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beks73 [17]
2 years ago
12

The Olympic record for a marathon is 2.00 hour 9.00 minute 21.0 seconds if the average speed of the runner achieving this record

is 5.436 m/s what is the marathon distance
Physics
1 answer:
iren2701 [21]2 years ago
3 0
(2.00 hours) x (3,600 seconds/hour)  =  7,200 seconds

(9.00 minutes) x (60 seconds/minute)  =  540 seconds

The record time = (7,200 + 540 + 21) = 7,761 seconds

Distance  =  (speed) x (time)

                 = (5.436 m/s) x (7,761 sec)  =<span>  42,188.8 meters
________________________________________________
</span>
The official length of the marathon run is  42,195 meters.
If we divide that by the record time in the question, we get

                       5.4368... m/s .

Rounded to the nearest thousandth, that's  5.437 m/s.

If the question had given the speed as  5.437  instead of  5.436 ,
then we would have calculated the distance to be 

            (5.437 m/s) x (7,761 sec)  =<span>  42,196.6 meters,

4.6 meters closer to the official distance than the answer we did get.
</span>
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Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena
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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

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Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

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<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .

When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.
 
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Answer:

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Explanation:

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