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aleksley [76]
2 years ago
5

What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​

Physics
1 answer:
Ksju [112]2 years ago
5 0
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
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What do work and energy have in common
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Energy and Work have the same unit of measurement which is Joules in SI units.

Explanation:

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4 years ago
You are watching a magic show. For one trick the magician rolls a ball down a hill. Suddenly the ball stops moving down the hill
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The forces (what causes the ball to accelerate) are gravity, friction, and the normal force. In this case, gravity is a downward force caused by the gigantic mass of the Earth and the mass of the ball. Keep in mind that a force is acceleration. Acceleration is a change in velocity. The ball speeds up. Than it stops speeding up at a certain point where the frictional force (along with air friction) equals the parallel component of gravity. 




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4 years ago
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A student is observing a pendulum swinging back and forth. Each time it swings, the pendulum bob reaches a lower maximum, and ev
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Answer:

the pendulum loses momentum and stops because of gravity and wind resistance. it does not violate the law of conservation of energy because it is not gaining any more momentum than what it had started with

Explanation:

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A colloidal liquid in a has is called
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Aerosol is the answer
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A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
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Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

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The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

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Q' = COP * W

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Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

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Δt = 1025.7 s

Δt ~ 1026 s

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