The use of different units in science and everyday life makes it important to be able to convert between different units.
Example 1: One kilometer is roughtly 58. What is a mile in kilometers?
Solution:
1km  =  58 mile  
8 km  =  5 miles  (multiply by 8) 
85km  =  1 mile  (divide by 5) 
Therefore, one mile is roughly ^{5} / _{8} of a kilometre.
Exercise 1: Express the following quantities in the units requested.
Solution: To calculate how many seconds there are in a year, recall that there are 365 days, each of which lasts 24 hour. Each hour has 60 minutes, each of 60 seconds duration. Thus, we get:
Number of seconds in a year  =  365 × 24 × 60 × 60 
=  31,536,000s 
Solution:
0.016miles  =  0.016 × 85 km 
=  0.0256miles 
Solution: A speed of 10 miles per hour is:
10 miles per hour  =  10miles 1hour 
=  10 × ^{8}/_{5}kilometres1hour  
=  16 kilometres per hour 
Solution: One million pounds in pennies is given by:
10^{6} × 100 = 10^{8} pennies
 which is one hundred million pennies.
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It is important to take care with powers of units:
Example 2: Consider the area of the square shown in the figure on the right. From the module on Dimensional Analysis, we know that this has dimensions of length squared (L^{2}). Its numerical value depends on the units used. Two possible ways of expressing the area are either:

In general, different units are linked by a conversion factor and an area given in one unit may be expressed in another unit by multiplying the area by the square of the conversion factor.
Since 1cm = 10mm, the conversion factor is 10. So an area of one square centimetre is 10^{2} = 100 square millimetres.
Now try these two short quizzes.
Quiz 1: Which of the following is the area of the square whose sides are 1cm long, in the SI units of square metres?
Explanation: We want to express an area of 1cm^{2} in square metres.
One metre is 100cm, so 1cm = 10^{−2}m. Thus the conversion factor is 10^{−2}.
∴ 1cm^{2}  =  (10^{−2})^{2}m^{2} 
=  1 × 10^{−4}m^{2}  
=  0.0001m^{2} 
A square centimetre is a ten thousandth part of a square metre.
Quiz 2: An area of a square mile is a mile times a mile. Recalling that 8km = 5miles, what is a square miles in square kilometres?
Explanation: We want to express an area of 1mile^{2} in square kilometres.
One mile is ^{8}/_{5}km, so the conversion factor is ^{8}/_{5}. We thus obtain:
1mile^{2}  =  ( ^{8}/_{5} )^{2}km^{2} 
=  ^{64}/_{25}km^{2} 
Note that the area is multiplied by the square of the conversion factor. This is illustrated in the following example:
Example 3: Consider the area of the rectangle shown in the figure on the right. Its area is given by the product of the sides. This is either:
Note the square of the conversion factor in the last line. 
In general, an area of Acm^{2} can be expressed in square millimetres as
Acm^{2}  =  A × (10mm)^{2} 
=  A × 10^{2}mm^{2}  
=  100Amm^{2} 
Similarly, a volume (whose dimensions are L^{3}) can be converted from one set of units to another by multiplying by the cube of the appropriate conversion factor.
Quiz 3: What is the volume of 3cm^{3} expressed in cubic millimetres?
Explanation: We want to express a volume of 3cm^{3} in cubic millimetres. One centimetre is 10mm, so the conversion factor is 10. We thus obtain:
3cm^{3}  =  3 × (10)^{3}mm^{3} 
=  3,000mm^{3} 
Quiz 4: A litre (l) is a cubic decimetre, where 1dm = 0.1dm. Express 0.5l in cubic metres.
Explanation: We want to express a volume of 0.5l in cubic metres. A litre is 1dm^{3} and 1dm = 0.1m, i.e. the conversion factor is 0.1. Thus:
0.5l  =  0.5dm^{3} 
=  0.5×(0.1)^{3}m^{3}  
=  5×10^{−1}×10^{−3}m^{3}  
=  5×10^{−4}m^{3} 
Similarly, with negative powers of a unit, we divide by the appropriate power of the conversion factor. An example of this follows.
Example 4: To convert a density (dimension ML^{−3}) of 1kg m^{−3} into kg dm^{−3} we use:
1kg m^{−3}  =  1 kgm^{3} 
=  1 kg(10dm)^{3}  
=  1 kg1000dm^{3}  
=  0.001kg dm^{−3} 
This can be practised in Quiz 5.
Quiz 5: Express a density of 3 × 10^{2}kg m^{−3} in terms of kilograms per cubic centimetre.
Explanation: We want to express a density of 3×10^{2}kg m^{−3} in terms of kilograms per cubic centimetre. A metre is 100cm, so the conversion factor is 100. One cubic metre is thus (100)^{3}=10^{6}cm^{3}. Therefore
3 × 10^{2}kg m^{−3}  =  3 × 10^{2}kg1m^{3} 
=  3 × 10^{2}kg1 × 10^{6}cm^{3}  
=  3 × 10^{2} × 10^{−6}kg cm^{−3}  
=  3 × 10^{−4}kg cm^{−3} 
Exercise 2: Perform the following unit conversions.
