Search results

Re: MX2 2015 Integration Marathon $For (b), let the integral be $I$. Splitting the integral yields $I=\int _{-1} ^0 \tan^{-1} \left(\mathrm{e}^{\sin x} \right)\text{ d}x + \int_0 ^1 \tan^{-1} \left(\mathrm{e}^{\sin x}\right)\text{ d}x$. For the first integral, let $u=-x$, so $\sin x = \sin...

re: HSC Physics Marathon Archive Impossible to understand at HSC level. I'd just memorise whatever the TB says for that (like Jacaranda). For a true explanation, one must have a look at the London Equations: https://en.wikipedia.org/wiki/London_equations

Re: why the mass of the planet plays no role in determining a satellite's orbital spe $Alternatively, it is can be derived via calculus. Let $x=x(t)$ be the displacement of the satellite from the planet's centre at time $t\geq 0$. We know from Newton that $\ddot{x}= -\frac{GM}{x^2}$,where $M$...

Re: why the mass of the planet plays no role in determining a satellite's orbital spe Well actually, we don't equate the kinetic energy to the potential energy (if we did, there'd be a negative, as you said). What we actually do is equate the total mechanical energy to 0. It is explained here:

Re: why the mass of the planet plays no role in determining a satellite's orbital spe Actually gravitational potential energy.

Re: why the mass of the planet plays no role in determining a satellite's orbital spe It's not the mass of the planet that plays no role in determining the satellite's orbital speed, it's the mass of the satellite.

re: HSC Physics Marathon Archive Maybe the one in (B) actually went below the t-axis for a tiny amount, but too small for our eyes to see. :P

$Basically, since the angle is constant, we have $\tau = nBIA \Rightarrow \tau \propto I$. When the motor speeds up, its back EMF increases by Lenz's Law, and so the net current $I$ flowing through the coil decreases as the motor picks up speed, so torque decreases.$

It's (D), as he said in his edit.

Answer's not actually (B), he misread it.

Back EMF.

Re: MX2 2015 Integration Marathon $Let $I_n =\int \left(\sin^{-1} x\right)^n \text{ d}x$, for $n\in \mathbb{N}$. We want to find $I_3$. Note that$ $$\begin{align*}I_1 &= \int \sin^{-1}x \text{ d}x \\ &= x\sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}}\text{ d}x\\ &= x \sin^{-1}x +...

Re: MX2 2015 Integration Marathon I think you made a mistake expanding when going from the top line here to the next.

$This is the auxiliary angle transformation. We can write $6\cos 2t + 8\sin 2t \equiv R\cos (2t-\alpha)$, for some $R>0$ and $\alpha \in \mathbb{R}$. Expanding with the compound angle formula, we have$ $$6\cos 2t + 8\sin 2t \equiv R \cos 2t \cos \alpha + R\sin 2t \sin \alpha$,$ $and...

$The key thing to remember therefore for calculating work functions is the equation $W_M=hf_{0,M}$, i.e. multiply the frequency intercept for that metal's $E$-$f$ graph by $h$.$

$Reading off the graph then, we have $ $$W_X = hf_{0,X}\approx 6.626 \times 10^{-34}\text{ J s}\times 4\times 10^{14}\text{ Hz}$ (since the horizontal intercept for metal $X$ is about 4 (based on a glance, use a ruler in the actual exam for more accurate value))$ $$\approx 2.65 \times...

$Recall that the lines are $E=hf-W_M$, where $W_M$ is the work function of the metal $M$. Note that all metals thus have the same slope for the line, and the only difference is the vertical intercept, which is determined by the metal's work function $W_M$.$ $For a metal $M$, let $f_{0, M}$ be...

$Note that$ $$\left( 2x^3 -\frac{1}{x}\right)^{12} = \left( 2x^3 -x^{-1}\right)^{12}$$ $$= \sum _{k=0}^{12} \binom{12}{k}\left(2x^3\right)^{12-k}\left(-x^{-1} \right)^k$$ $$= \sum _{k=0}^{12} \binom{12}{k}2^{12-k}x^{36-3k}\cdot (-1)^k \cdot x^{-k}$$ $$= \sum _{k=0}^{12}...

re: HSC Physics Marathon Archive The choice of positive direction is arbitrary. (And out of the given options, only (B) has a graph with constant non-zero acceleration, so (B) is the answer.)

I don't think students needed to simplify the binomial sum for the last question part (iii).