A wind with speed 50 ,m/s blows parallel to t | vedclass

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(C) According to Bernoulli's principle, for horizontal flow, $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$. Inside the house, the air

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$4.5 	imes 10^5 \,N$

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(C) According to Bernoulli's principle, for horizontal flow, $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$. Inside the house, the air is stationary, so $v_1 = 0$ and $P_1 = P_{atm}$. Outside the house, the wind speed is $v_2 = 50 \,m/s$ and the pressure is $P_2$. Thus, $P_{atm} = P_2 + \frac{1}{2} ho v_2^2$, which gives the pressure difference $\Delta P = P_{atm} - P_2 = \frac{1}{2} ho v_2^2$. Substituting the values: $\Delta P = \frac{1}{2} 	imes 1.2 \,kg/m^3 	imes (50 \,m/s)^2 = 0.6 	imes 2500 = 1500 \,N/m^2$. The force exerted on the roof is $F = \Delta P 	imes A = 1500 \,N/m^2 	imes 300 \,m^2 = 4.5 	imes 10^5 \,N$.

$A$ wind with speed $50 \,m/s$ blows parallel to the roof of a house. The area of the roof is $300 \,m^2$. Assume that the pressure inside the house is atmospheric pressure. The density of air is $1.2 \,kg/m^3$. The magnitude of the force exerted by the wind on the roof will be:

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