Answer:
strains for the respective cases are
0.287
0.318
0.127
and for the entire process 0.733
Explanation:
The formula for the true strain is given as:

Where
True strain
l= length of the member after deformation
original length of the member
<u>Now for the first case we have</u>
l= 1.6m

thus,


<u>similarly for the second case we have</u>
l= 2.2m
(as the length is changing from 1.6m in this case)
thus,


<u>Now for the third case</u>
l= 2.5m

thus,


<u>Now the true strain for the entire process</u>
l=2.5m

thus,

Answer:
1
Created on Nov 3, 2018 @author: ASLand
7import atexit
#Read, nanes of both files
Rrintll"Enter tvo files to be compared below
userliamel input ("Enter the nome of the first file: ")
userliame2 input("Enter the name of the second file: ")
ROpen each file
f1 - open(userNamel, r')
@17 f2 = opan(useriame 2, )
tread all the lines into a list
d1 f1.readlines ()
d2 f2.readlines()
re equivalent, print "Yes" else pri
oiterate, and conpare
#11
the
y
if dl == d2:
print("Yes")
atexit
elif for i in range(@, min(len (d1), len(d2))):
if di[i]!=d2[i]:
PCint("No")
print(d1[i])
pcint(d2[])
Answer:
Answer for the question:
1. A compressed air system consists of a compressor and receiver, 1500 ft of 4-in pipe, two gate valves, six standard elbows, and a manifold. Four rock drills requiring 200 cu-ft/min each are connected to the manifold by 1.25-in hoses 100 ft long. Pressure drop in the manifold is 3 psig and line leakage is 5%. Determine the pressure at the drill when all four drills are operating simultaneously and receiver pressure is 100 psig.
is explained in the attachment.
Explanation:
Answer:
8100 lbin²
Explanation:
Moment or inertia is expressed using the formula
I = mr²
M is the mass of the body
r is the radius of gyration
Given
W = 100lb
r = 9in
Required
Moment of inertia
I = Wr²
I = 100(9)²
I = 100×81
I = 8100lbin²
Hence the moment of inertia about the center of gravity for the rotator is
8100 lbin²