and marking for each other are two things that will really help you succeed in any subject regardless of whether or not you are doing distance ed/virtual tutoring.
Another source of motivation might be to maintain a journal where you reflect upon your study/results and in order to really evaluate your progress, what you are doing good, and what you might need to focus on. Regularly evaluating your performance Is something that (I think) is a valuable tool for improving (and when you don't have a teacher you interact directly to push you to improve as such, keeping a jounral can help you do this yourself). You could do this journal in a book/privately, or you could put your academic reflections in a journal on AN (see more about it here ). I personally wouldn't have been able to keep a journal privately through my year 12, but found that having a journal here was much easier to keep updated especially with people checking in a reminding you to update you journal from time to time. Having either a journal here about your endevours learning physics (or all your subjects) also provides a space for you to express any challenges/problems you are facing and for others to provide some wise advice!
Hopefully this might be of some help!
Good luck with physics next year and looking around to seeing you on the forums
Post by: EricAyl on December 27, 2019, 11:41:26 pm
Post by: milanander on December 28, 2019, 11:05:29 am
Golly, I'm glad I have someone with me on this journey then 😅, thank you for the advice! I do hope it all goes well otherwise RIP atar :P Thanks a lot pal, I shall keep your wise words close to heart and act on upon them accordingly!!!!!! Will update (probably) if things aren't too good :)
Ayyyy we distance people gotta stick together!
Also, just got told this recently - I’ve heard from a friend who did a subject through distance this year that it helps to get a schedule of when each topic will be covered from a high ranking school so that you can make sure you’re staying on track throughout the year.
Post by: redleafbun on January 10, 2020, 05:58:56 pm
I am having trouble understanding the questions, what does it mean by "approximate direction"??
Thanks :D
Post by: Erutepa on January 10, 2020, 06:32:42 pm
Hello, GUys!The question stem shows a diagram with dipole magnets
I am having trouble understanding the questions, what does it mean by "approximate direction"??
Thanks :D
dipole magnets produce magnetic fieilds like that shown below:
(https://www.researchgate.net/profile/Lokesh_Sharma7/publication/322069152/figure/fig7/AS:616329182724097@1523955736050/Magnetic-field-lines-of-a-dipole-10.png)
Here the arrows show the direction of the magnetic field. The North pole of a magnet will produce a magnetic acting outwards from the north pole, and the south pole produces a magnetic field acting inwards towards the south pole. The density of the lines (how close they are together) dictates the strength of the field.
For your question, there are two dipole magnets which produce magnetic fields that interact with eachother. For VCE physics you are going to have to know how to draw magnetic fields for a range of situations such as this one here. The magnetic field produced by these magnets is shown below
(https://upload.wikimedia.org/wikipedia/commons/9/95/VFPt_cylindrical_magnets_repelling.svg)
As can be seen, the north pole of each magnet produces a magnetic acting outwards from the north pole, and each south pole produces a magnetic field acting inwards towards the south pole, with the fields from each magnet interacting to result in the feild shown above. You should also note that at the point exactly between the two magnets, there is not feild due to the equal and opposite magnetic fields from each magnet interating.
Your question asks for the 'aproximate direction' meaning it wants you to aproximately/roughly indicate what the direction of the magnetic field is at the point referenced by applying your knowledge of the resultant magnetific field as shown in the diagram above. The inclusion of the compass in the diagram also indicates that the answer should be interms of those such directions (north, south, east, and west).
Hopefully this answers your question, but feel free to point out anything that I explained poorly or that you don't quite understand :)
Post by: redleafbun on January 12, 2020, 10:14:06 pm
The question stem shows a diagram with dipole magnets
dipole magnets produce magnetic fieilds like that shown below:
(https://www.researchgate.net/profile/Lokesh_Sharma7/publication/322069152/figure/fig7/AS:616329182724097@1523955736050/Magnetic-field-lines-of-a-dipole-10.png)
Here the arrows show the direction of the magnetic field. The North pole of a magnet will produce a magnetic acting outwards from the north pole, and the south pole produces a magnetic field acting inwards towards the south pole. The density of the lines (how close they are together) dictates the strength of the field.
For your question, there are two dipole magnets which produce magnetic fields that interact with eachother. For VCE physics you are going to have to know how to draw magnetic fields for a range of situations such as this one here. The magnetic field produced by these magnets is shown below
(https://upload.wikimedia.org/wikipedia/commons/9/95/VFPt_cylindrical_magnets_repelling.svg)
As can be seen, the north pole of each magnet produces a magnetic acting outwards from the north pole, and each south pole produces a magnetic field acting inwards towards the south pole, with the fields from each magnet interacting to result in the feild shown above. You should also note that at the point exactly between the two magnets, there is not feild due to the equal and opposite magnetic fields from each magnet interating.
Your question asks for the 'aproximate direction' meaning it wants you to aproximately/roughly indicate what the direction of the magnetic field is at the point referenced by applying your knowledge of the resultant magnetific field as shown in the diagram above. The inclusion of the compass in the diagram also indicates that the answer should be interms of those such directions (north, south, east, and west).
Hopefully this answers your question, but feel free to point out anything that I explained poorly or that you don't quite understand :)
Post by: ^^^111^^^ on January 14, 2020, 08:02:56 am
Post by: Bri MT on January 14, 2020, 08:38:09 am
Does anyone know if we have to learn about the derivations of the relativistic equations for units 3/4? Thanks.
You don't need to know this
Post by: milanander on January 14, 2020, 08:41:53 am
Does anyone know if we have to learn about the derivations of the relativistic equations for units 3/4? Thanks.
No. It’s not on the study design.
Post by: redleafbun on January 15, 2020, 07:55:28 pm
btw i struggle to understand chapter 2 (magnetic, electric fields) any resource recommendation?
Post by: Erutepa on January 16, 2020, 10:36:29 pm
Hey Guess! I am back with another question :DHey redleafbun :)
btw i struggle to understand chapter 2 (magnetic, electric fields) any resource recommendation?
You seem to have posted multiple questions - are you having trouble with all three, or is there one particular question that has you confused?
I am more than happy to answer your question, but before I do its important for you to have a go at it first and try to explain your current aproach to the question. Don't worry about being wrong as thats all part of the learning process.
Having a go at the question first and showing us your thinking process helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question.
Interms of recomendations to help develop your understanding of fields, If you are currently a bit confused as to what exactly you need to know it might be a good first step to have a look at the study design here which outlines precisely that (if there are any points which you don't quite understand, feel free to ask about it here :)).
If you see some content listed in the study design that you aren't confident on, it might be good to revisit your textbook and re-read over/make notes on those topics. In addition to reading over the textbook/making notes, I found that making your own diagrams when possible really helped. I think that a fair amount of feild questions (e.g. determining the direction of induced currents) requires the ability to visualise the problem - which I found drawing diagrams helped with.
