Alan

You might notice here that 22 + 81 = 103, so I suspect that what is wanted is 103!/(22!*81!) which is nCr(103,22)  (or nCr(103,81), which is the same).  This can be calculated using the website calculator using the ncr function:

$${\left({\frac{{\mathtt{103}}{!}}{{\mathtt{22}}{!}{\mathtt{\,\times\,}}({\mathtt{103}}{\mathtt{\,-\,}}{\mathtt{22}}){!}}}\right)} = {\mathtt{15\,197\,882\,802\,514\,693\,962\,310}}$$

$${\left({\frac{{\mathtt{103}}{!}}{{\mathtt{81}}{!}{\mathtt{\,\times\,}}({\mathtt{103}}{\mathtt{\,-\,}}{\mathtt{81}}){!}}}\right)} = {\mathtt{15\,197\,882\,802\,514\,693\,962\,310}}$$

1 eV = 1.60217657 × 10^-19 joules

8.5 eV = 8.5*1.60217657 × 10^-19 joules = 1.36185 x 10^-18 J

Look up Snell's law.  

This is what I get:

xvxvxv: In question number 1  why I can't use   P=V.I

You can, but you have to do it at the normal condition first: so 280=240*I_normal

I_normal = 280/240.

Ohms law says V = IR  so 240 = R*280/240 or R = 240²/280 

The resistance R, will be approximately the same at reduced voltage, so using Ohm's law again:

200 = I_reduced*240²/280

 I_reduced = 200*280/240²=0.972

(PS on question 16, go with thermal conductivity!).

In fact, there is a way to get the calculator here to calculate this (using the npr function):

$${\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}\left({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{335}}\right)){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{30}}}}} = {\mathtt{0.293\: \!683\: \!757\: \!280\: \!731\: \!3}}$$  

I entered the above as npr(365,(365-335))/365^30.

The result isn't as small as you might expect:

Re. 16).  You might well be right David.  I just thought in terms of a relatively small block of metal/wood for which one might write  m.c_p.dT/dt = heff.A(T_hand - T) where heff is an effective heat transfer coefficient that would include thermal conductivity as well as a contact heat transfer coefficient.  In this case the heat transfer is likely to be dominated by c_p when it transfers to the right-hand side.

If we are talking about large blocks, with effectively infinite heat capacity, then it is probably better to write a similar equation from the hand's point of view, in which case it will probably be the contact resistance that dominates.

Never really thought about it much before - an interesting question to ponder!

In the last part of the last image it looks to me as though B) A and E is circled.  This is correct here, as amplitude is usually mean-to-peak.  Think of the curve as y = a*sin(x).  a is generally thought of as the amplitude and y varies from +a to -a.

NB David is correct about question1.  I missed that!

The result of doing exactly what you specified is 1.069.

However, it looks to me as though you are trying to calculate the standard deviation of this data set, in which case you need to divide by 11 before taking the square root.  Also, the mean of the data (the value you are subtracting from each number) is 65.25 not 65.26.  If it is the standard deviation you really want then the result is 3.545.