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# ch22_p46 - 8.78 × 10 11 m/s 2 Thus Eq 4-21 gives t = x v o...

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46. Due to the fact that the electron is negatively charged, then (as a consequence of Eq. 22-28 and Newton’s second law) the field E pointing in the + y direction (which we will call “upward”) leads to a downward acceleration. This is exactly like a projectile motion problem as treated in Chapter 4 (but with g replaced with a = eE/m =
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Unformatted text preview: 8.78 × 10 11 m/s 2 ). Thus, Eq. 4-21 gives t = x v o cos 40 º = 3.00 m 1.53 x 10 7 m/s = 1.96 × 10 − 6 s. This leads (using Eq. 4-23) to v y = v o sin 40º − a t = − 4.34 × 10 5 m/s . Since the x component of velocity does not change, then the final velocity is v → = (1.53 × 10 6 m/s) i ^ − (4.34 × 10 5 m/s) j ^ ....
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