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Author Topic: HSC Physics Question Thread  (Read 1043152 times)  Share 

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jamonwindeyer

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Re: Physics Question Thread
« Reply #750 on: August 03, 2016, 10:06:58 pm »
+1
What is the action and reaction force(s) when it comes to running, swimming and rowing a boat?

Does the water affect this at all?

It's all pretty much the same I think! You apply a force on the ground/water, the reaction force propels you forward! ;D the difference being I suppose that for the water, it is the buoyancy of the water rather than the normal reaction force at play :)

MysteryMarker

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Re: Physics Question Thread
« Reply #751 on: August 03, 2016, 10:27:37 pm »
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Hey guys,

Could someone explain this question to me and how the answer is B? Wouldn't it be A as I = V/R and therefore the current is directly proportional to an increase in voltage?

Cheers.

jamonwindeyer

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Re: Physics Question Thread
« Reply #752 on: August 03, 2016, 10:43:20 pm »
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Hey guys,

Could someone explain this question to me and how the answer is B? Wouldn't it be A as I = V/R and therefore the current is directly proportional to an increase in voltage?

Cheers.

Hey there! I'm very happy you asked this question because it is an important principle to remember: Cathode/anode setups do not obey Ohm's Law! The relationship is more complicated, Ohm's Law only applies in certain situations and this isn't one of them ;D

That being said, we can approach this question by process of elimination.

Graph D is probably the most incorrect, since it implies that the current is infinite with no voltage applied. That is wrong on many levels, so no go.

Graph A is linear, which we would not expect in this situation.

We can explain that the answer is B pretty simply even without knowing that though. If there were no voltage source (that is, V=0), we would still expect some current flow. Why? Because of the photoelectric effect. The photoelectric effect causes a small photocurrent to flow through the circuit, completely independent of the applied voltage. The only graph that has a current flowing when V=0 is B, so, the answer must be B by default ;D

Aliceyyy98

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Re: Physics Question Thread
« Reply #753 on: August 04, 2016, 12:16:10 am »
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Hey Alice!

So they are applied ever so slightly differently. Are you okay with the Physics behind how the deflection plates actually work? Let me know if you aren't, but assuming you are, it's all about what external signal the horizontal and vertical deflection plates are aligned to.

In Oscilloscopes, the horizontal deflection is controlled by a time varying signal. This causes the beam to sweep horizontally across the screen, from left to right, at a constant pace. The vertical deflection is controlled by some other signal, a measurement (for example, a microphone). What this creates is a vertical axis dependent on signal strength of some stimuli, and a horizontal axis dependent on time. Think of your typical heartbeat monitor in the hospital drama shows, that is exactly what I'm talking about ;)

In Cathode Ray Televisions, both the horizontal and vertical deflection plates have time varying strengths. The idea here, put simply, is that it makes the electron beam sweep across the entire television screen. It sweeps along the top row of phosphor dots (kind of like pixels on your laptop), then the second row, then the third row, so quick that you just see the solid image.

There are 3 electron guns in Colour CRT's, each corresponding to either Red, Green, or Blue. Each is tasked with creating a specific colour on the screen (there are different phosphors, one for each colour). The intensity of each is controlled with a separate accelerating anode  :D

This explanation is definitely lacking a bit in certain areas, there is some pretty complex stuff involved, but this is a good overview. Is there any specific part of this that is a bit iffy for you?

(Image removed from quote.)

Thank you Jamon! I'm not quite sure how the signals in deflection plates work actually...  :'( could you clarify for me!

Cheers!

jamonwindeyer

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Re: Physics Question Thread
« Reply #754 on: August 04, 2016, 09:19:27 am »
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Thank you Jamon! I'm not quite sure how the signals in deflection plates work actually...  :'( could you clarify for me!

Cheers!

Sure!

Okay, so let's just assume we are working with electric field deflection (you can apply the same idea to magnetic). The idea is that by controlling the electric field strength and direction, we can change the path of the electron beam. In this case, this would be done by applying different voltages to the deflection plates  :)

Let's say we can apply a voltage between 100V and -100V to the plates (same max magnitude, just opposite polarity) to a pair of plates controlling vertical deflection. 0V would mean the electron beam is unaffected, and travels straight. If we apply 100V, the beam might (for example) be deflected such that it hits the very top of the screen. 99V means it hits JUST below that. 98V, just below that again. All the way to zero where we are back in the centre. Then, -1V, just below the centre. And the process continues, all the way to -100V where the beam hits the bottom of the screen.

If we do this for both horizontal and vertical deflection, we can hit ANY point on our two dimensional screen. The top right corner might be 100V applied to vertical plates, 100V applied to horizontal plates. A point in the bottom right would be -100V vertical, but still 100V horizontal. We adjust the voltages of each pair to move up/down, or left/right.

We can do creative things with this ideas (again, just assume electrical is used for everything to keep things numerically simple). For example, in an Oscilloscope, the horizontal plates have a time varying voltage. We start by applying -100V, and the beam is at the very left. We gradually increase this to 0, and then to 100, to cause the beam to sweep across the screen. Then, when we hit 100V, we immediately reset to -100V and start the sweep again (for prospective electrical engineers, this signal would be called a sawtooth wave).

