Answer: 
Explanation:
Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period 
of a planet is proportional to the cube of the semi-major axis 
of its orbit”:
(1)
Now, if is measured in years (Earth years), and is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
So, knowing 
and isolating from (2) we have:
![a=\sqrt[3]{T^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{(619.36 years)^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%28619.36%20years%29%5E%7B2%7D%7D)
(4)
Finally:
T
his is the distance between the dwarf planet and the Sun in astronomical units
Converting this to kilometers, we have:
Answer:
yes
Explanation:
please Mark my answer in brainlist and follow me
Your grade will probably go down to a D 68% or little higher than that
Explanation:
Below is an attachment containing the solution
Answer:
The horizontal component of her velocity is approximately 1.389 m/s
The vertical component of her velocity is approximately 7.878 m/s
Explanation:
The given question parameters are;
The initial velocity with which Margaret leaps, v = 8.0 m/s
The angle to the horizontal with which she jumps, θ = 80° to the horizontal
The horizontal component of her velocity, vₓ = v × cos(θ)
∴ vₓ = 8.0 × cos(80°) ≈ 1.389
The horizontal component of her velocity, vₓ ≈ 1.389 m/s
The vertical component of her velocity, 
= v × sin(θ)
∴ = 8.0 × sin(80°) ≈ 7.878
The vertical component of her velocity, ≈ 7.878 m/s.
