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Grade: 12th pass
        
A flat plate moves normally with a speed v` towards a horizontal jet of water of uniform area of cross section . The jet discharges water at the rate of volume V per second at a speed of v° . The density of water is ¶ .assume that water splashes along the surface of the plate at right angles to the original motion . The magnitude of the force acting on the plate due to the jet of water is 
one month ago

Answers : (1)

Vikas TU
8721 Points
							
Length of liquid jet hitting wall / sec = v1 + v2 
Vol of liquid cong out / sec 
V = Area of pipe * v2 
Vol of liquid hitting wall /sec = V/v2 *(v1 +v2)
Momentum of liquid hitting wall per second is P 
=> V/v2 *(v1 +v2)¶ * (v1+v2) = (v1+v2)^2 Vp /v2 
Finaal momentum perpendicular to wall =0 
Force is the change in momentum / sec 
F = (v+v2)^2 Vp /v2 
 
one month ago
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