Answer:
screw is the answer of the question
The answer is b, I hope this helps you
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :


J = 207.6 
So the shear stress at point A is :



Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
450,000m = 450km = 4.5E5
32,600,000W = 32.6MW = 3.26E7
59,700,000,000cal = 59.7Gcal = 5.97E10
0.000000083s = 83ns = 8.3E-8
35,000Ω = 35kΩ = 3.5E4
Explanation:
Giga = 1,000,000,000
Mega = 1,000,000
kilo = 1,000
unit = 1
deci = .1
centi = .01
milli = .001
micro = .000001
nano = .0000000001
pico = .000000000001
You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.