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klasskru [66]
2 years ago
8

Help!! Help!! Help!! Help!!

Physics
1 answer:
Mice21 [21]2 years ago
3 0

I think it's Quadratic. With all the rapid increase of the values of y.

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Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena
saul85 [17]

Answer:

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

<em><u>Translation:</u></em>

<em><u></u></em>

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

5 0
2 years ago
A stalled car is being pushed up a hill at constant velocity by three people. the net force on the car is
frosja888 [35]
I think you need more information like the force of gravity and the force of the three people. Once you combine the two, however, you should get the net force.
6 0
2 years ago
A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels
spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

7 0
3 years ago
An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?
never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

<span>Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.</span>

6 0
3 years ago
A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?
emmainna [20.7K]
Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω
8 0
3 years ago
Read 2 more answers
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