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Mariana [72]
3 years ago
7

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

lowly heated until it reaches its melting temperature, at which point it flattens into a uniform disk of thickness D 2 . By what factor is the angular velocity changed? (Give your answer as a factor of ωi.)
Physics
1 answer:
Leona [35]3 years ago
5 0

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

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3 years ago
A person with myopia (near-sightedness) has a far point of 45.0cm while their near point is 15.0cm. Upon wearing glasses, they h
Crank

Answer:

p = 22.5 cm

Explanation:

For this exercise we must use the equation of the constructor

       \frac{1}{f} =  \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm

since the image is on the same side as the object according to the sign convention the distance is negative

         \frac{1}{f} = \frac{1}{\infty }  + \frac{1}{-45}

          f = -45.0 cm

now let's use the near point (q = 15.0 cm) at what distance the object should be

          \frac{1}{p} = \frac{1}{f} - \frac{1}{q}

          \frac{1}{p} = \frac{1}{-45} - \frac{1}{-15}1 / p = 1 / -45 - 1 / -15

         \frac{1}{p} = - \frac{1}{45} + \frac{1}{15} 1 / p = -1/45 + 1/15

         \frac{1}{p} = 0.0444

          p = 22.5 cm

this is the closest distance you can see an object clearly

4 0
3 years ago
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