Step-by-Step Solution
Step 1
We are given the units ${\rm{m/ms}}$, ${\rm{\mu km}}$, ${\rm{ks/mg}}$ and ${\rm{km}} \cdot {\rm{\mu N}}$.
We are asked to estimate the correct SI form of the units.
Step 2
(a)
To find the SI form of unit ${\rm{m/ms}}$ we need to reduce the combinations of units.
Since $1\;{\rm{ms}} = {\rm{1}}{{\rm{0}}^{ – 3}}\,{\rm{s}}$, then\[\begin{array}{c} {\rm{m/ms}} = \left( {{\rm{m/ms}}} \right)\left[ {\frac{{\rm{m}}}{{\left( {{\rm{ms}} \times \frac{{{\rm{1}}{{\rm{0}}^{ – 3}}\,{\rm{s}}}}{{{\rm{1}}\;{\rm{ms}}}}} \right)}}} \right]\\ {\rm{m/ms}} = \left( {{{10}^3}\;{\rm{m/s}} \times \frac{{1\;{\rm{km/s}}}}{{{{10}^3}\;{\rm{m/s}}}}} \right)\\ {\rm{m/ms}} = 1\;{\rm{km/s}} \end{array}\]
Step 3
(b)
To find the SI form of unit ${\rm{\mu km}}$ we need to reduce the combinations of units.
Since $1\;{\rm{\mu m}} = {\rm{1}}{{\rm{0}}^{ – 6}}\,{\rm{m}}$, then\[\begin{array}{c} {\rm{\mu km}} = \left( {{\rm{\mu km}} \times \frac{{{{10}^{ – 6}}\,{\rm{km}}}}{{1\;{\rm{\mu km}}}}} \right)\left( {{\rm{km}} \times \frac{{{{10}^3}\,{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\\ {\rm{\mu km}} = \left( {{{10}^{ – 3}}\;{\rm{m}} \times \frac{{1\;{\rm{mm}}}}{{{{10}^{ – 3}}\;{\rm{m}}}}} \right)\\ {\rm{\mu km}} = 1\;{\rm{mm}} \end{array}\]
Step 4
(c)
To find the SI form of unit ${\rm{ks/mg}}$ we need to reduce the combinations of units.
Since $1\;{\rm{ks}} = {\rm{1}}{{\rm{0}}^3}\,{\rm{s}}$ and $1\;{\rm{mg}} = {\rm{1}}{{\rm{0}}^{ – 6}}\;{\rm{kg}}$, then\[\begin{array}{c} {\rm{ks/mg}} = \frac{{\left( {{\rm{ks}} \times \frac{{{{10}^3}\,{\rm{s}}}}{{1\;{\rm{ks}}}}} \right)}}{{\left( {{\rm{mg}} \times \frac{{{{10}^{ – 6}}\,{\rm{kg}}}}{{1\;{\rm{mg}}}}} \right)}}\\ {\rm{ks/mg}} = \left( {{{10}^9}\;{\rm{s/kg}} \times \frac{{1\;{\rm{Gs/kg}}}}{{{{10}^9}\;{\rm{s/kg}}}}} \right)\\ {\rm{ks/mg}} = 1\;{\rm{Gs/kg}} \end{array}\]
Step 5
(d)
To find the SI form of unit ${\rm{km}} \cdot {\rm{\mu N}}$ we need to reduce the combinations of units.
Since $1\;{\rm{km}} = {\rm{1}}{{\rm{0}}^3}\,{\rm{m}}$ and $1\;{\rm{\mu N}} = {\rm{1}}{{\rm{0}}^{ – 6}}\;{\rm{N}}$, then\[\begin{array}{c} {\rm{km}} \cdot {\rm{\mu N}} = \left( {{\rm{km}} \times \frac{{{{10}^3}\,{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {{\rm{\mu N}} \times \frac{{{{10}^{ – 6}}\,{\rm{N}}}}{{1\;{\rm{\mu N}}}}} \right)\\ {\rm{km}} \cdot {\rm{\mu N}} = \left( {{{10}^{ – 3}}\;{\rm{m}} \cdot {\rm{N}} \times \frac{{1\;{\rm{mm}} \cdot {\rm{N}}}}{{{{10}^{ – 3}}\;{\rm{m}} \cdot {\rm{N}}}}} \right)\\ {\rm{km}} \cdot {\rm{\mu N}} = 1\;{\rm{mm}} \cdot {\rm{N}} \end{array}\]