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Arisa [49]
1 year ago
14

If a runner exerts 457 j of work to make 321 w of power then how long did it take the runner to do the work

Physics
1 answer:
MA_775_DIABLO [31]1 year ago
5 0

The time taken by the runner to do the work is 1.42 seconds.

Given the data in the question;

  • Work done; W = 457J
  • Power; P = 321W
  • Time elapsed; t = \ ?

<h3>Power</h3>

Power can be simply referred to as the quantity of energy transferred per unit time.

It is expressed as;

P = \frac{W}{t}

Where W is work done and t is time elapsed.

To determine the time it took the runner to do the work, we substitute our given values into the expression above.

P = \frac{W}{t} \\\\t = \frac{W}{P} \\\\t = \frac{457J}{321W} \\\\t = \frac{457 kgm^2/s^2}{321 kgm^2/s^3}\\\\t = 1.42s

Therefore, the time taken by the runner to do the work is 1.42 seconds.

Learn more about Power and Work: brainly.com/question/2962104

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Answer:

u=speed, w=wavelenght, f=frequency

It's known that u=w*f =>  f=u/w

                                       u=20m/s     ==>  f=20/0,5  => f=40 Hz

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Explanation:

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1 year ago
What is being transferred as you do work?<br> A. Energy<br> B. Power<br> C. Heat<br> D. Strength
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\huge\bold{\purple{\bold{⚡A. Energy⚡}}}

\huge\underline\mathtt\colorbox{cyan}{Reason:}

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3 years ago
Read 2 more answers
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
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Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

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Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

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