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 × 8/5kilometres1hour | |
= | 16 kilometres per hour |
Solution: One million pounds in pennies is given by:
106 × 100 = 108 pennies
- which is one hundred million pennies.
Click on questions to reveal their solutions
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 (L2). Its numerical value depends on the units used. Two possible ways of expressing the area are either:
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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 102 = 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 1cm2 in square metres.
One metre is 100cm, so 1cm = 10−2m. Thus the conversion factor is 10−2.
∴ 1cm2 | = | (10−2)2m2 |
= | 1 × 10−4m2 | |
= | 0.0001m2 |
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 1mile2 in square kilometres.
One mile is 8/5km, so the conversion factor is 8/5. We thus obtain:
1mile2 | = | ( 8/5 )2km2 |
= | 64/25km2 |
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 Acm2 can be expressed in square millimetres as
Acm2 | = | A × (10mm)2 |
= | A × 102mm2 | |
= | 100Amm2 |
Similarly, a volume (whose dimensions are L3) 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 3cm3 expressed in cubic millimetres?
Explanation: We want to express a volume of 3cm3 in cubic millimetres. One centimetre is 10mm, so the conversion factor is 10. We thus obtain:
3cm3 | = | 3 × (10)3mm3 |
= | 3,000mm3 |
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 1dm3 and 1dm = 0.1m, i.e. the conversion factor is 0.1. Thus:
0.5l | = | 0.5dm3 |
= | 0.5×(0.1)3m3 | |
= | 5×10−1×10−3m3 | |
= | 5×10−4m3 |
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 kgm3 |
= | 1 kg(10dm)3 | |
= | 1 kg1000dm3 | |
= | 0.001kg dm−3 |
This can be practised in Quiz 5.
Quiz 5: Express a density of 3 × 102kg m−3 in terms of kilograms per cubic centimetre.
Explanation: We want to express a density of 3×102kg 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=106cm3. Therefore
3 × 102kg m−3 | = | 3 × 102kg1m3 |
= | 3 × 102kg1 × 106cm3 | |
= | 3 × 102 × 10−6kg cm−3 | |
= | 3 × 10−4kg cm−3 |
Exercise 2: Perform the following unit conversions.
Solution: To calculate the density 3 × 102kg m−3 in grams per cubic decimetre, we need to use the conversion factors:
1kg | = | 103g |
1m | = | 10dm |
So we have:
3 × 102kg m−3 | = | 3 × 102 × 103(10)3 |
= | 3×102g 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 = 0.14m2.
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 × 10−3J | = | 6 × 10−2J |
0.5μs | = | 0.5 × 10−6s | = | 5 × 10−7s |
We recall that:
Power = energytime
so the power is:
Power | = | 6 × 10−2(J)5×10−7(s) |
= | 1.2 × 10−2+7J s−1 | |
= | 1.2 × 10−5W |
- 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 × 106J
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 sub-atomic physics. It is related to the SI unit of energy, the Joule, by: 1eV = 1.6 × 10−19J. The energy of a photon is related to its frequency ν by E = hν, where h = 6.6 × 10−34J 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 × 10−19J. It follows that the photon's frequency is:
ν | = | Eh |
= | 3.52 × 10−19J6.6 × 10−34Js | |
= | 3.526.6 × 10−19+34s−1 | |
∴ ν | = | 5.3 × 1014s−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 (milli-Ampere) current is measured. In SI units, the current is I = 4 × 10−3A 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: vesc = √ 2gr. In SI units, g = 9.8m s−2 and r = 6380 × 1000 = 6.38 × 106m. Subsitituting these values into the equation gives:
vesc | = | √ 2 × 9.8m s−2 × 6.38 × 106m |
= | √ 1.25 × 108m2 s−2 | |
= | 1.1 × 104m 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 × 104kg m−3, and P=100 × 1000 = 105Pa. This gives:
H | = | 105Pa1.4 × 104kg m−3 × 9.8m s−2 |
= | 11.4 × 9.8 × 105−4 × N m−2 kg−1 m3 m−1 s2 | |
= | 0.73 N kg−1 s2 |
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 × 104g of a solid, what is ℓ?
Explanation: We are given E = mℓ and need to calculate ℓ given that E = 5MJ and m = 2 × 104g.
Rearranging the equation we obtain ℓ = E/m. We also need to express E and m in SI units:
E | = | 5 × 106J |
m | = | 2 × 104 × 10−3 = 20kg |
Thus we have:
Quiz 7: The area of a circle is πr2 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 πm2 and the smaller one has a radius of 50cm.
(a) 1.44 πm2 Correct - well done!
(b) 0.194 πm2 Incorrect - please try again!
(c) 2.5 πm2 Incorrect - please try again!
(d) 1.44×10−3 πm2 Incorrect - please try again!
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Explanation: The area of the disc between the two circles is given by the difference of their areas. The larger has an area of 1.69πm2, whilst the smaller has a radius of 50cm. In SI units, the radius is 0.5m. Such a circle has area πr2 = π(0.5)2m2 = 0.25πm2. Thus the difference is:
Choose the solutions from the options given