# 1. Introduction

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 5/8. What is a mile in kilometers?

Solution:

 1km = 5/8 mile 8 km = 5 miles (multiply by 8) 8/5km = 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.

(a) One year in seconds.

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
Therefore, there are 3.1536 × 106 seconds in a year (just over 30 million seconds).
(b) 0.016miles in kilometers

Solution:

 0.016miles = 0.016 × 8/5 km = 0.0256miles
(c) A speed of 10 miles per hour in kilometres per hour

Solution: A speed of 10 miles per hour is:

 10 miles per hour = 10miles / 1hour = 10 × 8/5kilometres/1hour = 16 kilometres per hour
(d)One million pounds in pennies.

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

# 2. Powers of Units

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: (1cm) × (1cm) = (1cm2) or: (10mm) × (10mm) = (100mm2) 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?

(a) 0.01m2 Incorrect - please try again!
(b) 100m2 Incorrect - please try again!
(c) 0.0001m2 Correct - well done!
(d) 10,000m2 Incorrect - please try again!

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?

(a) 25/64km2 Incorrect - please try again!
(b) 64/25km2 Correct - well done!
(c) 16/10km2 Incorrect - please try again!
(d) 10/16km2 Incorrect - please try again!

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: area = (2cm) × (1cm) = 2cm2 or: area = (20mm) × (10mm) = 2 ×102mm2 = 200mm2 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?

(a) 300mm3 Incorrect - please try again!
(b) 27,000mm3 Incorrect - please try again!
(c) 3 × 104mm3 Incorrect - please try again!
(d) 3,000mm3 Correct - well done!

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.

(a) 5 × 10−4m3 Correct - well done!
(b) 1.25 × 10−2m3 Incorrect - please try again!
(c) 0.0005m3 Incorrect - please try again!
(d) 5,000m3 Incorrect - please try again!

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 kg/m3 = 1 kg/(10dm)3 = 1 kg/1000dm3 = 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.

(a) 3 × 10−4kg cm−3 Correct - well done!
(b) 2.7 × 1011kg cm−3 Incorrect - please try again!
(c) 3 × 108kg cm−3 Incorrect - please try again!
(d) 3 × 105kg cm−3 Incorrect - please try again!

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 × 102kg/1m3 = 3 × 102kg/1 × 106cm3 = 3 × 102 × 10−6kg cm−3 = 3 × 10−4kg cm−3

Exercise 2: Perform the following unit conversions.

(a) Find the density 3 × 102kg m−3 in grams per cubic decimetre.

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
(b) Calculate the area of the triangle shown below in square metres. 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:

area = 1/2 base × height

So the area is 1/2 × 0.7 × 0.4 = 0.14m2.

(c) The power in Watts of a device that uses 60mJ (milli Joules) of energy in half a microsecond. (Power is the rate of conversion of energy with time. See the module on Units.)

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 = energy/time

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.

(d) A kilowatt hour is the energy used by a device with a power output of one kilowatt in one hour. Express this in Joules.

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

Click on questions to reveal their solutions

# 3. Equations and Units

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 = , 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 = E/h = 3.52 × 10−19J/6.6 × 10−34Js = 3.52/6.6 × 10−19+34s−1 ∴ ν = 5.3 × 1014s−1

Exercise 3: Put the quantities below into SI units to perform the necessary calculations:

(a) Ohm's law V = IR links the voltage V across a resistance R to the current I flowing through it. Calculate the voltage across a 4.3 Ω (Ohm) resistor if a 4mA (milli-Ampere) current is measured.

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:

V = 4 × 10−3A × 4.3 Ω = 1.72 × 102V

- where we use that 1A Ω = 1V.

(b) The escape speed, vesc, of a projectile from a planet is given by vesc = 2gr, where r is the radius of the planet and g is the acceleration due to gravity on the surface of the planet. The Earth's radius is 6380km and g = 9.8m s−2. Calculate the escape speed from Earth.

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
(c) Air pressure P at sea level is roughly 100kPa (kilo-Pascal). The height H of a mercury column in a barometer is related to the pressure by P = Hρg where ρ is the density of mercury (14 tonnes per cubic metre) and g is the acceleration due to gravity. Find the height of the column of mercury.

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 = P/ρg

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 = 105Pa/1.4 × 104kg m−3 × 9.8m s−2 = 1/1.4 × 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.

Click on questions to reveal their solutions

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?

(a)0.2J kg−1 Incorrect - please try again!
(b)250,000J kg−1 Correct - well done!
(c)2.5J kg−1 Incorrect - please try again!
(d)20J kg−1 Incorrect - please try again!

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:

ℓ = E/m = 5 × 106 (J)/20 (kg) = 2.5 × 105J kg−1
 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! 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:

area of difference = 1.69 π − 0.25 π = 1.44 πm2

# 4. Quiz on Converting Units

Choose the solutions from the options given

1.The sunspot cycle takes place over 11 years. Roughly how many cycles take place per millennium?
(a) 110
(b) 11
(c) 91
(d) 181
2.What is a cubic decimetre in terms of cubic centimetres?
(a) 1 × 10−3cm3
(b) 1 × 104cm3
(c) 1 × 106cm3
(d) 1 × 103cm3
3.Express the area of a square of side 5μm in square metres.
(a) 25μm2
(b) 2.5 × 10−12m2
(c) 2.5 × 10−11m2
(d) 2.5 × 10−13m2
4.Find the current in Amperes through a 10Ω resistor with a 2mV potential difference across it.
(a) 2 × 10−3A
(b) 500A
(c) 0.2A
(d) 2 × 10−4A