The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = 
d = 1.5

= 1.767 inches²
<h3>Solve for inertia</h3>

= 0.2485inches⁴
Solve for the tensile force from here

30x10³ = 
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
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Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
Answer:
Option (C) = 4 is correct.
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
A because i know thats it
Answer:
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The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.