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

The image mentioned is in the attachment

a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

a) Pressure is a force applied to a surface of an object or fluid per unit area.

The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P =

P =

P =

P = 2450 Pa

The pressure generated by the block is P = 2450 Pa.

b) A static liquid can also exert pressure and can be calculated as:

ρ.g.h

where

ρ is the density of the fluid

h is the depth of the fluid

g is acceleration of gravity

600.9.8.0.5

2940 Pa

The pressure in the fluid at 50 cm deep is 2940 Pa.

c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:

=

F =

Adjusting the units, = 0.002 m².

F =

F = 4.9 N

The force necessary to be equilibrium is F = 4.9 N.