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AQA A Physics Unit 4 24th Jan 2012

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Reply 60
Could someone explain gravitational potential to me please? I can do the questions via formal but I don't really get it...
Can anyone think of anything strange that could come up?. Also If I cover everything on the spec that should be enough right?
Reply 61
Original post by kimmey
thanks for this but can you please tell me how volume/area = v ? thanks


Well 2.0*10^-4 is how much volume of water is flowing past one point each second, so it is cross-sectional area * velocity, and you want just velocity, you divide it by the cross-sectional area to get velocity.

Plus you can just look at the units, to get the unit of velocity (ms^-1) using m^3s^-1 (meters cubed per second) and m^2 (meters squared), simply divide the first by the second and you're left with ms^-1 (meters per second).
Reply 62
After being rejected by Cambridge, I'm even more pumped to own this exam.
Original post by Femto

Original post by Femto
After being rejected by Cambridge, I'm even more pumped to own this exam.


Lol same here, although I got rejected from oxford :P
Original post by maths134

Original post by maths134
Could someone explain gravitational potential to me please? I can do the questions via formal but I don't really get it...
Can anyone think of anything strange that could come up?. Also If I cover everything on the spec that should be enough right?


Nothing strange should come up, but if it does we cant really prepare for it.
Reply 65
Does anyone have a graph for kinetic energy against velocity for SHM, if that exists?
Reply 66
Original post by --NWzD9--
Does anyone have a graph for kinetic energy against velocity for SHM, if that exists?


Surely it would take the shape of a parabola?

EDIT: Like this:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0008/KEgraphs.gif

But it would have to have a negative velocity as well surely as, with any SHM system, the velocity changes direction.
(edited 12 years ago)
Reply 67
Original post by Femto
Surely it would take the shape of a parabola?

EDIT: Like this:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0008/KEgraphs.gif

But it would have to have a negative velocity as well surely as, with any SHM system, the velocity changes direction.


Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?
Reply 68
Original post by --NWzD9--
Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?


You could draw a displacement time graph then find the gradient and that would be v against t and square it? Might seem too long though. :tongue:
Reply 69
Original post by --NWzD9--
Oops, sorry, what i wrote was wrong. I was meant to say kinetic energy against time, starting from maximum displacement?


Hmm, I don't think you'd be asked that. Although I'm not 100% sure about what I've just said - would somebody be able to clarify this?
Reply 70
http://www.has.vcu.edu/new-phy/WSG/CH16%20images/fig1.jpg

I was told it would be shaped like this, instead with the y-axis as KE
Reply 71
How do you work out the answers to Q.13 & 24 from the Multiple Choice section of the June 2010 paper? Mark scheme says answers are A & C respectively

Link to paper: http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JUN10.PDF
Reply 72
Original post by --NWzD9--
How do you work out the answers to Q.13 & 24 from the Multiple Choice section of the June 2010 paper? Mark scheme says answers are A & C respectively

Link to paper: http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JUN10.PDF


for 13 us v=1/4pi epsilon nought*(q/r)
then use v4=v-16
then you have 4/x=16/(120-x)
then working out you get x =24
or you could do it as a sort of ratio so 4+16=20
and 4 in proportion of that is 5
then 120/5=24

then for 24 it rotates twice as fast so cuts twice as much flux so there has got to be 2 waves in the graph space, and then as it is double omega then amplitude will double so its c

hope that helps :smile:
Reply 73
Original post by nexl
The 6 mark questions in past papers are:

Specimen - SHM
Jan 10 - Transformers
Jun 10 - Momentum
Jan 11 - Capacitors
Jun 11 - Electricity

There could be a question on fields this time, since it hasn't come up before.


is there anything else that could show up apart from fields???
Reply 74
Does anyone have the mark scheme for June 2011 cos I just did it as a mock and wanna I know how I did lol.
Reply 75
can any1 explain how to do this Question

19. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?
Original post by mk207

Original post by mk207
can any1 explain how to do this Question

19. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle?


Ok. W=2pie/T. V=wr so v=2rpie/T, square it and sub into Ek=1/2mv2
Reply 77
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?
Original post by mk207

Original post by mk207
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?


What are the options? Or link to paper please.
Reply 79
Original post by mk207
and another 1

23. Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time taken for one complete orbit?


do mw^2r=BQV dont forget w=2pi/T solve for T :smile:

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