An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.
<h3>What is an orbit?</h3>
An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.
In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.
Read more on orbit here: brainly.com/question/18496962
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Here is the full question
Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?
(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)
Answer:
1000 light-years (ly)
Explanation:
If we go by the hint; The area of the disk can be expressed as:

where D = 100, 000 ly
Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:



replacing d =
in the equation above; we have:




The distance (s) between each civilization = 
= 2 (500 ly)
= 1000 light-years (ly)
Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² =
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo =
=
- g t
t =
/ g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 =
² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J
b) PE=mgh=145*9.8*3.5=4973.5 J
c) ME=KE+PE=m(0.5v^2+gh)=62524 J