# OCR A Physics AS Breadth 24/5/16Watch

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#21
(Original post by sanchit117)
I disagree with question 8, the force constant changed so you can't use that formula, you need to use E=1/2Fx - since the force is the same, E is proportional to x and so it's E/3
Agreed - edited.
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#22
(Original post by julesaquilina)
for the very last question about number of photons could you have used e=pt and divided e by 1.6x10^-19? Cuz they gave you the power and they said it was one second
I don't think so - its photons not electrons
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3 years ago
#23
I have different opinion on one of the questions.

For question 27, the graph is current against voltage (If my memory still work). So the gradient of the graph should be 1/R. The first two bits of the answer is agreed, first the resistance is infinite and then it starts to decrease. However, after that the graph is a straight line going upwards for the last part. If it is a straight line, it suggests that the resistance is constant. Although the R value is less than before, but it is constant, not continue to decrease.
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#24
(Original post by kuashdfkus)
also for question 11 did people not get current clockwise?
No - it has to go + to - for the bigger emf
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3 years ago
#25
(Original post by NotGambit)
For Q1, would it not be correct to say "rate of work done" which i think was A ? :/
It asked for a definition of the watt not a definition of power
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3 years ago
#26
(Original post by kuashdfkus)
also for question 11 did people not get current clockwise?
It was anticlockwise because the EMF of the cell on the left was greater; and because they were facing opposite directions and conventional current goes from + to - you had to find the difference; which ended up being anticlockwise.
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3 years ago
#27
(Original post by wuhuapple001)
I have different opinion on one of the questions.

For question 27, the graph is current against voltage (If my memory still work). So the gradient of the graph should be 1/R. The first two bits of the answer is agreed, first the resistance is infinite and then it starts to decrease. However, after that the graph is a straight line going upwards for the last part. If it is a straight line, it suggests that the resistance is constant. Although the R value is less than before, but it is constant, not continue to decrease.
It's not constant because the scales for V and I were different I think; the fact that it was linear would have meant that it decreased at a constant rate.
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#28
(Original post by wuhuapple001)
I have different opinion on one of the questions.

For question 27, the graph is current against voltage (If my memory still work). So the gradient of the graph should be 1/R. The first two bits of the answer is agreed, first the resistance is infinite and then it starts to decrease. However, after that the graph is a straight line going upwards for the last part. If it is a straight line, it suggests that the resistance is constant. Although the R value is less than before, but it is constant, not continue to decrease.
This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.
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3 years ago
#29
Super happy as it turns out I got them right; probably dropped a couple marks here and there; especially with the asteroid question because I crossed out what I initially wrote about the force being in the opposite direction; but I did draw a reflection in the x axis. Not bad indeed.
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3 years ago
#30
(Original post by teachercol)
This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.
Does the linear gradient mean it decreases at a constant rate?
0
#31
(Original post by kuashdfkus)
are you sure about 11 + 20
Yes - you have to use angles to the normal and work out c for glass.
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3 years ago
#32
(Original post by teachercol)

x
Hi,

Thanks for posting these

Based on the questions in this Breadth paper, do you have any predictions for topics (especially practical questions) which are likely to come up in the Depth one?
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#33
(Original post by Parhomus)
Does the linear gradient mean it decreases at a constant rate?
I don't think so. Draw lines from the origin.
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#34
(Original post by voltz)
Hi,

Thanks for posting these

Based on the questions in this Breadth paper, do you have any predictions for topics (especially practical questions) which are likely to come up in the Depth one?
Moments torques and centre of gravity
Measuring Youngs Modulus.
LDRs thermistors and potential dividers
Resistivity and series / parallel
1
3 years ago
#35
(Original post by Parhomus)
Yes, that's exactly what they wanted people to think, at least in my opinion; because clearly if the same force is applied and extension is different then the value of the force constant will be different.
Yeah, so if the force constant is different, you can't use the equation EPE=1/2kx^2 as k is also different.
1
3 years ago
#36
where could we find the questions?
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#37
(Original post by kuashdfkus)
where could we find the questions?
You cant - until the paper is released in a years time.
1
3 years ago
#38
(Original post by teachercol)
This is a common mistake.
R is not 1 / gradient; its V/I.
Its only = 1 / gradient for straight lines through the origin.
Indeed R=V/I, but as the graph is plotted as current against voltage, gradient will be dI/dV, which means that it is equal to 1/R.

It does not matter whether the line is going through the origin or anywhere else, the gradient of the graph represents the change of current with repect to voltage. As dV/dI measures resistance, dI/dV represents 1/resistance.
0
3 years ago
#39
Judging from this I can only see possibly 3 marks lost, mostly on multiple choice, but I could almost guarantee the actual mark scheme will be very specific for some things, not allowing answers which are correct because of a keyword missing, or not stating a fairly obvious point, kind of seems to be the trend with the new spec :/

Fairly happy though, all calculations questions were fine, didn't completely screw up anything.
1
3 years ago
#40
(Original post by Parhomus)
It's not constant because the scales for V and I were different I think; the fact that it was linear would have meant that it decreased at a constant rate.
Difference in scale only affects how the graph looks, not the actual value of the gradient.
1
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