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(j) The two possibilities presented by a simple calculation for the angle between the
vector described in part (i) and the +
x
direction are tan
–1
[(–11 m)/(–2.0 m)] = 80°, and
180° + 80° = 260°. The latter possibility is the correct answer (see part (k) for a further
observation related to this result).
(k) Since
G
G
G
G
ab
ba
−=−
−
()
(
)
1
, they point in opposite (antiparallel) directions; the angle
between them is 180°.
28. Many of the operations are done efficiently on most modern graphical calculators
using their builtin vector manipulation and rectangular
↔
polar “shortcuts.” In this
solution, we employ the “traditional” methods (such as Eq. 36).
(a) The magnitude of
G
a
is
22
(4.0 m)
( 3.0 m)
5.0 m.
a
=+
−
=
(b) The angle between
G
a
and the +
x
axis is tan
–1
[(–3.0 m)/(4.0 m)] = –37°. The vector is
37°
clockwise
from the axis defined by
#
i
.
(c) The magnitude of
G
b
is
(6.0 m)
(8.0 m)
10 m.
b
=
(d) The angle between
G
b
and the +
x
axis is tan
–1
[(8.0 m)/(6.0 m)] = 53°.
(e)
ˆˆ
ˆ
ˆ
(4.0 m 6.0 m) i [( 3.0 m) 8.0 m]j
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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