A bus travels North for 10 miles then turns around and

During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced

frez [133]

Answer:

$r_f=137493m$

$v=8638940m/s$

Explanation:

During this process the mass $M=2\times10^{30}Kg$ will be considered constant. We start from a radius $r_i=7\times10^8m$ and a period $T_i=30\ days=(30)(24)(60)(60)s=2592000s$ . The final period is $T_f=0.1s$ .

Angular momentum <em>L</em> is conserved in this process. We can use the formula $L=I\omega$ , where I is the momentum of inertia (which for a solid sphere is $I=\frac{2mr^2}{5}$ ) and $\omega=\frac{2\pi }{T}$ is the angular velocity, so we can write the star's angular momentum as:

$L=I\omega=\frac{2mr^2}{5}\frac{2\pi }{T}=\frac{4\pi mr^2 }{5T}$

Since $L_f=L_i$ we have:

$\frac{4\pi mr_f^2 }{5T_f}=\frac{4\pi mr_i^2 }{5T_i}$

Which can be simplified as:

$\frac{r_f^2 }{T_f}=\frac{r_i^2 }{T_i}$

Which means:

$r_f=\sqrt{\frac{r_i^2 T_f}{T_i}}=r_i \sqrt{\frac{T_f}{T_i}}$

Which for our values is:

$r_f=r_i \sqrt{\frac{T_f}{T_i}}=(7\times10^8m) \sqrt{\frac{0.1s}{2592000s}}=137493m$

And we calculate the speed of a point on the equator by dividing the final circumference over the final period:

$v=\frac{C_f}{T_f}=\frac{2\pi r_f}{T_f}=\frac{2\pi (137493m)}{(0.1s)}=8638940m/s$

3 0

4 years ago

a 2.0 kg hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear spe

e-lub [12.9K]

Answer:

72 J

Explanation:

Total kinetic energy will be tge sum of rotational and translational energy.

Rotational kinetic energy is given by $0.5I\omega^{2}$ where I is moment of inertia which is given by $mr^{2}$ and here m is mass, r is radius. Also, $v=\omega r$ hence making \omega the subject then $\omega=\frac {v}{r}$ where v is the velocity.