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(b) The
y
component of
G
d
1
is
d
1
y
=
d
1
sin
θ
1
= –2.83 m.
(c) The
x
component of
G
d
2
is
d
2
x
=
d
2
cos
2
= 5.00 m.
(d) The
y
component of
G
d
2
is
d
2
y
=
d
2
sin
2
= 0.
(e) The
x
component of
G
d
3
is
d
3
x
=
d
3
cos
3
= 3.00 m.
(f) The
y
component of
G
d
3
is
d
3
y
=
d
3
sin
3
= 5.20 m.
(g) The sum of
x
components is
d
x
=
d
1
x
+
d
2
x
+
d
3
x
= –2.83 m + 5.00 m + 3.00 m = 5.17 m.
(h) The sum of
y
components is
d
y
=
d
1
y
+
d
2
y
+
d
3
y
= –2.83 m + 0 + 5.20 m = 2.37 m.
(i) The magnitude of the resultant displacement is
22
2
2
(5.17 m)
(2.37 m)
5.69 m.
xy
dd
d
=+
=
+
=
(j) And its angle is
= tan
–1
(2.37/5.17) = 24.6° which (recalling our coordinate choices)
means it points at about 25° north of east.
(k) and (l) This new displacement (the direct line home) when vectorially added to the
previous (net) displacement must give zero. Thus, the new displacement is the negative,
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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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