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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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Resisting force is
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An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
4 years ago
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a
cupoosta [38]

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point =  mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

7 0
3 years ago
Needddd helppppppp!!!
yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0
3 years ago
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