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# EDEXCEL A2 Physics EXAM Unit 4 Physics On The Move 20th June 2016 (NOT I-A-L) watch

1. (Original post by oShahpo)
I put that as d increases, C decreases, C= Q/V, V is constant so Q decreases too and there's a current.
Did you have to mention charge?
2. (Original post by target21859)
Did you have to mention charge?
I think this is what the question was about, but my memory is absolutely terrible so don't take my word for it.
3. I would say the speedometer will be overestimated.

if the tyre is under inflated, r will be less as well as v will be less.
As v=(2pi*r)/t. The r for the speedometer is constant for that specific bikeso actual speedometer will display higher v.
4. (Original post by qq1102141262)
I would say the speedometer will be overestimated.

if the tyre is under inflated, r will be less as well as v will be less.
As v=(2pi*r)/t. The r for the speedometer is constant for that specific bikeso actual speedometer will display higher v.
Same as I did. I said would be exaggerated due to lower radius. I also mentioned that the value itself wouldn't change as the angular velocity is constant.
5. (Original post by oShahpo)
Same as I did. I said would be exaggerated due to lower radius. I also mentioned that the value itself wouldn't change as the angular velocity is constant.
Wasnt this question related to something with pulses of emf?
6. (Original post by habiba_9)
Wasnt this question related to something with pulses of emf?
Was it? I don't remember the question exactly. Do you remember the wording of the question?
7. (Original post by oShahpo)
Was it? I don't remember the question exactly. Do you remember the wording of the question?
Nop but i wrote my answer in terms of magnitude of emf anyone else who did that aint sure about my answer tho
8. (Original post by oShahpo)
Yea I said the same thing, but I also that the value itself wouldn't change as the angular velocity stays the same, does that negate the first bit I said?
The angular velocity of the wheel remains constant and hence so does the speed reading.
9. (Original post by oShahpo)
Was it? I don't remember the question exactly. Do you remember the wording of the question?
I said that for a fixed speed V, the frequency of rotation of the wheel is V/2*pi*r and so for a smaller radius at a given speed the frequency of rotation would be greater and so the frequency of EMF pulses would be greater causing the speedometer to overestimate the speed.
10. (Original post by jtebbbs)
Resolve vertically: F*sin22 = 80 * 9.81 gives F = 846

Set horizontal component equal to mv^2/r, rearrange to get

r = (80 * 9^2) / (846 * cos22) which gives you 20.4m
I thought I didn't round but got 20.5m, did you use exact values in your calc?
11. (Original post by akiwicat)
I said that for a fixed speed V, the frequency of rotation of the wheel is V/2*pi*r and so for a smaller radius at a given speed the frequency of rotation would be greater and so the frequency of EMF pulses would be greater causing the speedometer to overestimate the speed.
V isn't constant, in fact, V is reduced when the tire is not inflated properly. It would overestimate the speed, but not because the frequency of emf pulses is greater, rather because of the calibration of the device.
12. Guess I was the only one who somehow got 40.5% for max decrease... oops
13. Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol
14. (Original post by AtomicAnal)
Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol
What was question 3 again???
15. (Original post by PhysicsIP2016)
Surely the speed is directly proportional? v=2pir/T?
Yeah you're right, looking back I'm not sure what I meant!

I think what I'm saying is as the radius decreases, the speed on the speedo relative to the true speed increases, but yeah, the true speed is directly proportional to the radius. Hoping I didn't make the same mistake in the exam, can't remember!
16. (Original post by jjjzzz)
Thank you!!!!!!!!! I insisted on 33.3% until I saw your comment. 👍
I agree on 33.3%
In fact I am pretty sure that 33.3 is more likely to be correct cause
This is what u get from the result of
{Max-Min} /Max *100%=33.3%
17. (Original post by AtomicAnal)
Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol
MeV was option C, and is a unit of energy, so it was C. D was Nm^-1s^2, in other words Newtons divided by metres times seconds squared. A newton is kgms^-2, or kilograms times metres divided by seconds squared, so the metre and second parts cancel out, leaving mass.
18. (Original post by AtomicAnal)
Question 3 was obviously D, Newtons aren't mass, can't believe you all got that wrong lol. My teacher said it was D too lol
I put MeV was not a unit of mass. I think D was Nm^-1s^2 which is a unit of mass.
19. (Original post by tkjoption)
I agree on 33.3%
In fact I am pretty sure that 33.3 is more likely to be correct cause
This is what u get from the result of
{Max-Min} /Max *100%=33.3%
I think the question asking for "max decrease" implies that you can either increase or decrease from the starting point. If you use 2.8mm to find your starting point, you're already at the max so you can't increase further, therefore 3.5mm must be the 'starting point'.
20. (Original post by jtebbbs)
MeV was option C, and is a unit of energy, so it was C. D was Nm^-1s^2, in other words Newtons divided by metres times seconds squared. A newton is kgms^-2, or kilograms times metres divided by seconds squared, so the metre and second parts cancel out, leaving mass.
No no no, no no no. It was D. All my teachers said D

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