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Black_prince [1.1K]
3 years ago
10

Which situation describes the highest rate of power?

Physics
1 answer:
Darya [45]3 years ago
8 0

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

P=\frac{w}{t}

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

P=\frac{200}{10}=20W

(B) A machine does 400 joules of work in 10 seconds.

 P=\frac{400}{10}=40W

(C) A machine does 200 joules of work in 5 seconds.

 P=\frac{200}{5}=40W

(D) A machine does 400 joules of work in 5 seconds.

P=\frac{400}{5}=80W

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

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A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?
NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

W =F*d

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

W = 294*5\\W = 1470 [J]\\

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%

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A student forms the following hypothesis if people wear bright colors, they will be happier. What is/are the problems with this
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In 1958 a large earthquake in Alaska produced a tsunami. What was the approximate height of the tsunami?
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6 0
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Read 2 more answers
You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.5 m out
Alinara [238K]

Answer:

W = 1562.5 J

Explanation:

Path 1:

W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J

Path 2:

W₂ = F₂*d₂ = 130 N * 10 m = 1300 J

Path 3:

W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J  (opposite to the motion)

We get

W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J

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BigorU [14]
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