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faltersainse [42]
2 years ago
5

A child with a weight of 430 N rides on a Ferris wheel, which has a radius of 17 m, and the linear velocity of the 3.5 m/s at an

y point on the circle. What is the child’s “apparent weight” at the lowest point of the circular motion? (Hint: The “apparent weight” is the normal force, or the force of the seat on the child at any point.)
Physics
1 answer:
Murrr4er [49]2 years ago
6 0

At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is

∑ F = ma

n - 430 N = (430 N)/g • a

where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is

a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²

and so

n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N

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Answer:

Coefficient of friction = 0.5

Explanation:

Given:

Mass of box = 5 kg

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Acceleration = 2 m/s²

Find:

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Friction force = Mass x Acceleration.

Friction force =  5 x 2

Friction force = 10 N

Coefficient of friction = Friction force / Force applied

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Answer:

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Explanation:

Applying,

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From the question,

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