If a cart of 2 kg mass has a force of 8 newtons exerted on it, what is its acceleration?
2 answers:
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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.
Answer:
a) equal 1, b) less than 1
Explanation:
a) the electric force is given by
Fe = q E
The charge of the electron and proton has the same value, that of the proton is positive and that of the electron is negative
Proton
Fp = qE
Electron
Fe = - q E
Fp / Fe = -1
If we do not take into account the sign the relationship is equal to one (1)
b) to calculate the force we use Newton's second law
F = ma
qE = m a
a = q E / m
The mass of the proton much greater than the mass of the electron
ap = q E /
ae = - q E /
ap / ae = /
=
/1600
=1/1600
It is much smaller than 1