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AysviL [449]
3 years ago
9

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

e entropy of the two reservoirs changes and determine if the second law is satisfied
Physics
1 answer:
vekshin13 years ago
6 0

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

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Answer:

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Explanation:

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The mechanical energy can also be transformed into electrical energy through a sort of dynamo system in vehicles. Stereo players use the electrical energy to produce sound.

We see that multiple energy conversions are common in a motor car.

6 0
3 years ago
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Vesnalui [34]
The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
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<span>F * 3.0 = 150 x 9.81 x 1.20 
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3 0
3 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
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F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
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Therefore, force is 22500 N

b) From newton's second law of motion;
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Thus; a = F/m
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But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
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3.9 divided by 15 gcuu
Vsevolod [243]

0.26 there you go buddy

8 0
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Ne4ueva [31]

1m per minute

or .5 m per 30 seconds

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3 years ago
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