Amineshaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. young's modulus of the cable is 10×10^10n/m2. when the cable is fully extended, the end of the cable is 900 m below the support. how much does the fully extended cable stretch when 1500 kg of ore is loaded? express your answer with the appropriate unit

ΔL = 0.2697 m = 26.97 cm

Explanation:

Given

∅ = 2.5 cm = 0.025 m

E = 10*10¹⁰ N/m²

L = 900 m

m = 1500 Kg

ΔL = ?

We get P = W as follows

W = m*g = (1500 Kg)(9.81 m/s²) =  14715 N

then we apply the equation of Hooke's Law

ΔL = P*L / (A*E)

⇒  ΔL = 14715 N*900 m / (π*0.25*(0.025 m)²*10*10¹⁰ N/m²)

⇒  ΔL = 0.2697 m = 26.97 cm

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..