A-level Physics/The SI System of Units

SI units are used throughout science in many countries of the world. There are seven base units, from which all other units are derived.

Base units

Every other unit is either a combination of two or more base units, or a reciprocal of a base unit. With the exception of the kilogram, all of the base units are defined as measurable natural phenomena. Also, notice that the kilogram is the only base unit with a prefix. This is because the gram is too small for most practical applications.

Quantity	Name	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

Derived units

Most of the derived units are the base units divided or multiplied together. Some of them have special names. You can see how each unit relates to any other unit, and knowing the base units for a particular derived unit is useful when checking if your working is correct.

Note that "m/s", "m s^-1", "m·s^-1" and $\frac{\text{m}}{\text{s}}$ are all equivalent. The negative exponent form is generally prefered, for example "kg·m^-1·s^-2" is easier to read than "kg/m/s²".

Quantity	Name	Symbol	In terms of other derived units	In terms of base units
Area	square metre	$m^2$		$m\times m$
Volume	cubic metre	$m^3$		$m\times m\times m$
Speed/Velocity	metre per second	$m$ $s^{-1}$		$\frac{m}{s}$
Acceleration	metre per second squared	$m$ $s^{-2}$		$\frac{\frac{m}{s}}{s}$
Density	kilogram per cubic metre	$kg$ $m^{-3}$	$\frac{kg}{m^3}$	$\frac{kg}{m\times m\times m}$
Specific Volume	cubic metre per kilogram	$m^3$ $k{g}^{-1}$	$\frac{m^3}{kg}$	$\frac{m\times m\times m}{kg}$
Current Density	ampere per square metre	$A$ $m^{-2}$	$\frac{A}{m^2}$	$\frac{A}{m\times m}$
Magnetic Field Strength	ampere per metre	$A$ $m^{-1}$		$\frac{A}{m}$
Concentration	mole per cubic metre	$mol$ $m^{-3}$	$\frac{mol}{m^3}$	$\frac{mol}{m\times m\times m}$
Frequency	hertz	Hz		$\frac{1}{s}$
Force	newton	N		$m$ $kg$ $s^{-2}$
Pressure/Stress	pascal	Pa	$\frac{N}{m^2}$	$m^{-1}$ $kg$ $s^{-2}$
Energy/Work/Quantity of Heat	joule	J	N m	$m^2$ $kg$ $s^{-2}$
Power/Radiant Flux	watt	W	$\frac{J}{s}$	$m^2$ $kg$ $s^{-3}$
Electric Charge/Quantity of Electricity	coulomb	C		s A
Electric Potential/Potential Difference/Electromotive Force	volt	V	$\frac{W}{A}$	$m^2$ $kg$ $s^{-3}$ $A^{-1}$
Capacitance	farad	F	$\frac{C}{V}$	$m^{-2}$ $k{g}^{-1}$ $s^4$ $A^2$
Electric Resistance	ohm	$\mathrm{\Omega }$	$\frac{V}{A}$	$m^2$ $kg$ $s^{-3}$ $A^{-2}$
Electric Conductance	siemens	S	$\frac{A}{V}$	$m^{-2}$ $k{g}^{-1}$ $s^3$ $A^2$
Magnetic Flux	weber	Wb	V s	$m^2$ $kg$ $s^{-2}$ $A^{-1}$
Magnetic Flux Density	tesla	T	$\frac{Wb}{m^2}$	$kg$ $s^{-2}$ $A^{-1}$
Inductance	henry	H	$\frac{Wb}{A}$	$m^2$ $kg$ $s^{-2}$ $A^{-2}$
Celsius Temperature	degree celsius	°C		K - 273.15
Luminous Flux	lumen	lm		cd sr
Illuminance	lux	lx	$\frac{lm}{m^2}$	$m^{-2}$ $cd$ $sr$
Activity of a Radionuclide	bequerel	Bq		$s^{-1}$

Prefixes

The SI units can have prefixes to make larger or smaller numbers more manageable. For example, visible light has a wavelength of roughly 0.0000005 m, but it is more commonly written as 500 nm. If you must specify a quantity like this in metres, you should write it in standard form. As given by the table below, 1nm = 1*10^-9m. In standard form, the first number must be between 1 and 10. So to put 500nm in standard form, you would divide the 500 by 100 to get 5, then multiply the factor by 100 (so that it's still the same number), getting 5*10^-7m. The power of 10 in this answer, i.e. -7, is called the exponent, or the order of magnitude of the quantity.

Prefix	Symbol	Factor	Common Term
peta	P	$10^{15}$	quadrillions
tera	T	$10^{12}$	trillions
giga	G	$10^9$	billions
mega	M	$10^6$	millions
kilo	k	$10^3$	thousands
hecto	h	$10^2$	hundreds
deca	da	$10^1$	tens
deci	d	$10^{-1}$	tenths
centi	c	$10^{-2}$	hundredths
milli	m	$10^{-3}$	thousandths
micro	µ	$10^{-6}$	millionths
nano	n	$10^{-9}$	billionths
pico	p	$10^{-12}$	trillionths
femto	f	$10^{-15}$	quadrillionths

Homogenous equations

Equations must always have the same units on both sides, and if they don't, you have probably made a mistake. Once you have your answer, you can check that the units are correct by doing the equation again with only the units.

For example, to find the velocity of a cyclist who moved 100 metres in 20 seconds, you have to use the formula $velocity=\frac{displacement}{time}$, so your answer would be 5$m$ $s^{-1}$.

This question has the units $m÷s$, and should give an answer in $m$ $s^{-1}$. Here, the equation was correct, and makes sense.

Often, however, it isn't that simple. If a car of mass 500kg had an acceleration of 0.2 $m$ $s^{-2}$, you could calculate from $F=ma$ that the force provided by the engines is 100N. At first glance it would seem the equation is not homogenous, since the equation uses the units $kg\times m{s}^{-2}$, which should give an answer in $kg$ $m$ $s^{-2}$. If you look at the derived units table above, you can see that a newton is in fact equal to $kg$ $m$ $s^{-2}$, and therfore the equation is correct.

Using the same example as above, imagine that we are only given the mass of the car and the force exerted by the engines, and have been asked to find the acceleration of the car. Using $F=ma$ again, we need to rearrange it for $a$, and we now have the formula: $a=\frac{m}{F}$. By inserting the numbers, we get the answer $a=5m{s}^{-2}$. You already know that this is wrong from the example above, but by looking at the units, we can see why this is the case:$m{s}^{-2}=\frac{kg}{kgm{s}^{-2}}$. The units are $m{s}^2$, when we were looking for $m{s}^{-2}$. The problem is the fact that $F=ma$ was rearranged incorrectly. The correct formula was $a=\frac{F}{m}$, and using it will give the correct answer of 0.2 $m$ $s^{-2}$. The units for the correct formula are $m{s}^{-2}=\frac{kgm{s}^{-2}}{kg}=m{s}^{-2}$.