A person riding a Ferris wheel moves through positions at the top, the bottom, and midheight. In the following questions, you will be rankin

Question

A person riding a Ferris wheel moves through positions at the top, the bottom, and midheight. In the following questions, you will be ranking these points on the path. If multiple points rank equally, use the same rank for each, then exclude the intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the ranking would be A:1, B:1, C:3). If all points rank equally, rank each as ‘1’.

If the wheel rotates at a constant rate, rank these three positions according to the magnitude of the person’s centripetal acceleration, greatest first.

Top
– -123
Bottom
– -123
Midheight
– -123

1. Greatest
2. Second greatest
3. Third greatest

If the wheel rotates at a constant rate, rank these three positions according to the magnitude of the net centripetal force on the person, greatest first.

Top

– -123
Bottom
– -123
Midheight
– -123

1. Greatest
2. Second greatest
3. Third greatest

If the wheel rotates at a constant rate, rank these three positions according to the magnitude of the normal force on the person, greatest first.

Top
– -123
Bottom
– -123
Midheight
– -123

1. Greatest
2. Second greatest
3. Third greatest

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Khải Quang 3 years 2021-08-30T10:58:01+00:00 1 Answers 56 views 0

Answers ( )

  1. danthu
    0
    2021-08-30T10:59:38+00:00
    Reply

    Answer:

    1) A:1, B:1, C:1

    2) A:1, B:1, C:1

    3)  B:1, C:2, A:3

    Explanation:

    1)

    a = \frac{V^{2}}{R}, where V – linear speed.

                          R – radius of ferris wheel.

    These values are constant, so at any position centripetal acceleration will be the same.

    2)

    F = ma = m\frac{V^{2}}{R},  where a – centripetal acceleration.

                                        V – linear speed.

                                        R – radius of ferris wheel.

    These values are constant, so at any position centripetal force will be the same.

    3)

    top: N = mg-m\frac{V^{2}}{R} \\ midheight: N = mg \\ bottom: N = mg+m\frac{V^{2}}{R}

    So, at bottom point normal force is maximume, at top point normal force is minimum.

    B:1, C:2, A:3 (bottom: Greatest, midheight: Second greatest, top: Third greatest)

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