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Maksim231197 [3]
2 years ago
15

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Physics
1 answer:
Mariana [72]2 years ago
7 0

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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2 1100 kg cars drive east; the first moving at 30 m/s the 2nd at 15 m/s what is the magnitude of the total momentum of the syste
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<h3>Answer:</h3>

49500 kgm/s

<h3>Explanation:</h3>

Data given;

  • First car; Mass = 1100 kg
  • Velocity = 30 m/s
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  • Velocity = 15 m/s

We are required to calculate the total momentum of the system.

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Momentum of the first car = 1100 kg × 30 m/s

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Momentum of the second car = 1100 kg × 15 m/s

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Therefore;

Total momentum = 33,000 kgm/s + 16,500 kgm/s

                             = 49500 kgm/s

Thus, the total momentum of the system is 49500 kgm/s

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A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance
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