What is the purpose for this experiment
1 answer:
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Answer:
stress = 50MPa
Explanation:
given data:
Length of strain guage is 5mm
displacement
stress due to displacement in structural steel can be determined by using following relation
where E is young's modulus of elasticity
E for steel is 200 GPa
stress = 50MPa
Answer:
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Explanation:
Given that
Heat loss from room (Q)= 60 KJ/min
Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s
We know that 1 W = 1 J/s
Sign convention for heat and work:
1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
So
Q = -60 KJ/min
In 10 min Q = -600 KJ
W = -1.2 KJ/s
We know that
1 min = 60 s
10 min = 600 s
So W = -1.2 x 600 KJ
W = -720 KJ
WE know that ,first law of thermodynamics
Q = W + ΔU
-600 = - 720 + ΔU
ΔU = 120 KJ
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean :
Standard deviation :
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, , the z-value corresponds to 2.39 will be :-
z-value corresponds to 2.60 will be :-
Using the standard normal table for z, we have
P-value =
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%