Question

A 20 kg load is suspended by a wire of cross section 0.4 mm^2 . The stress produced in n/m^2 is

Answer 1

Hey dear,

◆ Answer-
Stress = 4.9×10^8 N/m^2

● Solution-
# Given-
m = 20 kg
g = 9.8 m/s^2
A = 0.4 mm^2 = 0.4×10^-6 m^2

# Solution-
Stress produced in the wire is given by-
Stress = Force / Area
Stress = mg / A
Stress = 20×9.8 / 4×10^-7
Stress = 49×10^7
Stress = 4.9×10^8 N/m^2

Hope that is useful...

Answer 2

HEY BUDDY HERE IS YOUR ANSWER