Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
Answer:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Explanation:
(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.
(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.
Answer:
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Explanation:
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Answer:
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Explanation:
For steel bolt
Stress = 210 MPa or 210 N/mm2
Pressure = Stress* Area
Pbolt = 210 N/mm2 * 16^2 *(pi)/4
Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N
For Brass spacer
Pressure = 42201.6 N
Area of Brass spacer = Pressure/Stress
Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2
Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2
d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758
d^2 = 370.758 + 16^2
d^2 = 626.758
d = 25.03 mm
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
