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Licemer1 [7]
3 years ago
8

A train rolls past a stationary observer. To him, the train is moving at a speed of 23m/s west, and a woman on the train is movi

ng at a speed of 22.4m/s west. How long does it take the women to move 13 m relative to the train?
A. 24.1s
B.21.7s
C. 23.2s
D. 25.6s
Physics
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

21.7 seconds.

Explanation:

Woman's velocity relative to train (23 m/s - 22.4 m/s) = 0.6 m/s

Distance woman wants to travel = 13m

To find how long she will take to move 13m relative to the train, take the distance she wants to travel divided by her velocity relative to the train.

(13m)/(0.6 m/s) = 21.6667 seconds or 21.7 seconds.  

Therefore, it will take the woman 21.7 seconds to move 13m.

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3 years ago
An ambulance with a siren emitting a whine at 1800 hz overtakes and passes a cyclist pedaling a bike at 2.23 m/s. after being pa
Anna71 [15]
It's the Doppler Effect that "up and down sound." 

I was on Yahoo--- Brainly doesn't have an option for Credientials or Site credit, so I'll just put this in the quotes:


<span>The formula for doppler effect is always (s is speed and f is frequency): </span>
<span>f_perceived.by.observer = f_of.emitted.wave * (s_wave + s_observer) / (s_wave + s_source.of.wave) </span>

<span>And you should pay attention to the signs: </span>
<span>s_observer is positive if the receiver is moving towards the source, negative otherwise </span>
<span>s_source.of.wave is positive if the source is moving away from the observer, negative otherwise </span>

<span>Applying it to this case: </span>

<span>s_source.of.wave = ? (positive), speed of ambulance </span>
<span>s_observer = + 2.44 m/s speed cyclist </span>
<span>f_of.emitted.wave =1800 Hz frequency of whine </span>
<span>f_perceived.by.observer = 1760 frequency heard by cyclist </span>
<span>s_wave = 343 m/s speed of sound in air </span>

<span>Now you know every value in the equation for doppler effect except by s_source.of.wave, so you can solve for s_source.of.wave.</span>
5 0
3 years ago
Why must indirect evidence be used to study the structure of atoms?
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Indirect evidence must be used to study the structure of atoms, because the direct evidence affects the position and energy of the particles of the atomic structure.

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So indirect means would probably give more accurate answers about the atomic structure than that of the direct means.

An example of the indirect means would be for example setting up a mathematical equation or  model of the atomic structure and trying to provide solution for the model or equation.
8 0
3 years ago
Explain how the wave behave as the Frequency changes using the characteristic described in
Serggg [28]

Let's find the relationship

\\ \rm\Rrightarrow E=h\nu

\\ \rm\Rrightarrow \lambda=\dfrac{c}{\nu}

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2 years ago
La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0
3 years ago
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