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:
Total distance travelled = 210m
Explanation:
Distance travelled = 80m + 50m + 10m + 70m
= 210m
Answer:
The magnetic field in the System is 0.095T
Explanation:
To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.
By Faraday's law we know that

Where,
electromotive force
N = Number of loops
B = Magnetic field
A = Area
t= Time
For Ohm's law we now that,
V = IR
Where,
I = Current
R = Resistance
V = Voltage (Same that the electromotive force at this case)
In this system we have that the resistance in series of coil and charge measuring device is given by,

And that the current can be expressed as function of charge and time, then

Equation Faraday's law and Ohm's law we have,



Re-arrange for Magnetic Field B, we have

Our values are given as,





Replacing,


Therefore the magnetic field in the System is 0.095T