Answer:
Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;
The closest (minimum) distance of Neptune from the Earth is 29 AU
The farthest (maximum) distance of Neptune fro the Earth is 31 AU
Explanation:
The following parameters are given;
The distance from the Earth to the Sun = 1 AU
The distance of Neptune from the Earth = 30 AU
We have;
When the Sun is between the Earth and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 + 1 = 31 AU
When the Earth is between the Sun and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 - 1 = 29 AU
Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU
The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.
Answer:

Now when it will reach at point B then its normal force is just equal to ZERO


Explanation:
Since we need to cross both the loops so least speed at the bottom must be

also by energy conservation this is gained by initial potential energy


so we will have

now we have

here we have
R = 7.5 m
so we have


Now when it will reach at point B then its normal force is just equal to ZERO

now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)


now normal force at point C is given as



Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N