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ikadub [295]
3 years ago
8

Find the work (in ft-lb) required to pump all the water out of a cylinder that has a circular base of radius 7 ft and height 200

ft. Use the fact that the weight-density of water is 62.4 lb/ft3
Mathematics
1 answer:
Virty [35]3 years ago
7 0

Answer:

<em>work done is equal to 384279168 lb-ft</em>

<em></em>

Step-by-step explanation:

The cylinder has a circular base of 7 ft.

The height of the cylinder is 200 ft

The weight density of water in the cylinder is 62.4 lb/ft^3

First, we find the volume of the water in the cylinder by finding the volume of this cylinder occupied by the water.

The volume of a cylinder is given as \pi r^{2} h

where, r is the radius,

and h is the height of the cylinder.

the volume of the cylinder = 3.142* 7^{2}*200 = <em>30791.6 ft^3</em>

Since the weight density of water is 62.4 lb/ft^3, then, the weight of the water within the cylinder will be...

weight of water = 62.4 x 30791.6 = <em>1921395.84 lb</em>

We know that the whole weight of the water will have to be pumped out over the height of cylindrical container. Also, we know that the work that will be done in moving this weight of water over this height will be the product of the weight of water, and the height over which it is pumped. Therefore...

The work done in pumping the water out of the container will be

==> (weight of water) x (height of cylinder) = 1921395.84 x 200

==> <em>work done is equal to 384279168 lb-ft</em>

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Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

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Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

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