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Ivan
3 years ago
8

What frequency is received by the ambulance after reflecting from a wall near the person watching the oncoming ambulance (as in

the previous problem). The ambulance is moving at 109 km/h and emitting a steady 821-Hz sound from its siren
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

900.48925 Hz

979.9785 Hz

Explanation:

v_a = Relative velocity of ambulance =109\ km/h=\dfrac{109}{3.6}

v_w = Velocity of wall = 0

v = Velocity of sound in air = 343 m/s

From doppler effect we have

f=f'\dfrac{v+v_w}{v-v_a}\\\Rightarrow f=821\dfrac{343+0}{343-\dfrac{109}{3.6}}\\\Rightarrow f=900.48925\ Hz

The frequency of sound is 900.48925 Hz

When the wall acts like a source

f=f'\dfrac{v+v_a}{v-v_w}\\\Rightarrow f=900.48925\dfrac{343+\dfrac{109}{3.6}}{343-0}\\\Rightarrow f=979.9785\ Hz

The frequency of sound is 979.9785 Hz

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an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque
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Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

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v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

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= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0
3 years ago
A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :
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Explanation:

It is given that,

The acceleration of the toboggan, a=1.8\ m/s^2

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

s=ut+\dfrac{1}{2}at^2

At t = 1 s

s=\dfrac{1}{2}\times 1.8\times 1^2

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At t = 2 s

s=\dfrac{1}{2}\times 1.8\times 2^2

s_2=3.6\ m

At t = 3 s

s=\dfrac{1}{2}\times 1.8\times 3^2

s_3=8.1\ m

Hence, this is the required solution.

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3 years ago
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