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hjlf
3 years ago
6

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

illed with 42 lb of water and is pulled up at a rate of 1.5 ft/s, but water leaks out of a hole in the bucket at a rate of 0.15 lb/s. Find the work done in pulling the bucket to the top of the well. Show how to approximate the required work by a Riemann sum. (Let x be the height in feet above the bottom of the well. Enter xi* as xi.)
Mathematics
1 answer:
jonny [76]3 years ago
4 0

Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of \frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}

Work done required to pull the bucket to the top of the well is given by integral

W=\int_{a}^{b} F(x) dx

Here, function F(x) is the total weight of the bucket and water x feet above the bottom of the well. That is,

F(x)=4+(42-0.1 x)

=46-0.1x

a is the initial height and b is the maximum height of well. That is,

a=0 \text { and } b=60

Find the work done as,

W=\int_{a}^{b} F(x) d x

=\int_{0}^{60}(46-0.1 x) dx

&\left.=46x-0.05 x^{2}\right]_{0}^{60}

=(2760-180)-0[

=2580 \mathrm{ft}-\mathrm{lb}


Hence, the work done required to pull the bucket to the top of the well is 2580 \mathrm{ft}- \mathrm{lb}

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saw5 [17]

Answer: (a) 0.006

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Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

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and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

                                                       = P(AA) x P(D/AA)

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(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

                         = P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

                         = 0.3(0.02) + 0.70(0.03)

                         = 0.027

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