Suggested Solutions

[Alwin]

Alwin if you want a LaTeX template to get started with, I started making some solutions, until I realised I had to go out
So I'll put up what I got so far, hopefully it'll help..
https://docs.google.com/file/d/0B_pcvLkZRgy2OHV3U0xFTlNkaWc/edit?usp=sharing
Cheers

Haha thanks but I started mine already... up until I realised I hadn't read half the qs properly and was posting some wrong answers HAHA.

Thanks for the offer tho

[BananaPi]

Question 8b (motion) asked for the magnitude of the angle.  Is the negative sign still applicable, then?
edit: 17d: Was the order of A, B and C out (A-C-B-C-A)?  I seem to recall that one of the questions had strange ordering.
edit: 18c: Just check this, a number of high-scoring students that I know got 12% for this one (144/1200).

[Jtc]

questions 18.c for power should be 6^2x4 not 6x4^2 it should give a percentage of 12% not 8% --- fairly sure.

[lzxnl]

Hey guys, I'm at school now looking at a spare copy of the exam (which I think my teacher wants back). I'll scan it when I get home and whatnot
Here are my solns so far:
(will make some nice LaTeXed Worked solns later )

Electric Power
14.    Should look like two north poles facing each other

15.a) 1:6 ratio so 18.0 V
     b) Peak = root(2) x 18 = 25.5 V
     c) P=V^2/R = 3^2/1200 = 0.0075 W
     d) Explain general principles of how transformers work (one mark)
          Relate it to this situation here with the DC input and the on and off position (one mark)
          Explain what happens when it’s turned on using Lenz’s Law (one mark)

But RMS voltage was 18 V using the transformer...so wouldn't the power be 18^2/1200=0.27?

17.a) EMF=change in flux/time = (0.6-0.2)/0.5 = 0.8V
          I = 0.8/0.1=8A   …note that I don’t think you’d get this formula if you did the sin(2pif) rule…

     b) EMF zero when flux is max/min, think of derivatives if it helps. So t=0, 0,5, 1, 1.0, 1.5, 2.0, 2.5

Wasn't there a time domain given? Up to like 2.0s? And in those questions do we include the endpoints? The question didn't say.

     c) Moving towards a North Pole.
         To oppose this motion, a North Pole is induced on the side of the ring closest to the N Pole
         So, induced field direction is down towards the magnet.
         Using RH Grip rule, the current flows clockwise
         Or, the magnetic field strength is increasing so magnetic flux increases as the ring moves down
         To oppose this, the induced current will create a field in the opposite direction, etc etc
         As it doesn’t specify if the speed is constant and VCE doesn’t require you to know about
         strength of magnetic fields, I don’t think you would be penalised if you didn’t mention how
         the current changes over time (decreases for those interested)

     d) Flux is relative to the magnetic field strength/density. Technically strongest part is around the
          poles, 1/6th the total length I think I read somewhere last year, but we assume it’s at the
          middle of the magnet. When it’s far away, weaker field so less current induced (think of Lenz’s
          Law if it helps).
          So, first min of graph is A (given),the first max is B (middle of magnet), next min is C, then B etc
          Aka A: t=0, 2     B: t=0.5, 1.5, 2.5,     C: t=1.0

But C is the middle of the magnet...

18.a) R = Vloss/I = 24/6=4.0 ohm
     b) Voutput = 1200/6.0 = 200V
     c) Ploss = 6 x 4^2 =96.  So 96/1200 = 8%

Wait what? Ploss = I^2 = 6^2*4=144?

[Alwin]

But RMS voltage was 18 V using the transformer...so wouldn't the power be 18^2/1200=0.27?
Wasn't there a time domain given? Up to like 2.0s? And in those questions do we include the endpoints? The question didn't say.
But C is the middle of the magnet...

Wait what? Ploss = I^2 = 6^2*4=144?

frick.

I really should learn how to read questions . ty again (Y)

[Uberjew]

Q17.

The graph tells you the magnetic flux through the ring, not induced in the ring.

Does that mean that when you're doing 17.d, A and B are both a minimum flux whereas C is max, thus shouldn't you have it go through:

A at t=0, 2
C at t=0.5,1.5,2.5
B at t=1

[rany]

Do you guys think A+ cut off would be around 130/150 marks?

[~T]

frick.

I really should learn how to read questions . ty again (Y)

I've found that the most ridiculous subjects in terms of having to read the question carefully are definitely Methods and Physics. I'd blame it on the fact that you knocked both of them over last year

[randomposter]

What did you guys put for 'describe the intensity at the centre of the interference pattern'

I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.

[Robert123]

For the gain question in electronics (second last), what were we meant to do with the curvy parts to the clipping, should that be included for calculations?

[Chazef]

wait a minute we weren't expected to graph the voltage amplifier were we (just asking)

[~T]

What did you guys put for 'describe the intensity at the centre of the interference pattern'

I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.

I'm not sure it works exactly like that, but maybe. I just said that it would be a bright spot due to constructive interference, and that it would be the brightest of the bright spots due to it being closer to the source(s) than the others.

For the gain question in electronics (second last), what were we meant to do with the curvy parts to the clipping, should that be included for calculations?

Nope, just the linear region (the straight section). Pick two points in there that you can accurately take off the graph and then take the gradient.

[Apink]

Can someone explain 2b please?

[BananaPi]

Can someone explain 2b please?

I started with the entire system: mass=8kg, force causing acceleration=2*10N, Friction = 0N.
Next, calculate that the acceleration of the system a=F/m=20N/8kg=2.5ms^-1.
The tension is equal to the net force acting on mass 2, which is F=T=ma=6kg*2.5ms^-1, which is equal to 15N.

Hope that helps.

[joey7]

     b) Work function  = KE max / hf = 2.96e-19/(6.63e-34 x 1e15) = 0.446
   

Doesn't work function equal KE max - hf