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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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Explain the difference between si base units and derived units. give an example of each
34kurt
Base SI unit is the unit that used for simple quantity like time=second and length= meter. They only have one unit.
Derived unit is more complex because you multiply or divide at least two base SI, making it have more than 1 unit. The example could be velocity which was time/length = m/s
6 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
Where are the oldest rocks found ?
nignag [31]

Answer:

zircons

Explanation:

they are over 4.375 billion years old

7 0
3 years ago
Please help on this one someone
Nonamiya [84]

Valence electrons are the electrons in the outermost energy level of an atom — in the energy level that is farthest away from the nucleus.

I think it's A.

6 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

And the wave speed is given by:

v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0
3 years ago
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