D is the wrong answer. New information does often completely change the theory. Its hard to change something and leave the major theory intact.
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
The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:
1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)
1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)
<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω
Answer:
mass of ball 1=m1
mass of ball 2=m2
velocity of ball=r1w1
velocity of ball 2=r2w2
Total angular momentum=m1*v1+m2*v2
but
v1=r1*w1
v2=r2*w2
Substitute values in above equation
Total angular momentum of the system=m1*r1*w1+m2*r2*w2
Shale, sandstone, and limestone are the most commoc types of sedimentary rocks. They are formed by the most common mineral that is found on or near the surface of the Earth