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Anna007 [38]
3 years ago
5

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a

tmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Physics
1 answer:
s2008m [1.1K]3 years ago
6 0

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

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Given that

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K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

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a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

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Substituting

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A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
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