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lakkis [162]
3 years ago
5

Someone help with this also

Physics
1 answer:
Mumz [18]3 years ago
3 0
I’m pretty sure it’s average speed= total distance and total time which is A.
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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
3 years ago
An object elongates from a length of 45 cm to a length of 55 cm. The percent strain is
suter [353]

Answer:

22%

Explanation:

55 - 45 = 10

10/45 simplifies to 2/9

2/9 = 0.22222... so 22.22% (rounded 22%)

5 0
3 years ago
A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,
Alika [10]

Explanation:

Given that,

Displacement in ice block, d=14i-11j

Force exerted by water, F=158i-179j

To find,

Work done by the force during the displacement.

Solve,

We know that the product of force and displacement is called work done. It is also equal to the dot product of force and displacement as :

W=F.d

W=(158i-179j).(14i-11j)

We know that, i.i = j.j = k.k = 1

W=2212+1969=4181\ J

So, the work done by the force on the block during the displacement is 4181 Joules.

7 0
3 years ago
Determine weather or not the equation below is balanced.
Liula [17]

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

8 0
3 years ago
A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring
topjm [15]

Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

W_s = -0.05 J

so now plug in all data above

- 0.05 - \mu(0.98) = 0 - 2

\mu = 1.99

so above is the friction coefficient


4 0
4 years ago
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