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P_{loss}=I^2R. If we have a higher resistance, then we would have a greater power loss. So ideally, you'd want a low resistance for a low power loss. If the secondary coil has a lower voltage output, then it will have a higher current flowing through it. From the formula, you can see that a...

Have a read through this: http://www.purplemath.com/modules/rounding2.htm

In this case 0 is. The first significant figure is the first non-zero digit. So 0.000102 is rounded to 3 significant figures, not 2 or 6.

For i) they multiplied everything in the brackets by 1/2. \begin{align*}&=\frac{1}{2}s\sin{a}(2x-2l+\frac{l^2}{x}) \\&=s\sin{a}(\frac{2x}{2}-\frac{2l}{2}+\frac{1}{2}\times\frac{l^2}{x}) \\&=s\sin{a}(x-l+\frac{l^2}{2x})\end{align*} For ii) you aren't differentiating x, rather x^{-1}...

Re: HSC 2015 2U Marathon y=\sqrt{25-x^2}$ is in the form $y=\sqrt{r^2-x^2}$, which is the equation of a semicircle with radius $r$ units and centre $(0,0)$. So $y=\sqrt{25-x^2}$ is a semicircle with radius 5 units. For this function the range is: $0 \leq y \leq 5$. For $y=2\sqrt{25-x^2}$, we...

Read through this thread: http://community.boredofstudies.org/14/mathematics-extension-2/75261/if-you-year-11-read-first.html

Re: HSC 2015 3U Marathon

You can find some here: http://discoveryhshs.weebly.com/practice-exams.html

$The reaction is: FeS$_{(s)}$ + 2HCl$_{(aq)}$ $\to$ FeCl$_{2}$ + H$_{2}$S$_{(g)}$.$ $(a) n(H$_2$S) $=\frac{V}{V_m}=\frac{0.1225}{24.79}=0.00494$ moles. $(b) n(FeS) $=$ n(H$_{2}$S)$=0.00494$ moles. \\m(FeS) $=n \times M =0.00494 \times [55.85+32.07]=0.43446 $ grams. $(c) n(Reacted HCl) $=$...

It probably wouldn't explicitly state it. The question might say Copper (I) nitrate, or CuNO3, which shows that Cu+ is the ion you're dealing with.

Most of the time you would use Cu2+. I think it's because for aqueous solutions Cu2+ is more stable than Cu+. The reason why this is true seems to be beyond HSC Chemistry (http://www.chemicalforums.com/index.php?topic=3055.0)

How did you recognise that? I approached it like it was strictly rates of change, so I was stuck for a while on this one. Thanks for the help InteGrand!

Can someone help with this one too? $Two men P and Q walk along roads OA, OB at right angles to one another. P walks at 6 km/h along OA and Q at 5 km/h along OB; they start simultaneously from O. Show that the rate at which the distance between the 2 men is increasing is constant, and find this...

Ah okay, I misinterpreted the question. Thanks for the help!

$Sand is pouring from a chute at the rate of 12 m^3$/$min. The falling sand forms a conical pile on the ground, whose altitude is always equal to the diameter of the base. Show that when the altitude is $h$ m, the volume $V$ m$^3$ of sand in the pile is given by $V=\frac{\pi h^3}{12}$. I...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Thanks for the clarifications!

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread So I can't assume that y=Ae^{kx}+C, and then use that assumption to show that it works? My thinking was, if y is equal to that, and we sub it in to dy/dx and then integrate, and end up with y=Ae^{kx}+C again, then it is a solution.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread You differentiate y= Ae^{kx} + C to get dy/dx = k(y-C).

Nevermind, subbing in t=0 to x doesn't give x=0, so I'll assume the question is wrong.

$A moving particle has an acceleration $a$ metres per second per second at time $t$ seconds $(t\geq0)$. The acceleration of the particle is given by $a=2\pi^2\cos{(\pi t)}$. At time $t=0$, the object is at the point $x=0$ and has a velocity $v=\pi$ metres per second. \\(c) Show that the...