Answer:
Explanation:
Using equation of pure torsion

where
T is the applied Torque
is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
Outer Diameter
Inner Diameter

Applying values in the above equation we get
x 
Thus from the equation of torsion we get

Applying values we get

T =829.97Nm
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Answer:
Between 5 & 10 MPa in Table B.1.4
Explanation:
150F = 65°C
State 1:
T = 65°C , x = 0.0; Table B.1.1: v = 0,001020 m^3
/kg
Process: T = constant = 65°C
State 2:
T, v = 0.99 x v
f (65°C) = 0.99 x 0,001020 = 0.0010098 m^3
/kg
Between 5 & 10 MPa in Table B.1.4
Answer:
Explanation:
neither of the technicians is correct