Air is streaming past a horizontal aeroplane | vedclass

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(C) Using Bernoulli's theorem:$P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho gh_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho gh_{2}$Since the wing is h

Vedclass · Accepted Answer

$4095$

Vedclass · Answer

(C) Using Bernoulli's theorem:$P_{1} + \frac{1}{2} ho v_{1}^{2} + ho gh_{1} = P_{2} + \frac{1}{2} ho v_{2}^{2} + ho gh_{2}$Since the wing is horizontal,$h_{1} = h_{2}$,the equation simplifies to:$P_{1} + \frac{1}{2} ho v_{1}^{2} = P_{2} + \frac{1}{2} ho v_{2}^{2}$Rearranging to find the pressure difference $\Delta P = P_{2} - P_{1}$:$\Delta P = \frac{1}{2} ho (v_{1}^{2} - v_{2}^{2})$Given $ho = 1.3\, kg/m^3$,$v_{1} = 120\, m/s$ (upper surface),and $v_{2} = 90\, m/s$ (lower surface):$\Delta P = \frac{1}{2} 	imes 1.3 	imes (120^{2} - 90^{2})$$\Delta P = 0.65 	imes (14400 - 8100)$$\Delta P = 0.65 	imes 6300 = 4095\, N/m^2$.

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