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.
Answer:
10.1 N
Explanation:
Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
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Force required to pull one end at a constant speed</h3>
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;
F = ma
where;
- m is mass
- a is acceleration
At a constant speed, the acceleration of the object will be zero.
F = m x 0
F = 0
Thus, the force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
Learn more about constant speed here: brainly.com/question/2681210
Answer:
i/f = i/o + i/i f = focal, o = object, i = image
1 / i = 1 / f - 1 / o = (o - f) / o f
i = o * f / ( o - f) image distance
i = 12.5 * 22 / (12.5 - 22) = -28.9 cm
Image is real
Image is 28.9 cm to left of lens
M = - i / o = = 28.9 / 12.5 = 2.3 magnification (convex lens)
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again
it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something