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Vlad1618 [11]
2 years ago
10

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s .
Physics
1 answer:
vlada-n [284]2 years ago
6 0

frequency fapproach is heard by a passenger = 302.05 Hz

frequency frecede is heard by a passenger = 228.37433 Hz

Given :

speed of train to the ground = 30.0 m/s

frequency emitted by the train whistle = 262 Hz

speed of sound in air = 344 m/s

To Find :

(A) frequency fapproach (B) frequency frecede

Solution :

The frequency of approach is given by

f' = f <u>v + v(relative)</u>

v - v(air)

= 262x <u>344 + 18</u>

344 - 30

f' = 302.05 Hz

The frequency of approach is 302.05 Hz

The frequency of recede is given by

f' = f <u>v - v(relative)</u>

v + v(air)

= 262 x <u>344 - 18</u>

344 + 30

The frequency of recede is 228.37433 Hz

Learn more about Frequency here:

brainly.com/question/254161

#SPJ4

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65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

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Momentum after the collision:
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3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
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