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I cant seem to register. :confused:

I like this proof ... z = cosx + isinx dz/dx = -sinx + icosx = i(cosx+isinx) = iz dz/dx = iz dz/z = i dx ln[z] + c = xi x = 0, z = 1 thus c = 0 xi = lnz therefore z = cosx + isinx = e^(xi) cos[pi] + isin[pi] = e^(i Pi) = -1

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Yay! We love Vafa!

Re: Finished HSC!! Finished yesterday :rofl:

Why did you leave after 2 hours? I thought it was a ridiculously easy exam. I had around 15 minutes to go, wrote 22 full pages overall (used 8 booklets). I kinda wish it was harder, cause now its going to scale down. The third last question in developers view of hardware option was actually fun...

The fact that copper does not become a superconductor at very low temperatures is completely irrelevant. You are expected to know that -50C is too high for ANY material that exists today to reach a super-conductive state. They could have said 'modern high-temperature ceramic superconductor' and...

The answer could not be 0 EVEN if they did mean *after* it lands. They didn't mention frictional forces, but lets assume the ground does produce frictional forces on the ball. This frictional force would have to be infinite for the horizontal velocity to reduce down to 0m/s in 0 time.

This is a very simply question ... try to stick to a problem and solve it before asking here, it helps alot more when you realize how to solve something yourself.

You dont need equation for chord of contact Simply, dy/dx = x/4 tangent at point (t,t^2/8) y - t^2/8 = t/4(x-t) 8y - t^2 = 2t(x-t) goes through (m,n) 8n - t^2 = 2tm - 2t^2 t^2 - 2tm + 8n = 0 Since t is the x coordinate, x^2 - 2mx + 8n = 0 has roots that are the x-coordinates...

Man these exams are more about guessing about what the examiners want than answering the questions.

What does kinetic energy -> heat have to do with the forces experienced during re-entry and orbit? Why would that be part of the marking scheme?

Now way its 1MJ per 10,000KM, as I said, the field gets weaker. Thus it would take less energy further out.

Damn, you are right about 7. Now im not too confident.

Not really. Simply, further distance = weaker field.

8.A - because maximum FLUX goes through the loop when its perpendicular to field 9.B - Not too sure, but no way is it D? The conductor is always cutting flux at different parts, but its always at the same rate, so it had to be DC... 10.A Shouldn't the voltage be OPPOSITE? (ie. negative)...

Had half an hour left, went through it again and added random things here and there just to make sure their gay marking criteria doesnt fuck me up.

No, the weight was decreasing as the current was being increased. Thus attracted.

The disc is connected to a circuit...

I liked the exam, much better than 2005. 2005 was more like an English paper.