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(A) The force exerted is equal to the rate of change of momentum of the sand particles.First,we calculate the velocity $v$ with which a sand particle

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The force exerted is equal to the rate of change of momentum of the sand particles.First,we calculate the velocity $v$ with which a sand particle strikes the bottom of the hour-glass. Since the particle starts from rest $(u = 0)$ and takes $t = 2 \, s$ to reach the bottom,using the equation of motion $v = u + gt$ (taking $g = 10 \, m/s^2$):$v = 0 + 10 	imes 2 = 20 \, m/s$.The change in momentum of a single particle upon hitting the bottom (assuming it comes to rest) is:$\Delta p = m(v - u_{final}) = m(v - 0) = mv$.Given $m = 0.2 \, g = 0.2 	imes 10^{-3} \, kg$,we have:$\Delta p = 0.2 	imes 10^{-3} \, kg 	imes 20 \, m/s = 4 	imes 10^{-3} \, kg \cdot m/s$.Since $100$ particles fall per second,the total rate of change of momentum (which equals the average force) is:$F = n 	imes \Delta p = 100 	imes 4 	imes 10^{-3} \, N = 0.4 \, N$.Thus,the average force exerted is $0.4 \, N$.

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