I believe the answer is the mass of the object and the speed at which it is moving.
The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,

Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit,
: but as we saw from the previous equation, the orbital period of the Earth is proportional to
, therefore as
increases, T increases as well.
Therefore, the period of the orbit would increase.
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Answer:
Explanation:
Let the vertical height by which it descends be h . Let it acquire velocity of v .
1/2 mv² = mgh
v² = 2gh
As it leaves the surface of sphere , reaction force of surface R = 0 , so
centripetal force = mg cosθ where θ is the angular displacement from the vertex .
mv² / r = mg cosθ
(m/r )x 2gh = mg cosθ
2h / r = cosθ
cosθ = (r-h) / r
2h / r = r-h / r
2h = r-h
3h = r
h = r / 3
Good morning.
We have:

Where
j is the unitary vector in the direction of the
y-axis.
We have that

We add the vector
-a to both sides:

Therefore, the magnitude of
b is
47 units.
A parallel circuit exists when an electric charge flows in more than one path best describes it.
<h3>What is a Parallel circuit?</h3>
This type of circuit has branches in which the current divides and only part of it flows through any of the branch.
Parallel circuit having more than one branch therefore means that electric charge will flow in more than one path thereby making option A the most appropriate choice.
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