Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:

(1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun
Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so

so the radius of the orbit is

And if we re-arrange the equation (1), we can find the orbital period of Neptune:

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.
Explanation:
a) m1u1 + m2u2 = v(m1+m2)
1000×6 + 5000×0 = v(1000+5000)
6000 + 0 = 6000v
v = 6000/6000
v = 1 m/s
b) ½ ×(m1+m2)v²
= 0.5×6000×1²
=0.5×6000
=3000J
=3KJ
c) solve for b4 collision and compare
Goodluck
Answer:
42 degrees, virtual image, same size as the object (26 cm)
Explanation:
The law of reflection states that:
- When a ray of light is incident on a flat surface (such as the plane mirror), the angle of reflection is equal to the angle of incidence
So, since in this case the angle of incidence is 42 degrees, the angle of reflection is also 42 degrees.
Moreover, the image formed by a plane mirror is always:
- Virtual (on the same side as the object)
- The same size as the object
So in this case, since the object's size is 26 cm, the image's size is also 26 cm.