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Arisa [49]
2 years 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]2 years 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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194,400 joules of kinetic energy.

Explanation:

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We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
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The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

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\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

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\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

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