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Phantasy [73]
3 years ago
11

Water is leaking out of an inverted conical tank at a rate of 1.5 cm3 /min at the same time that water is being pumped into the

tank at a constant rate. The tank has height 10 cm and the diameter at the top is 6 cm. If the water level is rising at a rate of 1.0 cm/min when the height of the water is 2 cm, find the rate at which water is being pumped into the tank. (a) Draw a picture of the situation for any time t. (b) What quantities are given in the problem

Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

a) Check Explanation

b) Check Explanation

c) The rate at which water is being pumped into the tank = 2.631 cm³/min

Explanation:

Let the rate of flow of water into the tank be k cm³/min

a) The image of the conical tank is presented in the attached image

Note, the radius and height of a cone are related through the similar triangles principle.

As shown in the attached image, it is evident that

r/h = 3/10

r = 3h/10 = 0.3 h

b) The quantities given in the problem.

- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3

- total height of the tank, H = 10 cm

- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm

- rate at which water is leaking from the tank = 1.5 cm³/min

- water is being pumped into the tank at constant rate of k cm³/min

- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min

c) volume of the tank at any time = πr²h/3

Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)

dV/dt = k - 1.5

V = πr²h/3 and r = 0.3 h, r² = 0.09 h²

V = 0.03πh³

dV/dt = (dV/dh) × (dh/dt)

dV/dh = 0.09π h²

dV/dt = 0.09π h² (dh/dt)

dV/dt = k - 1.5

0.09π h² (dh/dt) = k - 1.5

But at h = 2 cm, (dh/dt) = 1.0 cm/min

0.09π h² (dh/dt) = k - 1.5

0.09π 2² (1) = k - 1.5

k - 1.5 = 1.131

k = 1.5 + 1.131 = 2.631 cm³/min

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We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

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By substituting these values into the formula, we have

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s cancels out. We are left with m. Thus,

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4 0
1 year ago
The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge
nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to 1.129m^2

(b) Charge on the capacitor is equal to 1.5\times 10^{-6}C

Explanation:

We have given spacing between the plates d = 0.05 mm = 5\times 10^{-5}m

Value of capacitance C=1\mu F=10^{-6}F

(A) Capacitance of a parallel plate capacitor is equal to C=\frac{\epsilon _0A}{d}

So 10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}

A=1.129m^2

So surface area of the plate will be equal to 1.129m^2

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to Q=CV

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The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction. Heavier atoms are products of a fusion reaction.

<h3 /><h3>What is nuclear fusion?</h3>

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

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How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

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Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

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3*t = t(40 - 10)/t        Combine. Cancel t's on the right

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