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Nataliya [291]
2 years ago
15

An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th

e rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 m/s.(a) What is the speed of the car after it has collected 3000 kg of rainwater?(b) Since the rain is falling downward, how is it able to affect the horizontal motion of the car?
Physics
1 answer:
skelet666 [1.2K]2 years ago
7 0

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

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Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

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The other expressions are incorrect, let’s prove it:

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Answer:

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