Objective
because it means<span> based on measurement, or reasoning free of bias.</span>
Explanation:
The tangential speed of Andrea is given by :

Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,

New angular speed is twice that of the Chuck's speed.
Answer:
Applying enough force in one direction to move the object, making kinetic energy.
Explanation:
Simpleness
Answer:
P = 2 pi R / v period of space station
F / m = v^2 / R centripetal force per unit of mass
So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2
Also, F / m = 9.8 m/s^2 earth's gravitational attraction
So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m
Or D = 2 R = 390 m the diameter required
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