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3 years ago

Calculate the power in kilowatts required to lift a 500 kg mass 12 meters in one second

1 answer:

8 0

The weight of the load is (mass) x (gravity).

The work done to lift it is (weight) x (distance) = (mass) x (gravity) x (distance)

The power is (work)/(time) =

 (mass) x (gravity) x (distance) / (time) =

 (500 kg) x (9.8 m/s²) x (12 m) / (1 sec) =

(500 x 9.8 x 12 / 1) (kg-m / sec²) (m) / (sec) =

 (500 x 9.8 x 12 / 1) newton - meter / sec =

(500 x 9.8 x 12 / 1) joule / sec =

 58,800 watts

 = 58.8 kW

 = about 78.8 horsepower

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Answer:

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Explanation:

The angular acceleration of the fan is given by:

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_f$ is the final angular speed

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$\Delta t$ is the time interval

For the fan in this problem,

$\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s$

Substituting,

$\alpha = \frac{180-0}{4}=45 rpm/s$

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

$\omega' = \omega_i + \alpha t$

where

$\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s$

Substituting,

$\omega' = 0 + (45)(5)=225 rpm$

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$\text {Acceleration of the object } \propto \frac{\text {Net force on the object}}{\text {Mass of the object}}$

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