Physics 112  HW #3 Solutions
329
(a)
Rotation period:
T
=
(24 h)
3600 s
h
=
8.64
×
10
4
s
v
=
2
π
R
T
=
2
π
(6.38
×
10
6
m)
8.64
×
10
4
s
=
4.01
×
10
7
m
8.64
×
10
4
s
=
464 m/s
a
rad
=
v
2
R
=
(464 m/s)
2
6.38
×
10
6
m
=
0.0337 m/s
2
=
0.0034 g
=
3.4
×
10
3
g
(b)
a
rad
=
v
2
R
>
g
⇒
v
>
gR
=
(9.8 m/s
2
)(6.38
×
10
6
m)
=
7910 m/s
⇒
T
=
2
π
R
v
<
2
π
(6.38
×
10
6
m)
7910 m/s
= 5070 s
=
1.41 h
334
(a)
a
rad
=
v
2
r
=
(3.00 m/s)
2
14.0 m
=
0.643 m/s
2
(b)
a
tan
=
0.500 m/s
2
a
=
a
rad
2
+ a
tan
2
=
(0.643 m/s
2
)
2
+ (0.500 m/s
2
)
2
=
0.814 m/s
2
tan
θ
=
a
tan
a
rad
=
0.500 m/s
2
0.643 m/s
2
=
0.778
⇒
θ
=
38°
a
rad
a
tan
a
v
θ
337
Let "w" = woman, "s" = sidewalk, and "g" = ground.
Using the textbook's notation:
v
w/g
=
v
w/s
+ v
s/g
(a)
v
w/g
=
1.5 m/s + 1.0 m/s
=
2.5 m/s
⇒
t
=
∆
x
v
w/g
=
35.0 m
2.5 m/s
=
14 s
(b)
v
w/g
=
 1.5 m/s + 1.0 m/s
=
 0.5 m/s
⇒
t
=
∆
x
v
w/g
=
35.0 m
0.5 m/s
=
70 s
(Now why would anyone want to do this?)
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381
Let "p" = plane, "a" = air, and "g" = ground.
Using the textbook's notation:
(a)
In xy (EN) components:
v
p/a
=
(220 km/h , 0)
v
p/g
=
120 km
0.50 h
,
20 km
0.50 h
=
(240 km/h , 40 km/h)
v
p/a
v
p/g
v
a/g
v
p/a
+
v
a/g
=
v
p/g
v
a/g
=
v
p/g

v
p/a
=
(240 km/h  (220 km/h) , 40 km/h)
=
(20 km/h , 40 km/h)
wind speed
=

v
a/g

=
(20 km/s)
2
+ (40 km/h)
2
= 45 km/h
tan
θ
=
20 km/h
40 km/h
=
0.50
⇒
θ
= 27° W of S
(b)
sin
θ
=
v
a/g
v
p/a
=
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 Fall '07
 LECLAIR,A
 Physics, mechanics, General Relativity, Velocity, m/s, km/h

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