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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.
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?