While going over content in this way, If there are any really confusing bits that you encounter which you can't understand on your own, myself (and others) are happy to help explain stuff to you, just try to be specific with what you do and don't understand so that we can help you more effectivley.
After going over the bits of content that you are stuggling with, I would recoment jumping back into answering questions (the most effective ones would be exam style questions if you have access to any - These are things like Atar notes topic tests or checkopints). :)
Post by: TheEagle on January 19, 2020, 12:16:54 am
I have a question regarding the area under the following graphs:
-gravitational force–distance
-gravitational field–distance
Why is it that we must multiply the area of the (gravitation field - distance) graph by mass, whereas, we don't for the gravitational force one? I have attached the textbooks 'explanation'
I have also attached an example in which the solution multiplies the area by mass
Thank you :)
Post by: Tau on January 19, 2020, 12:24:04 am
Hello everyone
I have a question regarding the area under the following graphs:
-gravitational force–distance
-gravitational field–distance
Why is it that we must multiply the area of the (gravitation field - distance) graph by mass, whereas, we don't for the gravitational force one? I have attached the textbooks 'explanation'
I have also attached an example in which the solution multiplies the area by mass
Thank you :)
Well gravitational field strength is the force divided by mass, and we know that the area under the force displacement graph is work (by definition). So to ‘recover’ the mass that’s missing from the field strength, we multiply by mass to obtain the force.
Post by: TheEagle on January 19, 2020, 12:34:01 am
Well gravitational field strength is the force divided by mass, and we know that the area under the force displacement graph is work (by definition). So to ‘recover’ the mass that’s missing from the field strength, we multiply by mass to obtain the force.
Sorry, I should have been more specific; I was asking in terms of energy (specifically kinetic energy).
Post by: Tau on January 19, 2020, 12:40:27 am
Sorry, I should have been more specific; I was asking in terms of energy (specifically kinetic energy).
I’m not quite sure what you mean. The area under a force-distance graph is work done, and by scaling a field strength-distance graph by mass the are is thus the same as under a force-distance,
Post by: TheEagle on January 19, 2020, 12:47:42 am
I’m not quite sure what you mean. The area under a force-distance graph is work done, and by scaling a field strength-distance graph by mass the are is thus the same as under a force-distance,
I don't get why we must scale a field strength - distance graph's area by mass to attain the change in energy. Whereas, the area under a force - distance graph is equal to energy without multiplying by mass?
Post by: Tau on January 19, 2020, 12:55:40 am
I don't get why we must scale a field strength - distance graph's area by mass to attain the change in energy. Whereas, the area under a force - distance graph is equal to energy without multiplying by mass?
By definition we have that \(W=\int F dx\), and for a Force-distance graph we can work directly with that. However, gravitational field strength is defined by \(g=F/m\) - force per unit mass - if we multiply \(g\) by \(m\) we get our \(F\) and can then use the first equation (well, counting squares really :) ) to find the work done.
Post by: redleafbun on January 20, 2020, 12:33:39 pm
Hey redleafbun :)
You seem to have posted multiple questions - are you having trouble with all three, or is there one particular question that has you confused?
I am more than happy to answer your question, but before I do its important for you to have a go at it first and try to explain your current aproach to the question. Don't worry about being wrong as thats all part of the learning process.
Having a go at the question first and showing us your thinking process helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question.
Interms of recomendations to help develop your understanding of fields, If you are currently a bit confused as to what exactly you need to know it might be a good first step to have a look at the study design here which outlines precisely that (if there are any points which you don't quite understand, feel free to ask about it here :)).
If you see some content listed in the study design that you aren't confident on, it might be good to revisit your textbook and re-read over/make notes on those topics. In addition to reading over the textbook/making notes, I found that making your own diagrams when possible really helped. I think that a fair amount of feild questions (e.g. determining the direction of induced currents) requires the ability to visualise the problem - which I found drawing diagrams helped with.
While going over content in this way, If there are any really confusing bits that you encounter which you can't understand on your own, myself (and others) are happy to help explain stuff to you, just try to be specific with what you do and don't understand so that we can help you more effectivley.
After going over the bits of content that you are stuggling with, I would recoment jumping back into answering questions (the most effective ones would be exam style questions if you have access to any - These are things like Atar notes topic tests or checkopints). :)
Post by: thisissara on January 20, 2020, 03:35:59 pm
Could someone help me establish the difference between Banked Tracks and Inclined Planes? I.e in regards to forces etc! Tysm!
Post by: DrDusk on January 20, 2020, 04:38:36 pm
Hey guys,Banked tracks are like those tilted race tracks that you see for professional bicycle races whereas an inclined plane is basically just a hill. Difference is that on a Banked track you are going in a circle and an inclined plane your just going up or down. Obviously this means the biggest difference in terms of forces is that a Banked track has a Centripetal force unlike an inclined plane.
Could someone help me establish the difference between Banked Tracks and Inclined Planes? I.e in regards to forces etc! Tysm!
Post by: puzzledstudent on January 26, 2020, 10:59:42 pm
I'm struggling so hard with this one question and I would be very thankful if anyone has the time to help me solve it and explain it.
A net force of 20.0 N acts on a 5.00 kg mass for 6 seconds. The original velocity of the mass is 100 m s-1 east. Calculate the velocity after the 6seconds if
a) The net force is to the east
b) The net force is to the west
Thank you soSO MUCH in advance!
Post by: Bri MT on January 26, 2020, 11:14:24 pm
So I am starting Year 11 and I haven't finished the holiday homework sent out :[
I'm struggling so hard with this one question and I would be very thankful if anyone has the time to help me solve it and explain it.
A net force of 20.0 N acts on a 5.00 kg mass for 6 seconds. The original velocity of the mass is 100 m s-1 east. Calculate the velocity after the 6seconds if
a) The net force is to the east
b) The net force is to the west
Thank you soSO MUCH in advance!
Hi! Welcome to the forums :)
Have you used the constant acceleration formulas before?
(Hint: the east vs west difference is about whether acceleration is in the same direction as initial velocity)
1. Define a direction to be positive
2. Write down your values including units
3. Find the magnitude of acceleration using f=ma
4. Identify the constant acceleration formula matching the info you have
5. Sub and solve
For part b,
1. Multiply the acceleration by -1
2. Sub this into the formula from before & solve
If any part of it doesn't make sense or you'd like to confirm your answer feel free to reply & I'm happy to help further :)
Post by: puzzledstudent on January 26, 2020, 11:35:48 pm
Ok let me assume East is positive.
F-20N M-5kg T-6s V-100ms-1
f=ma
20=5*a
a=4m/s
What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?
Uhh I think I've done this last year but I've completely forgotten about every single information LOL
x[
Post by: DrDusk on January 27, 2020, 12:29:04 am
What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?If a force acts towards the East on an object that is travelling East how will the object have the same velocity after 6 seconds? It's going to accelerate.
