Answer:
All 3 principal stress
1. 56.301mpa
2. 28.07mpa
3. 0mpa
Maximum shear stress = 14.116mpa
Explanation:
di = 75 = 0.075
wall thickness = 0.1 = 0.0001
internal pressure pi = 150 kpa = 150 x 10³
torque t = 100 Nm
finding all values
∂1 = 150x10³x0.075/2x0,0001
= 0.5625 = 56.25mpa
∂2 = 150x10³x75/4x0.1
= 28.12mpa
T = 16x100/(πx75x10³)²
∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]
= 1/2[84.37±√791.2969+5.827396]
= 1/2[84.37±28.33]
∂1 = 1/2[84.37+28.33]
= 56.301mpa
∂2 = 1/2[84.37-28.33]
= 28.07mpa
This is a 2 d diagram donut is analyzed in 2 direction.
So ∂3 = 0mpa
∂max = 56.301-28.07/2
= 14.116mpa
Answer:
1. Can you tell me something about yourself?
2. What are you weaknesses?
3. If you would describe yourself in one word?
Explanation: Those questions above 1, 2, and 3 are not harmful to ask your client. Bit the last two 4 and 5 are very harmful, because you don't need to be all up in they business and you don't want to put a lot of pressure on your client.
Hope this helps☝️☝☝
A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 * 
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Answer:
lead dioxide,sulfate and lead acid
