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timurjin [86]
3 years ago
13

Suppose you are standing at the exact center of a park surrounded by a circular road. An ambulance drives completely around this

road, with siren blaring. How does the pitch of the siren change as it circles around you?
Physics
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer: The pitch of the sound does not change.

Explanation: The pitch of sound wave is dependent on the frequency of the sound wave. The frequency of sound wave when their is a relative motion between an observer and a source is given by Doppler effect.

Doppler effect is a mathematical equation that gives the relationship between the observed frequency by an observer from any sound source and the relative motion between the observer and the source.

Mathematically,

f'= (v+v') /(v-vs) * f

Where f' = observed frequency of sound wave

v= speed of sound in air

v'= velocity of observer relative to sound source

vs= velocity of sound source relating to observer

f= frequency of sound produced by source

The observer is at the center, thus the distance between the observer and the source is constant (according to mensuration, the radius is constant for any given circle and since the car is moving along a circular path and the observer is at the center, thus the distance between them is constant), thus making the relative velocity between the observer and the source constant (vs=constant).

Also the frequency of sound wave produce by the source is a constant (f=constant)

The speed of sound in air is also a constant (v=336m/s)

The observer is standing at the center thus he is not moving, hence the relative motion between observer and source is also constant (v'=constant)

Since all parameters are constant, then the observed frequency will be constant too.



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Answer:

yes

Explanation:

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8 0
3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne
alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E =  U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

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