So we have an electron beam sweeping across the middle of the screen (maybe once a second). If we then connect the vertical deflection plates to some source we want to measure (for example, an electrode connected to the chest to measure heartbeat), that means the electron beams vertical position is dictated by what we are measuring. Thus, we then have your typical cardiograph, with the heartbeat visible as it fluctuates over time! The beam sweeps across, and jumps up and down as it sweeps in correspondence with the heartbeat.

Television screens are more complex, but basically, both the vertical and horizontal deflection plates are time varying. The periods are set up such that the beam does a sweep across the top row of the screen, then the next row in the reverse direction, then the next row, etc etc. Remember Donkey Kong? How you'd go all the way to the right, then jump, then all the way to the left, then jump, etc? That's what happens here (roughly speaking). It's a little tougher to picture, but let's say we start in the top left corner:

Horizontal = 100V, Vertical = 100V

Then we want the 'phosphor dot' to its right:

Horizontal = 99V, Vertical = 100V

We keep sweeping to the right until we get to the top right corner:

Horizontal = -100V, Vertical = 100V

Then we shift down:

Horizontal = -100V, Vertical = 99V

Then we begin sweeping left:

Horizontal = -99V, Vertical = 99V

And the process repeats in this fashion  ;D


Neutron

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Re: Physics Question Thread
« Reply #755 on: August 04, 2016, 10:01:06 am »
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Hey there! I'm very happy you asked this question because it is an important principle to remember: Cathode/anode setups do not obey Ohm's Law! The relationship is more complicated, Ohm's Law only applies in certain situations and this isn't one of them ;D

That being said, we can approach this question by process of elimination.

Graph D is probably the most incorrect, since it implies that the current is infinite with no voltage applied. That is wrong on many levels, so no go.

Graph A is linear, which we would not expect in this situation.

We can explain that the answer is B pretty simply even without knowing that though. If there were no voltage source (that is, V=0), we would still expect some current flow. Why? Because of the photoelectric effect. The photoelectric effect causes a small photocurrent to flow through the circuit, completely independent of the applied voltage. The only graph that has a current flowing when V=0 is B, so, the answer must be B by default ;D

Wait wait but just with this question, when the voltage is equal to the stopping voltage, shouldn't the current be zero since no photoelectrons actually reach the cathode? Why do none of the graphs account for this D:

Neutron

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Re: Physics Question Thread
« Reply #756 on: August 04, 2016, 10:13:19 am »
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Ahh it's the morning of the exam and I still have a crap tonne of questions (not a good sign, rip)

Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)

Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)

Neutron

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Re: Physics Question Thread
« Reply #757 on: August 04, 2016, 10:20:58 am »
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Here are the others, genuinely so sorry but freaking out and no way to ask teacher :/

Adios amigos

Rip me HAHAHA

Thank you, thank god my exams in the afternoon

7.The Earth’s radius at the equator is close to 6 380km, and its rotational period is equal to the rotational period of the Earth. If a new satellite launching station were established at Woomera in South Australia, which of the following would best describe the launch velocity advantage, given its intended orbit?

(A) 464 m s-1 Intended orbit from east → west

(B) more than 464 m s-1 Intended orbit from south → north

(C) 464 m s-1 Intended orbit from west → east

(D) less than 464 m s-1 Intended orbit from west → east


When you have a current carrying conductor that's travelling diagnoally to the right through a magnetic field coming out of a page, can it experience a force to the bottom right? Or only to the bottom? (won't elt me screenshot)

Ahh apparently the file size is too big, will just have to bathe in this ignorance hahah

jamonwindeyer

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Re: Physics Question Thread
« Reply #758 on: August 04, 2016, 10:38:56 am »
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Ahh it's the morning of the exam and I still have a crap tonne of questions (not a good sign, rip)

Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)

Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)

You'll be fine Neutron!! Just relax mate, you'll be sweet  ;D

First question, leave it with me, nothing strikes me immediately.

Next, the answer is definitely D. Orbital decay causes the velocity of the satellite to decrease, thus causing its orbital radius to drop. However, the drop in orbital radius causes an increase in orbital velocity (smaller orbit, needs to move faster to stay in orbit), so it actually ends up travelling faster than it did before the drop. The net effect? Orbital decay causes satellites to gradually spiral inwards towards earth, travelling faster and faster as it goes.

Next, there is definitely changing flux in the answer. However, the changing flux on opposite sides of the loop cancel each other out. One side moves down, the other moves up,, the net effect is no change in magnetic field strength and thus no induced current. Every change in flux on one side is opposite in sign and equal in magnitude to the change in flux directly opposite. Everything cancels, no induced current. The answer is therefore B. Every other example, there is some change in magnetic field strength in some part of the loop. The diagram is pretty bad, the magnet is at the dead centre of the loop, which is why this occurs. If the magnet was somewhere closer to one side of the loop the outcome would be different.

Question 11, I agree with you, I would say the answer is C not D (any takers? Anything I missed?)

