Apparent weight/weightlessness

[TonyHem]

Hey, I'm just a little confused over how it works.

What I know is: F normal = the force some surface exerts on someone, so that's the apparent weight and they feel apparent weightlessness if they move with an acceleration equal to the gravitational field strength right?

So EG:
2) Someone weights 25KG. They're holding a toy that ways 0.15KG. They're jumping on a trampoline. They place the toy on the palm of their outstretched hand and perform a jump.
What is the apparent weight of the toy during the jump?

A=0 - Is this because the toy is placed over the side and has nothing pushing against it?

4) A girl of mass 50kg is riding on a "Big Drop" thrill ride that is descending at 9.8MS^-2
Is the normal force 0 because  the acceleration of the girl going down ='s gravity?

3) During a space mission, an astronaut of mass 80KG initially accelerates at 30MS^-2 upwards, then travels in a stable circular orbit at an altitude where the gravitation field strength is 8.2 N/KG.

A) what is the apparent weight of the astronaut during lift-off?
Is this worked out by doing F = 80 x 30 = 2400N - Force traveling up
and gravity is F = 80 x 10 = 800N traveling down
[a = 3200N]
so I flip the direction of the gravity around and add them together to get 3200N?
Or is there like a normal force which is equal but opposite to the "force traveling up" to give a 2400 going down which I add to the gravity?

I've answered these questions but kinda unsure about how some parts work, so any help is appreciated

[Mao]

2) the apparent weight of the toy would be greater than its actual weight during the 'lift off', as the hand must provide N > g to accelerate the toy up. During descend, the toy would feel weightless as the hand does not need to provide any force as both the person and toy are in free fall.

4) correct

3) correct