Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
It will increase..........................
The action or process of magnifying something or being magnified, especially visually. Hope this helped
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) = 
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ = –K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
b) 4m/s/s
This is because you divide the speed you reach, by the time it takes to get to that speed:
12m/s ÷ 3s = 4m/s/s
The units come from what you divide, meters per second ÷ seconds this can be written as m/s/s or ms-²
