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krok68 [10]
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 wmax that the engine might have performed?
Physics
1 answer:
tino4ka555 [31]3 years ago
6 0

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it
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here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
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Answer:

r = 4.24x10⁴ km.  

     

Explanation:

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<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

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I hope it helps you!

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An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 1.9 times the mass of piece B.
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Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

<h3>What is the relation between the masses of A and B?</h3>
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Mass of piece B = Mb

  • Velocities of pieces A and B are Va and Vb respectively.
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Ma×Va = Mb×Vb

  • Here, Ma=1.9Mb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

<h3>What are the kinetic energy of piece A and B?</h3>
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  • Kinetic energy of piece B = 1/2 × Mb × Vb²
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=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

  • Kinetic energy of piece B = 7900 - 2724 = 5176 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

Learn more about the kinetic energy here:

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