Gravitational pull changes between each planet while its the "MASS" which doesnt change
Answer:

Explanation:
We could use the following suvat equation:

where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m

And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:

where the negative sign means the direction is downward.
Rotational speed would increase...
v = omega . r
which means it's directly proportional to radius...
Answer:
See the answers below.
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity = 10 [m/s]
Vo = initial velocity = 40 [m/s]
t = time = 5 [s]
a = acceleration [m/s²]
Now replacing:
![10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]](https://tex.z-dn.net/?f=10%3D40-a%2A5%5C%5C40-10%3Da%2A5%5C%5C30%3D5%2Aa%5C%5Ca%3D6%5Bm%2Fs%5E%7B2%7D%5D)
Note: The negative sign in the above equation means that the velecity is decreasing.
2)
To solve this second part we must use the following equation of kinematics.

where:
x = distance [m]
![(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]](https://tex.z-dn.net/?f=%2810%29%5E%7B2%7D%20%3D%2840%29%5E%7B2%7D%20-2%2A6%2Ax%5C%5C100%3D1600-12%2Ax%5C%5C12%2Ax%3D1600-100%5C%5C12%2Ax%3D1500%5C%5Cx%3D125%5Bm%5D)