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attashe74 [19]
2 years ago
14

A dancer moves around a path like the one

Physics
1 answer:
mash [69]2 years ago
7 0

Given what we know, despite not having the figure attached to the question, we can still confirm that the magnitude for the acceleration of the dancer will be zero.

<h3>Why is the dancer's acceleration equal to zero?</h3>

This has to do with how the question clarifies the speed of the dancer. Though it does not give us an exact value, we are told that the speed is constant. This is an indicator that the acceleration is zero because with any other value for acceleration the speed <u>cannot remain</u> constant.

Therefore, given that any value for acceleration will increase or decrease the speed of the dancer, but we are told that the dancer's speed is constant throughout the trip, we can confirm that the magnitude for the acceleration of the dancer is zero.

To learn more about acceleration visit;

brainly.com/question/12134554?referrer=searchResults

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A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
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Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

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The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

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