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valentina_108 [34]

3 years ago

9

A moving freight car collides with an identical one that is at rest. If momentum is conserved, what happens to the second car af

ter the collision? It attains the same speed as the first car. It moves at half the speed of the first car. It moves at twice the speed of the first car.

1 answer:

yKpoI14uk [10]3 years ago

4 0

For all elastic collisions we have

$e = \frac{v_2 - v_1}{u_1 - u_2}$

$e = 1 = \frac{v_2 - v_1}{u_1 - u_2}$

$v_2 - v_1 = u - 0$

also by momentum conservation we will have

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

since two cars are identical so we know that

$m_1 = m_2 = m$

$u + 0 = v_1 + v_2$

now by solving two equations we will have

$v_2 = u$

$v_1 = 0$

so the correct answer must be

It attains the same speed as the first car.

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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

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Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

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$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

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5 0

2 years ago

A 100 kg box sits on an incline held at rest by a very thin rope attached to another mass hanging as shown. The force of frictio

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If friction is acting along the plane upwards

then in this case we will have

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$mgsin\theta = F_f + T$

also for other side of hanging mass we have

$T = Mg = 50(9.8) = 490 N$

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In other case we can assume that friction will act along the plane downwards

so now in that case we will have

$mgsin\theta + F_f = T$

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$T = Mg = 50(9.8) N$

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$sin\theta = 0.397$

$\theta = 23.45 degree$

<em>So the range of angle will be 23.45 degree to 37 degree</em>

6 0

3 years ago