A copper wire has a diameter of 4.00 x 10-2 inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit

Complete question is;

A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.

F_max = 28.9 lbf

Explanation:

Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.

Thus;

Elastic limit = Max stress

Formula for max stress is;

Max stress = F_max/A

Thus;

Elastic limit = F_max/A

F_max is maximum load

A is area = πr²

We have diameter; d = 4 × 10^(-2) inches = 0.04 in

Radius; r = d/2 = 0.04/2 = 0.02

Plugging in the relevant values into the elastic limit equation, we have;

2.30 × 10⁴ = F_max/(π × 0.02²)

F_max = 2.30 × 10⁴ × (π × 0.02²)

F_max = 28.9 lbf

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.