A force acting on a body of mass 5 kg is (4 | vedclass

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(A) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \ N$,Initial velocity $\vec{u} = 0$,Time $t = 10 \ s$.Using Newton's

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$2 \sqrt{29}$

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(A) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \ N$,Initial velocity $\vec{u} = 0$,Time $t = 10 \ s$.Using Newton's second law,acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{4 \hat{i} - 2 \hat{j} + 3 \hat{k}}{5} = (0.8 \hat{i} - 0.4 \hat{j} + 0.6 \hat{k}) \ m/s^2$.Using the equation of motion $\vec{v} = \vec{u} + \vec{a}t$:$\vec{v} = 0 + (0.8 \hat{i} - 0.4 \hat{j} + 0.6 \hat{k}) 	imes 10 = (8 \hat{i} - 4 \hat{j} + 6 \hat{k}) \ m/s$.The magnitude of velocity is $|\vec{v}| = \sqrt{8^2 + (-4)^2 + 6^2} = \sqrt{64 + 16 + 36} = \sqrt{116} \ m/s$.$|\vec{v}| = \sqrt{4 	imes 29} = 2 \sqrt{29} \ m/s$.

$A$ force acting on a body of mass $5 \ kg$ is $(4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \ N$. If the body is initially at rest,then the magnitude of its velocity at the end of $10 \ s$ in $m/s$ will be:

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