Post by: Bri MT on January 27, 2020, 08:04:06 am
Hi thanks!
Ok let me assume East is positive.
F-20N M-5kg T-6s V-100ms-1
f=ma
20=5*a
a=4m/s
What I don't get is the net force to the east, west bit. Because it says the original velocity is 100ms-1 east. So doesn't this mean the velocity for east is 100ms-1?
Uhh I think I've done this last year but I've completely forgotten about every single information LOL
x[
No worries!
First up, your acceleration isn't 4 metres per second, it's 4 metres per second per second. Additionally, usually initial velocity would be indicated with u rather than v, and then we use v for final velocity.
As DrDusk said, the object is undergoing acceleration. By definition, this means that its velocity is changing. Because we've calculated the acceleration, we know that the velocity will change by 4 metres per second each second.
In part A, with acceleration and initial velocity in the same direction, we know that the speed will increase by 4 m s^-2 . We also know how many seconds this acceleration is being applied for & the initial velocity. This allows us to find the final velocity.
Post by: puzzledstudent on January 28, 2020, 10:53:59 pm
If anyone could take the time to make a worked solution for me, I would really appreciate it. I'm not lazy, I just do not understand anything you are saying etc "second per second". I'm new to Physics and I feel like I'm going to fail it at this pace. I hope if anyone has used Jacaranda 1/2 Physics, can they please tell me whether it could be so complicated? If it is, I'm definitely dropping it =[
Post by: Bri MT on January 28, 2020, 11:21:31 pm
I honestly still have no idea..
If anyone could take the time to make a worked solution for me, I would really appreciate it. I'm not lazy, I just do not understand anything you are saying etc "second per second". I'm new to Physics and I feel like I'm going to fail it at this pace. I hope if anyone has used Jacaranda 1/2 Physics, can they please tell me whether it could be so complicated? If it is, I'm definitely dropping it =[
You calculates the acceleration to be 4 m/s^2. Aside from getting the units wrong your working here was fine.
This means that each second, the velocity changes by 4 metres per second (i.e. change of 4 metres per second per second)
In part a we start off at 100 m/s east. This is the same direction as our acceleration. After 1 second, the new velocity will be 104 m/s east. After another second, it will be 108 m/s east. After the 3rd second, it will be 112 m/s east... until eventually after 6 seconds it will be 100 m/s +(4m/s^2)(6s) = 124 m/s
In part be we start off at 100 m/s west. This is the opposite direction to our acceleration. After 1 second the velocity will be 96 m/s. After another second it will be 92 m/s. After 3 seconds it will be 88 m/s.... after 6 seconds it will be 100 m/s - (4 m/s^2)(6) = 76 m/s west
In both cases, we can use the forumla v = u + at to arrive at our answers. In this approach, we can jump straight to
a)
v = u + at
= 100 + (4)(6)
= 124
v= 124 m/s east
b)
v = u + at
= -100 + (4)(6)
= -76
v = 76 m/s west
I hope this clarifies things a bit :)
Post by: BitcoinEagle on January 29, 2020, 12:03:52 pm
As a result of being hit from behind by a toy truck, a 250 g toy car, initially at rest rolls 10 m across a floor that applies a constant retarding force of 0.75N to it. The car stops 6 seconds after being hit.
a)If the truck was in contact with the car for 0.12 seconds, calculate the impulse given to the car.
b) Calculate the momentum of the car just after being hit.
c) Calculate the instantaneous speed of the car just after being hit.
d) Calculate the average force applied to the car during the collision.
Post by: SmartWorker on January 29, 2020, 12:43:47 pm
How does significant figures work, like how many sig figs are there in 13.07 and 13.0 and 13 and 2.00, 2.0, 3.03 x10^8
Thanks in advance😀😀
Post by: DrDusk on January 29, 2020, 07:23:27 pm
Hey everyone,I'll be completely honest here. I went through Mathematics Ext 2, hardcore University Maths like Complex Analysis and Vector Calculus and I still don't know how they work. I just let my calculator do sig figs for me lol
How does significant figures work, like how many sig figs are there in 13.07 and 13.0 and 13 and 2.00, 2.0, 3.03 x10^8
Thanks in advance😀😀
Post by: Evolio on January 29, 2020, 07:29:38 pm
Hey everyone,Hey SmartWorker!
How does significant figures work, like how many sig figs are there in 13.07 and 13.0 and 13 and 2.00, 2.0, 3.03 x10^8
Thanks in advance😀😀
13.07: 4 sig figs
13.0: 3 sig figs
13: 2 sig figs
2.00: 3 sig figs
2.0: 2 sig figs
3.03x10^8: 3 sig figs.
Rules:
- Following zeros (after non-zero numbers) are always significant
- Leading zeros (before non-zero numbers) are never significant
- Zeros in between non-zero numbers count as a sig fig
- All numbers except 0 count as a sig fig
This is specified by VCAA here: https://www.vcaa.vic.edu.au/curriculum/vce/vce-study-designs/Physics/advice-for-teachers/Pages/MeasurementinScienceQuantitativeAnalysisofUncertainty.aspx
Please correct me if I'm wrong!
Post by: milanander on January 29, 2020, 08:37:40 pm
I'll be completely honest here. I went through Mathematics Ext 2, hardcore University Maths like Complex Analysis and Vector Calculus and I still don't know how they work. I just let my calculator do sig figs for me lol
What type of calculator figured out sig figs for you?
Post by: DrDusk on January 29, 2020, 08:53:22 pm
What type of calculator figured out sig figs for you?Casio fx-100 AU PLUS. It's a standard calculator that most people use in NSW.
Post by: Erutepa on January 30, 2020, 09:44:48 pm
Hi, I was wondering if I could have some help with this physics motion question. Thanks! ;DHey BitcoinEagle
As a result of being hit from behind by a toy truck, a 250 g toy car, initially at rest rolls 10 m across a floor that applies a constant retarding force of 0.75N to it. The car stops 6 seconds after being hit.
a)If the truck was in contact with the car for 0.12 seconds, calculate the impulse given to the car.
b) Calculate the momentum of the car just after being hit.
c) Calculate the instantaneous speed of the car just after being hit.
d) Calculate the average force applied to the car during the collision.
myself and others are more than happy to answer your question, but before we do its important for you to have a go at it first and try to explain your current aproach to the question. Are you having difficulty with all of these parts (a,b,c,d) or is there one particular part that you are not understanding?
Having a go at the question first and showing us your thinking process helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question. And don't worry about being wrong as every mistake is an opportunity to learn and improve!
