If there wasn't any battery before, then there was no current
in the circuit before, and there IS one now. That's just about
the greatest change possible.
If there WAS a battery there before and you added another one
in series with it, then there are a few different possibilities for the
effect on the current in the circuit:
-- If the new battery has the same voltage as the original one,
AND you connect the new one so that they're both in the same
direction, then the current in the circuit will become double the
original current (twice as much as it was before).
-- If the new battery has the same voltage as the original one, AND
you connect the new one so that they're in opposite directions, then
the two batteries cancel each other, the total voltage becomes zero,
and the current in the circuit completely disappears.
-- If the voltage of the two batteries is different AND you connect
the new one so that they're both in the same direction, then the
current in the circuit increases, by a factor of
(sum of the two battery voltages)
divided by
(voltage of the original battery alone).
-- If the voltage of the two batteries is different AND you
connect the new one so that they're in opposite directions,
then the current in the circuit decreases, by a factor of
(difference of the two battery voltages)
divided by
(voltage of the original battery alone)
and the current flows in the direction of whichever battery has
the greater voltage. If the new battery has greater voltage than
the original one alone, then the current reverses, and flows in
the opposite direction.
I think that covers all the possibilities.
The equation we can use here is:
v^2 = v0^2 + 2 a d
where v is final velocity, v0 is intial velocity, a is
acceleration and d is distance
14^2 = 8^2 + 2 a (44)
<span>a = 1.5 m/s^2</span>
Answer:
option C
Explanation:
Given,
Refractive index of medium 1 = n₁
Refractive index of medium 2 = n₂
For total internal reflection to take place light should move from denser medium to the rarer medium.
Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2
n₁ > n₂
The correct answer is option C
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>
<h3>B )
The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>
C ) the direction of motion can be changed without changing the direction of the applied force .</h3>
D ) the system would continue to move without any further, after and initial force has set in motion.