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Bond [772]
3 years ago
15

A 328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at 13.0 m/s in the same direction. If th

e second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision
Physics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

14.04 m/s

Explanation:

To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

We have the following data:

m1 = m1' = 328,

m2 = m2' = 790,

v1 = 19.1,

v2 = 13,

v2' = 15.1.

Using this data, we can find v1' (final velocity of the first car):

328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1

16534.8 = 328 * v1' + 11929

328 * v1' = 4605.8

v1' = 14.04 m/s

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Answer:

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Explanation:

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Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

v = \frac{2 \pi r}{T} (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite (1.50x10^{7} m) and the Earth radius (6.38x10^{6} m) :

r = 1.50x10^{7} m+6.38x10^{6}m

r = 21.38x10^{6}m

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a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}    

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Answer with Explanation:

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Let \theta be the angle .

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30 cos\theta=12.6

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