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Llana [10]
2 years ago
12

A motor takes 38 amperes on a 220-volt circuit. Find the horsepower output (hp) of the motor shown with an efficiency of 90%. Ex

press the answer to the nearest hundredth.
Engineering
1 answer:
IrinaK [193]2 years ago
7 0

This question involves the concepts of power and current.

The power output of the motor is "10.08 hp".

<h3>Power Output</h3>

The power output of the motor can be found using the following formula:

P = ηVI

where,

  • P = Power = ?
  • η = efficiency = 90 % = 0.9
  • I = current = 38 A
  • V = potential difference = 220 V

Therefore,

P = (0.9)(38 A)(220 V)

P = 7524 watt

converting this to horsepower:

P = (7524 watt)(\frac{1\ hp}{746\ watt})

P = 10.08 hp

Learn more about power here:

brainly.com/question/13948282

#SPJ1

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By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

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y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

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dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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4 0
2 years ago
Technician A says test lights are great for performing simple tests. Technician B says you can use a test light to check SRS cir
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The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.

<h3>Which Technician is Correct?</h3>

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Now, the  two values ​​of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.

Thus, we can conclude that Technician A is correct.

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The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit
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Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

T = \pi \sqrt{\frac{l}{g} }

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Then the equation will be in the form

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Now putting the dimensional formula:

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Equating the powers gives:

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so;

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Therefore:

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Answer:

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