Answer:
6 light years = 57 million km
Explanation:
Given;
A light year = 9.5 million km
To calculate how far is 6 light years;
6 light years = 6 × 1 light year = 6 × 9.5 million km
6 light years = 57 million km
(1) You must find the point of equilibrium between the two forces,
<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2
So,
x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.
</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and
R - x = 3.46*10^8 m
from the center of the Earth.
(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
Answer:

Explanation:


If the sun considered as x=0 on the axis to put the center of the mass as a:

solve to r1


Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

where T = orbital period
then L'(x',y') = L(x) by conservation of angular momentum. So that means

Since
then

Most likely it would dislodge the coconut on the way down due to gravity because on the way up the gravity would slow down the rock but on they down the gravity pulls the rock
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.