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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. W area (boxed with red outline) is represented as the hot reservoir while Z area 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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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

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A stretched rubber band is storing elastic potential energy. (A)

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Lerok [7]

Answer:

Yes

Explanation:

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2 years ago

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The basic methods and techniques used by psychologists to investigate human behavior?

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3 years ago

Sally and Sam are in a spaceship that comes to within 17,000 km of the asteroid Ceres. Determine the force Sally experiences, in

MAXImum [283]

Answer:

0.01606 Newtons

Explanation:

r = Distance between the asteroid and Sally = 17000000 m

m₁ = Mass of the asteroid = 8.7× 10²⁰ kg

m₂ = Mass of Sally = 80 kg

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From Newton's Universal law of gravity

$F=G\frac{m_1m_2}{r^2}\\\Rightarrow F=6.67\times 10^{-11}\times \frac{8.7\times 10^{20}\times 80}{17000000^2}\\\Rightarrow F=0.01606\ N$

The force Sally experiences is 0.01606 Newtons

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3 years ago