It would be 4 atm, because the way to figure out the final pressure is that (P1)(V1)=(P2)(V2)
meaning that the original pressure x original volume is equal to the final pressure x final volume. This gas law is called Boyle's law if you'd like to learn more about it.
But (1 atm)(40 mL)=(4 atm)(10 mL)
So it would be the second choice.
The system is located in the HairyFunyon aka near the floor
This interaction is known as <em>constructive interference</em>. It's a result of linear superposition.
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find
Answer:
19.2 m/s
Explanation:
The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added
18 + 1.2 = 19.2
If you were walking backwards (west) your velocity with respect to the ground would be
18 - 1.2 = 16.8