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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×[tex]10^{8}[/tex] N
B. 5.0×[tex]10^{7}[/tex] N
C. 8.0×[tex]10^{6}[/tex] N
D. 4.0×[tex]10^{8}[/tex] N
E. 800 N

Respuesta :

Answer:

800 N

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

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