2.1. Physical Quantity, Magnitude, and Units


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Physical quantity

A physical quantity is something you can measure.


  • Length, \(l\)
  • Weight (mass), \(m\)
  • Speed, \(v\)
  • Voltage, \(U\)
  • Cost


A unit is a definite magnitude of a certain physical quantity.


  • Length, \(l\) – meter/metre, \(\text{m}\)
  • Weight (mass), \(m\) – gram, \(\text{g}\)
  • Speed, \(v\) – meters per second, \(\text{m/s}\)
  • Voltage, \(U\) – volt, \(\text{V}\)
  • Cost – € (Euro) etc.

SI units

Get used to them!

  • Mass: 1 kg = 1000 g
  • Volume: 1 L = 1 dm3 = 1000 mL

Also: Learn the prefixes:

Prefix  Base 10
Name Symbol
mega M 106
kilo k 103
deci d 10–1
milli m 10–3
micro μ 10–6

Numerical magnitude

The amount (magnitude) of the unit you’re measuring.


Phys. quant.
Example Numerical magnitude
Length \(l = 1.93\text{m}\) 1.93
Mass \(m = 250\text{g}\) 250
Speed \(v = 25\text{m/s}\) 25
Voltage \(U = 12\text{V}\) 12

The relation between physical quantity, magnitude, and unit

The relation between physical quantity magnitude and unit.

Note: There is a multiplication sign between "\(250\)" and "\(\text{g}\)": \(m = 250 \times \text{g}\)

  • This is similar to algebraic notation, e.g. \(y = 250x\).

How to use physical quantity, magnitude, and unit

Example 1

If I dissolve 25g salt in 0.5dm3 water, which is the salt concentration? Give your answer in the unit g/dm3.


Since the answer is to be given in the unit \(\frac {\text{g}}{\text{dm}^3}\), I have to divide the mass \(m\) by the volume \(V\). We write the concentration \(c\):

\[c = \frac {m}{V} = \frac {250\text{g}}{0.5\text{dm}^3} = 50 \frac {\text{g}}{\text{dm}^3}\]

Answer: \(c = 50\text{g/dm}^3\)

↑ Note: Both physical quantity, magnitude, and unit in the answer!

Example 2

A salt solution has a concentration of 50g/dm3. From this solution I pour 0.100dm3 in a glass. What is the mass of the salt in the glass?


I want to know the mass \(m\), which is measured in \(\text{g}\).

I know the concentration \(c\), which is measured in \(\frac {\text{g}}{\text{dm}^3}\).

How do we go from the unit \(\frac {\text{g}}{\text{dm}^3}\) to \(\text{g}\)? We must multiply \(\frac {\text{g}}{\text{dm}^3}\) with \(\text{dm}^3\):

\[\frac {\text{g}}{\text{dm}^3} \times \text{dm}^3 = \text{g}\]

Thus, we can write:

\[m = cV = 500\frac {\text{g}}{\text{dm}^3} \times 0.100\text{dm}^3 = 50.0\text{g}\]

Answer: \(m = 50.0\text{g}\)

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