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3 years ago

13

An electric train operates on 800 V. What is its power consumption when the current flowing through the train's motor is 2,130 A

?

1 answer:

Musya8 [376]3 years ago

5 0

Answer:

<h2>1704 kW</h2>

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

<em>Power= current * voltage</em>

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800= 1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed by the trains motor is 1704 kW

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If a dog is accelerating at a constant rate of 6.57 m/s2, what is the mass of the dog if 25 N of force is being applied to it?

Oliga [24]

Answer:

<h3>The answer is 3.81 kg</h3>

Explanation:

The mass of the dog can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{25}{6.57} \\ = 3.80517503$

We have the final answer as

<h3>3.81 kg</h3>

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3 years ago

Band of which of these colours are not seen in a spectrum

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3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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How much power is required to do 40 J of work on an object in 5 seconds?

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so it depends on the phase

</span>

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3 years ago

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