Un móvil posee un movimiento uniformemente acelerado, con una velocidad inicial de 20 m/s y aceleración 1,5 m/s. ¿Qué velocidad tendrá cuando hayan transcurrido 2 minutos? ¿Qué espacio habrá recorrido durante ese tiempo?

Answers

Answer 1

Answer:

La velocidad que tendrá el móvil cuando hayan transcurrido 2 minutos es 200 m/s.

El espacio que habrá recorrido el móvil durante 2 minutos es 13200 metros.

Explanation:

El movimiento rectilíneo uniformemente variado o MRUV es un movimiento que ocurre sobre una línea recta con aceleración constante. En otras palabras, un cuerpo realiza un  movimiento rectilíneo uniformemente variado cuando su trayectoria es una línea recta y su aceleración es constante y distinta de 0. Esto implica que la velocidad aumenta o disminuye su módulo de manera uniforme.

La velocidad que tendrá un móvil luego de un tiempo t será:

v= v0 + a*t

donde v0 es la velocidad inicial y a es la aceleración.

En este caso v0= 20 m/s, a=1,5 m/s² y t=2 minutos= 120 segundos. Reemplazando y resolviendo se obtiene:

v= 20 m/s + 1.5 m/s²* 120 segundos

v= 200 m/s

La velocidad que tendrá el móvil cuando hayan transcurrido 2 minutos es 200 m/s.

La expresión de la  posición en función del tiempo para el movimiento uniformemente variado es:

x= x0 + v0*t + [tex]\frac{1}{2}[/tex] *a*t²

donde x0 es la posición inicial, v0 es la velocidad inicial y a es la aceleración.

En este caso x0= 0 m, v0= 20 m/s, a=1,5 m/s² y t=2 minutos= 120 segundos. Reemplazando y resolviendo se obtiene:

x= 20 m/s* 120 s +  [tex]\frac{1}{2}[/tex] * 1.5 m/s² *(120 s)²

x= 2400 m + 10800 m

x= 13200 m

El espacio que habrá recorrido el móvil durante 2 minutos es 13200 metros.


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Answers

Answer:

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Answers

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Answers

Answer:

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no files.

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Answer:

Following are the responses to the given choices:

Explanation:

The maximum pressure is 10 V from across condenser.

The highest charge mostly on condenser.

[tex]vc = 10\times 1\ mf\\\\[/tex]

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The time it takes for the condenser to be 5 V different.

[tex]time = ln_2 \times 1000 \times 1mf[/tex]

        [tex]=0 .69 \times 10^{-3}\ sec[/tex]

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Answer:

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