Question

A vertical jet of water of density leaves a nozzle at a speed of . Area of cross section of nozzle is . It suspends a plate having a mass of 1000 kg as shown in the figure. if the vertical distance h is meters then x=? (take g = 10 m/s2 ) (assume water after striking starts moving parallel to plane of the plate and at every point of vertical stream pressure remains same)

Answer 1

Answer:

assume

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Answer 2

Given:

$Density, \rho = 997kg/m^{3} \\Mass, M = 1000kg\\Acceleration due to gravity, g = 9.8m/s^{2}$

Find: We have to calculate the velocity

Step by Step Solution:

So, we know the relation

Force by liquid is equal to the product of the mass and the acceleration due to gravity,

$\rho gh = mg\\\\997\times10\times h = 1000\times 10\\\\\\h = \dfrac{997}{1000} m\\\\h = 997\times 10^{-3} m$