The entire population of a given community is examined, and all who are judged to be free from bowel cancer are questioned exten
1 answer:
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Answer:
I. Tension (cable A) ≈ 6939 lbf
II. Tension (cable B) ≈ 17199 lbf
Explanation:
Let's begin by listing out the data that we were given:
mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,
g = 32.17405 ft/s²
The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.
Mathematically represented thus:
T = mg + ma
where:
T = tension, m = mass, g = gravitational force,
a = acceleration
I. For Cable A, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-20)]
T = 18339.2085 - 11400 = 6939.2085
T ≈ 6939 lbf
II. For Cable B, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-2)]
T = 18339.2085 - 1140 = 17199.2085
T ≈ 17199 lbf
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
Answer:
Detailed solution is given in the attached diagram
