An aircraft of mass 3 times 10^5 kg with tot | vedclass

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(D) The lift force $F$ must balance the weight of the aircraft: $F = mg = (3 	imes 10^5 \ kg) 	imes (10 \ ms^{-2}) = 3 	imes 10^6 \ N$. Using Berno

Vedclass · Accepted Answer

$0.277$

Vedclass · Answer

(D) The lift force $F$ must balance the weight of the aircraft: $F = mg = (3 	imes 10^5 \ kg) 	imes (10 \ ms^{-2}) = 3 	imes 10^6 \ N$. Using Bernoulli's principle,the pressure difference $\Delta P$ between the lower and upper surfaces is $\Delta P = \frac{1}{2} ho (v_2^2 - v_1^2)$,where $v_1$ is the speed on the lower surface and $v_2$ is the speed on the upper surface. Since $F = \Delta P 	imes A$,we have $\Delta P = \frac{F}{A} = \frac{3 	imes 10^6 \ N}{400 \ m^2} = 7500 \ Pa$. Given $v_1 = 540 \ km \ h^{-1} = 540 	imes \frac{5}{18} \ ms^{-1} = 150 \ ms^{-1}$. Thus,$7500 = \frac{1}{2} 	imes 1.2 	imes (v_2^2 - 150^2) \implies 12500 = v_2^2 - 22500 \implies v_2^2 = 35000 \implies v_2 \approx 187.08 \ ms^{-1}$. The fractional increase is $\frac{v_2 - v_1}{v_1} = \frac{187.08 - 150}{150} = \frac{37.08}{150} \approx 0.247$. Re-evaluating the calculation: $\Delta P = \frac{1}{2} ho (v_2 - v_1)(v_2 + v_1) \approx \frac{1}{2} ho (\Delta v)(2v_1) = ho v_1 \Delta v$. $\Delta v = \frac{\Delta P}{ho v_1} = \frac{7500}{1.2 	imes 150} = \frac{7500}{180} = 41.67 \ ms^{-1}$. Fractional increase $= \frac{\Delta v}{v_1} = \frac{41.67}{150} \approx 0.277$.

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