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A body of mass $0.6 \mathrm{~kg}$ is moving along a circular path of radius $1 \mathrm{~m}$. If the body moves with $\frac{900}{\pi}$ revolutions per minute, its kinetic energy
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Verified Answer
The correct answer is:
$270 \mathrm{~J}$
We have
$$
\begin{aligned}
& \mathrm{V}=\omega \mathrm{r} \\
& =2 \pi \mathrm{fr} \\
& =2 \pi \times \frac{15}{\pi} \times 1 \\
& =30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, Kinetic energy $=\frac{1}{2} \times 0.6 \times 900$
$$
=270 \mathrm{~J}
$$
$$
\begin{aligned}
& \mathrm{V}=\omega \mathrm{r} \\
& =2 \pi \mathrm{fr} \\
& =2 \pi \times \frac{15}{\pi} \times 1 \\
& =30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, Kinetic energy $=\frac{1}{2} \times 0.6 \times 900$
$$
=270 \mathrm{~J}
$$
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