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.
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Explanation:
Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)
The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)
Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current
I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches
R1 = (8 V)/(3 A) = 8/3 Ω
R2 = (8 V)/(4 A) = 2 Ω
R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω
V1 = V2 = V3 = 8 V
Answer:
Expressions are made up of terms.
A term is a product of factors.
Coefficient is the numerical factor in the term
Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
Explanation:
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