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faltersainse [42]
3 years ago
14

A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce

s off the warehouse door and travels back to the police officer in the moving car. Using 340 m/s as the speed of sound, what is the beat frequency the officer hears
Physics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = (\frac{c}{c+Vs}) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = \frac{340}{340+30} · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

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≈

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A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
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Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

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Pi = Pf

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where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

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m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

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V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

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