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

hey , the answer is c

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s