A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What
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Answer:
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Explanation:
Energy conservation law: In isolated system the amount of total energy remains constant.
The types of energy are
- Kinetic energy.
- Potential energy.
Kinetic energy
Potential energy =
Here, q₁= +5.00×10⁻⁴C
q₂=-3.00×10⁻⁴C
d= distance = 4.00 m
V = velocity = 800 m/s
Total energy(E) =Kinetic energy+Potential energy
+
=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E +
m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, 
:
First we need to find the net charge flowing at a certain point of the wire in one second,
. Using I=0.92 A and re-arranging the previous equation, we find
Now we know that each electron carries a charge of
, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
First we need to find the net charge flowing at a certain point of the wire in one second,
Now we know that each electron carries a charge of