An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
Are these estimates of physical quantities accurate?
The error $\Delta Z$ of the quantity $Z=\frac{A}{B}$ where $A$ and $B$ are independent satisfies $\left(\frac{\Delta Z}{Z}\right)^2 = \left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2$.
The error $\Delta Z$ of the quantity $Z=A+B$ where $A$ and $B$ are independent satisfies $(\Delta Z)^2 = (\Delta A)^2 + (\Delta B)^2$.
The error $\Delta Z$ of the quantity $Z=kA$ satisfies $(\Delta Z)^2 = (|k|\Delta A)^2 $.
$$\begin{align*}\left(\frac{\Delta r'}{r'}\right)^2 &= \left(\frac{8.3\times10^{-35}}{1.67...\times10^{-27}}\right)^2 + \left(\frac{4.5\times10^{-38}}{9.10...\times10^{-31}}\right)^2 \\&= 4.90\times^{-15}.\\\Rightarrow \Delta r' &= 1.29\times10^{-4}\end{align*}$$
Therefore, $r' = 1836.15267(13)$.
The proton/electron mass ratio, $r$, is $1836.152\, 672\, 4718(80)$. These values are consistent, as the given value for $r$ is within the error of $r'$. It appears the mass ratio is known to much greater accuracy than the individual masses.
The atomic mass of oxygen-16 is 15.99491461956(16)u, and the atomic mass of hydrogen-1 is 1.00782503207(10)u. The atomic mass of a water molecule is therefore $18.0104536837(26)u = 2.99072411(15)\times10^{-26}kg$. Therefore 1 mole of water weighs (mulitplying by Avogadro's constant) $1.80105647(13)\times10^{-2}kg$.