That being said, it might be good to start with the formula:
where 'I' here is the impulse
Post by: Gogurt on February 20, 2020, 08:23:46 pm
Post by: DrDusk on February 20, 2020, 09:22:43 pm
Can anyone explain why the answer is A? I've spent a good amount of time on this, and I haven't got a clue. I'm mostly confused on why the friction force is facing the same direction that the car is accelerating in.Think about it this way. Suppose a car is moving to the right. This means its tires are rotating clockwise. The bottom of the tire therefore exerts a force to the left on the road in order for the car to go forward. Friction needs to oppose this motion force which means it will point to the right.
When you think about these kind of questions conceptually you must think about exactly what the force is acting on or else you will get confused. This is because the car is moving to the right but the tires are actually rotating in the 'opposite' direction of the cars acceleration so to speak. So isolate the component that the force is actually acting on and think about it that way instead of just the whole car itself. Friction is NOT acting on the car but rather the tires and the tires follow a separate motion to the car in a sense..
Post by: Erutepa on February 20, 2020, 09:39:59 pm
Can anyone explain why the answer is A? I've spent a good amount of time on this, and I haven't got a clue. I'm mostly confused on why the friction force is facing the same direction that the car is accelerating in.Welcome to the forums Gogurt!
In addition to what Dr Dusk has said above,
In diagram A, the wheel rotates clockwise and applie a force on the ground acting to the left, consequentially the ground applied an equal and opposite frictional force acting towards the right. This causes the car to accelerate.
In diagram B, think of the wheel is slowing down/stopped as it applies a force on the ground acting to the right (this is essentially the same as imagining a stationary wheel staring to reverse - as it rotates anticlockwise, it will apply a force on the ground acting to the right. As a result, an equal and opposite frictional force will be exerted by the ground on the wheel in the left direction.
This kinda explains why the force is acting as it is, but the best way to get to the answer is by simply considering the directions of motion and the directions of the force. In diagram A, the frictional force acts in the direction of motion, causing the car to accelerate to the right, while in diagram B, the frictional force acts opposite to the direction of motion, causing the car to deccelerate.
Hopefully this helps, but feel free to point out anything you still don't unserstand/anything I may not have explained clearly. :)
Post by: Gogurt on February 23, 2020, 09:56:05 am
Think about it this way. Suppose a car is moving to the right. This means its tires are rotating clockwise. The bottom of the tire therefore exerts a force to the left on the road in order for the car to go forward. Friction needs to oppose this motion force which means it will point to the right.
When you think about these kind of questions conceptually you must think about exactly what the force is acting on or else you will get confused. This is because the car is moving to the right but the tires are actually rotating in the 'opposite' direction of the cars acceleration so to speak. So isolate the component that the force is actually acting on and think about it that way instead of just the whole car itself. Friction is NOT acting on the car but rather the tires and the tires follow a separate motion to the car in a sense..
In diagram A, the wheel rotates clockwise and applied a force on the ground acting to the left, consequentially the ground applied an equal and opposite frictional force acting towards the right. This causes the car to accelerate.
In diagram B, think of the wheel is slowing down/stopped as it applies a force on the ground acting to the right (this is essentially the same as imagining a stationary wheel staring to reverse - as it rotates anticlockwise, it will apply a force on the ground acting to the right. As a result, an equal and opposite frictional force will be exerted by the ground on the wheel in the left direction.
This kinda explains why the force is acting as it is, but the best way to get to the answer is by simply considering the directions of motion and the directions of the force. In diagram A, the frictional force acts in the direction of motion, causing the car to accelerate to the right, while in diagram B, the frictional force acts opposite to the direction of motion, causing the car to deccelerate.
Hopefully this helps, but feel free to point out anything you still don't unserstand/anything I may not have explained clearly. :)
Ah that makes so much sense, thanks a lot!
Post by: TheEagle on May 03, 2020, 07:16:01 pm
It says, "when the elevator is accelerating upwards, you feel heavier due to a larger normal force, however, I thought you will feel lighter due to gravity being less, thus less drag downwards compared to the normal force.
Post by: ashmi on May 03, 2020, 07:33:30 pm
Why is the normal force considered as your apparent weight?Hey TheEagle! ;D
It says, "when the elevator is accelerating upwards, you feel heavier due to a larger normal force, however, I thought you will feel lighter due to gravity being less, thus less drag downwards compared to the normal force.
If you think about it realistically on Earth, your weight (mg) won't change regardless of what is happening to you, however, your Normal force can change to reflect net forces. Let's say someone was 70kg and in an elevator. Their weight would be 868 Newtons ( 70* 9.8 ) and that will never change so long as they stay within the surface of Earth.
Now let's take that scenario of an elevator accelerating upwards (Fnet is up). Your weight (mg) is always constant and will never change, and due to Newton's Second Law, F = ma, for you to be accelerating upwards, your normal force will actually increase to reflect this scenario. I've drawn a diagram below to help explain this visually. Notice how the arrow length of the Normal force changes? This is to account for the situation someone in an accelerating/decelerating frame. Your acceleration on Earth is always 9.8ms^-2 DOWN no matter where you are (so long as you don't go past the surface), your mass doesn't change and as such your weight doesn't change either.
When the normal force is less than mg, you feel lighter and when it is bigger than mg, you feel heavier. You FEEL the effect of the force. Hope that helps!
(https://i.imgur.com/JNq2c0r.png)
Post by: thisissara on June 02, 2020, 01:37:28 pm
I was just wondering is there ever a time where you do use Fleming's left-hand rule in VCE physics? I've always thought that the left-hand rule was used for motors and the right-hand rule for generators? Could someone clarify this for me? if I am wrong could you also explain why? cheers! ;D
Post by: Ryanyeltrom on July 26, 2020, 08:55:37 am
I feel like I always spend so long writing out notes and stuff on the topics, but then when it comes to using the formulas and applying concepts I just can't remember how to do it or how to use the equations. My teacher isn't helping the stress levels when it comes to this subject as he has been piling on SAC's and has jumped around from topic to topic without any real demonstrations on how to do what it's asking of us.
I don't want to fail physics or get a really bad mark for the exam all because my teacher wasn't sufficient with his teaching methods or resources he has supplied us with. And I'm struggling to figure out how to get into a good study routine for physics, because it's not a subject that comes very naturally or easily for myself.
Any replies or help will be greatly appreciated, and I know this is a bit late in the year but I'm just dumb and didn't realise how little I've comprehended so far.
Thanks.
Post by: Owlbird83 on July 26, 2020, 10:18:10 am
Okay so my physics teacher at my school is new (previously retired) and he pretty much has no idea how to teach my class, even though we're an extremely small class. I'm not complaining or anything because I can't really do much about it, but I was wondering if anybody would be able to help me with tips for studying physics?
I feel like I always spend so long writing out notes and stuff on the topics, but then when it comes to using the formulas and applying concepts I just can't remember how to do it or how to use the equations. My teacher isn't helping the stress levels when it comes to this subject as he has been piling on SAC's and has jumped around from topic to topic without any real demonstrations on how to do what it's asking of us.
