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AnnZ [28]
3 years ago
15

How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?

Physics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W

So, the required power is 70 W.

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A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?
EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by d = \frac { m } { v }.

Here in this problem, we are given a mass of 565 g which occupies a volume of 50 cm^3.

So plugging the data in the above formula to find the density:

Density = \frac { 565 } { 50 } = 11.3

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

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An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
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Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

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Answer:

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