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dem82 [27]
3 years ago
13

A ballet student with her arms and a leg extended spins with an initial rotational speed of 0.90rev/s . As she draws her arms an

d leg in toward her body, her rotational inertia becomes 0.80 kg⋅m2 and her rotational velocity is 4.1rev/s .
Determine her initial rotational inertia.
Physics
1 answer:
Rainbow [258]3 years ago
5 0
Initial inertia = 3.6 kg . m^2

final inertia = 0.80 kg.m^2rotational velocity final = 4.1 rev/s
rotational velocity initial = 0.90 rev/s 
lo= lf(final inertia) *  Wf(rotational velocity final)/ Wo (rotational velocity initial)

lo =  0.8*4.1 / 0.90 = 3.6 kg m2
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4 years ago
Greg is giving a slide show presentation to a large audience. How might a laser best help him with the presentation?
gizmo_the_mogwai [7]

Explanation :

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3 years ago
Read 2 more answers
A 75 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each
shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9  m/s²,  

g = 9.81 m/s² : acceleration due to gravity

W= 75 N :  Block weight

W= m*g  

m =  W/g = 75/9.8= 7.65 kg :  Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax    

20-Ff= m*ax    

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

μk = 0.26885

4 0
3 years ago
Which law does this diagram illustrate?
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In short, Your Answer would be Option D

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