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Makovka662 [10]
3 years ago
5

A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outst

retched hands are 0.795 m from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to 40.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in
Physics
1 answer:
USPshnik [31]3 years ago
6 0

Answer:

0.3950m

Explanation:

Use conservation of angular momentum:

Let L be the angular momentum(a vector).

We know that:p=mv, L=l\omega, \ \ \ l=\sum (m_ir_i^2)} and that the two masses have the same radius:

L_o=2mr_o^2\omega_o=L_f=2mr_f^2\omega_f\\\\r_o=0.795m,\omega_o=10\times2\pi \ rad/m , \omega_f=40.5\times 2\pi \ rad/m\\\\\therefore r_f=r_o\sqrt{(w_o/w_f)}\\\\r_f=0.795\sqrt{(10.0/40.5)}\\\\r_f=0.3950m

Hence, the weights are 0.3950m away .

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20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

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u = 1/5.0 m⁻¹

u = 0.2 m

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Then the acceleration of this object will be equal to the gravitational acceleration:

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Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

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