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A body of mass $0.05 \mathrm{~kg}$ is observed to fall with an acceleration of $9.5 \mathrm{~ms}^{-2}$. The opposing force of air on the body is $\left(g=9.8 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$0.015 \mathrm{~N}$
From, Newton's laws of motion,
or
$\begin{aligned}
\mathrm{mg}-\mathrm{F}_{\mathrm{air}} &=\mathrm{ma} \\
\mathrm{F}_{\mathrm{air}} &=\mathrm{m}(\mathrm{g}-\mathrm{a}) \\
&=0.05(9.8-9.5) \\
&=0.015 \mathrm{~N}
\end{aligned}$
or
$\begin{aligned}
\mathrm{mg}-\mathrm{F}_{\mathrm{air}} &=\mathrm{ma} \\
\mathrm{F}_{\mathrm{air}} &=\mathrm{m}(\mathrm{g}-\mathrm{a}) \\
&=0.05(9.8-9.5) \\
&=0.015 \mathrm{~N}
\end{aligned}$
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