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LUCKY_DIMON [66]
3 years ago
11

A student was asked to push a large box down the hallway about 35 meters. the task took the student five minutes and a force of

150 newtons. at what rate was the student pushing the box?
Physics
1 answer:
zaharov [31]3 years ago
5 0
Rate of work or Power done by the student can be computed with below formulas.
1. Formula for Work
Work = Force x distance
Force or F = Force applied to the object 
distance or d = displacement

2. Formula for Power
P = Work done / Time it was done or completed

Substituting the values we have,
Work or Work done = 150 N x 35 meters = 5250 J (unit of work).

Lastly computing for the rate we have,
Power or rate of work = 5250 J / 300 seconds (or 5 min) = 17.5 Joule/S or Watts

Answer is 17.5 watts

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Equation 1 :
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Placing in  equation 1

we get T = 218.18N
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If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

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On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

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Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.
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