As mentioned, SI-system informs basic units. For example, the basic unit for mass in one kilogram. However, this unit is exaggerated then talking about the mass of a spider. Furthermore, the measurements of a box are usually given in centimetres but volume in litres. [br][br]Prefix of a unit express its magnitude compared to a basic unit. For example,[br][br]  [math]\Large 1\text{ km} = 1\cdot 10^3\text{ m}=1000 \text{ m}\\[br]\Large 1\text{ cm}=1\cdot 10^{-2}\text{ m}=0.01\text{ m}[/math][br][br] [br]If volume is solved with units of length but the context requires solution in litres, transformation can be done with [br][br]  [math]\LARGE \textcolor{blue}{1\text{ l}=1\text{ dm}^3.}[/math] [br][br]

Let us study situation, where one metre is transformed to centimetres. As known, the result is 100 cm. If we use the previous table, the idea is to[color=#0000ff] subtract power of end case from power of starting case[/color]. In our examples, the starting case is [math]\Large 1\text{ m} = 10^\textcolor{blue}{0} \text{ m}[/math] and the end case is [math] \Large 1 \text{ cm} = 10^{\textcolor{purple}{-2}} \text{ m}[/math]. Thus,[br][br]   [math] \Large 1\text{ m}=1\cdot 10^{\textcolor{blue}{0}-(\textcolor{purple}{-2})}\text{ cm}=1\cdot 10^2\text{ cm}=100\text{ cm}.[/math] [br][br]The same idea is usable also with harder cases. See the examples below.

[u]Lengths:[/u][br][br][math]\Large 2.3\text{ cm}=2.3\cdot 10^{-2-0}\text{ m}=2.3\cdot 0.01 \text{ m}=0.023 \text{ m}\\[br]\Large 0.5\text{ km}=0.5\cdot 10^{3-(-1)}\text{ dm}=0.5\cdot 10^4 \text{ dm}=5000 \text{ dm}[/math][br][br][u]Area:[/u][br][br][math]\Large 1\text{ m}^2=1\cdot (10^{0-(-2)}\text{ cm})^2=1\cdot (10^2)^2\text{ cm}^2=10^4\text{ cm}^2=10000\text{ cm}^2\\[br]\Large 0.01\text{ ha}=0.01\cdot (10^{2-0}\text{ m})^2=0.01\cdot 10^4 \text{ m}^2=1\cdot 10^2 \text{ m}^2=100\text{ m}^2\\[br]\Large 4.7\text{ dm}^2=4.7\cdot (10^{-1-(-2)}\text{ cm})^2=4.7\cdot 10^2 \text{ cm}^2=470\text{ cm}^2[/math][br][br][u]Volume:[/u][br][br][math]\Large 1\text{ m}^3=1\cdot (10^{0-(-1)}\text{ dm})^3=1\cdot (10^1)^3\text{ dm}^3=10^3\text{ dm}^3=1000\text{ dm}^3=1000\text{ }\cal l\\[br]\Large 2\text{ }\cal l=2 \text{ dm}^3=2\cdot (10^{0-1}\text{ m})^3=2\cdot 10^{-3} \text{ m}^3=2\cdot 0.001\text{ m}^3=0.002\text{ m}^3\\[br]\Large 5.2\text{ dl}=5.2\cdot 10^{-1-(-3)}\text{ ml}=5.2\cdot 10^2 \text{ ml}=520\text{ ml}\\[br]\Large 3.1\text{ dl}=0.31 \text{ }\cal l=0.31 \text{ dm}^3=0.31\cdot 1000 \text{ cm}^3=310\text{ cm}^3[br][br][/math]

[color=#0000ff]Example 2[/color]. Wind speed is informed to be 18 m/s. What speed of a car does it correspond?[br][br]Speed of a car is usually given as km/h. Let us transform the given wind speed to this unit:[br][br]  [math]\Large 18 \text{ m/s} = 18 \frac{10^{-3}\text{ km} }{\frac{1}{60\cdot 60}\text{ h}}=18\cdot \frac{3600}{10^3} \text{ km/h}\approx 64.8 \text{km/h}.[/math][br][br][br][br][color=#0000ff]Example 3[/color]. It is noticed, that there is dropping water from a pipe with a speed of 60 litres in two hours. Calculation need units in SI system, so let us transform it to m[sup]3[/sup]/s.   [br][br]  [math]\Large \frac{60\text{ l}}{2\text{ h}}= \frac{60\text{ dm}^3}{2\cdot 60\cdot 60\text{ s}}=\frac{(10^{-1})^3\text{ m}^3}{2\cdot 60 \text{ s}}\approx 8.3\cdot 10^{-6}\text{ m}^3\text{/s}.[/math][br] [br][br]