Answer: 2.86 m
Explanation:
To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,
ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)
In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have
mgh + 0 = 0 + KE(f)
To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have
mgh = 1/2mv² + 1/2Iw²
To get the inertia of the bodies, we use the formula
I = [m(R1² + R2²) / 2]
I = [2(0.2² + 0.1²) / 2]
I = 0.04 + 0.01
I = 0.05 kgm²
Also, the angular velocity is given by
w = v / R2
w = 4 / (1/5)
w = 20 rad/s
If we then substitute these values in the equation we have,
0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)
4.9h = 4 + 10
4.9h = 14
h = 14 / 4.9
h = 2.86 m
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.
While ice melts, it remains at 0 °C, and the liquid water that is formed with the latent heat of fusion is also at 0 °C. The heat of fusion for water at 0 °C is approximately 334 joules per gram, and the heat of vaporization at 100 °C is about 2,230 joules per gram. So it will be C
Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
