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2 years ago

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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James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

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.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

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

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