I don't want to fail physics or get a really bad mark for the exam all because my teacher wasn't sufficient with his teaching methods or resources he has supplied us with. And I'm struggling to figure out how to get into a good study routine for physics, because it's not a subject that comes very naturally or easily for myself.
Any replies or help will be greatly appreciated, and I know this is a bit late in the year but I'm just dumb and didn't realise how little I've comprehended so far.
Thanks.
Hey, personally I think for physics, think that spending a lot of time writing notes isn't the most helpful, and you would be better to make a cheat sheet and spend that time practice questions. I know it might seem daunting to do exam questions if you don't feel confident on understanding the content but, it will get easier and you'll start to understand it better, (I had a topic that finally 'clicked' for me while I was doing the sac haha). I know it may seem oversimplified to say 'do' the questions if you don't know where to start, but you want to sort of wing the questions until they start to seem familiar. Have you made a cheat sheet?
-I recomend scanning your notes for all the formulas and writing them down concisely and try to avoid excess info, (you can add the units, and the meaning of each of the letters/symbols too)
(here's an example of how I set out part of my cheat sheet)
Spoiler
(https://i.imgur.com/BnJA5ox.jpg)
-Then scan your cheat sheet for a formula with each of the pieces of info (eg, look through them all until you see F=nIlB)
- Check to see if they are in the correct units as they should be imputted into the formula, then sub the numbers in (rearrange if necessary), then type it exactly into your calculator.
-Remember to add units
Sorry if what I've written is something that's obvious, but I completely believe that for majority of the questions with formulas it's not necessary to understand what's going on to be able to do them. I relied heaps on my cheat sheet all the time.
For the questions that relate to understanding, it might be a good idea to familiarise yourself with the common ones that come up. Most of the worded questions that get you to explain things have been asked in slightly different ways throughout the years on past exams. If you practice them you'll start to feel more confident in knowing the certain points they want you to make each time. For example a common one is Len's law where you have to explain which way the current flows and why.
Spoiler
-flux will be increasing in coil towards right
-to oppose this, a flux is created in coil towards left
-therefore the induced current is upwards in the wire
(basically always these three points, but change the directions depending on question)
-to oppose this, a flux is created in coil towards left
-therefore the induced current is upwards in the wire
(basically always these three points, but change the directions depending on question)
Since you've mentioned it being confusing with the jumping around different topics, maybe when you are studying, only focus on one AOS at a time until you get confident with it. Physics has super separate areas, so you want to be able to identify what areas the questions are from.
In summary, making cheat sheet and doing past questions is going to be the most helpful.
Don't worry if you don't understand everything right now, you aren't alone, (I never fully understood the last area of study, and blindly subbed the numbers into the formulas and it worked well for me).
Hope this helps, good luck :)
Post by: a weaponized ikea chair on August 24, 2020, 09:31:25 pm
I am not doing VCE, but am struggling to understand the mathematical formula for average velocity. Is it the change in displacement over the change in time, or is it the total displacement over the total time taken, or is it the displacement at the position in question over the total time taken? I have heard all three and am confused.
Any help would be greatly appreciated.
Post by: Chocolatepistachio on August 24, 2020, 09:38:26 pm
Post by: rozmaaate on September 06, 2020, 07:45:46 pm
Post by: Owlbird83 on September 06, 2020, 08:30:27 pm
Hey guys I’m super confused with this question as I thought in order to calculate change in velocity in 2 dimensions you are suppose to add the 2 vectors like this, v+(-u), However this isn’t shown in the solution for this questions, as the drawing shown should be drawn head to tail , so 16m/s south west then -u which is 20m/s north west.
If the ball moved in a straight line forwards or back you could do do the calculations like you said, but because it's going in the south west then north west directions you have to draw out the vectors in a triangle, and calculate the resultant vector (the one connecting from the start of the initial to end of the final) using pythagoras to find the change in velocity. And because velocity is a vector you need the direction too, so to find the direction of the resultant vector/change in velocity, you must find the angle in the triangle in the picture in the answers, then add 45deg, because you know the initial velocity is 45deg from south (south-west).
I hope that answers your question? Let me know if you are still confused.
Post by: rozmaaate on September 06, 2020, 09:08:31 pm
If the ball moved in a straight line forwards or back you could do do the calculations like you said, but because it's going in the south west then north west directions you have to draw out the vectors in a triangle, and calculate the resultant vector (the one connecting from the start of the initial to end of the final) using pythagoras to find the change in velocity. And because velocity is a vector you need the direction too, so to find the direction of the resultant vector/change in velocity, you must find the angle in the triangle in the picture in the answers, then add 45deg, because you know the initial velocity is 45deg from south (south-west).
I hope that answers your question? Let me know if you are still confused.
Thanks for your time , unfortunately I’m still confused here’s what I have so far
Post by: Owlbird83 on September 06, 2020, 09:38:38 pm
Thanks for your time , unfortunately I’m still confused here’s what I have so far
All good!
Sorry, when I referred to the angle in the triangle in my last post, I muddled it up a little.
Although the answers suggest calculating the angle on the right side of the triangle, I reckon it's easier to do it like this:
(also I hope this isn't confusing, I'm calculating a different theta than they are in the answers, cos it seems more simple to me).
my drawing
(https://i.imgur.com/BTA4zPf.jpg)
You want to find the direction of the final velocity, and you can find the angle that I found in my picture, and add this (38.7) to 45 which equals S83.7W.
Spoiler
Alternatively, you have found the angle on the right side correctly, so since you know there's 180deg in a triangle you can do 180-90-51.3=38.7 which will be the angle on the left side
Let me know if you want more clarification! :)
Edit: oh no I did this the wrong way around, ahhh I'm sorry, I hope I didn't make you more confused? :-\
I feel bad, ignore what I said cos I did it all wrong and am also confused
Post by: ImamB1 on September 07, 2020, 07:02:13 am
Does anyone have a list of the questions removed from the past VCAA Physics examinations. I’ve only found a list from 2017 onwards but I’m looking to go much prior to that.
Post by: Newton is Nice on September 24, 2020, 09:09:16 pm
Anyone's help would be greatly appreciated.
Post by: Bri MT on September 24, 2020, 10:21:28 pm
Hello, I was just curious to know how an AC generator reverses the current of the coil even though it only has slip rings? I though only a split ring commutator can reverse the current of the coil.
Anyone's help would be greatly appreciated.
Hey!
In an AC generator the reason you have current going in the opposite direction sometimes is because the change in flux is in the opposite direction & therefore the induced current is in the other direction so that the change is still opposed; the change in direction is definitely not because of slip rings "swapping it" or anything. I would recommend not saying "an AC generator reverses the current of the coil" because then the person marking you might think you're getting confused with how split ring commutators work.
