baltazar (kmb2869) – HW1 – ditmire – (58216)
1
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printout
should
have
21
questions.
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before answering.
001
10.0 points
A certain corner of a room is selected as the
origin of a rectangular coordinate system.
If a fly is crawling on an adjacent wall at a
point having coordinates (3
.
2 m
,
2
.
4 m), what
is the distance of the fly from the corner of
the room?
Correct answer: 4 m.
Explanation:
Let :
Δ
x
= 3
.
2 m
and
Δ
y
= 2
.
4 m
.
Using the Pythagorean Theorem,
d
=
radicalBig
(Δ
x
)
2
+ (Δ
y
)
2
=
radicalBig
(3
.
2 m)
2
+ (2
.
4 m)
2
=
4 m
.
002
10.0 points
When an object falls through air, there is
a drag force
(
with dimension M
·
L
/
T
2
)
that
depends on the product of the surface area of
the object and the square of its velocity;
i.e.
,
F
air
=
C A v
2
,
where
C
is a constant.
What is the dimension for constant
C
?
1.
[
C
] =
M
T
·
L
2
2.
[
C
] =
T
·
L
M
3.
[
C
] =
T
M
4.
[
C
] =
M
L
3
correct
5.
[
C
] =
M
T
2
·
L
2
6.
[
C
] =
M
T
7.
[
C
] =
T
2
·
L
M
8.
[
C
] =
T
2
·
L
2
M
9.
[
C
] =
M
L
2
10.
[
C
] =
T
·
L
2
M
Explanation:
[
F
] = M
·
L
/
T
2
,
[
A
] = L
2
,
and
[
v
] = L
/
T
,
so
[
C
] =
[
F
]
[
A
] [
v
]
2
=
M
·
L
/
T
2
L
2
·
(L
/
T)
2
=
M
·
L
T
2
·
T
2
L
4
=
M
L
3
.
003
10.0 points
Consider a cube of soft, spongy material.
Which piece below has the larger density?
1.
cutting out a piece of the cube that has
oneeighth the volume
2.
Unable to determine
3.
compressing the cube until it has one
eighth the volume
correct
4.
Densities are the same.
Explanation:
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baltazar (kmb2869) – HW1 – ditmire – (58216)
2
ρ
1
=
m
V
Compressing the cube results in a denser ma
terial.
Compared to the piece cut out, the
compressed piece has density
ρ
2
=
m
1
8
V
= 8
m
V
= 8
ρ
1
.
004
(part 1 of 3) 10.0 points
There are roughly 10
59
neutrons and protons
in an average star and about 10
11
stars in a
typical galaxy. Galaxies tend to form in clus
ters of (on the average) about 10
3
galaxies,
and there are about 10
9
clusters in the known
part of the Universe.
Approximately how many neutrons
and
protons “#” are there in the known Universe?
1.
#
≈
10
43
2.
#
≈
10
87
3.
None of these
4.
#
≈
10
52
5.
#
≈
10
47
6.
#
≈
10
82
correct
Explanation:
Let :
N
n
= 10
59
,
N
s
= 10
11
,
N
g
= 10
3
and
N
c
= 10
9
.
The number of particles in the observable
Universe equals the product of the numbers
of particles in each astrophysical unit
N
nU
=
N
n
N
s
N
g
N
c
=
(
10
59
) (
10
11
) (
10
3
) (
10
9
)
= 10
82
neutrons and protons
.
005
(part 2 of 3) 10.0 points
Suppose
all
this
matter
were
compressed
into a sphere of nuclear matter such that
each nuclear particle occupied a volume of
1
.
401
×
10
−
45
m
3
(which is approximately the
“volume” of a neutron or proton).
What would be the radius of this sphere of
nuclear matter?
1.
R
≈
10
12
m
correct
2.
R
≈
10
14
m
3.
R
≈
10
23
m
4.
R
≈
10
35
m
5.
R
≈
10
25
m
6.
None of these
Explanation:
Let :
V
p
= 1
.
401
×
10
−
45
m
3
The volume of the sphere would equal the
product of the number of protons and neu
trons in the observable Universe and the vol
ume of such particles;
i.e.
,
N
nU
V
p
=
4
3
π r
3
r
=
parenleftbigg
3
N
nU
V
p
4
π
parenrightbigg
1
3
=
bracketleftBigg
3
(
10
82
) (
1
.
401
×
10
−
45
)
4
π
bracketrightBigg
1
3
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 Spring '10
 Ditmire
 Cos, Correct Answer, Observable universe

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