Answer:
A) equal to the battery's terminal voltage.
Explanation:
When the capacitor is fully charged after long hours of charging , its potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and current becoming zero . There is no charging current when the capacitor is fully charged .
Answer:
Explanation:
Let the potential difference between the middle point and one of the plate be ΔV .
electric potential energy will be lost and it will be converted into kinetic energy .
Electrical potential energy lost = Vq , where q is charge on charge particle .
For proton
ΔV× q = 1/2 M V² ( kinetic energy of proton )
where M is mass and V be final velocity of proton .
For electron
ΔV× q = 1/2 m v² ( kinetic energy of electron )
where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .
As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same
1/2 M V² = 1/2 m v²
(V / v )² = ( m / M )
(V / v ) = √ ( m / M )
In other words , their velocities are inversely proportional to square root of their masses .
"Light year" is a distance, not a speed. It's the distance light travels in one year, at the speed of 299,792,458 meters per second.
Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.