Step-by-Step Solution
Step 1
The given quantities are (a) ${{354\;{\rm{mg}}\left( {45\;{\rm{km}}} \right)} \mathord{\left/ {\vphantom {{354\;{\rm{mg}}\left( {45\;{\rm{km}}} \right)} {\left( {0.0356\;{\rm{kN}}} \right)}}} \right. } {\left( {0.0356\;{\rm{kN}}} \right)}}$, (b) $\left( {0.00453\;{\rm{Mg}}} \right)\left( {201\;{\rm{ms}}} \right)$, and (c) ${{435\;{\rm{MN}}} \mathord{\left/ {\vphantom {{435\;{\rm{MN}}} {23.2\;{\rm{mm}}}}} \right. } {23.2\;{\rm{mm}}}}$
We are asked to determine the value of given quantities in SI units in three significant figures.
Step 2
(a)
The mass $1\;{\rm{mg}}$ is equal to ${10^{ – 3}}\;{\rm{kg}}$, the length $1\;{\rm{km}}$ is equal to ${10^3}\;{\rm{m}}$, and weight $1\;{\rm{kN}}$ is equal to ${10^3}\;{\rm{N}}$ in SI units.
Step 3
Now, convert the units and evaluate.\[\begin{array}{c} \frac{{354\;{\rm{mg}}\left( {45\;{\rm{km}}} \right)}}{{\left( {0.0356\;{\rm{kN}}} \right)}} = \frac{{\left( {354\;{\rm{mg}} \times \left( {\frac{{{{10}^{ – 3}}\;{\rm{kg}}}}{{1\;{\rm{mg}}}}} \right)} \right) \times \left( {45\;{\rm{km}} \times \left( {\frac{{{{10}^3}\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)} \right)}}{{\left( {0.0356\;{\rm{kN}} \times \left( {\frac{{{{10}^3}\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right)} \right)}}\\ = 4.47 \times {10^2}\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{N}}}} \right. } {\rm{N}}} \end{array}\]
Step 4
(b)
The mass $1\;{\rm{Mg}}$ is equal to ${10^3}\;{\rm{kg}}$, and time $1\;{\rm{ms}}$ is equal to ${10^{ – 3}}\;{\rm{s}}$ .
Step 5
Now, convert the units and evaluate.\[\begin{array}{c} \left( {0.00453\;{\rm{Mg}}} \right)\left( {201\;{\rm{ms}}} \right) = \left( {0.00453\;{\rm{Mg}} \times \left( {\frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)} \right) \times \left( {201\;{\rm{ms}} \times \left( {\frac{{{{10}^{ – 3}}\;{\rm{s}}}}{{1\;{\rm{ms}}}}} \right)} \right)\\ = 91.0 \times {10^{ – 2}}\;{\rm{kg}} \cdot {\rm{s}} \end{array}\]
Step 6
(c)
The weight $1\;{\rm{MN}}$ is equal to ${10^6}\;{\rm{N}}$, and length $1\;{\rm{mm}}$ is equal to ${10^{ – 3}}\;{\rm{m}}$.
Step 7
Now, convert the units and evaluate.\[\begin{array}{r} \frac{{435\;{\rm{MN}}}}{{23.2\;{\rm{mm}}}} = \frac{{435\;{\rm{MN}} \times \left( {\frac{{{{10}^6}\;{\rm{N}}}}{{1\;{\rm{MN}}}}} \right)}}{{23.2\;{\rm{mm}} \times \left( {\frac{{{{10}^{ – 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)}}\\ = 1.87 \times {10^{10}}\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} \end{array}\]
Step 8
Convert the units again.\[\begin{array}{c} \frac{{435\;{\rm{MN}}}}{{23.2\;{\rm{mm}}}} = 1.87 \times {10^{10}}\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}\\ = 18.7 \times {10^9}\;\left( {\frac{{\rm{N}}}{{\rm{m}}}} \right) \times \left( {\frac{{1\;{\rm{GN}}}}{{{{10}^9}\;{\rm{N}}}}} \right)\\ = 18.7\;{{{\rm{GN}}} \mathord{\left/ {\vphantom {{{\rm{GN}}} {\rm{m}}}} \right. } {\rm{m}}} \end{array}\]