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Ok I finally got a copy of the exam so I can now comment. :) In my opinion the numerical approximation of 0.1 (one significant digit) should be fine. The data you were given, t1~=0.56, was itself only accurate to 2 significant digits, and in general the series of calculations required would...

What! I voted up your "rep power" for nuffin'. :wink: Ok, I see it now. Dang, it was buried somewhere in the middle of the "general thoughts" thread. :)

At last, someone has uploaded. Thanks ocatal. :cool2:

Yes, if you used the population standard deviation (sigma) then you should have ended up with a Z-score of about -1.25 and a rounded mark of 59 in part ii. If however you used the sample standard deviation (S) then you should have ended up with a Z-score of about -1.18 and a rounded mark of 60...

No Q24 was definitely "c" (43 degrees).

Thanks hatake and D94. Yeah I thought he only needed about 50% (60/120) raw to get a half decent mark. It looks like he must have only got about 40% or less in that exam. Oh well, he did spend more time playing guitar hero than studying. :haha:

I was looking at my nephew's (2012) HSC results today and was a bit surprised how low was his 2U math score. He got all Bands 4 and 5 in his other subjects but only scored a pretty lowly 60 in 2u maths. I was just wondering, roughly what sort of raw exam mark would have netted an aligned mark of...

B I L \sin \theta_1 L = L_{perp}/\sin \theta_2 OP (lolcakes) has fallen for a classic trap here in assuming that "theta1" is equal to "theta2" and hence the sin(theta) terms cancel out. In fact theta2 = 30 degrees while theta1 (the angle between the current and the B field) is 90 degrees. So...

Solutions. I've only just got handed a copy of the exam tonight, but these are what I believe to be the correct answers. (* = up for debate) 1. d 2. a 3. c 4. b 5. c 6. c 7. b 8. a 9. b 10. d 11. c 12. a 13. b 14. d 15. d 16. c 17. b 18. c 19. c* 20. b * Question 19 could be debated. Too...

Q20. Sorry I haven't read all the replies, but just in case there's still any uncertainty over this one the correct answer is "B" (InSb). - Si and GaN (C and D) have too higher bandgap, so few if any carriers will be optically excited across the bandgap by the given infrared signature. -...

Re Q16. That's true, but for the operating conditions given (constant rotational speed) then back EMF equals applied voltage for the IDEAL dc motor. For any real motor the answer *would* have been "A", but for this question the answer is "C".

Wow it's not really all that confusing mate. What you propose is certainly easy for anyone who is doing the course, but pretty difficult for anyone else who isn't.

Do you mean given that c > 1/2 show that r > 1/2? Rearrange 4c = 1 + 4 r^2 to, $r = \sqrt{c-0.25}$. And note that when c=0.5 that r=0.5, and that $$\sqrt{\cdot}}$$ is an increasing function. Edit: Re-reading your question I'm not sure I really answered what you asked. Anyway, it's easiest...

The question states that the circle touches at two points located symmetrically, so it really can't touch at y=0 (or it would at least touch at an odd number of points). Of course it wouldn't hurt to state an explicit justification, but in a 2 unit paper I highly doubt that it would be necessary...

So for those of us that have to wait to see it, when does BOS officially make these available for download? I've never paid too close attention to the timing, is it usually feb/march/april ???

A bit late but I've only just got to look at this exam today. Using discriminates was a good solution, but another way to tackle 16c is to equate both the gradients and the coordinates at the point of contact. One advantage of this approach is that it makes the second part basically trivial...

Nonsense, those two forms are exactly the same thing. Negative divided by negative is positive so it was fine as it was. AusBluesMan, just make sure that you use brackets on that denominator, Sn=a(r^n-1)/(r-1), when using an ordinary divide symbol (as opposed to a "fraction bar" which has...

Hey it looks pretty similar to Terry Lee's solutions. Which would of course be as expected if all the answers were correct, but what if they had an identical answer to Terry and it was one he actually got wrong (well partially wrong anyway). :p I'm referring to question 3 part (a) (iii) with...

In the annuity question of Q27 (a) part (ii), what exactly was it asking for? Was it asking for the amount per year needed to be invested in the annuity ($48395.15) or was it asking for the amount of an equivalent lump sum investment having the same value after 7 years ($289318.37) ??? I...

No doubt the correct answer was "Juan is correct, experiment 2 was closer to theoretical" It seems that most people who managed to get the theoretical probabilities correct could also get the right answer "by inspection" but I'm just wondering if anyone tried anything quantitative? Something...