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Setler [38]
2 years ago
11

Assuming a mass of 0.04 g, what is the power exerted by the electrical forces on a fragment of fiber if the fragment has a veloc

ity of 200 mm/s, 10 ms after ejection? For this problem, I used the formula P= Force * velocity = mass* (velocity/time)*velocity = (mass * velocity^2) /time. However I got the problem wrong. In the answer, they used the formula P= W/t = 1/2* (mass*velocity^2)/time. I am confused.When do you use P = Force * velocity = (mass * velocity^2) /time and when do you use P= W/t = 1/2* (mass*velocity^2)/time?
Physics
1 answer:
schepotkina [342]2 years ago
8 0

The power exerted by the electrical forces on a fragment of fiber is 8.0× 10^-5 Watts

<h3>Power and Energy or Work done</h3>

Power is the rate of doing work.

  • Power = work done / time

Energy is the ability to do work.

Work done and energy are used interchangeably.

  • Work done = Force × distance

In the example above, the energy possessed by the mass is kinetic energy.

Kinetic energy is the energy a body has by virtue of its motion.

  • Kinetic energy, KE = 1/2 mv^2

Calculating kinetic energy of the body:

mass of the body = 0.04 g = 4.0 × 10^-5 kg

velocity of the body = 200mm/s = 0.2 m/s

Kinetic energy of the body = 1/2 × 4.0 × 10^-5 × 0.2^2

Kinetic energy of the body = 8.0× 10^-7

Thus, work done = 8.0× 10^-7 J

Calculating power exerted:

time taken = 10 ms = 0.01 s

Power = work done/ time

Power exerted = 8.0× 10^-7/0.01

Power exerted = 8.0× 10^-5 Watts

Therefore, the power exerted by the electrical forces on a fragment of fiber is 8.0× 10^-5 Watts

Learn more about about electrical energy and power at: brainly.com/question/60890

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for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

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The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

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