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Aleks [24]
3 years ago
11

22.0-kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is

1.25 m from its center? (b) What centripetal force is exerted if the merry-go-round rotates at 3.00 rev/min and he is 8.00 m from its center? (c) Compare each force with his weight.
Physics
1 answer:
Likurg_2 [28]3 years ago
8 0

Answer:

a) Centripetal force is 482.79 N towards center.

b) Centripetal force is 17.37 N towards center.

c) First centripetal force is 482.79 N which is greater than weight of 215.82 N

   First centripetal force is 17.37 N which is less than weight of 215.82 N

Explanation:

Centripetal force is given by the expression, F = mrω², m is the mass, r is the radius and ω is the angular velocity,

a) Mass, m = 22 kg

   Radius, r = 1.25 m

   Angular speed , ω = 40 rev /min = 4.19 rad/s

  Substituting

                     F = mrω²

                    F = 22 x 1.25 x 4.19² = 482.79 N

   Centripetal force is 482.79 N towards center.

b) Mass, m = 22 kg

   Radius, r = 8 m

   Angular speed , ω = 3 rev /min = 0.314 rad/s

  Substituting

                     F = mrω²

                    F = 22 x 8 x 0.314² = 17.37 N

   Centripetal force is 17.37 N towards center.

c) Weight = mg = 22 x 9.81 = 215.82 N

   First centripetal force is 482.79 N which is greater than weight of 215.82 N

   First centripetal force is 17.37 N which is less than weight of 215.82 N

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A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10
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Explanation:

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B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

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B = ma / qvSinΘ

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B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

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Answer:

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