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konstantin123 [22]
2 years ago
12

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Physics
1 answer:
Triss [41]2 years ago
4 0

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

                              ρA₁V₁ = ρA₂V₂

                                 A₁V₁ = A₂V₂

making V₂ the subject of the equation;

                               V_{2} = \frac{A_{1}V_{1}}{A_{2}}

the pipe is widened to twice its original radius,

                                r₂ = 2r₁          

then the cross-sectional area A₂ = 4A₁

                           ⇒  V_{2}= \frac{A_{1}V_{1}}{4A_{1}}

                                  V_{2}= \frac{V_{1}}{4}

This implies that the water speed will drop by a factor of  \frac{1}{4} because of the increase the pipe cross-sectional area.  

 The Bernoulli Equation;

     Energy per unit volume before = Energy per unit volume after    

        p₁ + \frac{1}{2}ρV₁²  + ρgh₁ = p₂ + \frac{1}{2}ρV₂²  + ρgh₂  

Total pressure is constant and P_{T} = P = \frac{1}{2}ρV₂²ρV²  

        p₁ + \frac{1}{2}ρV₁²  = p₂ + \frac{1}{2}ρV₂²

Making p₂ the subject of the equation above;

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρV₂²

But V_{2}  = \frac{V_{1}}{4} so,

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρ\frac{V_{1}^{2}}{4^{2}}      

       p₂ = 3.0 x 10⁴ + (\frac{1}{2} × 1000 × 12²) - ( \frac{1}{2} × 1000 × 12²/4² )

      P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³    

       P₂ = 9.79 × 10⁴Pa      

      P₂ = 9.8 × 10⁴Pa                      

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Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

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As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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