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re: HSC Chemistry Marathon Archive What is the marking criteria?

Realised you probably just want a numerical answer lol. According to http://www.wolframalpha.com/widgets/view.jsp?id=78baf4f3a070cc5b9b226664d2ce80ec and http://wolframalpha.com/input/?i=sum+from+k+%3D+168+to+240+of+%28240+choose+k%29%281%2F3%29%5E%28k%29+*%282%2F3%29%5E%28240-k%29&x=0&y=0 ...

I've heard rumours that there actually is Options scaling, but haven't investigated it much.

Re: HSC 2015 4U Marathon This isn't correct. The second line of the working is an incorrect conclusion from the first line; something we could conclude from the first line is that -1/z = r.cis(-2pi/3), where r is the modulus of -1/z (which is the reciprocal of the modulus of z).

Re: HSC 2015 4U Marathon And of course the physical interpretation is that k represents (something proportional to) some sort of resistance force (air resistance), and as that goes to 0, the ball dropped under gravity has its motion approach that of no resistance.

Re: HSC 2015 4U Marathon Yeah it is. It's asking to find x(t), just integrate basically. But the original Q. involved solving a differential equation to find v(t) I think. And the last part involves Maclaurin series.

Re: HSC 2015 4U Marathon haha thought so

Re: HSC 2015 4U Marathon Is this a typo? Is the dt just meant to be t? (Or should the dt just not be there?)

If the mgh formula were valid (i.e. if the altitude were low), then the Change in potential energy would indeed by mgh, since: $$\Delta U = mg(\Delta h) = mgh_2 - mgh_1 = mgh_2$, as $h_1 = 0$ (Earth's surface), where $h$ refers to height above Earth's surface in this formula.$

Should the U = -Gmm/r one be used instead here? Since h is quite significant haha.

That would be the case if (and only if) θ is in quadrants 1 or 2, which it does not need to be.

Actually no-one did get "over" 70% lol.

Nothing can guarantee it, you still need to do well in the actual HSC exam and internally of course.

$The angle $\theta$ may be in any quadrant. To prove (i), note that $\mathrm{cis}(-\theta)=\cos (-\theta) + i\sin (-\theta) = \cos \theta - i\sin \theta$ (using even and odd properties of cosine and sine).$

Didn't sit the paper.

$Assuming you have a 1 in 3 chance of getting an answer right on a given guess, the answer is $\sum _{k=168}^{240}\binom{240}{k}\left(\frac{1}{3} \right)^k \left(\frac{2}{3} \right)^{240-k}$, which you should not be expected to have to calculate without use of a computer.$

Wow first 70! Congratulations to that person. Does someone owe someone lunch now? :P

HSC Physics is pointless for uni physics and engineering. The most important HSC subject for uni engineering and physics would be maths – generally, the higher level of maths you do, the better. This is because university physics is mathematical and problem-solving based, whereas the HSC Physics...

Re: HSC 2015 4U Marathon $My method is simply converting rpm's (which is the units we are given the speed of rotation in) to radians per second:$ $ $45$ rpm $= 45\text{ }\frac{\text{rev}}{\text{min}}$$ $$=45\times \frac{2\pi\text{ rad}}{60\text{ s}}$ (since one revolution is $2\pi$...

Re: MX2 2015 Integration Marathon $Just from the bounds, I would guess a substitution of $u=\frac{1}{x}$.$