A-level Physics

1.1 SI units and prefixes — Learn

Prerequisites: decimal place value and standard form; substitution into a formula; the idea that a quantity is a number with a unit.

By the end of this section you can: name the seven SI base units; build a derived unit from the base units; change a prefix accurately, including for areas and volumes; and check whether an equation is homogeneous.

1. Base quantities and base units

A physical quantity is a number together with a unit. The number alone means nothing: "the wire is 2.5" is not a measurement, but "the wire is 2.5 mm across" is.

Physics builds every quantity from seven base quantities. Each has one base unit, fixed by international agreement.

Base quantity Base unit Symbol
Length metre m
Mass kilogram kg
Time second s
Electric current ampere A
Thermodynamic temperature kelvin K
Amount of substance mole mol
Luminous intensity candela cd

A derived unit is a combination of base units. Speed is distance divided by time, so its unit is metres per second, written m s−1\text{m}\,\text{s}^{-1}. The newton is a derived unit built from the base units by Newton's second law:

N=kg m s−2\text{N} = \text{kg}\,\text{m}\,\text{s}^{-2}

The kelvin is the base unit of temperature, but a temperature change is also often written in degrees Celsius. One degree Celsius is the same size as one kelvin, so a change of 10 ∘C10\,^\circ\text{C} is a change of 10 K10\,\text{K}.

2. Prefixes change the size of a unit

A prefix multiplies a unit by a power of ten. This lets one unit cover very large and very small quantities. The prefixes you will meet most often are below.

Prefix Symbol Factor
tera T 101210^{12}
giga G 10910^{9}
mega M 10610^{6}
kilo k 10310^{3}
centi c 10−210^{-2}
milli m 10−310^{-3}
micro μ\mu 10−610^{-6}
nano n 10−910^{-9}
pico p 10−1210^{-12}
femto f 10−1510^{-15}

To change into base units, replace the prefix by its factor. A capacitance of 4.7 μF4.7\,\mu\text{F} is 4.7×10−6 F4.7 \times 10^{-6}\,\text{F}. A force of 1.2 MN1.2\,\text{MN} is 1.2×106 N1.2 \times 10^{6}\,\text{N}.

Worked example 1. Write 3.6 kN3.6\,\text{kN} and 25 ms25\,\text{ms} in base units.

3.6 kN=3.6×103 N=3600 N25 ms=25×10−3 s=0.025 s\begin{aligned} 3.6\,\text{kN} &= 3.6 \times 10^{3}\,\text{N} = 3600\,\text{N} \\ 25\,\text{ms} &= 25 \times 10^{-3}\,\text{s} = 0.025\,\text{s} \end{aligned}

Worked example 2. A volume is 250 cm3250\,\text{cm}^3. Write it in m3\text{m}^3.

1 cm=1×10−2 m1 cm3=(1×10−2)3 m3=1×10−6 m3250 cm3=250×10−6 m3=2.5×10−4 m3\begin{aligned} 1\,\text{cm} &= 1 \times 10^{-2}\,\text{m} \\ 1\,\text{cm}^3 &= (1 \times 10^{-2})^3\,\text{m}^3 = 1 \times 10^{-6}\,\text{m}^3 \\ 250\,\text{cm}^3 &= 250 \times 10^{-6}\,\text{m}^3 \\ &= 2.5 \times 10^{-4}\,\text{m}^3 \end{aligned}

Watch out. A prefix applies to the whole unit, and a power applies to the prefix as well. So 1 cm21\,\text{cm}^2 is (10−2)2 m2=10−4 m2(10^{-2})^2\,\text{m}^2 = 10^{-4}\,\text{m}^2, not 10−2 m210^{-2}\,\text{m}^2. A student who writes 250 cm3=2.5 m3250\,\text{cm}^3 = 2.5\,\text{m}^3 has divided by 10210^{2} instead of 10610^{6}: the centimetre was not cubed. Check by asking whether the answer should be much smaller or much larger than the original number.