Solution: To calculate the density 3 × 10^{2}kg m^{−3} in grams per cubic decimetre, we need to use the conversion factors:
1kg  =  10^{3}g_{ } 
1m  =  10dm 
So we have:
3 × 10^{2}kg m^{−3}_{ }  =  3 × 10^{2} × 10^{3}(10)^{3}_{ } 
=  3×10^{2}g dm^{−3}_{ } 
Solution: We need to calculate the area of a triangle with base 70cm and height 400mm. In SI units, these lengths are 0.7m and 0.4m respectively.
The formula for the area of a triangle is:
So the area is ^{1}/_{2} × 0.7 × 0.4 = 0.14m^{2}.
Solution: We want to find the power in Watts of a device that uses 60mJ (milli Joules) of energy in half a microsecond. In SI units:
60mJ  =  60 × 10^{−3}J_{ }  =  6 × 10^{−2}J_{ } 
0.5μs  =  0.5 × 10^{−6}s_{ }  =  5 × 10^{−7}s_{ } 
We recall that:
Power = energytime
so the power is:
Power  =  6 × 10^{−2}(J)5×10^{−7}(s) 
=  1.2 × 10^{−2+7}J s^{−1}_{ }  
=  1.2 × 10^{−5}W_{ } 
 where we recall that a Watt is a Joule per second.
Solution: To express a kilowatt hour in Joules, note that a kilowatt hour is the energy used in one hour by a device with a power consumption. In SI units, an hour is 3,600s and a kilowatt is 1,000W. So we have
energy = 1,000 × 3,600 = 3.6 × 10^{6}J
where we again use that 1J = 1W s
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Generally it is best in equations to express all quantities in SI units before performing calculations.
Example 5: The electron Volt (eV) is a widespread unit of energy in atomic and subatomic physics. It is related to the SI unit of energy, the Joule, by: 1eV = 1.6 × 10^{−19}J. The energy of a photon is related to its frequency ν by E = hν, where h = 6.6 × 10^{−34}J s. What is the frequency of a photon with energy 2.2eV?
Solution: The energy of the photon in SI units is E = 2.2 × 1.6 × 10^{−19} = 3.52 × 10^{−19}J. It follows that the photon's frequency is:
ν  =  Eh 
=  3.52 × 10^{−19}J6.6 × 10^{−34}Js  
=  3.526.6 × 10^{−19+34}s^{−1}  
∴ ν  =  5.3 × 10^{14}s^{−1}_{ } 
Exercise 3: Put the quantities below into SI units to perform the necessary calculations:
Solution: We want to find the voltage across a 4.3 Ω (Ohm) resistor if a 4mA (milliAmpere) current is measured. In SI units, the current is I = 4 × 10^{−3}A and the resistance is R = 4.3 Ω. From Ohm's law, V=IR, so:
 where we use that 1A Ω = 1V.
Solution: We need to find the escape speed of a projectile from the Earth: v_{esc} = √ 2gr. In SI units, g = 9.8m s^{−2} and r = 6380 × 1000 = 6.38 × 10^{6}m. Subsitituting these values into the equation gives:
v_{esc}  =  √ 2 × 9.8m s^{−2} × 6.38 × 10^{6}m_{ } 
=  √ 1.25 × 10^{8}m^{2} s^{−2} _{ }  
=  1.1 × 10^{4}m s^{−1}_{ } 
Solution: We wish to calculate the height of a column of mercury in a barometer if the air pressure, P, is 100kPa. We use
P = Hρg or H = Pρg
where g=9.8m s^{−2}, and ρ=14 tonnes per cubic metre. In SI units, ρ = 14 × 1000 = 1.4 × 10^{4}kg m^{−3}, and P=100 × 1000 = 10^{5}Pa. This gives:
H  =  10^{5}Pa1.4 × 10^{4}kg m^{−3} × 9.8m s^{−2}_{ } 
=  11.4 × 9.8 × 10^{5−4} × N m^{−2} kg^{−1} m^{3} m^{−1} s^{2}_{ }  
=  0.73 N kg^{−1} s^{2}_{ } 
Now 1N = 1kg m s^{−2}, so therefore the height is H = 0.73m.
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Here are two further quizzes to practise on.
Quiz 6: The energy E required to melt a mass m of a substance is given by E=mℓ where ℓ is the specific latent heat of fusion (in J kg^{−1}). If 5MJ is required to melt 2 × 10^{4}g of a solid, what is ℓ?
Explanation: We are given E = mℓ and need to calculate ℓ given that E = 5MJ and m = 2 × 10^{4}g.
Rearranging the equation we obtain ℓ = ^{E}/_{m}. We also need to express E and m in SI units:
E  =  5 × 10^{6}J 
m  =  2 × 10^{4} × 10^{−3} = 20kg 
Thus we have:
Quiz 7: The area of a circle is πr^{2} where r is its radius. Calculate the area of the region between the two circles illustrated in the figure on the right if the larger one has an area of 1.69 πm^{2} and the smaller one has a radius of 50cm.
(a) 1.44 πm^{2} Correct  well done!
(b) 0.194 πm^{2} Incorrect  please try again!
(c) 2.5 πm^{2} Incorrect  please try again!
(d) 1.44×10^{−3} πm^{2} Incorrect  please try again!


Explanation: The area of the disc between the two circles is given by the difference of their areas. The larger has an area of 1.69πm^{2}, whilst the smaller has a radius of 50cm. In SI units, the radius is 0.5m. Such a circle has area πr^{2} = π(0.5)^{2}m^{2} = 0.25πm^{2}. Thus the difference is:
Choose the solutions from the options given