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?
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Ответ:
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64 JuliosExplicación:
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
Variación de la energía cinética = 64 Julios
Ответ:
The maximum speed is![u = 22 \ m/s](/tpl/images/0694/2973/fef68.png)
Explanation:
From the question we are told that
The distance covered by the debris is![R = 50 \ m](/tpl/images/0694/2973/c2a2b.png)
The maximum range of the debris projectile is mathematically represented as
At maximum![\theta = 90 ^o](/tpl/images/0694/2973/b5b70.png)
Now making u which is the maximum speed at which debris was blown outward by the explosion.
we have
substituting values