Answer:
Check the explanation
Explanation:
Given
1) CU traixial compression test
2) Devatoric stress at failure = бd = 50 kN/
3) Confining pressure at failure = бd = 48 kN/
4) Pore pressure at failure = u = 18 kN/
5) Unconfined compression stress = q = 20 kN/
6) Undrained cohesion = q/2 = 20 kN/
To find:
1) Effective and total stress strength failure envelope
Kindly check the attached image below
.
Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = 
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/(
) = 
m^2
recall that area = 
= 
= 
= 
= 
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
Answer:
#Initialise a tuple
team_names = ('Rockets','Raptors','Warriors','Celtics')
print(team_names[0])
print(team_names[1])
print(team_names[2])
print(team_names[3])
Explanation:
The Python code illustrates or printed out the tuple team names at the end of a season.
The code displayed is a function that will display these teams as an output from the program.
Answer: 0.93 mA
Explanation:
In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.
We can express the resistance as follows:
R = ρ.L/A
In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:
A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2
Replacing by the values, we get R as follows:
R = 1.4 1010 Ω
Applying Ohm’s Law, and solving for the current I:
I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
