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NISA [10]
3 years ago
5

A Carnot engine operates with a cold reservoir at a temperature of 474 K and a hot reservoir at a temperature of 628 K. What is

the net entropy change (in J/K) as it goes through a complete cycle? slader
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

zero

Explanation:

A Carnot engine is an ideal engine in which the process is cyclic and reversible.

A Carnot engine is the engine whose efficiency is 100 % and the entropy in is equal to entropy out. That means the net change in entropy is zero.

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Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 years ago
Simple physics (Final) (Pic provided)
Igoryamba

Answer:

15m/s

Explanation:

Divide distance by time

3 0
3 years ago
A lever is being used to move a heavy stone from a garden. The stone weighs 250 newtons. The board used as a lever is 6 meters l
DerKrebs [107]

Answer:

Force = 125 [N]

Explanation:

In the attached image we can see a sketch of the lever system.

And if we make a sum of moments at the point O equal to zero (0).

In the equation showed in the image, we can determinate the force that we need

7 0
3 years ago
For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.
Lelechka [254]

Answer:

B. 161.5 J

Explanation:

n = Number of moles = 0.37

\Delta T = Rise in the temperature of the oxygen gas = 15 K

Q = heat added in order to raise the temperature

c_{p} = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J

6 0
2 years ago
Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do
Kaylis [27]

Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work  is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J  .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .

4 0
2 years ago
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