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solmaris [256]
3 years ago
13

Sam is walking through the park. He hears a police car coming down the street toward him. What happens to the sound of the siren

as the police car approaches and then passes Sam?
Sound waves from the siren remain constant. As the police car drives past Sam, the sound waves decrease in frequency.
The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency.
The frequency of sound waves from the siren decrease. As the police car drives past Sam, the sound waves increase in frequency.
The frequency of the sounds waves from the siren increase and continues to increase as the police car passes and drives away from Sam.
Physics
2 answers:
geniusboy [140]3 years ago
8 0
I believe it is the first one.
This is because of the Doppler Effect. 
The Doppler Effect is when the frequency of the wave changes for an observer moving relative to the wave source.
Basically, as the police car moves away from you, the distance between you and the car increases, which makes the sound waves spread out. 
Damm [24]3 years ago
5 0

Answer:

The correct answer is "The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency."

Explanation:

The correct answer is "The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency."

This is due to the effect called "Doppler Effect".

The Doppler effect is defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave movement when the sender and the receiver, or observer, move relative to each other.

According to this effect, the frequency perceived by the receiver will increase when the receiver and sender increase their separation distance and will decrease whenever the separation distance between them is reduced. As mentioned, then when the police car approaches Sam, the wave "compresses", that is, the wavelength is short, and the frequency is high. On the other hand, when the police car moves away, the wave "decompresses", that is, the wavelength is long and the frequency low.

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Ultrasound with a frequency of 4.257 MHz can be used to produce images of the human body. If the speed of sound in the body is t
shutvik [7]

Answer:

2.49 * 10^(-4) m

Explanation:

Parameters given:

Frequency, f = 4.257 MHz = 4.257 * 10^6 Hz

Speed of sound in the body, v = 1.06 km/ = 1060 m/s

The speed of a wave is given as the product of its wavelength and frequency:

v = λf

Where λ = wavelength

This implies that:

λ = v/f

λ = (1060) / (4.257 * 10^6)

λ = 2.49 * 10^(-4) m

The wavelength of the sound in the body is 2.49 * 10^(-4) m.

4 0
3 years ago
If you increase the length ? of a pendulum by a factor of 9, how will the period t increase?
garik1379 [7]
T = 2 * pie √(L/g)

so, if length is increased by 9

then time period is increased by √9 = 3

hope it helped :)
3 0
3 years ago
Read 2 more answers
Which of the following best describes a galaxy?
valentinak56 [21]

Answer:

B-Large groups of stars held together by gravity

Explanation:

4 0
3 years ago
A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0
4 years ago
If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now
Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g)       ................equation 1

where T is the period on earth

gravitational acceleration on the moon is g/6

T1 = 2π√[L/(g/6)]

T1=2π√(6L/g)               ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

5 0
4 years ago
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