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June 2011 G485-Fields, Particles and Frontiers of Physics

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Reply 920
Avatar for apo1324
apo1324
11
Anyone notice the error? :P Oh nvm.
Original post by Pheylan
What was the potential difference across the 150 uF capacitor? (I got 1.5V)


Same.
Reply 922
Avatar for Raimu
Raimu
I thought it was a pretty ****ty paper overall. A lot of mechanics and not enough stuff on stars/quarks/decay. I messed up the angle question and the midpoint question, but all of the describe questions were really easy marks. I was kind of lucky and didn't notice 0.05 cm and wrote in meters instead, so that was cool. What did everyone put for the KE graph? Also how did you do the question about the decrease in diameter? I said the current would decrease by a factor of 2^2 and just used that in the equation.
Reply 923
Avatar for anshul96
anshul96
I sucked a cock or two
(edited 12 years ago)
Reply 924
Avatar for Pheylan
Pheylan
Original post by supervernagirl
Same.


And 4.5V across the 450 uF one? (I'm trying to figure out how I got 1/81 for the ratio :s-smilie: )
Reply 925
Avatar for emlath
emlath
Original post by OneInSolidarity
Someone I know got the ratio as 3. I got the ratio as 1/27 however (I'm sure this is wrong xD)


The majority of the people taking the exam at our school got 1/27, only two people got 3. One didn't revise.
Reply 926
Avatar for Pheylan
Pheylan
Original post by Raimu
Also how did you do the question about the decrease in diameter? I said the current would decrease by a factor of 2^2 and just used that in the equation.


I used I=nAvq=nvq*pi*(d/2)^2 so I is proportional to d^2. If d is halved, I is quartered
Original post by Oh my Ms. Coffey
Am I the only one disappointed with the lack of Quarks/Fundamental particles.


Yes :biggrin:
Original post by Pheylan
And 4.5V across the 450 uF one? (I'm trying to figure out how I got 1/81 for the ratio :s-smilie: )


Sorry, can't remember my answer..
Did any1 else not finish the paper or was that just me
Original post by Raimu
I thought it was a pretty ****ty paper overall. A lot of mechanics and not enough stuff on stars/quarks/decay. I messed up the angle question and the midpoint question, but all of the describe questions were really easy marks. I was kind of lucky and didn't notice 0.05 cm and wrote in meters instead, so that was cool. What did everyone put for the KE graph? Also how did you do the question about the decrease in diameter? I said the current would decrease by a factor of 2^2 and just used that in the equation.


Synoptic question. R = (roe x l)/A.

If diameter was halved, area was quatered, therefore resistance was 4x greater.

V=IR, so the Current which was 4.0A is quatered and becomes 1.0A.

End force = 1 * 0.08Tesla * length

Basically a quarter of your previous force, like you said, but thats the explanation.
Original post by apo1324
Anyone notice the error? :P Oh nvm.


Let me just grab my 0.30 cm ruler..
Reply 932
Avatar for Raimu
Raimu
Original post by Pheylan
I used I=nAvq=nvq*pi*(d/2)^2 so I is proportional to d^2. If d is halved, I is quartered


Sweet, that's what I've done then too. You just used F = BIL then right, with I being 1?
Reply 933
Avatar for ziigmund
ziigmund
my answers (not sure if they're all correct so quote/edit if you disagree :tongue:)

Q1di)219 =~220 turns
ii)1)34.3A
2)6.86W

Q2) bi)1.5v
ii)2.25*10^4 C
iii)1.125*10^-4 F
ciii) 1/27 =0.037

Q3ii)1.4*10^9 s
v)4.18*10^7 m/s
vi)7.94*10^-16 J

Q4a)6.6*10^-6 N/C towards B (think it was equal to towards the right)
c)18.8 degrees

Q5c)i)1.6*10^-2 N
ii) 4.484N
d)0.064N

Q6c)i)1.4*10^22 kg/m^3
ii) mass of electrons was negligible

Q7b) H=2.1*10^-18 /s p0=7.9*10^-27 kg/m^3
I realised it was a quarter after finishing the exam >< - I guessed and said halved.

And wtf is this mistake you're on about, I saw a 0.05m in the question and used that, I didn't bother reading the text... None of the examiners said anything either!
Original post by emlath
The majority of the people taking the exam at our school got 1/27, only two people got 3. One didn't revise.


Ooh i got 1/27, thought it was absurd. I got different answers with W = 0.5QV and W= 0.5 CV^2!
Reply 936
Avatar for scotzbhoy
scotzbhoy
Original post by Pheylan
I used I=nAvq=nvq*pi*(d/2)^2 so I is proportional to d^2. If d is halved, I is quartered

I should have been halved. I=nAvq, so if A is halved, I is halved.
The way I looked at it was if we take V to be 1, the initial resistance is R=V/I=1/4=0.25. The diameter is then halved, hence the resistance is doubled, so I=V/R=1/0.5=2.
Reply 937
Avatar for VDX
VDX
Original post by JohnnySAFC19
Did any1 else not finish the paper or was that just me

I just about finished on time but some of my friends finished way earlier :dontknow:
(edited 12 years ago)
Original post by ViralRiver
I realised it was a quarter after finishing the exam >< - I guessed and said halved.

And wtf is this mistake you're on about, I saw a 0.05m in the question and used that, I didn't bother reading the text... None of the examiners said anything either!


In the text it said 0.05cm. The correct value was 0.05m, as on the diagram.
Original post by ziigmund
was the voltage 1.5v or 2v? idk if it was a ratio of 1:3 ie 1/4 or 1/3


I got 1.5v on 4.5capacitor and 4.5V on the 1.5capacitor.

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