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15 December, 17:28

A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is initially at rest. A manually operated plunger provides a force of 397 N in the pipe. Assuming that the other end of the pipe is open to the air, with what speed does the water emerge from the pipe?

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  1. 15 December, 17:45
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    v₂ = 97.4 m / s

    Explanation:

    Let's write the Bernoulli equation

    P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

    Index 1 is for tank and index 2 for exit

    We can calculate the pressure in the tank with the equation

    P = F / A

    Where the area of a circle is

    A = π r²

    E radius is half the diameter

    r = d / 2

    A = π d² / 4

    We replace

    P = F 4 / π d²2

    P₁ = 397 4 / π 0.058²

    P₁ = 1.50 10⁵ Pa

    The water velocity in the tank is zero because it is at rest (v1 = 0)

    The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

    Since the pipe is horizontal y₁ = y₂

    We replace on the first occasion

    P₁ = P₂ + ½ ρ v₂²

    v₂ = √ (P1-P2) 2 / ρ

    v₂ = √ [ (1.50-1.013) 10⁵ 2/1000]

    v₂ = 97.4 m / s
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