Step-by-Step Solution
Step 1
We are given the values $45320\;{\rm{kN}}$, $568\left( {{{10}^5}} \right)\;{\rm{mm}}$ and $0.00563\;{\rm{mg}}$.
We are asked to represent the number between 0.1 and 1000 using an appropriate prefix.
Step 2
(a)
To represent the number between 0.1 and 1000 we will use the following relation.
Since $1000\;{\rm{kN}} = 1\,{\rm{MN}}$, then\[\begin{array}{c} 45320\;{\rm{kN}} = \left( {45320\;{\rm{kN}}} \right)\left( {\frac{{1\;{\rm{MN}}}}{{1000\;{\rm{MN}}}}} \right)\\ = 45.320\;{\rm{MN}} \end{array}\]
Step 3
(b)
To represent the number between 0.1 and 1000 we will use the following relation.
Since ${10^6}\;{\rm{mm}} = 1\,{\rm{km}}$, then\[\begin{array}{c} 568\left( {{{10}^5}} \right)\;{\rm{mm}} = \left( {56800000\;{\rm{mm}}} \right)\left( {\frac{{1\;{\rm{km}}}}{{{{10}^6}\;{\rm{mm}}}}} \right)\\ = 56.8\;{\rm{km}} \end{array}\]
Step 4
(c)
To represent the number between 0.1 and 1000 we will use the following relation.
Since $1000\;{\rm{mg}} = 1\,{\rm{g}}$ and $1\;{\rm{\mu g}} = {10^{ – 6}}\;{\rm{g}}$, then\[\begin{array}{c} 0.00563\;{\rm{mg}} = \left( {0.00563\;{\rm{mg}}} \right)\left( {\frac{{1\;{\rm{g}}}}{{1000\;{\rm{mg}}}}} \right)\\ = 5.63 \times {10^{ – 6}}\;{\rm{g}} \times \left( {\frac{{{\rm{1}}\;{\rm{\mu g}}}}{{{\rm{1}}{{\rm{0}}^{ – 6}}\;{\rm{g}}}}} \right)\\ = 5.63\;{\rm{\mu g}} \end{array}\]