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pantera1 [17]
3 years ago
5

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

e tank at a constant rate. the tank has height 6 m and the diameter at the top is 4 m. if the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (round your answer to the nearest integer.)

Physics
1 answer:
satela [25.4K]3 years ago
4 0

The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}

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Answer:

Let's see the similarities between the two forces

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* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

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Explanation:

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* They are inversely proportional to the square of the distance

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The differences in them

* The electric force in much greater than the gravitational force

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