Answer:
28,400 N
Explanation:
Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

On the lower part of the hatch, there is a pressure equal to

So, the net pressure acting on the hatch is

which acts from above.
The area of the hatch is given by:

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

The gravitational force between two objects is given by:

where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of

, while the second "object" is the Earth, with mass

. The distance of the object from the Earth's center is

; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Answer: I would say it would be 3.9
but i believe there is something missing shorty.