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AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

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Reply 680
These answers are making me feel rubbish :facepalm: I found the paper really really difficult and looks like I haven't done very well :frown: Annoyed cause I actually tried really hard for this exam all year :frown:
Reply 681
Original post by Dirtybit
Probably said the sattelite was near the earths surface


they actually said that? :O

wow, I must've completed missed it

I guess when I saw the word satellite and instinctively thought it was in space, oh ****
Reply 682
Original post by fizzbizz
What did people get for the final transformers multiple choice question? 460 turns and less than 0.26A?


NO WAY is this right? I guessed last minute
Original post by MSI_10
1500?


Yh thats what I got. Then what are people talking about in the first q saying they got 1500 in q1?

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Positive sine curve yeah? Last

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Reply 685
Original post by Dirtybit
Probably said the sattelite was near the earths surface


Yeah it said low polar orbit
Reply 686
Original post by Jack93o
oh yeah, for the induced emf graph, everyone started off at zero, right?

but did you draw it sloping up (to the positive part of the y-axis) in the first quarter of a cycle?


I did this but it's wrong, it should've been sloping down to start with (minus sine)
Original post by saba146
section a: kgms-2, t root2 energy one i got 2E? ...

1a)define shm:
acceleration is proportinal to displacement, and acceleration acts toward equilibrium (2)
b)acceleration of pendulum = 0.503 3s.f (3)
c)frequency= 1500 revs per minute (2)
d) time for oscilations to be in phase again = 38 (3)

2) voltage = 30,000 (1)
b) (3)
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)

3) 6 marker on fields?

4) 2 conditions when no force is exerted on the partcle (2)
field is parallel to velocity
particle is stationary
b)
c)time taken for partile to go through 1 dee
d)show the time taken is independent of radius = 2pim/bq (3)
kinetic energy im MEV = 2.44 (3)

angle for max emf = (2)
gradient of graph = emf (1)

drawing a graph (2)
peak voltage (3)
calculate flux density = 0.26 (2)


what questions has this person missed out?? this doesn't add up to 50 and I can't remember :/
Reply 688
For satellite P and Q, P had a greater KE but lower PE so answer was C?
Reply 689
Original post by cooldudeman
Yh thats what I got. Then what are people talking about in the first q saying they got 1500 in q1?

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Oh lol someone mixed it up for the list of Qs posted..

The freq for the 3 marker on Q1 was 0.503 guys the acceleration was -0.4 ish!

Not 1500 lol that was for the last q
Reply 690
Original post by MSI_10
For satellite P and Q, P had a greater KE but lower PE so answer was C?


Yup
Reply 691
Original post by fizzbizz
I did this but it's wrong, it should've been sloping down to start with (minus sine)


you sure? why?
Reply 692
Original post by cooldudeman
Positive sine curve yeah? Last

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Started at 0, EMF was max when linkage was 0, probs wrong though..
Reply 693
What did everyone get for question 24 on the multi choice about the rod moving in a magnetic field?
Reply 694
shouldn't it be -sinx?
Original post by MSI_10
Oh lol someone mixed it up for the list of Qs posted..

The freq for the 3 marker on Q1 was 0.503 guys the acceleration was -0.4 ish!

Not 1500 lol that was for the last q


Theres me thinking I lost 6 marks.
I got 0.503 too. Thats 3sf right?

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Reply 696
Original post by cooldudeman
Theres me thinking I lost 6 marks.
I got 0.503 too. Thats 3sf right?

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I also put that
Reply 697
Original post by Jack93o
you sure? why?


Because of Lenz's law - induced emf always in the opposite direction

You could also think of it as being the gradient of the flux linkage curve which was -cos:

ddx[cos(x)]=sin(x) -\frac{d}{dx}[-cos(x)]=-sin(x)
Reply 698
Original post by Serpentine111
I think it would be brighter because there is more energy stored?


i said that !!!!!!!!1
Original post by The H
shouldn't it be -sinx?


yes, I'm pretty sure it should be

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