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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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3 0
3 years ago
sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N
yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

4 0
3 years ago
Why can the atmosphere hold on to heat
Naddika [18.5K]

Answer:

A. Air

Explanation:

4 0
2 years ago
A crate is acted upon by a net force of 100N. An acceleration of 5.0 m/s^2 results. What is the rate of the crate?
andreev551 [17]
Is there any other information given? I don't think you can solve this without a time
6 0
3 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
3 years ago
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