Square ABCD has a side length of 10 centimeters. Find the area of the square. Type a numerical answer in the space provided. Do
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Answer:
The total work done in pulling the bucket to the top of the well is approximately 3,139.1 ft·lb
Step-by-step explanation:
The given parameters are;
The mass of the bucket, W = 3.6 pounds
The depth of the well, h = 78 feet deep
The mass of water in the bucket = 38 ponds
The rate at which the water is pulled up = 2.9 feet per second
The rate at which water is leaking from the bucket, = 0.1 pounds per second
We separate and find the work done for lifting the bucket and the water individually, then we add the answers to get the solution to the question as follows;
The work done in lifting bucket empty from the well bottom, = W × h
∴ = 3.6 pounds × 78 feet = 280.8 ft-lb
The work done in lifting bucket empty from the well bottom, = 280.8 ft-lb
The time it takes to lift the bucket from the well bottom to the top, 't', is given as follows;
Time, t = Distance/Velocity
The time it takes to pull the bucket from the well bottom is therefore;
t = 78 ft./(2.9 ft./s) ≈ 26.897
The time it takes to pull the bucket from the well bottom to the top, t ≈ 26.897 s
The mass of water that leaks out from the bucket before it gets to the top, m₂, is therefore;
m₂ = × t
∴ m₂ = 0.1 lbs/s × 26.897 s = 2.6897
The mass of the water that leaks, m₂ = 2.6897 lbs
The mass of water that gets to the surface m₃ = m - m₂
∴ m₃ = 38 lbs - 2.6897 lbs ≈ 35.3103 lbs
Given that the water leaks at a constant rate the equation representing the mass of the water as it is lifted can b represented by a straight line with slope, 'm' given as follows;
The slope of the linear equation m = (38 lbs - 35.3103 lbs)/(78 ft. - 0 ft.) = 0.03448 lbs/ft.
Therefore, the equation for the weight of the water 'w' can be expressed as follows;
w = 0.03448·y + c
At the top of the well, y = 0 and w = 38
∴ 35.3103 = 0..03448 × 0 + c
c = 35.3103
∴ w = 0.03448·y + 35.3103
The work done in lifting the water through a small distance, dy is given as follows;
(0.03448·y + 38) × dy
The work done in lifting the water from the bottom to the top of the well, , is given as follows;
= (0.034483/2 × 78^2 + 35.3103 × 78) - (0.034483 × 0 + 38 × 0) ≈ 2,859.1
The work done in lifting only the water, ≈ 2,859.1 ft-lb
The total work done, in pulling the bucket to the top of the well, W = +
∴ W = 2,859.1 ft.·lb + 280.8 ft.·lb ≈ 3,139.1 ft·lb
The total work done, in pulling the bucket to the top of the well, W ≈ 3,139.1 ft·lb.