According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
<h2>
(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>
(2)
</h2>
Then:
<h2>
(3)
</h2>
Which is the same as:
<h2>
</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
Answer:
1.
Firstly removing off one strip and it leaves electrons behind, so the strip becomes positively charged.
2. The roll however is not negatively charged because it is "earthed " by the hand holding it, thus excess negatives repel each other away through the hand.
3.Tearing off the next strip and once more it leaves electrons behind, the new strip is also positively charged and will repel the first strip.
4. Then, tear two strips apart and one will leave electrons behind on the other. Meaning that one strip is positive and the other is negative and they will attract each other.
Okay! Thank you for this knowledge
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
= Angle at which the force is being applied = 
Horizontal component of force is given by
The horizontal component of the force acting on the crate is 19.01 N.
