Answer:
0.0006091222 m
Explanation:
q = Charge = 42 pC
V = Voltage = 620 V
= Permittivity of free space = 
Electric potential is given by (at r = R)

The radius of the drop is 0.0006091222 m
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
Read more about Electrostatics at; brainly.com/question/18108470
The
balloon’s potential energy was transformed to kinetic energy as it drops.
Before
the balloon was drop it possessed potential energy but at the moment the
balloon was dropped the potential energy it possessed was transformed into
kinetic energy.
<span>Hope
this answer will be a good h<span>elp for you.</span></span>
Answer:
1525 meters above ground
Explanation:
So to do this you will need to write this in slope intercept form or
. So 650 would be the b, 175 would be the m, and the x would be 5 so the equation would be
so if you solve or simplify the equation you will get 1525 meters above the ground and that would be our final answer.
Answer:
The net torque is 0.4962 N m
Explanation:
please look at the solution in the attached Word file