I recommend looking at diagrams of flux with the coil diagrams and trying to figure out what the direction of current should be at different points to consolidate this info.
I hope this helps, please feel free to reply with any further questions :)
Post by: Newton is Nice on September 25, 2020, 11:13:25 am
Hey!Thanks for replying. So essentially, it s the magnets that are reversing, right?
In an AC generator the reason you have current going in the opposite direction sometimes is because the change in flux is in the opposite direction & therefore the induced current is in the other direction so that the change is still opposed; the change in direction is definitely not because of slip rings "swapping it" or anything. I would recommend not saying "an AC generator reverses the current of the coil" because then the person marking you might think you're getting confused with how split ring commutators work.
I recommend looking at diagrams of flux with the coil diagrams and trying to figure out what the direction of current should be at different points to consolidate this info.
I hope this helps, please feel free to reply with any further questions :)
Post by: Newton is Nice on October 01, 2020, 03:26:37 pm
Post by: Bri MT on October 01, 2020, 03:29:21 pm
Thanks for replying. So essentially, it s the magnets that are reversing, right?
No worries. The relative position of the magnets and coil are cycling yeah.
- Why is the frictional force equal to accelerative force?
More context is needed for this question
Post by: Newton is Nice on October 02, 2020, 09:22:05 pm
No worries. The relative position of the magnets and coil are cycling yeah.
Hi thank you once again for replying. So, if commutators reverse the current in the coil, but AC generators do not use commutators, so they have the reversion for current due to the change of direction of magnets (relative) right? If so, why do DC generators require split ring commutator for reversion of current and AC generators do not?
Post by: Newton is Nice on October 09, 2020, 09:40:01 pm
Post by: S200 on October 09, 2020, 11:16:20 pm
So, if commutators reverse the current in the coil, but AC generators do not use commutators, so they have the reversion for current due to the change of direction of magnets (relative) right?An AC motor really only works because the magnetic field changes in sync with the phase of the AC current.
Because the magnetic field flips 180 degrees you dont have the deceleration issue that affects DC motors.
Challenge
Once you have that locked away, try to work out how to do an AC motor with permenant magnets.
If so, why do DC generators require split ring commutator for reversion of current and AC generators do not?DC Motors require split ring commutators due to the constant magnetic field given by the permenant magnets.
Under Lenz's law, the rotor would end up completely perpendicular to the magnetic field if you didn't have a split-ring commutator.
(https://upload.wikimedia.org/wikipedia/commons/7/73/Ejs_Open_Source_Direct_Current_Electrical_Motor_Model_Java_Applet_%28_DC_Motor_%29_80_degree_split_ring.gif)
Post by: PhysicsS on November 11, 2020, 09:40:35 pm
I have a question about Luminosity I'm hoping to get some help with.
Q is: stars of Deneb, and Mimosa, "appear to be the same brightness", with Deneb at 3230 light years away, and Mimosa at 353 light years. Which star has the greater luminosity, and by how much?
Clearly Deneb's intrinsic brightness or luminosity would be stronger, if they appear similarly bright, despite Deneb's much further distance. The second part of the Q is the area we're unsure of.
An example compares the luminosity of Sirius, at a distance of 8.61 light years, to that of the sun. It states Sirius's apparent brightness as 8.8x10^-11. So in a worked example, it coverts 8.61 to AU (x 63 240), to determine the solar distance ratio (R/R⊙), = 5.44x10^5. The question states "we know the brightness ratio, b/b⊙" which is 8.8x10^-11". They then work it all out: L/L⊙ = b/b⊙ x (R/R⊙)^2 =8.8x10^-11 x (5.44x10^5)² = 26. This makes sense.
However, my query is, how exactly did they determine Sirius's apparent brightness?
In previous pages, info is given on apparent magnitude scales (i.e. the sun at -26.7), which prompted us to find Deneb's (+1.25) and Mimosa's (+1.25), there is mention but no example in the book, of the logarithmic scale it's all based on. The book also says "the brightness ratio can be worked out from the apparent magnitudes, but that is a little messy so we shall just quote the brightness ratio directly" ??? :-\
There's another worked example of the sun's luminosity, which uses it's watts per square metre on earth (1370), to calculate it's luminosity: L = b x 4πR² (b as brightness in watts, and 4πR² being the area of a sphere, R as the distance in AU (1) yet converted to metres: 1.5x10^11), it's also described as b = L/4πR².
The page also states the inverse law, or brightness declining as radiation increases with distance, written as a ratio of y₁/y₂ = (x₂/x₁)², or propto of y ∝ 1/x².
Hence, do we use the inverse law to compare the stars? if so, which values do we substitute for x and y (some online examples use AU, most American resources quote in parsec), from the question we're not sure all or the correct info is given, or is it again just a matter of conversions?
Is it best to use the calculation in the Sirius example to determine each stars' brightness ratio to the sun, then compare them to each other? (bearing in mind no watts radiation values are given, and internet searches yield varied results, but most tend to state info like "Deneb is 200,000x brighter than the sun", which I think is only the calculation result.)
Thus, which calculations are used to find each stars' luminosity, or brightness, especially if the equation with Sirius, is used?
Sorry again for all the info, just wasn't sure where exactly to look or which steps to use when (the question's driving me nuts, and has been for 6+ hours!)
Thanks so much again!
Post by: a weaponized ikea chair on November 12, 2020, 03:00:57 pm
Redshift is when the wavelength of light has been shifted to the red part of the electromagnetic spectrum.
A) True
B) False
I put false, but I'm not sure.
Post by: LE-0130 on January 12, 2021, 02:25:15 pm
I was just hoping to get help for this physics question. I'm sort of new to mechanical waves and I don't quite understand the answer for this question. Any help would be appreciated! Thanks :)
Post by: Bri MT on January 12, 2021, 02:46:46 pm
Welcome to the forums :)
This question is testing your understanding of how transverse waves work. Imagine that the graph is showing an ocean wave, you if you picked one location in the ocean, the water level would go up and down and up and down as the wave moves along. For this question, the locations you are interesting in are u and v - where are those particles going?
It might help you to watch a video or look at a gif explaining the basics of transverse waves if you're finding this very confusing.
Post by: Cupcake2423 on February 10, 2021, 06:53:11 pm
i know for the end of year exam i'm allowed to type up my notes (and its 2 double-sided A4 paper right??)
But for my SACs can i also type up my notes (like the summary sheet)??
Post by: Owlbird83 on February 10, 2021, 08:47:44 pm
hey!!Hey!
i know for the end of year exam i'm allowed to type up my notes (and its 2 double-sided A4 paper right??)
