1) A lid is put on a box that is 17 cm long, 11 cm wide, and 8.0 cm tall and the

Question

1) A lid is put on a box that is 17 cm long, 11 cm wide, and 8.0 cm tall and the box is then evacuated until its inner pressure is 8.9 104 Pa.
(a) How much force is required to lift the lid at sea level?

(b) How much force is required to lift the lid in Denver on a day when the atmospheric pressure is 67.5 kPa (2/3 the value at sea level)?

2) A nozzle of inner radius 1.04 mm is connected to a hose of inner radius 7.95 mm. The nozzle shoots out water moving at 25.0 m/s.

(a) At what speed is the water in the hose moving?

(b) What is the volume flow rate?

(c) What is the mass flow rate?

Explanation / Answer

(1) The area of the lid A = 0.17m * 0.11m = 0.0187 m^2 the height of the lid h = 0.08m The inner pressure Pi = 8.9*10^4 Pa              Force   = PA                          = (1.01*10^5 - 8.9*10^4 )(0.17*0.11)                          = 224.4 N (b) When the pressure P = 67.5 kPa               F = (67.5*10^3 - 89*10^3 Pa)(0.17)(0.11)                  = - 0.4*10^3 = - 400 N (2) The radius of the nozzle r = 1.04*10^-3 m        The radius of the hose R = 7.95*10^-3 m The velocity of the fluid through nozzle v1 = 25m/s from equation of continuity            A1v1 = A2v2     Therefore the velocity            v2 = (A1/A2) v1                = (r^2/R^2) (25m/s)                = (1.04/7.95)^2 (25)                = 3.27 m/s The volume rate of flow         V = Av = (1.04*10^-3)^2 (25)                      = 8.5*10^-5 m^3 /s The mass rate of flow             m = V* = (8.5*10^-5)(10^3)                           = 0.085 kg/s

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