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:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Answer:
2.5 ohm
Explanation:
R' and R''' are parallel
So,
1/R1= 1/R' + 1/R'''
1/R1 = 1/2 + 1/2
1/R1 = 1
so,
R1= 1 ohm
Now R1 and R'' are in series
so,
R= R1 + R''
R= 1 + 1.5
R= 2.5 ohm
