This is a very valid hypothesis for many reasons. One is that solar systems form from massive amounts of dust, ice, and debris that eventually form into planets and such. This means it is very possible for this 'excess material' if you will to have moved into orbit behind Neptune.
Answer:
Definimos:
Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.
En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:
RM = 400m/30s = 13.33 m/s
La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.
El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:
VM = 40m/30s = 1.33 m/s
Answer:

Explanation:
From the question we are told that:
Altitude 
Mass 
Radius 
Generally the equation for Satellite Speed is mathematically given by



Therefore
Period T is Given as



It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = 
where
f = focal length
Thus
f = 
f =
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:

where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,



Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm