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?
2 answers:
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:
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
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