How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?
1 answer:
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Answer:
v = 5.88 10⁷ m / s
Explanation:
For this exercise we use the relation
E = m c²
also indicate that all energy is converted into kinetic energy
E = K = ½ (M-2m) v²
where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined
we substitute
m c² = ½ (M-2m) v²
v² =
let's calculate
v =
v =
v = 5.88 10⁷ m / s