A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
1 answer:
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Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
In order to accelerate the dragster at a speed 
, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
:
So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
And since 1 horsepower is equal to 746 W, we can rewrite the power as
however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
And since 1 horsepower is equal to 746 W, we can rewrite the power as