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Alborosie
2 years ago
5

A refrigerator draws 4.5 A of current while operating on a 120-V power line. If the refrigerator runs 50% of the time and electr

ic power costs $0.15/kWh, how much does it cost to run this refrigerator for a 30-day month
Physics
1 answer:
Murljashka [212]2 years ago
3 0

Answer:

$ 29.16

Explanation:

The following data were obtained from the question:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 30 days

1 KWh = $ 0.15

Cost for 30 days =.?

From the question given above, we were told that the refrigerator runs 50% of the time.

Thus, the time of operation of the refrigerator will be:

Time (t) = 30 × 50% = 30 × 50/100

= 15 days.

Next, we shall convert 15 days to hours. This can be obtained as follow:

1 day = 24 h

15 days = 15 days × 24 h / 1 day

15 days = 360 h

Next, we shall determine the energy consumed in KWh This can be obtained as follow:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 360 h

Energy (E) =?

E = ivt

E = 4.5 × 120 × 360

E = 194400 Wh

Next, we shall convert 194400 Wh to KWh. This can be obtained as follow:

1000 Wh = 1 KWh

Therefore,

194400 Wh = 194400 Wh × 1 KWh/ 1000 Wh

194400 Wh = 194.4 KWh

Thus the energy consumed is 194.4 KWh

Finally, we shall determine the cost of energy consumed. This can be obtained as follow:

From the question given above,

1 KWh cost $ 0.15

Therefore, 194.4 KWh will cost = 194.4 KWh × $ 0.15 = $ 29.16

Therefore, the cost of running the refrigerator for 30 days is $ 29.16

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3 years ago
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
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Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
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<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
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  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

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At a 10° tilt, the arctic circle and antarctic circles would be a little less than half the distance from the poles as they are today.

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