A particle is moving at a speed less than c / 2 . If the speed of the particle is doubled, what happens to its momentum?
1 answer:
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Answer:
charge on each
Q1 = 2.06 × C
Q2 = 7.23 × C
when force were attractive
Q1 = 1.07 × C
Q2 = -1.39 × C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 × C
and
According to Coulomb's law force between two sphere is
Force F = .........1
Q1Q2 =
here F is force and r is apart distance and k is 9 × N-m²/C² put all value we get
Q1Q2 =
Q1Q2 = 1.49 × C²
and
we have Q2 = 93 × C - Q1
put here value
Q1² - 93 × Q1 + 1.49 ×
= 0
solve we get
Q1 = 2.06 × C
and
Q1Q2 = 1.49 ×
2.06 × Q2 = 1.49 ×
Q2 = 7.23 × C
and
if force is attractive we get here
Q1Q2 = - 1.49 × C²
then
Q1² - 93 × Q1 - 1.49 ×
= 0
we get here
Q1 = 1.07 × C
and
Q1Q2 = - 1.49 ×
2.06 × Q2 = - 1.49 ×
Q2 = -1.39 × C
Answer:
The equivalent capacitance will be
Explanation:
We have given two capacitance
They are connected in parallel
So equivalent capacitance
This equivalent capacitance is now connected in series with
In series combination of capacitors the equivalent capacitance is given by
So the equivalent capacitance will be