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A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at one end and a closed nozzle at the other end. As you might suspect, water pressure builds up and water squirts vertically out of the puncture to a height of 0.45 m. Determine the pressure inside the hose. (Enter your answer to the nearest 1000 Pa.)

Respuesta :

Answer:

104.5 x 10³ N

Explanation:

If v be the velocity of water at the point of puncture

v₂ = √2gh

= √ 2 x 9.8 x .45

= 3  m/s

We  shall apply Bernoulli's theorem to solve the problem. Water becomes stagnant (v₁ =0) inside the pipe as water accumulates and pressure( P₁) is increased . We water comes out , it is at atmospheric pressure (P₂) and its velocity (v₂) increases

For horizontal flow of water

P₁ - P₂ = 1/2 ρv₂²

P₂ atmospheric pressure

P₁ - 100 x 10³  = 1/2 ρv₂²

P₁ = 100 x 10³ + .5 x 10³x 9

= 104.5 x 10³ N

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