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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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Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

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The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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