3. Try it: the size of a prefix

A prefix is a step on a scale, and the steps are not all the same size.

Predict first. A human hair is about 70 µm across. Do not move the slider yet. Write down your answers.

  1. Which prefix marks a single micrometre?
  2. Where on the ruler do you expect the mark for a micrometre to sit: near the water molecule, near the bacterium or near the person?
  3. Giga is two steps of a thousand above kilo. How many times larger than kilo is giga?

Test. Move the slider through the prefixes. Watch the marker and the object that labels it.

Explain. Two neighbouring marks are usually a factor of 1000 apart, but milli and centi are only ten apart. Explain why, in two or three sentences. Then write the factor for micro and for nano as powers of ten.

4. Checking an equation with base units

An equation is homogeneous when both sides have the same base units. Every valid physical equation is homogeneous, so this is a quick check on a formula you have rearranged or remembered.

Write each quantity with its base units, then simplify. The constant numbers, such as the 12\tfrac{1}{2} in kinetic energy, carry no units and can be ignored.

Worked example 3. Show that Ek=12mv2E_k = \tfrac{1}{2}mv^2 is homogeneous.

left side=J=kg m2 s−2right side=kg×(m s−1)2=kg m2 s−2\begin{aligned} \text{left side} &= \text{J} = \text{kg}\,\text{m}^2\,\text{s}^{-2} \\ \text{right side} &= \text{kg} \times (\text{m}\,\text{s}^{-1})^2 \\ &= \text{kg}\,\text{m}^2\,\text{s}^{-2} \end{aligned}

Both sides are kg m2 s−2\text{kg}\,\text{m}^2\,\text{s}^{-2}, so the equation is homogeneous.

Worked example 4. The drag force on a small sphere moving slowly through a liquid is F=6πηrvF = 6\pi\eta rv. Find the base units of the viscosity η\eta.

η=F6πrv=kg m s−2m×m s−1=kg m−1 s−1\begin{aligned} \eta &= \frac{F}{6\pi rv} \\ &= \frac{\text{kg}\,\text{m}\,\text{s}^{-2}}{\text{m} \times \text{m}\,\text{s}^{-1}} \\ &= \text{kg}\,\text{m}^{-1}\,\text{s}^{-1} \end{aligned}

The number 6π6\pi is a pure number, so it has no unit and does not appear in the base units.

5. Using units to predict a unit

Homogeneity works in the other direction too. If you know an equation, you can find the unit of any one quantity in it.

Worked example 5. The period of a pendulum is T=2πLgT = 2\pi\sqrt{\dfrac{L}{g}}, where LL is the length. Find the base units of gg.

T2=4π2Lgg=4π2LT2=ms2=m s−2\begin{aligned} T^2 &= \frac{4\pi^2 L}{g} \\ g &= \frac{4\pi^2 L}{T^2} \\ &= \frac{\text{m}}{\text{s}^2} = \text{m}\,\text{s}^{-2} \end{aligned}

So gg has base units m s−2\text{m}\,\text{s}^{-2}, which is the unit of an acceleration.

Progress check

Return to the chapter problem. The pendulum length is 0.800 m, and the time for 20 oscillations is 35.90 s.

  1. Write 0.800 m in millimetres and in micrometres.
  2. Write the time 35.90 s in milliseconds.
  3. The period is T=1.795 sT = 1.795\,\text{s}. Use g=4π2LT2g = \dfrac{4\pi^2 L}{T^2} to show that gg has base units m s−2\text{m}\,\text{s}^{-2}.

If question 1 or 2 goes wrong, revisit Section 2 and check each prefix factor. If question 3 goes wrong, write the base units of the length and of T2T^2 before you divide.

You should now be able to

Settings

Colour scheme

Applies to every book you read here.

Reading theme

Applies to every book you read here.