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kipiarov [429]
3 years ago
12

An ambulance is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the ambulance d

raws closer? (2 points)
A) The frequency increases, and the pitch increases
B) The frequency increases, and the pitch decreases
C) The frequency decreases, and the pitch decreases
D) The frequency decreases, and the pitch increases

Physics
2 answers:
vagabundo [1.1K]3 years ago
5 0

Answer:

A )

Explanation:

This change in frequency observation occur due to doppler effect

if the wave source moves,In the time between one wave peak being emitted and the next, the source will have moved so that the shells will no longer be concentric. The wavefronts will get closer together in front of the source as it travels and will be further apart behind it.  (see the graph)

when the person standing still in front of the ambulance, he will observe a <em>higher frequency </em>than before as the source travels towards them.

f_(observed)=\frac{V_(wave)}{V_(wave)-V_(source)} *f_(original)

The pitch we hear depends on the frequency of the sound wave.

A high frequency corresponds to a high pitch

as we hear a higher frequency , it makes the <em>pitch higher</em> too

vesna_86 [32]3 years ago
3 0

Answer:

A I got it right on my quiz

Explanation:

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sweet-ann [11.9K]

Answer:

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8 0
3 years ago
A box at rest has the shape of a cube 3.5 m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship t
guapka [62]

Answer:

V_o=25.725\ m^3

Explanation:

Given:

sides of the cube, a=3.5\ m

speed of the cube with respect to the observer, v=0.8c

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a_o=a\div\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

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a_o= observed length of the side along the direction of velocity

a_o=3.5\div\frac{1}{\sqrt{1-\frac{(0.8c)^2}{c^2} } }

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So the observed volume will be:

V_o=a\times a\times a_o

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V_o=25.725\ m^3

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