But for my SACs can i also type up my notes (like the summary sheet)??
yep for size of exam notes
I'm sure you would be able to type and print out sac cheat sheets, I don't see why teachers wouldn't accept that unless they've told people it must be handwritten. I guess it depends on your school/teacher, but my school was fine with whatever for sacs so long as it was the size they told us
Post by: Roger Luo on March 26, 2021, 06:40:16 pm
Post by: Coolmate on March 26, 2021, 07:06:22 pm
I am quite ahead for physics, and don't know what to do, right now I am done in Unit 3 for physics, but our school is only halfway through it, should I start revising practice exams for unit 3, or get the other jacaranda book and do the questions of that. tbh, I don't know how to juggle Unit 3 review with stronger SAC prep.
Hey Roger Luo!
Welcome to the forums :)
It is great that you are ahead for VCE Physics, I didn't do VCE Physics, but I do strongly suggest starting practice questions early. The thing that really helped me during my HSC was applying the content to questions, rather than straight memorisation as you can get any question on the day.
If you run out of questions from past VCE papers, you could go through past HSC papers here and here and sort out any relevant questions to the VCE syllabus.
Good luck with everything :)
Coolmate 8)
Post by: ErnieTheBirdi on May 25, 2021, 02:16:18 pm
Post by: Samueliscool223 on July 18, 2021, 09:40:13 pm
Post by: Rachelrachel on July 25, 2021, 05:34:02 pm
"a railway wagon of mass 2.5 tonnes moving along a horizontal track at 2 m/s runs into a stationary engine and is coupled to it. After the collision, the engine and the wagon move off at 0.3 m/s. What is the mass of the engine alone?"
I used this formula:
m1u1 + m2u2 = m3v3
to obtain:
2.5*2+0 = (2.5+m)*0.3
This resulted in m=14.17 tonnes. However, my textbook says that 4.2 tonnes is correct. I have no idea how to obtain this answer.
Can anyone offer any insight?
Post by: ArtyDreams on July 25, 2021, 07:00:41 pm
Hey guys, I'm completing some Unit 1&2 conservation of momentum questions and am really confused by this one.
"a railway wagon of mass 2.5 tonnes moving along a horizontal track at 2 m/s runs into a stationary engine and is coupled to it. After the collision, the engine and the wagon move off at 0.3 m/s. What is the mass of the engine alone?"
I used this formula:
m1u1 + m2u2 = m3v3
to obtain:
2.5*2+0 = (2.5+m)*0.3
This resulted in m=14.17 tonnes. However, my textbook says that 4.2 tonnes is correct. I have no idea how to obtain this answer.
Can anyone offer any insight?
Hi! Your working out is correct but you have to make sure that you convert the tonnes to kilograms. In the formula p=mv, mass is in kilograms and v is in m/s, to get your momentum being kg/m/s. Hope this helps!
Post by: Rachelrachel on July 25, 2021, 07:57:54 pm
you have to make sure that you convert the tonnes to kilograms.
1 tonne = 1000 kg. Therefore converting to kg doesn't change the actual digits of the answer, only its magnitude (the mass works out to be 14167 kg instead of 14.17 tonnes). I'm still stuck on how the textbook gets an answer of 4200 kg.
Post by: ashmi on July 25, 2021, 09:41:25 pm
Hey guys, I'm completing some Unit 1&2 conservation of momentum questions and am really confused by this one.
"a railway wagon of mass 2.5 tonnes moving along a horizontal track at 2 m/s runs into a stationary engine and is coupled to it. After the collision, the engine and the wagon move off at 0.3 m/s. What is the mass of the engine alone?"
I used this formula:
m1u1 + m2u2 = m3v3
to obtain:
2.5*2+0 = (2.5+m)*0.3
This resulted in m=14.17 tonnes. However, my textbook says that 4.2 tonnes is correct. I have no idea how to obtain this answer.
Can anyone offer any insight?
First things first, I don't see anything wrong with your working out here. Just as Arty has mentioned above, it would be preferable if you kept the momentum to kg/m/s (just in case you lose track of units). I believe your answer (14.17 tonnes) is correct.
Post by: parieeelol on August 11, 2021, 12:29:21 pm
The angle of incidence was calculated to be 48 degrees, and angle of refraction as it enters the prism was 28.6 degrees. I used Snell's Law for both. But other than that, I have no idea how to tackle this question and would appreciate any help.
Post by: SmartWorker on August 13, 2021, 12:00:47 pm
Hey guys - I've been stuck on this question for a while now, and I believe it's asking for the angle of emergence, but I have no idea how to calculate it and the solutions aren't making sense. (Unit 4 Physics, AOS1)
The angle of incidence was calculated to be 48 degrees, and angle of refraction as it enters the prism was 28.6 degrees. I used Snell's Law for both. But other than that, I have no idea how to tackle this question and would appreciate any help.
Hi, i tried the question, i don't know if my solution is correct but i hope this at least helps a bit.
So yep the angle of refraction = 28.6 degrees.
Then notice: the triangle is equilateral. Make a right angled triangle (should be able to do this as the dotted line is the normal line) and then you can solve from there.
(https://i.imgur.com/KEHTH9h.png)
Post by: miyukiaura on August 14, 2021, 04:36:21 pm
Thanks!
Post by: mcpunjavu on September 07, 2021, 09:04:18 am
worth 3 marks
Explain with reference to quantised energy levels why only photons of certain wavelengths can be emitted by the atom.
thanks in advance
Post by: Billuminati on September 07, 2021, 09:37:47 am
Just about to do our final sac in physics, and no quite sure how to answer this question:
worth 3 marks
Explain with reference to quantised energy levels why only photons of certain wavelengths can be emitted by the atom.
thanks in advance
I can answer your question based on the knowledge I learned in my 2nd year uni physical + analytical chem unit, with the caveat that the required explanation in VCE physics may be different. Basically what they mean by quantised is that in a chemical species, when electrons are excited by energy and move up to an orbital of higher energy, the associated quantity of energy absorbed (we call this a transition) is highly specific. Hence, this specificity translates to a very specific wavelength emitted given you already know the speed (c) and frequency (f) of the wave. The energy of each energy transition is usually plotted in what's known as a Grotian diagram.
That's why atomic absorption spectroscopy is highly selective, because each chemical species has distinct absorption bands that rarely overlaps with others so you'll almost never get a false positive or overestimation of concentration because some other chemical species with a similar absorption band is present. Basically how AAS works is a lamp containing the element of interest is used to emit light of the same wavelength of absorption, this light source is shined onto a burner which converts the element of interest in your sample into gaseous, ground state atoms which can absorb this electromagnetic radiation. This absorbance is detected by computers, but not before being wavelength-restricted to the intended wavelength by a monochromator
Post by: mcpunjavu on September 09, 2021, 07:38:17 pm
I can answer your question based on the knowledge I learned in my 2nd year uni physical + analytical chem unit, with the caveat that the required explanation in VCE physics may be different. Basically what they mean by quantised is that in a chemical species, when electrons are excited by energy and move up to an orbital of higher energy, the associated quantity of energy absorbed (we call this a transition) is highly specific. Hence, this specificity translates to a very specific wavelength emitted given you already know the speed (c) and frequency (f) of the wave. The energy of each energy transition is usually plotted in what's known as a Grotian diagram.
