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OCR Physics A G485 - Frontiers of Physics - June 2016

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Reply 20

Mihael_Keehl

Original post by flamingoking

The average density is the average density of the universe, since this value varies depending on where you are in in the universe and the critical density is the density required for the universe to have a strong enough gravitation pull to stop its expansion.

interesting thank you.

Reply 21

Fox Corner

Original post by Mihael_Keehl

I won't be revising for a while but if I can help, I'll try, means less revision for me in the summer if I help people earlier on (or if people explain to me!)

I put this thread in the A Level Exam Directory Mega Thread for you which you can find here :parrot:

use that thread to find discussions about your other exams too :smile:

....

Can anyone explain for me the relationship of Magnetic flux and Induced E.m.f ?I know E.m.f = rate of change of Magnetic flux. But even when Magnetic flux double, the range of magnetic flux still the same ?

Reply 24

lai812matthew

Original post by Xuancong

magnetic flux doubles. time doubles. so no change in rate of change of magnetic flux. only if one of the magnetic flux/ time is fixed then an emf would change.

Reply 25

ukcolBehT

I need help on this, how do you do c) and d)?

Reply 26

lai812matthew

Original post by ukcolBehT

I need help on this, how do you do c) and d)?

i'm not sure
but i think c) using c=q/v, where c= answer in part b, and q= answer in part a
d) e=1/2qv=1/2cv2, 1/2v2/q whatever you like

Reply 27

ukcolBehT

Original post by lai812matthew

i'm not sure
but i think c) using c=q/v, where c= answer in part b, and q= answer in part a
d) e=1/2qv=1/2cv2, 1/2v2/q whatever you like

Do they count as being in series or in parallel?

And when I do it either way, how can it be that the final energy stored is different from the original energy stored? ( Higher)

Reply 28

lai812matthew

Original post by ukcolBehT

Do they count as being in series or in parallel?

And when I do it either way, how can it be that the final energy stored is different from the original energy stored? ( Higher)

series? it would be different because you are using different values of c, q and v.

Reply 29

ukcolBehT

Original post by lai812matthew

series? it would be different because you are using different values of c, q and v.

Yeah but the energy went up. Is it possible?

Reply 30

lai812matthew

Original post by ukcolBehT

Yeah but the energy went up. Is it possible?

ah sorry in this case, capacitors are connected in parallel.

Reply 31

lai812matthew

Original post by ukcolBehT

Yeah but the energy went up. Is it possible?

e=1/2cv2, where c= answer in part b, v= answer in part c.

Reply 32

Jmedi

Hello,
Is anyone here able to upload the G485 2015 paper and ms?
Thanks

Reply 33

ukcolBehT

Original post by lai812matthew

e=1/2cv2, where c= answer in part b, v= answer in part c.

The energy goes up wether I do it in series or in parallel.

Reply 34

lai812matthew

Original post by ukcolBehT

The energy goes up wether I do it in series or in parallel.

? the energy would definitely go down

Reply 35

ukcolBehT

Original post by lai812matthew

? the energy would definitely go down

Do you think its possible for you to do the question? Because I don't seem to get the results your talking about.

Reply 36

lai812matthew

Original post by ukcolBehT

Do you think its possible for you to do the question? Because I don't seem to get the results your talking about.

a) Q= 4.7x10^-6 x12=5.64x10^ -5 C
E=1/2 x 4.7x10^-6 x12^2 =3.38x10^-4 J
b) c=4.7+2.2=6.9 microfarads
c) V=5.64x10^-5 / 6.9x10^-6 =8.17V
d) E=1/2 x6.9x10^-6 x 8.17^2 =2.31x10^-4 J

Reply 37

ukcolBehT

Original post by lai812matthew

a) Q= 4.7x10^-6 x12=5.64x10^ -5 C
E=1/2 x 4.7x10^-6 x12^2 =3.38x10^-4 J
b) c=4.7+2.2=6.9 microfarads
c) V=5.64x10^-5 / 6.9x10^-6 =8.17V
d) E=1/2 x6.9x10^-6 x 8.17^2 =2.31x10^-4 J

Thank you.

Ok so where I went wrong is I thought that for any capacitance question in parallel pd stays the same and in series charge stays the same, but it makes sense what you did.

What is the general rule why you don't use the pd staying the same here?

Reply 38

lai812matthew

Original post by ukcolBehT

uh i just memorise it .the thing which stays constant is charge as the voltage supply stays constant. c=q/v, then pd across won't be the same with changing capacitance. (addition of capacitor)