SI units and unit conversion

Every physical quantity you report has two parts: a number and a unit. The International System of Units (SI) fixes seven base units — metre ($\mathrm{m}$), kilogram ($\mathrm{kg}$), second ($\mathrm{s}$), ampere ($\mathrm{A}$), kelvin ($\mathrm{K}$), mole ($\mathrm{mol}$), candela ($\mathrm{cd}$) — and every other unit is built from them.

Prefixes scale a unit by powers of ten: $1\ \mathrm{km} = 10^{3}\ \mathrm{m}$, $1\ \mathrm{cm} = 10^{-2}\ \mathrm{m}$, $1\ \mathrm{mg} = 10^{-3}\ \mathrm{g}$.

To convert, multiply by a conversion factor — a fraction equal to one — arranged so that the unwanted unit cancels:

$$250\ \mathrm{cm}\times\frac{1\ \mathrm{m}}{100\ \mathrm{cm}} = 2.5\ \mathrm{m}.$$

Keeping units in every line lets you check the result before you trust the number. If the final unit is wrong, the arithmetic is wrong.

Convert $250\ \mathrm{cm}$ to metres and $3.5\ \mathrm{km}$ to metres.

A pump delivers $1.5\ \mathrm{L}$ of water per $1\ \mathrm{min}$. Express this flow rate in $\mathrm{m^{3}/s}$.

A block measures $20\ \mathrm{cm}\times10\ \mathrm{cm}\times5\ \mathrm{cm}$ and has a mass of $2.7\ \mathrm{kg}$. Find its density in $\mathrm{kg/m^{3}}$ and in $\mathrm{g/cm^{3}}$, and state whether it sinks in water (water density $1.00\ \mathrm{g/cm^{3}}$).

A safe measurement task. Take a book, a plastic bottle, and a bag of rice.

1. Measure the three edges of the book in centimetres with a ruler and convert each value to $\mathrm{m}$ and $\mathrm{mm}$, writing the chain of fractions with units cancelling.
2. Copy the net mass in grams from the bottle or bag label and convert it to $\mathrm{kg}$ and $\mathrm{mg}$.
3. Find the volume of the book in $\mathrm{cm^{3}}$ and in $\mathrm{m^{3}}$, then compute its density in $\mathrm{kg/m^{3}}$ (weigh the book on a kitchen scale).

Do not measure anything hazardous yourself: mains voltage, hot cookware, or a pressurised gas cylinder. For such quantities take a published or simulated value (for example, the PhET "Measurement" / "Unit Rates" simulation) and convert only its units.

Finish by checking your work: run every answer through the prefix ladder and write one sentence about which conversion was hardest and why.

Make the links between SI prefixes visible as powers of ten, so the learner can choose the right conversion factor and immediately see the expected order of magnitude.

A prefix-ladder diagram: a vertical staircase where each step is a factor of $10$; the steps are labelled from nano ($\mathrm{n}$, $10^{-9}$) to giga ($\mathrm{G}$, $10^{9}$) with the prefix name, symbol, and exponent, and an arrow shows the direction of conversion (upward — divide, downward — multiply). Beside it, a small inset shows the fraction $\frac{1\ \mathrm{m}}{100\ \mathrm{cm}}$ with the unit $\mathrm{cm}$ struck through as it cancels, as a reminder of the conversion-factor method.

A runner covers $100\ \mathrm{m}$ in $9.58\ \mathrm{s}$. Average speed in SI: $v=\frac{100\ \mathrm{m}}{9.58\ \mathrm{s}}\approx 10.44\ \mathrm{m/s}.$ To get the value in $\mathrm{km/h}$, use $1\ \mathrm{m/s}=3.6\ \mathrm{km/h}$, so $10.44\ \mathrm{m/s}\times 3.6\approx 37.6\ \mathrm{km/h}$.

The scenario uses the SI base units: metre for length and second for time. The derived unit $\mathrm{m/s}$ comes from division, and conversion to $\mathrm{km/h}$ uses the factor $3.6$ because $1\ \mathrm{m/s}=3.6\ \mathrm{km/h}$.

A car travels at $90\ \mathrm{km/h}$ on a highway. To substitute this into $s=vt$ in SI units, convert: $90\ \mathrm{km/h}=\frac{90}{3.6}\ \mathrm{m/s}=25\ \mathrm{m/s}.$ In time $2\ \mathrm{h}=7200\ \mathrm{s}$, the distance is $s=25\ \mathrm{m/s}\times 7200\ \mathrm{s}=180000\ \mathrm{m}=180\ \mathrm{km}.$

The kilo prefix is used ($1\ \mathrm{km}=1000\ \mathrm{m}$), as is conversion of the non-SI hour to seconds ($1\ \mathrm{h}=3600\ \mathrm{s}$). Consistent SI units allow speed and time to be multiplied without extra factors.

A smartphone case states a capacity of $4000\ \mathrm{mAh}$. This is not an SI unit. Convert: $1\ \mathrm{mAh}=10^{-3}\ \mathrm{A}\times 3600\ \mathrm{s}=3.6\ \mathrm{C},$ so $4000\ \mathrm{mAh}=4000\times 3.6\ \mathrm{C}=14400\ \mathrm{C}.$ The charger supplies $5\ \mathrm{V}$ at a current of $2\ \mathrm{A}$, so the power is $P=UI=5\ \mathrm{V}\times 2\ \mathrm{A}=10\ \mathrm{W}.$

This applies the SI units ampere and second, the milli prefix, and the derived units coulomb and watt. Converting $\mathrm{mAh}$ to coulombs shows how the product of units $\mathrm{A}\times\mathrm{s}$ gives the new unit $\mathrm{C}$.