Question

A steel cable has a diameter of 0.16 cm. The tensile strength of the steel is 4 × 108 N/m2. What is the approximate tension force that will snap the cable?

A. 8.0×
`10^(8)` N
B. 5.0×
`10^(7)` N
C. 8.0×
`10^(6)` N
D. 4.0×
`10^(8)` N
E. 800 N

Answer 1

800 N

Step-by-step explanation:

The formula to solve this equation is stress = Force/ Area.

The first step is to rearrange this equation for force: Force = stress * area

The second step is to determine the area: A = pi/4 * d^2

A = pi/4 * (0.0016m)^2=0.00000201 m^2

The third step is to substitute the area back into the equation:

Force = (4 * 10^8)(0.00000201) = 804 N

804 N is approximately 800 N