A plane is in level flight at constant speed | vedclass

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(D) The total area of the wings of the plane,$A = 2 	imes 25 = 50 \; m^{2}$.Speed of air over the lower wing,$V_{1} = 180 \; km/h = 180 	imes \frac{

Vedclass · Accepted Answer

$4400$

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(D) The total area of the wings of the plane,$A = 2 	imes 25 = 50 \; m^{2}$.Speed of air over the lower wing,$V_{1} = 180 \; km/h = 180 	imes \frac{5}{18} = 50 \; m/s$.Speed of air over the upper wing,$V_{2} = 234 \; km/h = 234 	imes \frac{5}{18} = 65 \; m/s$.Density of air,$ho = 1 \; kg \; m^{-3}$.Using Bernoulli's equation,the pressure difference $(P_{1} - P_{2})$ between the lower and upper surfaces is:$P_{1} - P_{2} = \frac{1}{2} ho (V_{2}^{2} - V_{1}^{2})$.The upward lift force $F$ is given by $(P_{1} - P_{2}) A$:$F = \frac{1}{2} ho (V_{2}^{2} - V_{1}^{2}) A$.Substituting the values:$F = \frac{1}{2} 	imes 1 	imes (65^{2} - 50^{2}) 	imes 50$.$F = 25 	imes (4225 - 2500) = 25 	imes 1725 = 43125 \; N$.Since the plane is in level flight,the lift force balances the weight of the plane:$F = mg \implies m = \frac{F}{g}$.$m = \frac{43125}{9.8} \approx 4400.51 \; kg$.Rounding to the nearest value,the mass of the plane is $4400 \; kg$.

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