Question 7, you would normally need to calculate the velocity of the earth's surface at the equator. We can do this by considering the distance travelled (one circumference, the length of the equator) divided by the time taken (24 hours). However, the question doesn't really need it because it's the same number everywhere. If we launch from Woomera, the surface of the earth isn't moving as quickly as the equator.

To demonstrate this, pick up a ball of any kind. Put black dot on the very edge of the ball (where the equator would be), and then put one lower down. If you spin it, the black dot on the equator rotates faster. Why? Because it has a greater distance to travel in the same amount of time. The 'circumference' circumnavigated from Woomera is lower. Thus, the velocity is lower. We won't get as much benefit there. Therefore, the answer must be D.



jamonwindeyer

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Re: Physics Question Thread
« Reply #759 on: August 04, 2016, 10:42:44 am »
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Question 15, I have no idea how a fluorescent screen can be a particle? That makes no sense to me aha ;D not sure about this one either  :P


jakesilove

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Re: Physics Question Thread
« Reply #760 on: August 04, 2016, 10:46:32 am »
+1
Ahh it's the morning of the exam and I still have a crap tonne of questions (not a good sign, rip)

Question 3: How on Earth do you do it? The answer is C
Question 5: The answer is D, but is it correct to say that the satellite 'gains velocity' cause I thought the point was that they lost velocity and therefore could not maintain orbit and started falling.
Question7: Again, no idea how to do :/ Answer is D
Question 9: Answer is B :O But isn't that the set up of electromagnetic braking?? So eddy currents are induced on either sides? Is there a difference to the disk rotating relative to the magnet and the magnet rotating relative to the disk? :/
Question 11: The answer is D, does this mean you can't have a force in the sideways direction? Cause I thought it would go bottom right..
Question 13: Can't it technically be the torque as well? Answer is C :/
Question 15: I thought it was C but the answer is D (how tf is it the fluorescent screen, am I just dumb?)

Sorry for the bombardment, it's okay if you guys can't get through all of them! (i'll attach the other screenshots of the questions in a new post below)

Do I have a story about that first question: Gather round children.

I topped my Physics Trial by one mark, and it was because of this multiple choice question. How do you answer it? It's something about calculating the change in potential and kinetic energy, and subtracting terms. Just calculate the initial potential energy, the final kinetic+potential energy, and find that some energy is missing. By conservation of energy, that had to go somewhere (ie. heat and light).

However, Jake wasn't that smart in the exam. Instead, I noticed that D-A=C. By Multiple Choice theory, I assumed that D and A were tricks, something that could be calculated, but that the LOSS would be some sort of 'difference'. So, I put C, the correct answer. Round of applause.

What should you take from this? Be smart with you multiple choice questions. Even if an answer doesn't come to mind, you can logic it out almost every time. You'll be fine mate
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jamonwindeyer

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Re: Physics Question Thread
« Reply #761 on: August 04, 2016, 10:48:11 am »
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Better working in my next post thanks to Jake's suggestion, but this shows how far off you are when you neglect potential energy  ;)

Work is equal to the force applied over the distance. We need the gravitational force on the daredevil (which for this question, we can approximate with W=mg) applied over 10km, to figure out how much work is done accelerating the guy.



However, if we calculate the kinetic energy of the daredevil:



Edit: Indeed Jake! I think the calculation above is close, because the difference between the two is the same as C, off by a factor of 1000.
« Last Edit: August 04, 2016, 11:00:18 am by jamonwindeyer »

jakesilove

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Re: Physics Question Thread
« Reply #762 on: August 04, 2016, 10:53:39 am »
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Question 15, I have no idea how a fluorescent screen can be a particle? That makes no sense to me aha ;D not sure about this one either  :P

For 15, I think the answer has to be A right? First of all, it is clear that the difference between X and Y is either than one is more charged than the other, or one is faster than the other. That already only really leaves us with A. Gamma particles have no charge, so Z makes sense, and Beta particles will experience a force in the opposite direction to the Alpha particles (X,Y) so that makes sense to.
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jamonwindeyer

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Re: Physics Question Thread
« Reply #763 on: August 04, 2016, 10:57:22 am »
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Let me have another go and integrate potential energy this time, based on Jake's suggestion.



Now we take the potential energy and kinetic energy after the 10km drop:



The difference between those two results gives the answer of C, that's better!!  ;D

Ignore the above working, this is more correct and yields the correct answer, but if you did happen to do the above then that could also lead you to a correct guess!  ;D


jamonwindeyer

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Re: Physics Question Thread
« Reply #764 on: August 04, 2016, 10:59:05 am »
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For 15, I think the answer has to be A right? First of all, it is clear that the difference between X and Y is either than one is more charged than the other, or one is faster than the other. That already only really leaves us with A. Gamma particles have no charge, so Z makes sense, and Beta particles will experience a force in the opposite direction to the Alpha particles (X,Y) so that makes sense to.

I definitely agree!! Not sure why the answer would read D, or, how the hell a fluorescent screen is a particle?

I'll add that these are tricky questions Neutron, I don't blame you for getting stuck!  :o