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Aleksandr-060686 [28]
3 years ago
11

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

ll 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
Physics
1 answer:
Ivahew [28]3 years ago
7 0

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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Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

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Let's write the equations on each axis

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Y Axis  

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We calculate the acceleration

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b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

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