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tatuchka [14]
3 years ago
8

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 in

itially 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?
Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

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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The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

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The  cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of  the occurrence of plasmain nature,

It should be noted that this  thermonuclear power with plasmabe  posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.

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