# An airplane has a mass of 2.00 x 10^6 kg, and each wing has an area of 1200 m^2. The air flows past the lower surface of the wings at 95 m/s. How fast must the air flow over the upper surface of the wing if the plane has to stay in the air? Consider Bernoulli's principle and buoyancy. The density of the air is 1.26 kg/m^3. As an airplane flies, the speed of air passing over the top of the wing is higher than the speed at which air flows below the wing. This creates a differential in the pressure above and below the wing. The weight of the airplane is supported by this pressure differential; this allows the aircraft to fly.

Here, the mass of the aircraft is 2*10^8 kg and the area of each wing is 1200 m^2. The speed at which air flows past the lower surface of the wing is 95 m/s. The density of air is 1.26 kg/m^3.

Using the Bernoulli equation, `P_l + (1/2)*rho*v_l^2 = P_u + (1/2)*rho*v_u^2`

`P_l - P_u = (1/2)*rho*(v_u^2 - v_l^2)`

To balance the weight of the aircraft, `P_l - P_u = (2*10^6)/2400`

`(2*10^6)/2400 = (1/2)*1.26*(v_u^2 - 95^2)`

=> `v_u^2 = 10347.75`

=> `v_u = 101.72`

The speed of air flowing above the wing is approximately 101.72 m/s