A pressure-pump has a horizontal tube of cross sectional area $10 \mathrm{~cm}^{2}$ for the outflow of water at a speed of $20 \mathrm{~m} / \mathrm{s}$. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :
[given: density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$]
Solution
<p>${F_w} = \rho A{v^2}$</p>
<p>$= {10^3} \times 10 \times {10^{ - 4}} \times 20 \times 20$</p>
<p>$= 400\,N$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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