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umka2103 [35]
3 years ago
6

Describe how the doppler effect explains why a siren has a lower pitch as a fire engine moves away from you. How does the freque

ncy of a wave relate to the energy of the wave? Write your answers out in your own words please :))

Physics
2 answers:
Elena L [17]3 years ago
5 0
The Doppler effect is the change in sound or light that occurs whenever there is motion between the source and its observer. The siren of the fire engine has a lower pitch as it moves away because the waves are now spread out causing a lower frequency and a lower pitch. High-frequency sound waves have a higher pitch and higher energy than low-frequency waves. 
Anna71 [15]3 years ago
5 0

Answer:

<u>Sirens lower pitch when the fire engine moves away:</u>

When the fire engine(the source) moves away from the listener, there is a change or decrease in frequency, f of the sound produced by the fire engine.

Because, when there is a change in the distance, d between the listener and the source of the sound there is always a change in the frequency or the pitch of the sound. As the source of the sound comprises a given field in which the intensity of the sound produced is much more effective, then the person standing outside that field.

<u>Source of the Sound, when the source is leaving:</u>

  • f₁=a/a+U,
  • And the sound pitch will be low,  f₁∠f.

<u>Sound source approaching towards the listener:</u>

  • fa=f×(a/a-U),
  • And the pitch of the sound will be of high value as, f∠fa.

Explanation:

<u>Doppler effect: </u>

"The Doppler effect is the change in the frequency,f level between the listener and the source. As the difference between the intensity or level of frequency,f is due to the increase and decrease of the distance between the listener and the source."

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6 0
4 years ago
Water waves with a frequency of 4.5 Hz and a wavelength of 2.0m are traveling across a small harbor that is 200 m wide. How long
Crazy boy [7]

Answer:

22.2 seconds

Explanation:

recall that for a regular wave

v = fλ

where v = wave velocity (we need to find this to solve the next part)

f = frequency = 4.5 Hz

λ = wavelength = 2.0m

Substituting these into the equation above,

v = fλ

v = (4.5)(2)

v = 9.0m/s

Also recall that Distance travelled = velocity x time

in our case, we found velocity above (= 9.0 m/s) and the distance across the harbor is given as 200m, hence

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6 0
3 years ago
Two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has
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A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N

B. The force is repulsive

<h3>A. How to determine the magnitude of the electric force</h3>

From the question given above, the following data were obtained:

  • Charge 1 (q₁) = 7.03 nC = 7.03×10¯⁹ C
  • Charge 2 (q₂) = 4.02 nC = 4.02×10¯⁹ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 1.72 m
  • Force (F) =?

The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²

F = 8.60×10⁻⁸ N

<h3>B. How to determine whether the force is attractive or repulsive</h3>

From the question given, we were told that:

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Since both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts

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It's simple bro
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