Answer:
480J
Explanation:
Using the formula:
Delta U = Q - W
Q:Heat (J)
Delta U: Changes in internal Energy (J)
W:Work (J)
We can plug in the give numbers, Q and W.
Delta U = 658J - 178J = 480J
Answer:
The positively charged ball moves between both charged plates till the plates and the ball all become neutral.
Check Explanation for more.
Explanation:
Let the ball be in square brackets, and the plates in normal brackets.
(+) [+] (-)
From the law that like charges repel and unlike charges attract.
The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.
Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.
So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.
Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.
This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.
The ball hanging on the insulated thread becomes neutral too at this point.
Hope this Helps!!!
Answer:
calm down please its not that serious maybe no one saw it yet
Explanation:
It is gravity¿ what is the question?
With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating
at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating
at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:
