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konstantin123 [22]
3 years ago
6

Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar

m to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action?
Physics
1 answer:
liberstina [14]3 years ago
4 0

Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

 

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

          ∑ τ =0

                L₀ = L_{f}

              I₀ w₀ = I w

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Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

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m=1 kg

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The maximum acceleration of upper block can be μ g.

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The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
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1. una masa oscila a la frecuencia de 3Hz y con una amplitud de 6cm. ¿cuáles serán sus posiciones en los tiempos t=0 y t= 3.22 s
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3 years ago
If a book is knocked off a desk that is .75 m tall at a rate of 2.0 m/s, how far away from the desk does it fall
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Answer: 0.8 m

Explanation:

In the vertical direction, the speed is zero, u = 0.

Distance covered in the vertical direction, s = 0.75 m.

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From the equation of motion,

s = u t + 0.5 a t²

Substituting the above values, we will find out the time taken for the book to hit the ground.

⇒0.75 m=0+0.5×9.8 m/s²×t²

⇒t = √0.153 = 0.39 s ≈ 0.40 s

Now, the horizontal distance covered,

d = v×t ⇒d= 2.0 m/s × 0.40 s =0.8 m

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7 0
3 years ago
A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57
ololo11 [35]

Answer:

47.76°

Explanation:

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rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

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1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

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