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crimeas [40]
3 years ago
14

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

m/s, the speed of surface B is 62.2 m/s, and the speed of sound is 334 m/s. The source emits waves at frequency 1110 Hz as measured in the source frame. In the reflector frame, what are (a) the frequency and (b) the wavelength of the arriving sound waves? In the source frame, what are (c) the frequency and (d) the wavelength of the sound waves reflected back to the source?
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

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pochemuha

Answer:

Si logra alcanzar el bus.

Explanation:

Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:

x_{f} = x_{o}+(v*t)

donde:

Xf = Ubicacion del punto donde se encuentra el bus [m]

Xo = Ubicacion desde donde esta Alicia [m]

v = velocidad constante = 5 [m/s]

t = tiempo [s]

Xf - Xo = 15 [m]

15 = 5*t

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Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:

v_{f} = v_{i}+(a*t)

donde:

Vf = velocidad del bus despues de los 3 [s]

Vi = velocidad inicial = 0

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Vf = 0 + (0.5*3)

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La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.

7 0
3 years ago
A horizontal force of 50 N is applied to the object.
USPshnik [31]

Answer:

Mass of object (m) = 5.102 kg

Explanation:

Given:

Horizontal Force (F) = 50 N

Find:

Mass of object (m) = ?

Computation:

We know that, acceleration due to gravity (g) = 9.8 m/s²

⇒ Horizontal Force (F) = mg

⇒ 50 N = m (9.8 m/s²)

⇒ Mass of object (m) =  50 / 9.8

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<u>Explanation:</u>

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n\lambda =2d\sin \theta

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n = order of diffraction = 3

\lambda = wavelength of the light = 580nm=5.80\times 10^{-7}m    (Conversion factor:  1m=10^{9}nm )

d = spacing between the crystal planes = 0.100 mm = 1.0\times 10^{-4}       (Conversion factor:  1 m = 1000 mm)

\theta = angle of diffraction = ?

Putting values in above equation:

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