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

Data Matching

Use your skill and judgement to match the sets of random data.

More or Less?

Are these estimates of physical quantities accurate?

Constantly Changing

Age 14 to 16 Challenge Level:

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 $.

The error of $r' = \frac{m_p}{m_e} = 1836.15267...$ obeys:

$$\begin{align*}\left(\frac{\Delta r'}{r'}\right)^2 &= \left(\frac{8.3\times10^{-35}}{1.67...\times10^{-27}}\right)^2 +  \left(\frac{4.5\times10^{-38}}{9.10...\times10^{-31}}\right)^2 \\&= 4.90\times^{-15}.\\\Rightarrow \Delta r' &= 1.29\times10^{-4}\end{align*}$$

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) $1.80105647(13)\times10^{-2}kg$.

The energy in the mass of a mole of water is therefore $1.61870882(11)\times10^{15}J$.
Suppose a cup of tea has a volume of $200$ml and we need to raise its temperature from $20^{\circ}\mathrm{C}$ to $100^{\circ}\mathrm{C}$. Supposing there's no loss of energy in our heating system, the amount of energy in a mole of water could make $2.4\times10^{10}$ cups of tea!