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One student has left the Saturday class :( 3 hours of travelling was a BIT TOO MUCH in the end! Another student is joining this week, from the same agricultural high school :) 1 more spot only! :)

Question was "The electrical resistance, R, of a piece of wire was measured at different temperatures, T. Near room temperature, the resistacne can be modelled by the equation R=mT+b (b) Assess the validity of using the data from this experiment to estimate the electrical resistacne at -100 degC"

I don't tutor him, I teach him at my school :)

Yep, 2 marks

Agreed. However, when I talked to Margot Braithwaite (Board of Studies Liason Officer), she said the more the students write, the better! As long as they do not contradict themselves, the markers do not look for reasons to 'take marks off', which kind of surprised me.

13 is A, Arsenic is type 5, therefore more free electrons

Not too sure about concise ever being marked. I know coherent and logical are

Totally agree. It's upto the marking guidelines, which they create AFTER marking a few sets of papers apparently?

We did this exact question too, haha. It's from a past paper. It was contrast, so differences only. I would talk about the differences in individual energies of the photons of red/blue light and then the fact that there must be more photons in the red light as the power is the same, but...

oic, but was the constant thrust answering this guys q? Constant thrust was Q16, but thrust has nothing to do with a stable orbit? the answer for 16 is B.

This is true! He is my top student, I sometimes thinks he knows too much. It's funny though, he can write a page for a 4 mark answer, showing a logical progression of ideas and communicating ideas clearly, hahaha.

I went through pretty much this EXACT question with my class, except it was a 1mW laser with wavelength 700nm. Most of them remembered :) YAYYYY

?? Which MC was this?

Energy\; of\, a \, single\, photon=\frac{hc}{\lambda } \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{450\times 10^{-9}}=4.42\times 10^{-19}J Number\, of\, photons=\frac{Total\, Energy}{Energy\, of\, photon}=\frac{1\times 10^{-3}}{4.42\times 10^{-19}}=\, 2.26\times 10^{15} There u go...

You should say something about the forces on the sides of the coil are equal, but in opposite direction. Therefore net force = 0

I reckon 82ish, maybe even higher

I think it was about 7692ms-1? 29(a) would be 2.26x10^15 photons

That would be KEmax vs frequency It was asking for KEmax vs wavelength

One of my students wrote 6 extra booklets for the Q2Q option, hahaha. and another one wrote 4

It's not linear, its hyperbolic KEmax=\frac{hc}{\lambda }-\phi KEmax \; \alpha \; \frac{1}{\lambda }