smith (js57698) – HW01 – criss – (024809)
1
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printout
should
have
18
questions.
Multiplechoice questions may continue on
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before answering.
001
(part 1 of 2) 10.0 points
A person walks 27
.
0
◦
north of east for 3
.
98
km. Another person walks due north and due
east to arrive at the same location.
a) How large is the east component of this
second path?
Correct answer: 3
.
54621 km.
Explanation:
Δ
y
Δ
x
3
.
98 km
27
◦
Note:
Figure is not drawn to scale.
Basic Concept:
Δ
x
=
d
(cos
θ
)
Given:
θ
= 27
.
0
◦
d
= 3
.
98 km
Solution:
Δ
x
= (3
.
98 km)(cos 27
◦
)
= 3
.
54621 km
002
(part 2 of 2) 10.0 points
b) How large is the north component of this
second path?
Correct answer: 1
.
80688 km.
Explanation:
Basic Concept:
Δ
y
=
d
(sin
θ
)
Solution:
Δ
y
= (3
.
98 km)(sin 27
◦
)
= 1
.
80688 km
003
(part 1 of 2) 10.0 points
A submarine dives 138
.
0 m at an angle of
17
.
0
◦
below the horizontal.
a) What is the horizontal component of the
submarine’s displacement?
Correct answer: 131
.
97 m.
Explanation:
Δ
y
Δ
x
138 m
−
17
◦
Note:
Figure is not drawn to scale.
Basic Concept:
Δ
x
=
d
(cos
θ
)
Given:
d
= 138
.
0 m
θ
=
−
17
.
0
◦
Solution:
Δ
x
= (138 m)[cos(
−
17
◦
)]
= 131
.
97 m
004
(part 2 of 2) 10.0 points
b) What is the vertical component of the sub
marine’s displacement?
Correct answer:
−
40
.
3473 m.
Explanation:
Basic Concept:
Δ
y
=
d
(sin
θ
)
Solution:
Δ
y
= (138 m)[sin(
−
17
◦
)]
=
−
40
.
3473 m
005
(part 1 of 2) 10.0 points
A descent vehicle landing on the moon has
a vertical velocity toward the surface of the
moon of 29 m
/
s. At the same time, it has a
horizontal velocity of 54
.
8 m
/
s.
a) At what speed does the vehicle move
along its descent path?
Correct answer: 62
.
0003 m
/
s.
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smith (js57698) – HW01 – criss – (024809)
2
Explanation:
v
h
v
v
v
θ
The speeds act at right angles to each other,
so
v
=
radicalBig
v
2
h
+
v
2
v
=
radicalBig
(54
.
8 m
/
s)
2
+ (29 m
/
s)
2
=
62
.
0003 m
/
s
.
006
(part 2 of 2) 10.0 points
b) At what angle with the vertical is its path?
Correct answer: 62
.
1123
◦
.
Explanation:
v
h
is the side opposite and
v
v
is the side
adjacent to the angle, so
tan
θ
=
v
h
v
v
,
so
θ
= arctan
bracketleftbigg
v
h
v
v
bracketrightbigg
= arctan
bracketleftbigg
(54
.
8 m
/
s)
(29 m
/
s)
bracketrightbigg
=
62
.
1123
◦
.
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 Spring '08
 CRISS
 Physics, Vector Space, Acceleration, Force, Velocity, Euclidean vector, Antilock braking system

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