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# Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

• View Poll Results: The last question - Does resistance increase or decrease?
It increases ( using V=IR or some other method)
70.73%
It decreases using the 'lattice vibrations' theory
29.27%

1. (Original post by x0x)
Mine was a straight line with a negative gradient
the y intercept is the e.m.f
the negative of the gradient is the internal resistance
i guess there were 3 marks for the graph and 2 for explaning how to find e.m.f and r so i think you'll get some credit
3 marks for a straight line ? :P doubt it
2. 2 marks for the line
1 mark to identify EMF
1 mark to identify r
1 mark to write the equation of line: V=E-Ir
3. (Original post by Alexsk)
2 marks for the line
1 mark to identify EMF
1 mark to identify r
1 mark to write the equation of line: V=E-Ir
That is more like it
4. do you think if you got the gradient of the line correct but not the shape you would get mark, as for some reason i drew it curving downwards thinking it would be exponential because of heat or some stupid idea.
5. (Original post by gkweic)
do you think if you got the gradient of the line correct but not the shape you would get mark, as for some reason i drew it curving downwards thinking it would be exponential because of heat or some stupid idea.
I still think it was curved, a few of us have done that so it might be right!

EDIT: Probably not though
6. hahahaahahaha! Of course it is wrong! It was a line
7. (Original post by Alexsk)
hahahaahahaha! Of course it is wrong! It was a line
Hahahahahaha. Don't be so sure about yourself until you get the results.

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8. (Original post by Freddy-Francis)
On the last question. Wasnt it a series circuit. Cause of it was in series the. Of 1 bulb Is switched off the other one dont light up as it is an incomplete Circuit. :O

Am i wrong?

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yes .. It is series circuit , because if you saw the next three questions the current was always the same .. so if the current is the same the circuit is series
9. (Original post by StephenNaulls)
I still think it was curved, a few of us have done that so it might be right!

EDIT: Probably not though
Exactly what I done. But mine was a non zero start.

I thought the gradient represented R. Then I rearrange Emf= V+ Ir

and to make r the subject.

Hope i get some credit
10. (Original post by Jaydude)
Exactly what I done. But mine was a non zero start.

I thought the gradient represented R. Then I rearrange Emf= V+ Ir

and to make r the subject.

Hope i get some credit
Mine was a non zero start too

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11. The same VI graph is in the Hodder physx AS book pg155...so yeah the graph is straight and has a negative gradient...
12. i think it has to be non zero start as emf is when current is zero and will then decrease though i think it does have to be constant gradient as it was a variable resistor?
13. (Original post by hamzeh h)
but V was constant so u mean lifting it 200 Meters and lifting it 1 meter would give same usefull power output ?
Yes it would, remember power is joules per second, not how many joules in the time taken to lift, lifting it one meter would still have the same power output but will be just used for less time, so like 10w, power output, lifting for ten seconds will be 100j, but for one second it'd be 10j, the power will be the same no matter how high you lift it..

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14. (Original post by Jaydude)
Exactly what I done. But mine was a non zero start.

I thought the gradient represented R. Then I rearrange Emf= V+ Ir

and to make r the subject.

Hope i get some credit
Sorry, I don't take Maths so forgive me. How can you have it starting on the y intercept and decreasing but have a non-zero start; is it a quadratic curve or something like that?

It's times like these I wish I took Maths, or was good at it! I wrote E = IR+Ir and rearranged it, but crossed it out. Do you think they'll credit it?
15. (Original post by JoshThomas)
Yes it would, remember power is joules per second, not how many joules in the time taken to lift, lifting it one meter would still have the same power output but will be just used for less time, so like 10w, power output, lifting for ten seconds will be 100j, but for one second it'd be 10j, the power will be the same no matter how high you lift it..

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you use mgh I have a text book and it says it in there
16. (Original post by StephenNaulls)
Sorry, I don't take Maths so forgive me. How can you have it starting on the y intercept and decreasing but have a non-zero start; is it a quadratic curve or something like that?

It's times like these I wish I took Maths, or was good at it! I wrote E = IR+Ir and rearranged it, but crossed it out. Do you think they'll credit it?
I'm sure my answer is wrong, but worthy of some credit. If you look at the other posts, they did a negative straight line, non zero start. I did the same but with a positive gradient. Initially my graph was straight. It it curved at the top.

I done a curve at top because I thought as voltage increases, eventually resistance will become so small, current will be much greater. I can't explain this properly but I did this cause I saw it in past papers.

Y intercept is emf. That is worthy of credit.
I said th gradient represented R (not sure if this is correct)
I then said to get r, I rearrange the equation you just wrote to get r. ( not sure if correct)

As to if you would get some marks, did you write anything else? Such as what I did and other people did who did the negative gradeint graph? Or something similar?
If so, then it must likely has gotten a minimum of 2 marks.

Finger crossed for results we want!

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17. Is the poll for the last 4 mark question on the paper? If so then the answer is that the resistance does not change, when the current decreases it only means that the total voltage output will not reach what it needs to be for the bulb to light up, right? Obviously I put the resistance increases but my teacher said that is not correct :/
18. (Original post by chriselle15)
Is the poll for the last 4 mark question on the paper? If so then the answer is that the resistance does not change, when the current decreases it only means that the total voltage output will not reach what it needs to be for the bulb to light up, right? Obviously I put the resistance increases but my teacher said that is not correct :/
Yes it is for that question
And now that it has been confirmed by your teacher i guess almost everyone on TSR got it wrong :P
19. (Original post by ALevel96)
you use mgh I have a text book and it says it in there
I know its mgh:') i was just saying the persons justification for it was wrong

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20. (Original post by x0x)
Yes it is for that question
And now that it has been confirmed by your teacher i guess almost everyone on TSR got it wrong :P
lol that was a terrible question plus I was short on time so I just put down anything

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