Question

Evaluate the power in Watts that a 300-Kg rocket uses to overcome gravity when it travels 20 m at a 60°-angle above the horizontal line in 0.6 s.

Answer 1

The power generated by the rocket is 84870.48 W.

Explanation:

To solve this problem we must apply power property that is:

P = W / t

The power is work between the time, then we proceed to calculate the work:

W = 300 kg · (9.8 m / s ^ 2) · Sen (60º) · (20 m)

W = 50922.29 J

Now, we get the power, which tells us that:

P = 50,922.29 J / 0.6s

P = 84870.48 W

Therefore, the power generated by the rocket, in the horizontal component, is 84870.48 W.

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La potencia generada por el cohete es de 84870.48 W.  

Explicación:  

Para resolver este problema debemos aplicar propiedad de potencia es decir:  

P = W/t  

La potencia es trabajo entre el tiempo, entonces procedemos a calcular el trabajo:  

W = 300 kg·(9.8 m/s^2)·Sen(60º)·(20 m)

W = 50922.29 J  

Ahora, obtenemos la potencia, la cual nos indica que:  

P =50922.29 J/0.6s  

P = 84870.48 W  

Por tanto, la potencia que genera el cohete, en la componente horizontal, es de 84870.48 W.  

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