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sergey [27]
2 years ago
13

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a

cceleration. Is a person in this ride experiencing apparent weightlessness, true weightlessness, or neither? Explain.
Physics
2 answers:
o-na [289]2 years ago
5 0
An object experiences true weightlessness when the net force of all gravitational forces acting upon the object is zero. In this case, the gravitational force exerted by the earth on the people that are on the park ride while it's free falling never ceases to act on the people. If the person on the ride were in a case of true weightlessness then they would not fall in any direction in the first place. The answer is the apparent weightlessness.
blsea [12.9K]2 years ago
3 0

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.

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Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

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    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

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           t_1 =  \frac{d_2 }{v_2 }

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The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

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Answer:

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