A wind with speed 40 m/s blows parallel to t | vedclass

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(C) Applying Bernoulli's theorem just above and just below the roof:$P + \frac{1}{2}\rho v^2 = P_0 + 0$Here,$P$ is the pressure just above the roof,$P

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$2.4 	imes 10^5 \ N$,upwards

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(C) Applying Bernoulli's theorem just above and just below the roof:$P + \frac{1}{2}ho v^2 = P_0 + 0$Here,$P$ is the pressure just above the roof,$P_0$ is the atmospheric pressure inside the house,and $v = 40 \ m/s$ is the wind speed.The pressure difference is $\Delta P = P_0 - P = \frac{1}{2}ho v^2$.The force exerted on the roof is $F = \Delta P \cdot A = \frac{1}{2}ho A v^2$.Substituting the given values:$F = \frac{1}{2} 	imes 1.2 \ kg/m^3 	imes 250 \ m^2 	imes (40 \ m/s)^2$$F = 0.6 	imes 250 	imes 1600 = 2.4 	imes 10^5 \ N$.Since the pressure inside the house $(P_0)$ is greater than the pressure just above the roof $(P)$,the net force acts in the upward direction.

$A$ wind with speed $40 \ m/s$ blows parallel to the roof of a house. The area of the roof is $250 \ m^2$. Assuming that the pressure inside the house is atmospheric pressure,the force exerted by the wind on the roof and the direction of the force will be $(\rho_{air} = 1.2 \ kg/m^3)$.

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