Physics, 09.07.2020 01:01 ronniem

Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena está funcionando con una frecuencia de 360 Hz, ¿con qué frecuencia percibe la persona
el sonido de la sirena?

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Answer from: bnbjj

f_o=331.046Hz

Explanation:

Use Doppler effect equation:

The Doppler effect is a physical phenomenon where an apparent change in wave frequency is presented by a sound source with respect to its observer when that same source is in motion. The general equation is given by:

$f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Where:\\\\f_s=Actual\hspace{3}frequency\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

When the observer moves towards the source v_o

is positive.When the observer moves away from the source v_o

is negative.When the source moves towards the observer v_s

is negative.When the source moves away from the observer v_s

is positive.

Since the problem don't give us aditional information let's assume:

v=343m/s

Which is the speed of sound in air.

And using the information provided by the problem:

$v_s=30m/s\\v_o=0\\f_s=360Hz$

$f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz$

The frequency perceived by the person is 331.046Hz

Translation:

Usa la ecuación del Efecto Doppler:

El efecto Doppler es un fenómeno físico en el que una fuente de sonido presenta un cambio aparente en la frecuencia de onda con respecto a su observador cuando esa misma fuente está en movimiento. La ecuación general viene dada por:

$f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Donde:\\\\f_s=Frecuencia\hspace{3}real\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\f_o=Frecuencia\hspace{3}observada(percibida)\\v=Velocidad\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\v_o=Velocidad\hspace{3}del\hspace{3}observador\\v_s=Velocidad\hspace{3}de\hspace{3}la\hspace{3}fuente$

Cuando el observador se mueve hacia la fuente v_o

es positivo.Cuando el observador se aleja de la fuente es v_o

negativo.Cuando la fuente se mueve hacia el observador v_s

es negativa.Cuando la fuente se aleja del observador v_s

es positiva.

Como el problema no nos da información adicional, supongamos que:

v=343m/s

La cuál es la velocidad del sonido en el aire.

Y utilizando la información proporcionada por el problema:

$v_s=30m/s\\v_o=0\\f_s=360Hz$

$f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz$

La frecuencia percibida por la persona es 331.046Hz

Answer from: Quest

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

Answer from: Quest

answer: a. same unit

explanation:

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Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena está funciona...

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