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An electron has an initial velocity of 7.88*10^6 m/s in a uniform 3.31*10^5 N/C strength electric...

An electron has an initial velocity of 7.88*10^6 m/s in a uniform 3.31*10^5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.

a) How far does the electron travel before coming to rest?

b) How long does it take the electron to come to rest?

c) What is the electron’s velocity when it returns to its starting point?

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Answer #1

(a) x 3.31x15 16 acceleration, aa-qE -1.602xioly 9.11x1031 → a=-5.82066x10 m/ree? uz u²+2 as - 0 2 17.88 x 10) = 2x5.82066x10

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