contestada

A 1500kg space probe is moving at constant velocity of 4200m/sec. In a course change maneuver the rocket engines fire for 5 minutes increasing the probe's velocity to 5000m/sec. What is the probe's change in kinetic energy? A. 3.68x10^6 J. B. 5.52x10^9 J C. 1.1x10^10 J D. not enough info.

Respuesta :

fabianb4235

Answer:

B)5.52x10^9 J

Explanation:

The equation for calculating kinetic energy is as follows

E=[tex]\frac{1}{2} mv^{2}[/tex]

so what we have to do is calculate the kinetic energy of the space probe in each state and calculate the difference

E=[tex]\frac{1}{2} m.V2^{2}-\frac{1}{2} m.V1^{2}=E[/tex]

E=[tex]\frac{1}{2} m(V2^{2} -V1^{2})=E[/tex]

E=(1500)(5000^2-4200^2)/2

E=5.52x10^9 J

Otras preguntas