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Flura [38]
3 years ago
5

A 1000-kg car is traveling east at 20\:m.s^{-1}20 m . s − 1 and a 1200-kg car is traveling west at 22\:m.s^{-1}\:22 m . s − 1. W

hat is the total kinetic energy of the two-car system in the center of mass reference frame? (i)\:\:\:\:\:3.92\:\times\:10^5\:J
Physics
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

490,400 J

Explanation:

Mass of first car, m = 1000 kg

Mass of second car, M = 1200 kg

velocity of first car, u = 20 m/s east

velocity of second car, U = 22 m/s west

The formula for the kinetic energy is

k = \frac{1}{2}mv^{2}

where, m is the body and v be the velocity of the body.

Total kinetic energy is given by

k = \frac{1}{2}mu^{2}+\frac{1}{2}MU^{2}

k = \frac{1}{2}\times1000\times20^{2}+\frac{1}{2}\times1200\times22^{2}

k = 200000 + 290400

k = 490,400 J

Thus, the total kinetic energy of the system is 490,400 J.

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Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

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