As light shines from air to another medium, i = 30.0o, the light bends toward the normal and refracts at 22.0o. What is the inde
2 answers:
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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
The momentum of block B afterwards is .
Explanation:
We can solve this problem by applying the principle of conservation of momentum.
In fact, the total momentum before and after the collision must be conserved. Therefore, we can write:
where:
is the initial momentum of block A
is the initial momentum of block B
is the final momentum of block A
is the final momentum of block B
Solving for , we find:
So, the momentum of block B afterwards is .
Learn more about momentum:
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