Unit conversions and scientific notation

It is early October and in Year 1 of Italian high school physics has just started: quantities, measurements, the International System of Units. The first test will arrive in four to six weeks, in November, and experience says that the first failing grade of the year almost always starts here, not from real physics but from equivalences. The teacher explained the units in a couple of lessons, the textbook exercises look trivial, and then in the test $350 cm^{3}$ becomes $0.35 m^{3}$ and the whole problem is wrong by three orders of magnitude.

This post is the method I use for conversions and for scientific notation. If you learn it now, you are a step ahead: when the teacher reaches the test exercises, you will already have done the mechanical part.

The conversion-factor method

A conversion is not a trick to remember by heart. It is a multiplication by $1$ written cleverly. We know that $1 km = 1000 m$ , so the fraction

\frac{1000 m}{1 km}

equals $1$ . Multiplying by $1$ doesn’t change the quantity, it only changes how we write it. You choose the fraction so that the unit to eliminate sits on the opposite side and cancels:

5.2 km \cdot \frac{1000 m}{1 km} = 5200 m

The “km” at the top and bottom cancel, and the metres remain. If you get the fraction the wrong way up, the units don’t cancel and you notice at once: that is the check that makes the method safe.

The prefixes of the International System

Before conversions you need the prefixes, which are simply powers of ten:

Prefix	Symbol	Factor
giga	G	$1 0^{9}$
mega	M	$1 0^{6}$
kilo	k	$1 0^{3}$
centi	c	$1 0^{- 2}$
milli	m	$1 0^{- 3}$
micro	$μ$	$1 0^{- 6}$
nano	n	$1 0^{- 9}$

So $1 mm = 1 0^{- 3} m$ and $1 kg = 1 0^{3} g$ . Careful with the kilogram: the SI base unit for mass is already the kg, not the gram.

km/h and m/s: why you divide by 3.6

The conversion that appears in every test. Let’s apply the method:

1 \frac{km}{h} = \frac{1000 m}{3600 s} = \frac{1}{3.6} \frac{m}{s}

From this comes the rule: from km/h to m/s you divide by 3.6; from m/s to km/h you multiply by 3.6. One way not to swap them: the number in m/s is always “smaller” than the number in km/h, because the metre is small and the second is short. If the number grows when you go to m/s, you multiplied instead of dividing.

Two examples:

72 \frac{km}{h} = \frac{72}{3.6} \frac{m}{s} = 20 \frac{m}{s} 15 \frac{m}{s} = 15 \cdot 3.6 \frac{km}{h} = 54 \frac{km}{h}

Squared and cubed units: the mistake that costs the most

This is where most of the class goes wrong. If $1 cm = 1 0^{- 2} m$ , then for area and volume the factor has to be raised to the same power as the unit:

1 cm^{2} = (1 0^{- 2} m)^{2} = 1 0^{- 4} m^{2} 1 cm^{3} = (1 0^{- 2} m)^{3} = 1 0^{- 6} m^{3}

The typical mistake is writing $1 cm^{3} = 1 0^{- 2} m^{3}$ : you remember “centi equals $1 0^{- 2}$ ” and forget that the exponent 3 of the unit also applies to the factor. The result is wrong by a factor of ten thousand.

A common-sense check: a cube of $1 cm$ on a side is tiny, a cubic metre is a cube as big as a washing machine. A million of those little cubes fit in a cubic metre, so $1 cm^{3}$ can’t be $0.01 m^{3}$ (only a hundred would fit).

Other volume equivalences to know: $1 dm^{3} = 1 L = 1 0^{- 3} m^{3}$ , so $1 m^{3} = 1000 L$ .

Density has its compound units too

1 \frac{g}{cm ^{3}} = \frac{1 0 ^{- 3} kg}{1 0 ^{- 6} m ^{3}} = 1 0^{3} \frac{kg}{m ^{3}}

So $0.8 g/cm^{3}$ is $800 kg/m^{3}$ (water is $1 g/cm^{3} = 1000 kg/m^{3}$ ). You go from g/cm³ to kg/m³ by multiplying by $1000$ .

Scientific notation

A number is in scientific notation when it is written as $a \cdot 1 0^{n}$ with $1 \leq a < 10$ and $n$ an integer. To write $45000$ you move the decimal point until there is only one digit before it: $4.5 \cdot 1 0^{4}$ . For $0.0032$ you move it to the right: $3.2 \cdot 1 0^{- 3}$ . If the point moves to the left the exponent is positive, to the right it is negative.

Multiplying and dividing

Powers of ten follow the rules of powers: in a product the exponents are added, in a quotient they are subtracted. The numbers in front are treated separately.

