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Mariulka [41]
3 years ago
13

Calculate the power expended by peter who carries a 11 kg box 6m high in 3s

Physics
1 answer:
yawa3891 [41]3 years ago
3 0

Given that,

Mass of a box, m = 11 kg

It is lifted to a height of 6 m

Time, t = 3s

To find,

The power experienced by the Peter.

Solution,

Power = rate of doing work

P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{11\times 9.8\times 6}{3}\\\\P=215.6\ W

So, the power experienced by the Peter is 215.6  Watts.

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Answer:

5.714 hours / day

Explanation:

<u>Calculate the hours used in that week </u>

120000 / 3000 = 120 / 3 = 40 hours a week

<u>Calculate the amount it is used in one day</u>

40 / 7 = 5.71428571 hours or 5.714 hours/day

5 0
2 years ago
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The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor
Mekhanik [1.2K]

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

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So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

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mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

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4 years ago
Use this free body diagram to help you find the magnitude of the force F1 needed to keep this block in static equilibrium 15.3 N
bija089 [108]

Do you have a picture of the diagram?

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3 years ago
Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

7 0
3 years ago
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According to Bode’s Law a planet is missing between Jupiter and Saturn. True False
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