Step-by-Step Solution
Step 1
We are given the units $\left( {684\;{\rm{\mu m}}} \right)/\left( {43\;{\rm{ms}}} \right)$, $\left( {{\rm{28}}\;{\rm{ms}}} \right)\left( {{\rm{0}}{\rm{.0458}}\;{\rm{Mm}}} \right){\rm{/}}\left( {348\;{\rm{mg}}} \right)$ and $\left( {2.68\;{\rm{mm}}} \right)\left( {426\;{\rm{Mg}}} \right)$.
We are asked to evaluate each of the following significant figure and SI units.
Step 2
(a)
To find the SI unit we need to use the following relation.
Since ${\rm{1 \mu m}} = {\rm{1}}{{\rm{0}}^{ – 6}}\,{\rm{m}}$ and ${\rm{1 ms}} = {\rm{1}}{{\rm{0}}^{ – 3}}\,{\rm{s}}$, then\[\begin{array}{c} \left( {684\;{\rm{\mu m}}} \right)/\left( {43\;{\rm{ms}}} \right) = \frac{{\left( {684\;{\rm{\mu m}} \times \frac{{{{10}^{ – 6}}\;{\rm{m}}}}{{1\;{\rm{\mu m}}}}} \right)}}{{\left( {43\;{\rm{ms}} \times \frac{{{{10}^{ – 3}}\;{\rm{s}}}}{{1\;{\rm{ms}}}}} \right)}}\\ = 0.015\;{\rm{m/s}} \end{array}\]
Step 3
(b)
To find the SI unit we need to use the following relation.
Since ${\rm{1 mg}} = {\rm{1}}{{\rm{0}}^{ – 6}}\,{\rm{kg}}$, ${\rm{1 Mm}} = {\rm{1}}{{\rm{0}}^6}\,{\rm{m}}$ and ${\rm{1 ms}} = {\rm{1}}{{\rm{0}}^{ – 3}}\,{\rm{s}}$, then\[\begin{array}{c} \left( {{\rm{28}}\;{\rm{ms}}} \right)\left( {{\rm{0}}{\rm{.0458}}\;{\rm{Mm}}} \right){\rm{/}}\left( {348\;{\rm{mg}}} \right) = \frac{{\left( {28\;{\rm{ms}} \times \frac{{{{10}^{ – 3}}\;{\rm{s}}}}{{1\;{\rm{ms}}}}} \right)\left( {0.0458\;{\rm{Mm}} \times \frac{{{{10}^6}\;{\rm{m}}}}{{1\;{\rm{Mm}}}}} \right)}}{{\left( {348\;{\rm{mg}} \times \frac{{{{10}^{ – 6}}\;{\rm{kg}}}}{{1\;{\rm{mg}}}}} \right)}}\\ = 3.685 \times {10^6}\;{\rm{m}} \cdot {\rm{s/kg}} \end{array}\] Step 4
(c)
To find the SI unit we need to use the following relation.
Since ${\rm{1 mm}} = {\rm{1}}{{\rm{0}}^{ – 3}}\,{\rm{m}}$ and ${\rm{1 Mg}} = {\rm{1}}{{\rm{0}}^3}\,{\rm{kg}}$, then\[\begin{array}{c} \left( {2.68\;{\rm{mm}}} \right)\left( {426\;{\rm{Mg}}} \right) = \left( {2.68\;{\rm{mm}} \times \frac{{{{10}^{ – 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\left( {426\;{\rm{mg}} \times \frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{mg}}}}} \right)\\ = \left( {1141.68\;{\rm{m}} \cdot {\rm{kg}} \times \frac{{1\;{\rm{km}} \cdot {\rm{kg}}}}{{1000\;{\rm{m}} \cdot {\rm{kg}}}}} \right)\\ = 1.14\;{\rm{km}} \cdot {\rm{kg}} \end{array}\]