(3.0 \cdot 1 0^{5}) \cdot (4.0 \cdot 1 0^{- 2}) = 12 \cdot 1 0^{3} = 1.2 \cdot 1 0^{4}

Note the last step: $12 \cdot 1 0^{3}$ is not yet in scientific notation, because in front there is $12$ and not a number between $1$ and $10$ . You write $12 = 1.2 \cdot 10$ and add the exponents: $10 \cdot 1 0^{3} = 1 0^{4}$ . It is an oversight the teacher will mark.

For division:

\frac{6.0 \cdot 1 0 ^{3}}{2.0 \cdot 1 0 ^{- 4}} = 3.0 \cdot 1 0^{3 - (- 4)} = 3.0 \cdot 1 0^{7}

Watch the sign: $3 - (- 4) = 7$ , not $- 1$ .

Three worked exercises

Exercise 1: a speed conversion

A car travels at $108 km/h$ . What is its speed in m/s and what distance does it cover in $4.0 s$ at constant speed?

108 \frac{km}{h} = \frac{108}{3.6} \frac{m}{s} = 30 \frac{m}{s}

s = v \cdot t = 30 \frac{m}{s} \cdot 4.0 s = 120 m

Check: $108/3.6$ is $30$ because $3.6 \cdot 30 = 108$ . The result is already in metres because the seconds cancel.

If you want to practise this kind of question against the clock, the Year 1 test page collects the practice tests for the first two years. I’ll tell it as it is: they are general-purpose tests on Year 1 physics and maths, not a practice test built only on conversions; they serve as self-assessment tests, not as a substitute for the exercises in this post. The correct answer to each question is free, the guided step-by-step solution unlocks with your first lesson.

Exercise 2: volume, conversion and mass

A bottle contains $750 cm^{3}$ of water. Express the volume in $m^{3}$ in scientific notation and calculate the mass, knowing that the density of water is $1000 kg/m^{3}$ .

Conversion, with the correct factor $1 0^{- 6}$ :

750 cm^{3} = 750 \cdot 1 0^{- 6} m^{3} = 7.5 \cdot 1 0^{- 4} m^{3}

Mass from density, $m = d \cdot V$ :

m = 1000 \frac{kg}{m ^{3}} \cdot 7.5 \cdot 1 0^{- 4} m^{3} = 7.5 \cdot 1 0^{- 1} kg = 0.75 kg

Check: a bottle of three quarters of a litre of water weighs three quarters of a kilo. It works. If you had used $1 0^{- 2}$ you would have got $7.5 m^{3}$ and a mass of $7500 kg$ , which common sense rejects at once.

Exercise 3: scientific notation in a problem

Sunlight takes a certain time to reach the Earth. The distance is about $1.5 \cdot 1 0^{11} m$ and the speed of light is about $3.0 \cdot 1 0^{8} m/s$ . How long does the journey take?

t = \frac{s}{v} = \frac{1.5 \cdot 1 0 ^{11} m}{3.0 \cdot 1 0 ^{8} m/s} = 0.5 \cdot 1 0^{3} s = 5.0 \cdot 1 0^{2} s

The numbers in front: $1.5/3.0 = 0.5$ . The powers: $11 - 8 = 3$ . Then $0.5 \cdot 1 0^{3}$ is rewritten with one digit before the point: $5.0 \cdot 1 0^{2} s$ , that is $500 s$ , about $8$ minutes and $20$ seconds.

The mistakes I see most

Exponent not raised in area or volume. The case $1 cm^{3} = 1 0^{- 2} m^{3}$ seen above. If the unit has an exponent, so does the factor.
Dividing by 3.6 instead of multiplying (or vice versa). Always check the order of magnitude: $90 km/h$ is about $25 m/s$ , not $324$ .
Leaving the result out of form. $12 \cdot 1 0^{3}$ or $0.5 \cdot 1 0^{3}$ are not scientific notation: the number in front has to be brought between $1$ and $10$ .
Signs of the exponents in division. A negative exponent in the denominator means the exponent of the result grows.
Changing units halfway through a calculation. First you bring all the data to SI units, then you calculate. Mixing km and m in one formula is the quickest road to a wrong result.

Significant figures and when they are needed

In the test you are often also asked to give the result with the right number of significant figures: the result of a multiplication or division has as many as the data point that has the fewest. The full discussion, with the rules for sums and differences, is in the post on measurements and significant figures (written for a different school track but the rule is the same). When you get to measurement errors, you will find the method in measurement errors and error propagation; the first relationships between quantities, with graphs, are instead in direct, inverse and quadratic proportionality.

How to use the coming weeks

You don’t need to study a lot, you need to repeat a little and often. Every evening, for a week, do six conversions: two of speed, two of area or volume, two of scientific notation. Always write the conversion fraction out in full, even when you “already know it”: it is the habit that saves you in a test when nerves kick in.

If you want to know where you stand, the free introductory call is for this: you send me a test or an exercise that doesn’t work out for you and I tell you where the error is. The physics programme for the first two years, with the work on measurements, is here; lessons are online and one to one.