Answer:
a) omax=125 MPa
omin=75 MPa
b) Temperature = 1289.6°F
Explanation:
a) The mean stress is:
![mean stress=\frac{o_{max}+o_{min} }{2} =100 MPa](https://tex.z-dn.net/?f=mean%20stress%3D%5Cfrac%7Bo_%7Bmax%7D%2Bo_%7Bmin%7D%20%20%7D%7B2%7D%20%3D100%20MPa)
omax+omin=200 MPa
stress amplitude=50 MPa
omax-omin=50MPa
2omax=250 MPa
omax=125 MPa
omin=75 MPa
b) For copper, the creep is important at 0.65 of the melting point, thus:
Temperature = 0.65*1984°F=1289.6°F
Answer:
channel width = 2.621 ft
Explanation:
Given data :
Decreasing Factor = 2
subcritical value = 0.4
super critical value = 2.5
width = 13 ft
attached below is a detailed solution and
Explanation:
A ductile material can convert into brittle material due to following reasons
1.At very low temperature
2.Due to presence of notch
In titanic ,the base of ship strike to the large ice cube and lower part of titanic ship material was made of steel .We know that steel is ductile material and when steel came with very low temperature of ice due to this ductile material converted in to brittle material and titanic ship failed.Brittle material does not show any indication before failure.
Answer:
the required diameter is 0.344 m
Explanation:
given data:
flow is laminar
flow of carbon dioxide Q = 0.005 Kg/s
for flow to be laminar, Reynold's number must be less than 2300 for pipe flow and it is given as
![\frac{\rho VD}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20VD%7D%7B%5Cmu%20%7D%3C2300)
arrange above equation for diameter
\frac{\rho Q D}{\mu A }<2300
dynamic density of carbon dioxide = 1.47×
Pa sec
density of carbon dioxide is 1.83 kg/m³
![\frac{1.83\times 0.0056\times D}{1.47\times 10^{-5}\times \frac{\pi}{4} \times D^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.83%5Ctimes%200.0056%5Ctimes%20D%7D%7B1.47%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20D%5E%7B2%7D%20%7D%3C2300)
![\frac{1.83\times 0.0056}{1.47\times 10^{-5}\times \frac{\pi}{4} \times 2300}= D](https://tex.z-dn.net/?f=%5Cfrac%7B1.83%5Ctimes%200.0056%7D%7B1.47%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%202300%7D%3D%20D)
D = 0.344 m
Answer:
Option D. is correct
Explanation:
Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment, he is not able to reach the new network segment from his workstation.
The most problem is that the technician used a straight-through cable.
Option D. is correct.