Question

It is known that a cable with a​ cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the cable is proportional to its​ cross-sectional area, what size cable is needed to hold 70007000 ​lb?

Answer 1

Answer:

0.84 square in.

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area.

Mathematically,

C = k * A

A1 = 0.3 sq

C1 = 2500 lb

C2 = 7000 lb

C1/A1 = C2/A2

2500/0.3 = 7000/C2

= 7000 / 8333.33

= 0.84 square in.

Answer 2

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in