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Marta_Voda [28]
3 years ago
14

Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, are stuck together and move off af

ter the collision. How does the final velocity of train cars A and B after the collision compare to the initial velocity of train car B before the collision? The final velocity is double train car B’s initial velocity. The final velocity is the same as train car B’s initial velocity. The final velocity is half of train car B’s initial velocity. The final velocity is zero since train car B will stop.
Physics
2 answers:
olga_2 [115]3 years ago
7 0

the answer is c. The final velocity is half of train car B’s initial velocity.


k0ka [10]3 years ago
4 0
The final velocity is half of train car B's initial velocity. 
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The dipole moment of a molecule and its boiling point are directly varied. If the dipole moment is large, the attraction between the positive and negative charges of a molecule is strong thus, requiring stronger forces to break them. Hence, they will have higher boiling points. 
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a fixed mass of gas occupies a volume of 1000 CM3 at 0 degree celsius if it is heated at constant pressure of 100 degree celsius
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Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the c
tigry1 [53]

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let<em> t</em> represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

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The answer would be A) True.
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