Explanation:
Given that,
Electrostatic force, 
Distance, 
(a) 
, q is the charge on the ion
(b) Let n is the number of electrons are missing from each ion. It can be calculated as :
n = 2
Hence, this is the required solution.
Explanation:
Only few supernova are observed in our galaxy -
Type II supernovae ( i.e. the explosions of the massive stars ) occurred in the Milky Way, and they might be hidden by the intervening dust if they are located in the more distant parts of our Galaxy .
Type Ia supernovae , which need a white dwarf star in the binary star system , are brighter than the type II supernovae , but some of them could also happen in the older parts of Galaxy which are hidden due to the buildup of the dust and gas .
(a) 1200 rad/s
The angular acceleration of the rotor is given by:
where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for 
, we find the final angular speed after 10.0 s:
(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:
If we re-arrange it for t, we get:
where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,
The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
