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Goryan [66]
3 years ago
13

A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

0.6 m/s. What is the frequency heard by the observer
Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

f_o = \frac{v - v_o}{v}f

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz

The frequency heard by the observer is 993.52 Hz.

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Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

  1. The horizontal speed
  2. the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

x(t)=8.8 m/s *t

y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}

Now we calculate the time the hammer takes to get to the floor

17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s or t=-2.38s

Now, we keep the positive time result and calculate the horizontal distance travelled

x(1.47s)=8.8m/s*1.47s=12.94m

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