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AVprozaik [17]
3 years ago
5

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

diameter. Water is flowing in the inflow line at a velocity of 5 ft/s and in the outflow line at a velocity of 2 ft/s. Is the level in the tank rising or decreasing? With what speed (ft/sec)?
Physics
1 answer:
neonofarm [45]3 years ago
6 0

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

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Explanation:

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Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove
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Answer:

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Explanation:

Conceptual analysis

We apply the formula to calculate uniform moving distance[

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v: speed in miles/hour

Development of problem

The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:

travel data to the mountains: t₁= 8 hours ,  v=v₁

d= v₁*t₁=8*v₁ Equation (1)

data back home : t₂=4hours ,  v=v₂=v₁+45

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Equation (1)=Equation (2)

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