The figure below shows a gas contained in a vertical piston– cylinder assembly, where d = 10 cm. a vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. the total mass of the piston and shaft is 25 kg. while the gas is slowly heated, the internal energy of the gas increases by 0.1 kj, the potential energy of the piston–shaft combination increases by 0.2 kj, and a force of f = 1668 n is exerted on the shaft as shown in the figure. the piston and cylinder are poor conductors, and friction between them is negligible. the local atmospheric pressure is 1 bar and g = 9.81 m/s2. determine the work done by the shaft, in kj.

The work done by the force on the moving shaft is the product of the

height moved by the shaft and the magnitude of the force.

Correct response:

Given parameters are;

Diameter of the piston, D = 10 cm

Mass of the piston and shaft = 25 kg

The increase in the internal energy of the gas = 0.1 kJ

Potential energy of the piston-shaft combination = 0.2 kJ

Force of applied = 1,668 N

Cross-sectional area of the shaft = 0.8 cm²

Local atmospheric pressure = 1 bar

Required:

The work done by the shaft

Solution:

Potential energy, P.E. = ρ·g·h

Which gives;

Work done = Force × Distance

Therefore;

Work done, W ≈ 1,668 N × 0.82 m = 1,367.76 J

Learn more about the work done by a force here:

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they start with the same

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.