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A cable stretches by an amount d as it supports a crate of mass M. The cable is cut in half. What is the mass of the load that can be supported by either half of the cable if the cable stretches by an amount d?

Respuesta :

Answer:

2M

Explanation:

Let's say the initial length of the cable is L.

Now, if it is cut in two, it will have two lengths of L/2 each.

We are told that the cable stretches by d.

Now, young's modulus is;

E = stress/strain.

Stress = Force/Area = Mg/A

Strain = change in length/original length = d/(L/2)

Since g, A and L are constant, thus;

E α M/(L/2)

E α 2M/L

Thus, mass of the load that can be supported by either half of the cable if the cable stretches by an amount d is 2M

okpalawalter8

Answer:

  • The mass of the load that can be supported by either half of the cable = [tex]2M[/tex]

Explanation:

For the same stress as the full cable, the new cable would stretch by only [tex]\frac{d}{2}[/tex].

Since it has stretched by d, the stress must now be two times the original. Therefore, the mass must also be two times the original which is [tex]2M[/tex]

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