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Sindrei [870]
3 years ago
8

Un cuerpo de m=0,5 Kg se desplaza horizontalmente con v=4m/s y luego de un lapso de tiempo se mueve con v=20 m/s. ¿cual ha sido

la variación de la energÍa cinética?
Physics
1 answer:
Lilit [14]3 years ago
7 0

Responder:

<h2>64 Julios </h2>

Explicación:

La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;

m es la masa del cuerpo

v es la velocidad del objeto

Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.

Parámetros dados

masa del objeto m = 0,5 kg

Variación de velocidad = 16 m / s

Variación de la energía cinética = 1/2 * 0,5 * 16²

Variación de la energía cinética = 1/2 * 0,5 * 256

Variación de la energía cinética = 0,5 * 128

<em>Variación de la energía cinética = 64 Julios</em>

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Applying,

v = λf.................... Equation 1

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f = v/λ................ Equation 2

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From the question,

Given: λ = 3 cm = 0.03 m

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