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viva [34]
2 years ago
5

When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates i

ts own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop
Physics
1 answer:
Shkiper50 [21]2 years ago
7 0

Answer:

Len's law

Explanation:

We can explain this exercise using Len's law

when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.

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Claudia throws a baseball to her dog. Which free-body diagram shows the
chubhunter [2.5K]

Answer:

only the weight of the ball will act on the ball

Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown

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3 years ago
State the name in words and the symbol for the following:
BartSMP [9]

Answer:

<h2>Kilometer (km) and micrometer (um) respectively</h2>

Explanation:

<h3>One thousand meters is equal to one kilometer represented as km. </h3>

and

<h3>One thousandth of a meter mean 1/1000 m which implies one thousands part of a meter which is equal to micro meter and represented as um.</h3>
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What is physics mean and branch​
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3 years ago
In complete sentences, describe what causes the phases of the moon. (4 points)
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7 0
3 years ago
A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.
ruslelena [56]

Answer:

C. W = 115.12\,Btu

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process (\eta), no unit, is:

\eta = \left(1-\frac{T_{L}}{T_{H}} \right) (1)

Where:

T_{L} - Temperature of the cold reservoir, measured in Rankine.

T_{H} - Temperature of the hot reservoir, measured in Rankine.

If we know that T_{H} = 1809.67\,R and T_{L} = 584.67\,R, then the energy efficiency of the ideal thermal process is:

\eta = 0.678

By First Law of Thermodynamics, we calculate the work output:

W = Q_{H}-Q_{L}

W = \frac{W}{\eta} -Q_{L} (By definition of efficiency)

Q_{L} = \frac{W}{\eta}-W

Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W(2)

Where:

Q_{H} - Heat received by the engine, measured in Btu.

Q_{L} - Heat rejected by the engine, measured in Btu.

W - Work output, measured in Btu.

If we know that \eta = 0.678 and Q_{L} = 55\,Btu, then the work output of the Carnot engine is:

W = \frac{Q_{L}}{\frac{1}{\eta}-1 }

W = 115.807\,Btu

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

5 0
3 years ago
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