One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied perpendicularly to the end face (uniformly across the area) at the other end, pulling directly away from the vise. What is the stress on the rod?

1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?