Answer
a) \[\underline{28.9\times {{10}^{3}}\ \mu \text{m}}\]
b) \[\underline{1.432\ \text{L}}\]
c) \[\underline{\text{1}\text{.211}\times \text{1}{{\text{0}}^{6}}\ \text{Gm}}\]
Work Step by Step
(a)
\[\underline{28.9\times {{10}^{3}}\ \mu \text{m}}\]
\[\begin{align}
& 28.9\ nm=\left( 28.9\ nm \right)\left( \frac{{{10}^{-3}}\ \mu m}{1\ \text{nm}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
& =28.9\times {{10}^{-3}}\ \mu m
\end{align}\]\[\,\left[ \begin{align}
& \because 1nm\,\,=\,\,{{10}^{-9}}m \\
& \,\,\,\,\,1m\,\,\,\,=\,\,{{10}^{6}}\mu m\Rightarrow 1nm\,\,\,\,=\,\,{{10}^{-3}}\mu m\, \\
\end{align} \right]\]
The given quantity \[\text{28}\text{.9}\ \text{nm}\]in micrometres is \[\underline{28.9\times {{10}^{3}}\ \text{ }\!\!\mu\!\!\text{ m}}\].
(b)
\[\begin{align}
& 1432\ c{{m}^{3}}=\left( 1432\ c{{m}^{3}} \right)\left( \frac{0.001\ \text{L}}{1\ c{{m}^{3}}} \right) \\
& =1.432\ \text{L}
\end{align}\]
The given quantity\[\text{1432}\ \text{c}{{\text{m}}^{3}}\text{ }\]in litres is \[\underline{1.432\ \text{L}}\].
(c)
\[\begin{align}
& 1211\ Tm=\left( 1211\ Tm \right)\left( \frac{{{10}^{3}}\ \text{Gm}}{1\ Tm} \right) \\
& =1211\times {{10}^{3}}\ \text{Gm} \\
& =1.211\times {{10}^{6}}\ \text{Gm}
\end{align}\]
The given quantity\[\text{1211}\ \text{Tm}\] in Giga meters is \[\underline{\text{1}\text{.211}\times \text{1}{{\text{0}}^{6}}\ \text{Gm}}\].
