A body of mass m = 10^{-2} ; kg is moving in | vedclass

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(C) Let $v_f$ be the final speed of the body after $t = 10 \; s$.Given that the final kinetic energy is $\frac{1}{8} mv_0^2$,we have:$\frac{1}{2} m v_

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$10^{-4} \; kg \cdot m^{-1}$

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(C) Let $v_f$ be the final speed of the body after $t = 10 \; s$.Given that the final kinetic energy is $\frac{1}{8} mv_0^2$,we have:$\frac{1}{2} m v_f^2 = \frac{1}{8} m v_0^2 \Rightarrow v_f^2 = \frac{1}{4} v_0^2 \Rightarrow v_f = \frac{v_0}{2} = 5 \; m/s$.The equation of motion is $F = m \frac{dv}{dt} = -kv^2$.Rearranging the terms,we get $\frac{dv}{v^2} = -\frac{k}{m} dt$.Integrating both sides from $t = 0$ to $t = 10 \; s$ and $v = 10 \; m/s$ to $v = 5 \; m/s$:$\int_{10}^{5} v^{-2} dv = -\frac{k}{10^{-2}} \int_{0}^{10} dt$.$[-v^{-1}]_{10}^{5} = -100k [t]_{0}^{10}$.$-(\frac{1}{5} - \frac{1}{10}) = -100k(10)$.$-(\frac{1}{10}) = -1000k$.$k = \frac{1}{10000} = 10^{-4} \; kg \cdot m^{-1}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

$A$ body of mass $m = 10^{-2} \; kg$ is moving in a medium and experiences a frictional force $F = -kv^2$. Its initial speed is $v_0 = 10 \; m/s$. If,after $10 \; s$,its energy is $\frac{1}{8} mv_0^2$,the value of $k$ will be:

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