A cable is 50.0 m long and has a diameter of 2.50 cm. A force of 10,000 N is applied to the end of the cable. If the maximum stretch allowed in the cable is 2.00 mm, then what is the minimum elastic modulus allowed?

To solve this problem, we need to use the formula for stress and strain in the context of Hooke's Law.

1. First, we need to calculate the stress (σ) in the cable. Stress is defined as the force (F) divided by the cross-sectional area (A) of the object. The cross-sectional area of a cable (which is a cylinder) can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

   The diameter of the cable is given as 2.50 cm, which is 0.025 m. So, the cross-sectional area A = π*(0.025/2)^2 = 0.00049087 m^2.

   Now, we can calculate the stress: σ = F/A = 10000 N / 0.00049087 m^2 = 2.037 x 10^7 Pa.

2. Next, we calculate the strain (ε) in the cable. Strain is defined as the change in length (ΔL) divided by the original length (L0). The change in length is given as 2.00 mm, which is 0.002 m. The original length is given as 50.0 m.

   So, the strain ε = ΔL / L0 = 0.002 m / 50.0 m = 0.00004.

3. Finally, we can find the elastic modulus (E), which is defined as the stress divided by the strain. So, E = σ / ε = 2.037 x 10^7 Pa / 0.00004 = 5.0925 x 10^11 Pa.

Therefore, the minimum elastic modulus allowed for the cable is 5.0925 x 10^11 Pa.