Answer:
<h2>hope it helps you see the attachment for further information .....✌✌✌✌✌</h2>
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:
for a reverse carnot cycle T-S diagram is a rectangle which i have shown
net work for a complete cycle must be equal to net heat interaction.
Kindly check attachment for the step by step solution of the given problem.
Answer:
note:
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm
Required:
•). The magnitude of current= ?
•••••••••••••SOLUTION•••••••••••••
We can find the relation ship between current, voltage and resistance with the help of Ohms law.
According to ohms law;
V= IR.
Rearranging the above equation;
I= V/ R
Putt the values in the above equation; we get
I= 15V/ 1000ohm
I = 0.015 A( ampere)
••••••••••••••• CONCLUSION•••••••
The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.