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This leads (using Eq. 423) to
61
1
2
6
00
5
sin
(2.00 10 m/s)sin40.0
(8.78 10 m/s )(1.96 10
s)
4.34 10 m/s .
y
vv
a
t
θ
−
=−
=
×
°
−
×
×
×
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
^
.
50. Due to the fact that the electron is negatively charged, then (as a consequence of Eq.
2228 and Newton’s second law) the field
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This note was uploaded on 07/07/2010 for the course PHYSICS 244 taught by Professor Ds during the Fall '10 term at The Petroleum Institute.
 Fall '10
 DS
 Charge, Acceleration

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