## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 363: 54

#### Answer

(a) The weight of the plane is $10^5~N$ (b) The air speed above the wings is $85.1~m/s$

#### Work Step by Step

(a) We can find the total force that the pressure difference exerts on the wings: $F = \Delta P~A$ $F = (500~N/m^2)(2)(100~m^2)$ $F = 10^5~N$ Since the airplane flies on a level path, this upward force is equal in magnitude to the plane's weight. Therefore, the weight of the plane is $10^5~N$ (b) Let $P_1$ be the pressure below the wings and let $P_2$ be the pressure above the wings. We can use Bernoulli's equation to find the air speed $v_2$ above the wings: $P_2 + \frac{1}{2}\rho~v_2^2 = P_1+\frac{1}{2}\rho~v_1^2$ $\frac{1}{2}\rho~v_2^2 = P_1-P_2 +\frac{1}{2}\rho~v_1^2$ $v_2^2 = \frac{2(P_1-P_2)}{\rho} +v_1^2$ $v_2 = \sqrt{\frac{2(P_1-P_2)}{\rho} +v_1^2}$ $v_2 = \sqrt{\frac{2(500~N/m^2)}{1.3~kg/m^3} +(80.5~m/s)^2}$ $v_2 = 85.1~m/s$ The air speed above the wings is $85.1~m/s$

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