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