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never [62]
3 years ago
9

A sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A st

ationary observer hears a sound whose frequency is appreciably different from f because:________
a. interference effects set up a standing-wave pattern that alters the frequency
b. the wavelength established in the medium is not the same as it would be if the source were at rest.
c. the equation that relates velocity of propagation, frequency, and wavelength of a sound traveling through a medium does not apply in this situation
d. the sound wave travels through the medium with a velocity different from that which it would have if the source were at rest
Physics
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

The correct answer is d

Explanation:

In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.

The expression for the Doppler effect of a moving source is  

f ’= (v / (v- + v_s) f  

From this expression we see that if the speed the sound source is different from zero feels a change in the  frequency.

The correct answer is d

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the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

final angular velocity, \omega_f = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2

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3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

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