That's why atomic absorption spectroscopy is highly selective, because each chemical species has distinct absorption bands that rarely overlaps with others so you'll almost never get a false positive or overestimation of concentration because some other chemical species with a similar absorption band is present. Basically how AAS works is a lamp containing the element of interest is used to emit light of the same wavelength of absorption, this light source is shined onto a burner which converts the element of interest in your sample into gaseous, ground state atoms which can absorb this electromagnetic radiation. This absorbance is detected by computers, but not before being wavelength-restricted to the intended wavelength by a monochromator
thats super helpful, thanks so much!
Post by: Newton is Nice on September 26, 2021, 03:14:29 pm
2017 NHT- asked why beyond a certain voltage (1V), there is no increase in photocurrent, and here is what they said:
"At V = +1.0 V all of the available photoelectrons are being collected. Since there are no more
photoelectrons to be collected, increasing the voltage will not result in an increase in photocurrent."
However, if I were to increase the intensity of the light, then there would be an increase in the number of electrons emitted, thereby implying that there are still photoelectrons available in the metal that have not been emitted. This would contradict them saying "there are no more photoelectrons to be collected".
It would be great if anyone could kindly let me know if my reasoning is wrong, or if I am simply misinterpreting what they are saying.
Cheers.
Post by: eman27_hc on October 29, 2021, 04:13:16 pm
Hi guys - got a question regarding saturation current:
2017 NHT- asked why beyond a certain voltage (1V), there is no increase in photocurrent, and here is what they said:
"At V = +1.0 V all of the available photoelectrons are being collected. Since there are no more
photoelectrons to be collected, increasing the voltage will not result in an increase in photocurrent."
However, if I were to increase the intensity of the light, then there would be an increase in the number of electrons emitted, thereby implying that there are still photoelectrons available in the metal that have not been emitted. This would contradict them saying "there are no more photoelectrons to be collected".
It would be great if anyone could kindly let me know if my reasoning is wrong, or if I am simply misinterpreting what they are saying.
Cheers.
"if I were to increase the intensity of the light, then there would be an increase in the number of electrons emitted, thereby implying that there are still photoelectrons available in the metal that have not been emitted."
I think that's where your logical reasoning goes wrong. I don't believe that jump in logic is sound. Just because more electrons are being emitted from the light does not mean there are still more available electrons on the cathode that are able to be ionized.
Post by: james.358 on October 29, 2021, 06:21:23 pm
I just want to quickly add that the problem with Newton is Nice's answer isn't that it is necessarily wrong, it just doesn't answer what the question asks for. The question says why an increase in voltage doesn't lead to an increase in photocurrent. This is because all the available photoelectrons are ejected, so an increase in voltage (which basically makes the receiving plate more "attractive") wouldn't do anything.
However, an increase in intensity would increase the photocurrent. Remember that this is a circuit, so if you increase the intensity, the rate of incident photons will increase, so more electrons will be ejected and hence a higher current will be achieved.
Quote
are still more available electrons on the cathode that are able to be ionized.
This is a very minor detail, but note that the electrons are not being ionised. Ionisation would imply that the metal becomes positively charged after losing the electron. However, as the photelectric effect takes place in a circuit, the lost electrons are replenished and hence no ions are formed.
Hope this helps,
James
Post by: miyukiaura on November 05, 2021, 06:46:55 pm
Post by: eman27_hc on November 05, 2021, 08:13:33 pm
Hey guys, I just wanted to know what you think was the hardest exam, and the most poorly done topics in physics?
Hi, from the past VCAA exams that I've done, I don't think there's any particular exam that stands out to be clearly 'harder' than the others. In terms of the big practical investigation problem that they include every year, the 2019 VCAA paper is probably the hardest/most different from other years - just because it comprises of two strings which are concurrently used, and so has the potential to throw some people off.
In terms of which topics are done most poorly, I'm not sure with 100% certainty which topic it is. Though from what I can recall, I believe the chief assessor for physics said it was special relativity. In my opinion, the most poorly done problems are generally the explanation type questions. This is from looking at the chief examiner's report.
Hope it helps
Post by: eman27_hc on November 05, 2021, 09:01:44 pm
Why is answer A here wrong (q. 12)? isnt the velocity function a linear equation for objects falling under the influence of gravity, and hence isnt the speed of da object increasing at a constant rate? i wouldve thought both A and C are correct but apparently its just C
Hi, for physics, you have to break down the velocity into its vertical and horizontal components. Expanding upon your statement "isnt the velocity function a linear equation for objects falling under the influence of gravity", this assertion only applies to the vertical component of the velocity. This will be important to note.
Secondly, the speed of the object is a combination of the vertical and horizontal component of the velocity. Hence, you have to apply Pythagoras theorem to find the speed of the object. I've attached the mathematics behind how this works.
If you look at the final speed equation that I arrive at, it is clear that the speed does not increase at a constant rate - it is actually quadratic expression that is all under a square root!
Hopefully this helps!
Post by: miyukiaura on November 06, 2021, 12:30:59 pm
Thanks so much!
Post by: james.358 on November 06, 2021, 12:38:19 pm
Post by: TnGn74 on June 01, 2022, 06:21:04 pm
Post by: nillyadis on June 20, 2022, 08:40:27 pm
Post by: nillyadis on June 20, 2022, 09:20:05 pm
Is is because:
it uses a vertical filter
the vertical filter stops every transverse wave that doesn't oscillate vertically and only allows a vertically oscillating wave to pass through
a longitudinal wave oscillates parallel to the direction the wave propagates
it is not longitudinal because we do not observe all waves passing through the filter,
Post by: Bri MT on June 22, 2022, 08:33:44 pm
How does polarisation support the theory that light is a transverse wave and not longitudinal?
Is is because:
it uses a vertical filter
the vertical filter stops every transverse wave that doesn't oscillate vertically and only allows a vertically oscillating wave to pass through
a longitudinal wave oscillates parallel to the direction the wave propagates
it is not longitudinal because we do not observe all waves passing through the filter,
Polarisation selects a particular direction (not necessarily vertical) of oscillation. Evidence for this is indicated by light being able to pass through one polarisation filter, but not a second one that selects for the perpendicular direction of oscillation. In longitudinal waves, oscillation is parallel to the direction of wave movement whereas in transverse waves particles oscillate perpendicular to the direction the wave is moving in. Thus, only transverse waves undergo polarisation, and light being able to be polarised suggests that it is a transverse wave.