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:
Aluminum
Explanation:
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Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
The mean of the velocities.
(finalVelocity + initialVelocity)/2
Answer:
Energy will be 
Explanation:
We have given that dielectric constant k = 5
Area of the plate 
Distance between the plates 
Electric field E = 200 kN/C
We know that capacitance is given by 
We know that electric field is given by 
So 
So energy will be 
