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WARRIOR [948]
3 years ago
14

A Explain why the sound produced by the horn of an approaching car

Physics
1 answer:
aleksklad [387]3 years ago
5 0
When a car approaches you, the sound waves that reach you have a shorter wavelength and a higher frequency. You hear a sound with a higher pitch. When the car moves away from you, the sound waves that reach you have a longer wavelength and lower frequency.

?? ⬇️

An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.

The Doppler effect is an effect observed in light and sound waves as they move toward or away from an observer. One simple example of the Doppler effect is the sound of an automobile horn. Picture a person standing on a street corner. A car approaches, blowing its horn.

Comparing two waves of the same wavelength, a higher frequency is associated with faster movement. Comparing two waves of different wavelengths, a higher frequency doesn't always indicate faster movement, although it can. Waves of different wavelengths can have the same frequency.

The pitch of a sound is our ear's response to the frequency of sound. Whereas loudness depends on the energy of the wave. ... The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.
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Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

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