The Student Room Group

AQA A Physics Unit 4 24th Jan 2012

Scroll to see replies

Original post by maths134

Original post by maths134
If you guys want to spend a day working it out, you welcome to...


Oh come on, its about helping each other out. I've helped people on this thread with questions for example and on the ocr chemistry F324 thread I spent ages doing the same (collating questions) for everyone else to use as well-check if you don't believe me. But you spent the time doing it, so it's your choice. Good luck either way
Reply 101
Hi all, I hope someone can help me out... The question is regarding capacitance.

A capacitor is discharges through a fixed resistor, how long does it take the pd across the capacitor to fall from V_0 from V? (Assuming all values are known, aside the time)

I thought you would use the equation below, but I can't rearrange for t :'( (Never did maths)

V=V_0 e^(-t⁄RC)

Thank you in advance :smile:
Reply 102
Original post by schenker
Hi all, I hope someone can help me out... The question is regarding capacitance.

A capacitor is discharges through a fixed resistor, how long does it take the pd across the capacitor to fall from V_0 from V? (Assuming all values are known, aside the time)

I thought you would use the equation below, but I can't rearrange for t :'( (Never did maths)

V=V_0 e^(-t⁄RC)

Thank you in advance :smile:


Rearranging V=V_0 e^(-t⁄RC) for t
is simple if you do A-level maths, but since you don't here it is

t=RC In(V_0/V), In is the inverse of the exponential function.
Reply 103
Original post by albus
Rearranging V=V_0 e^(-t⁄RC) for t
is simple if you do A-level maths, but since you don't here it is

t=RC In(V_0/V), In is the inverse of the exponential function.


Gah, damn maths I was leaving the e in :P

Thank you!
Reply 104
jus done the june 2011 paper. i think i notices a couple of errors. anyone whos done it notice any ?
Original post by kimmey
jus done the june 2011 paper. i think i notices a couple of errors. anyone whos done it notice any ?


have u got the markscheme for it?
Original post by kimmey
jus done the june 2011 paper. i think i notices a couple of errors. anyone whos done it notice any ?


no not really... we did it as a mock and have been through it and i don't remember any errors...
Reply 107
Original post by schizopear
no not really... we did it as a mock and have been through it and i don't remember any errors...


i noticed 1 error. q14 it says work out the electic potential energy

now to to this u would usually do v=q/4piepsilon0r then sub v into w=qv to find the energy.

so this question should actually be work out the electric potential. the word energy should not be there. also another less relevant point is that energy is a scalar so there is no need for the +/- signs.
Reply 108
Original post by handsome7654
have u got the markscheme for it?


no my teacher marked it
Original post by kimmey
no my teacher marked it


Ohhh! :smile:
Does anybody know the grade boundaries from previous papers? I hear they are particularly low.
Reply 111
Original post by LastHype
Does anybody know the grade boundaries from previous papers? I hear they are particularly low.


http://web.aqa.org.uk/UMS/ :smile:
Original post by kimmey
i noticed 1 error. q14 it says work out the electic potential energy

now to to this u would usually do v=q/4piepsilon0r then sub v into w=qv to find the energy.

so this question should actually be work out the electric potential. the word energy should not be there. also another less relevant point is that energy is a scalar so there is no need for the +/- signs.


i didn't take into consideration the multiple choice paper so there may be mistakes in the paper then... :redface:
Reply 114
Mistake on qu 23 of circular motions posted on previous page I think
Just done Jan 2011 Mult. choice.

Can anyone help me out on question 2? I've tried everything! It'll be something simple probably.

Also for question 21, I used BANcosθ and got 3.8x10^-3 Wb turns, so I assumed the flux linkage would increase by this amount. The answer is in fact B, can anyone explain why?

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF
Reply 116
Original post by lee_vassallo
Can anyone help me out on question 2? I've tried everything! It'll be something simple probably.


Did you divide the diameter by 2 to get the radius?
Reply 117
Original post by lee_vassallo
Also for question 21, I used BANcosθ and got 3.8x10^-3 Wb turns, so I assumed the flux linkage would increase by this amount. The answer is in fact B, can anyone explain why?

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF


BANcos50 = 3.78x10^-3
BANcos0 = 5.88x10^-3

5.88x10^-3 - 3.78x10^-3 = 2.1x10^-3

Flux linkage is maximum when θ=0 because cos0=1, so it increases by 2.1x10^-3 Wb turns.
Oh thanks I get it.

For question 2 I tried to find angular speed using w=θ/t and then for the avg. speed v=rw (and divided by pi) but to no avail :frown:
Original post by lee_vassallo
Oh thanks I get it.

For question 2 I tried to find angular speed using w=θ/t and then for the avg. speed v=rw (and divided by pi) but to no avail :frown:


For Q2, note that the disc turns through an angle of 30 degrees in 20ms. Hence, it will turn through 360 degrees in 240ms.

Then, you apply the formula for velocity (not angular): v = 2pir / t.

=> v = 2pi(60x10-3) / 240x10-3

=> v = 0.25(2pi) = 0.5pi = A
(edited 12 years ago)

Quick Reply

Latest

Trending

Trending