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In short, you should expect the uncertainties for the volume to be around $0.5-1\%$ and the density to be around $2-5\%$.   \phantom{1}\\  For the last question of part 1, the key thing for students to realise is that if 10 coins were used then each relative uncertainty would decrease by a factor of 10, which would ultimately propagate to 10 times the precision. Writing out the whole derivation would probably take too long, so I wouldn’t be picky about that.\\  Basically, if $r_i$ is the relative uncertainty of one measurement, then with $x$ coins, this will become $\frac{r_i}{x}$. From there, a quick derivation:  \begin{align*} & 10=\dfrac {r_{old}} {r_{new}}\\ & =\dfrac {\sqrt {\sum r_{i}^{2}}} {\sqrt {\sum \left( \dfrac {r_{i}} {x}\right) ^{2}}}\\ & =x\end{align*} thus x=10 coins.\\  \phantom{1}\\  I think I found a good example for this question:  Example  (a) If 10 is added to each number in a set, what is the change in the mean? What is the change in the standard deviation?  (b) If each number in a set is multiplied by 10, what is the change in the mean? What is the change in the standard deviation?  Solution  (a) Since each value increases by 10, the mean increases by 10.   Adding 10 to each number does not change the distances between the numbers and the mean. Since the standard deviation is dependent on those distances, the standard deviation does not change.   If you are not sure about the answer, you can Plug In some numbers and try it.  The mean of {1, 2, 3} is 2. The mean of {11, 12, 13} is 12. The mean increased by 10.  The distances between {1, 2, 3} and their mean are 1. The distances between {11, 12, 13} and their mean are also 1. The standard deviation did not change.  (b) Since each value is multiplied by 10, the mean is multiplied by 10.  Multiplying each number by 10 changes the distances between the numbers and the mean by a factor of 10. Since the standard deviation is dependent on those distances, the standard deviation changes by a factor of 10.   If you are not sure about the answer, you can Plug In some numbers and try it.  The mean of {1, 2, 3} is 2. The mean of {10, 20, 30} is 20. The mean increased by a factor of 10.  The distances between {1, 2, 3} and their mean are 1. The distances between {10, 20, 30} and their mean are 10. Since the distances changed by a factor of 10, the standard deviation also changed by a factor of 10.\\  \phantom{1}\\  In part 3 (measuring reaction time), the free-fall distance associated with a reaction time of 0.18s is 16.2cm. This should be what people aim for, but it is difficult to test this in a reliable manner in the lab, so expect results to vary wildly - results will most likely be within the range of 10cm - 40cm.\\  \phantom{1}\\