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zaharov [31]
3 years ago
10

A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

00 s, how much power was developed, to the nearest watt?
Physics
2 answers:
Vika [28.1K]3 years ago
6 0
You need to calculate the work needed. This is force x distance in this case.
After you do that, divide the work by the time.
Brilliant_brown [7]3 years ago
3 0

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

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a = 4.76 cm

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mixas84 [53]

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Explanation:

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          U₁ =- k \frac{e^2 }{r+1}

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starting point.

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final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

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          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

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therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

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Answer:

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Explaination:

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<h2> and</h2>

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