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lesya [120]
3 years ago
5

A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold

reservoir. During that time, what is the maximum amount of work Wmaxthat the engine might have performed?Express your answer in terms of Qc, Th, and Tc.
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Answer: Qc * (1 - Th/Tc)

Explanation:  

Maximum Work done = Qh * η

                                Wmax  = Qh * (1 - Tc/Th)

For reversible cycle of engine: Qc/Tc + Qh/Th = 0

                                                   Qh = -(Qc*Th)/Tc

Substituting Qh back into equation for work:

Wmax = -(Qc*Th)/Tc *  (1 - Tc/Th)

Wmax = -(Qc*Th)/Tc + Qc

Wmax = Qc * (1 - Th/Tc)

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3 years ago
The diagram shows parts of a wave. A series of waves with an arrow passing through their centers. The highest point of one wave
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Explanation:

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In a city grid, each block on the east-west streets is 100 meters long. Each block on the north-south streets is 20 meters long.
alex41 [277]

Answer:

72.1 m

Explanation:

Hello!

When the walker walks west, each block he walks will be of 100 m, so the walker walked 4*100m = 400 m west.

Similarly, when the walker walks south, he walks 20 meters per block, therefore, the walker walked 4*20 m = 80 m south.

Since the directions west and south are perpendicular, the distance between the start ad end point is:

d = √(400^2 + 80^2) m = 407.92 m

However the walker traveled 480 m

Therefore, the walker traveled 480 - 407.9 m = 72.1 m farther than the actual distance

8 0
3 years ago
A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.
natima [27]

Answer:

(a)  24.025 m/s. downward.

(b)  31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish  = 29.41+1.25

Distance between the pelican and fish  = 30.66

Distance between the pelican and fish ≈ 31 m

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A cargo spacecraft has been launched to rendezvous with the International Space Station. The cargo ship must attain the same spe
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