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Anna35 [415]
3 years ago
9

An air track car with a mass of 0.55 kg and velocity of 5.8 m/s to the right collides and couples with a 0.45 kg car moving to t

he left with a velocity of 3.9 m/s. What is the combined velocity of the cars just after the collision?
Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Explanation:

Given:

m1 = 0.55 kg

m2 = 0.45 kg

vi1 = +5.8 m/a

vi2 = -3.9 m/a

Solve for V

m1vi1 + m2vi2 = (m1 + m2)V

(0.55 kg)(+5.8 m/s) + (0.45 kg)(-3.9 m/s) = (0.55 + 0.45)V

V = +1.46 m/a

This means that the combined mass after the collision is moving to the right at 1.46 m/s

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A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
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(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

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P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

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in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m
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<h3>Principle of conservation of linear momentum</h3>
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m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s

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