An aeroplane of mass 4.5 times 10^4 ,kg and t | vedclass

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(D) For an aeroplane to travel at a constant height,the upward lift force must balance the downward gravitational force (weight) of the aeroplane.Let

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$750 \,N \,m^{-2}$

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(D) For an aeroplane to travel at a constant height,the upward lift force must balance the downward gravitational force (weight) of the aeroplane.Let $m$ be the mass of the aeroplane,$A$ be the total wing area,$g$ be the acceleration due to gravity,and $\Delta P$ be the pressure difference between the lower and upper surfaces of the wings.The lift force $F_L$ is given by $F_L = \Delta P 	imes A$.The weight of the aeroplane is $W = m 	imes g$.Equating the two forces for constant height: $\Delta P 	imes A = m 	imes g$.Substituting the given values: $\Delta P 	imes 600 = (4.5 	imes 10^4) 	imes 10$.$\Delta P 	imes 600 = 4.5 	imes 10^5$.$\Delta P = \frac{4.5 	imes 10^5}{600} = \frac{450000}{600} = 750 \,N \,m^{-2}$.Thus,the pressure difference is $750 \,N \,m^{-2}$.

An aeroplane of mass $4.5 \times 10^4 \,kg$ and total wing area of $600 \,m^2$ is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

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