28384 *x soít cos estematema
Answer:
We would need background context,
Explanation:
Then I would be happy to help!
The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
Answer:
Detailed solution is given in the attached diagram
Answer:
3A
Explanation:
Using Ohms law U=I×R solve for I by I=U/R