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Evgen [1.6K]
3 years ago
13

g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.

After this head-on collision, the lighter ball bounces straight back with a momentum P’A=2.0 kg∙ m/s in negative x-direction. What is the magnitude of the heavier ball B’s momentum change?
Physics
1 answer:
Svet_ta [14]3 years ago
3 0

Answer:

7kgm/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

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zhuklara [117]
Let d =  distance that the fugitive travels to get on the train.

Let t =  the time to travel the distance d.
The fugitive starts from rest accelerates at a = 3.8 m/s².
Therefore
(1/2)*(3.8 m/s²)*(t s)² = (d m)
1.9 t² = d                      (1)

The train travels at constant speed  5.0 m/s.
Therefore
(5.0 m/s)*(t s) = d    
5t = d                          (2)

If the fugitive successfully boards the train, then equate (1) and (2).
1.9t² = 5t
t = 0 or t = 2.6316 s
Ignore t = 0, so t = 2.6316 s.

The speed of the fugitive after 2.6316 s, is
v = (3.8 m/s²)*(2.6316 s) = 10 s

This speed exceeds the maximum speed of the fugitive, therefore the fugitive fails to get on the train.

Answer: The fugitive fails to get on the train.
8 0
3 years ago
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a person hears an echo from the top of a tower , 2.2 second after the sound is produced.how far away is the tower from the perso
bekas [8.4K]

Answer:

730.4 m

Explanation:

The sound waves travels with a uniform motion (=constant velocity), therefore we can calculate the distance it travels using the formula:

d=vt

where

d is the distance

v is the speed of the sound wave

t is the time taken

In this problem we have:

v = 332 m/s is the speed of sound in air

t = 2.2 s is the time elapsed

Therefore, the distance between the tower and the person is

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5 0
3 years ago
Remaining Time: 1 hour, 49 minutes, 34 seconds.
ohaa [14]

Explanation:

i think C . it is twice the size of the object

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A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the fo
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Answer:

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

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5 0
3 years ago
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Which types of numbers does scientific notation best describe?
Travka [436]
The correct answer is
<span>c) very small and very large

Let's see this with a few examples:
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<span>2) Similarly, if we have a very large number:
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<span>we see that we can write it easily by using again the scientific notation:
</span>1 \cdot 10^{10}<span>
</span>
4 0
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