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3 years ago

The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Physics

2 answers:

Marat540 [252]3 years ago

7 0

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

Minchanka [31]3 years ago

4 0

The answer is Z is unusable

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4 years ago

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A 1.8-kg object moves in the x direction according to the following function:

DanielleElmas [232]

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3 years ago

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Fynjy0 [20]

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3 years ago

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a g

Gre4nikov [31]

Answer:

<h3>After 3seconds</h3>

Explanation:

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer

Using the formula for calculating speed expressed as;

Speed = Distance/Time

Given;

Distance = 3km = 3000m

Speed = 1000m/s

Required

How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)

From the formula;

Time = Distance/speed

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3 years ago

A 1.0-kilogram rubber ball traveling east at 4.0 meters per second hits a wall and bounces back toward the west at 2.0 meters pe

Mrrafil [7]

Answer:

8 J and 2 J

Explanation:

Given that,

Mass of the rubber ball, m = 1 kg

Initial speed of the rubber ball, u = 4 m/s (in east)

Final speed of the rubber ball, v = -2 m/s (in west)

We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.

Initial kinetic energy,

$K_i=\dfrac{1}{2}mv^2\\\\K_i=\dfrac{1}{2}\times 1\times (4)^2\\\\K_i=8\ J$

Final kinetic energy,

$K_f=\dfrac{1}{2}mv^2\\\\K_f=\dfrac{1}{2}\times 1\times (2)^2\\\\K_f=2\ J$

So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.

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3 years ago