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:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
Answer:
cultivation - preparing and planting crops
domestication - capturing, taming, and breeding animals
hunting and gathering - obtaining food from the wild
Explanation:
moo
I attached a photo that explains and gives the answer to your questions. Had to add a border because the whole picture didn’t fit.
Answer:
116.3 electrons
Explanation:
Data provided in the question:
Time, t = 2.55 ps = 2.55 × 10⁻¹² s
Current, i = 7.3 μA = 7.3 × 10⁻⁶ A
Now,
we know,
Charge, Q = it
thus,
Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)
or
Q = 18.615 × 10⁻¹⁸ C
Also,
We know
Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C
Therefore,
Number of electrons past a fixed point = Q ÷ q
= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]
= 116.3 electrons
