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Leya [2.2K]
2 years ago
5

Xander reached a final velocity of 4.5 m/s for 3.5 seconds. Finley reached a final velocity of 3.6 m/s for 4.2 seconds. Max reac

hed a final velocity of 7.3 m/s for 1.2 seconds. They all started at the same location from rest. Which lists them from least to most acceleration? Max Finley Xander Max Xander Finley Xander Finley Max Finley Xander Max
Physics
2 answers:
ANEK [815]2 years ago
6 0

Answer:

Finley, Xander and Max

Explanation:

v = Final velocity

t = Time

u = Initial velocity = 0

a = Acceleration

From kinematic equations we get

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}=\dfrac{v-0}{t}\\\Rightarrow a=\dfrac{v}{t}

Xander

v = 4.5 m/s, t = 3.5 s

a=\dfrac{4.5}{3.5}\\\Rightarrow a=1.29\ \text{m/s}^2

Finley

v = 3.6 m/s, t = 4.2 s

a=\dfrac{3.6}{4.2}\\\Rightarrow a=0.86\ \text{m/s}^2

Max

v = 7.3 m/s, t = 1.2 s

a=\dfrac{7.3}{1.2}\\\Rightarrow a=6.083\ \text{m/s}^2

The required list is Finley, Xander and Max.

pshichka [43]2 years ago
5 0

Answer:

Finley, Xander and Max

Explanation:

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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

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