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Edexcel Physics Unit 2 "Physics at work" June 2013

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Reply 760
Avatar for x0x
x0x
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7
Original post by ambbs
I put C too. I think we're right, because I remembered seeing this in the textbook ImageUploadedByStudent Room1370514521.634454.jpgImageUploadedByStudent Room1370514527.631827.jpg


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I think it is D
If it was C then Y should start from the origin and go below the X axis
i traced the wave to check :P
Reply 761
Avatar for Alexsk
Alexsk
Can anyone upload the Doppler? What was the answer?
Reply 762
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x0x
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7
Original post by vinhvu95
Did anyone see through the trick question in question 17? It wasn't a negative gradient graph, because that's when the voltmeter is measuring the load resistance, but in this question the voltmeter was measuring the Pd at the battery, so it would've been a positive graph


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Are you 100% sure?
I saw a few books with exactly the same circuit as the one we were shown in the exam and they used the same method?
Reply 763
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Alexsk
Upload the books!!!!!
Reply 764
Avatar for a10
a10
19
Original post by vinhvu95
Did anyone see through the trick question in question 17? It wasn't a negative gradient graph, because that's when the voltmeter is measuring the load resistance, but in this question the voltmeter was measuring the Pd at the battery, so it would've been a positive graph


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it was definitely a negative gradient graph with EMF as the intercept and gradient r
Reply 765
Avatar for Jaydude
Jaydude
11
So what did the graph look like?


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Reply 766
Avatar for ambbs
ambbs
12
Original post by x0x
I think it is D
If it was C then Y should start from the origin and go below the X axis
i traced the wave to check :P


I'm still convinced it's C :dontknow:
Reply 767
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x0x
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7
Original post by Alexsk
Upload the books!!!!!


Dude calm down! I am not saying you're wrong :P
I borrowed that book and i don't have it anymore
and i remember doing a paper 3 question about internal resistance. I drew the same arrangement and my teacher told me that it would work as there is no difference. Did your physics teacher tell you that your reasoning is right? If so then you're likely to be right :smile:
there were 5 marks for that part .. how did you find the internal resistance and the e.m.f?
Reply 768
Avatar for Jaydude
Jaydude
11
Original post by a10
it was definitely a negative gradient graph with EMF as the intercept and gradient r


I put positive gradient, but with a non zero start that curved at the end...am I completely wrong? Lol. I agree with how to work out emf and r though.


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Reply 769
Avatar for krisshP
krisshP
14
Original post by a10
it was definitely a negative gradient graph with EMF as the intercept and gradient r


Yes, that's what I did:smile:

But I stopped my graph when it intercepted the horizontal axis and I stopped it when it intercepted the vertical axis, is that right?
Reply 770
Avatar for vinhvu95
vinhvu95
Original post by a10
it was definitely a negative gradient graph with EMF as the intercept and gradient r


I'm not 100% sure but if you imagine that the load resistor had infinite resistance, Pd at the battery would be very low and overall current would be very small, if the load resistor had a very low resistance, then pd at the battery would be very high and overall current would be high because there is less resistance

Edit: So my graph looked like a positive gradient graph where the maximum y value was emf and gradient is r rather than -r

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(edited 10 years ago)
Reply 771
Avatar for krisshP
krisshP
14
What's the answer everyone put for 17b?
Reply 772
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x0x
OP
7
Original post by ambbs
I'm still convinced it's C :dontknow:


Can you explain why you think its C?
half a radian is equal to 90 degrees
90 degrees is a quarter of a full single cycle
if you move Y backwards 90 degrees its new position is over X hence that is why i think its D but again i might be wrong :P
Reply 773
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x0x
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7
Original post by krisshP
Yes, that's what I did:smile:

But I stopped my graph when it intercepted the horizontal axis and I stopped it when it intercepted the vertical axis, is that right?


yeah ... did the same
Reply 774
Avatar for a10
a10
19
Original post by Jaydude
I put positive gradient, but with a non zero start that curved at the end...am I completely wrong? Lol. I agree with how to work out emf and r though.


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it should be negative gradient because from the equation, but I think u can get away with drawing a positive graph but mentioning r as negative

E=V+IrE = V + Ir

in the form y= c + m.x

V=E−IrV=E - Ir

so the graph was V against I hence gradient should be negative, and the value is r
Reply 775
Avatar for Alexsk
Alexsk
Original post by x0x
Dude calm down! I am not saying you're wrong :P
I borrowed that book and i don't have it anymore
and i remember doing a paper 3 question about internal resistance. I drew the same arrangement and my teacher told me that it would work as there is no difference. Did your physics teacher tell you that your reasoning is right? If so then you're likely to be right :smile:
there were 5 marks for that part .. how did you find the internal resistance and the e.m.f?


HAHA I am calm and I agree with you for this question!
Reply 776
Avatar for ambbs
ambbs
12
Original post by x0x
Can you explain why you think its C?
half a radian is equal to 90 degrees
90 degrees is a quarter of a full single cycle
if you move Y backwards 90 degrees its new position is over X hence that is why i think its D but again i might be wrong :P


If you look at the pictures I had in my original post you'll see why I think it's C :smile: I saw a very similar question in the textbook
Reply 777
Avatar for a10
a10
19
Original post by vinhvu95
I'm not 100% sure but if you imagine that the load resistor had infinite resistance, Pd at the battery would be very low and overall current would be very small, if the load resistor had a very low resistance, then pd at the battery would be very high and overall current would be high because there is less resistance

Edit: So my graph looked like a positive gradient graph where the maximum y value was emf and gradient is r rather than -r

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that should still be correct by the sounds of it
Reply 778
Avatar for krisshP
krisshP
14
Original post by x0x
yeah ... did the same


E=V+Ir
V=-rI+E

Gradient is negative, but I still don't get how we are supposed to know that the graph stops at I=0 and at V=0.
Reply 779
Avatar for vinhvu95
vinhvu95
Original post by a10
it should be negative gradient because from the equation, but I think u can get away with drawing a positive graph but mentioning r as negative

E=V+IrE = V + Ir

in the form y= c + m.x

V=E−IrV=E - Ir

so the graph was V against I hence gradient should be negative, and the value is r


It wasn't Big V, it was small v that the voltmeter was measuring. Because the voltmeter was across the battery


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