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sleet_krkn [62]
3 years ago
6

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperatu

re of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you fill in the following table. P V T A 1 400 kPa 10.0 L 720 K B C 24 L D 15 L (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A ? B, B ? C, C ? D, and D ? A. Process Q (kJ) W (kJ) ?Eint (kJ) A ? B B ? C C ? D D ? A (c) Calculate the efficiency Wnet / |Qh|. % (d) Show that the efficiency is equal to 1 ? TC / TA, the Carnot efficiency. (Do this on paper. Your instructor may ask you to turn in this work.)

Physics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

Find attached the solution

Explanation:

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A fish tank is a cube of size L × L × L, where L = 1 m, filled with water. Find
Murljashka [212]

At the bottom of the tank :

P = ρgH

P = (1000 kg/m³)(10 m/s²)(1 m)

P = 10000 N/m²

F = P • A

F = (10000 N/m²)(1 m²)

F = 10000 N

At the side of the tank :

Pav = ½ρgH

Pav = ½(1000 kg/m³)(10 m/s²)(1 m)

Pav = 5000 N/m²

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F = 5000 N

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2 years ago
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7 0
3 years ago
A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft \omega _1=420\ rev/min

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment  of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

L_1=L_2

I_1\omega _1=I_2\omega _2

I\times (420)=7 I\times (\omega _2)

\omega _2=\frac{420}{7}

\omega _2=60\ rev/min  

8 0
3 years ago
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