schroeder (cs39963) – Resource I: Units, Dimensions, Conversions – Balasubramanya – (1401310) 1
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001
10.0 points
What would equal 5 meters?
1.
50 cm
2.
50 dam
3.
0.5 km
4.
50 dm
correct
5.
0.5 hm
6.
50 mm
Explanation:
(50 dm)
·
1 m
10 dm
= 5 m
002
10.0 points
A school bus would have what relationship to
a meter?
1.
the same as
2.
shorter than
3.
longer than
correct
Explanation:
“Waist high” on an average person is a
fairly reasonable approximation for a meter.
003
(part 1 of 2) 10.0 points
How many atoms are in a water molecule?
1.
1
2.
3
correct
3.
2
4.
4
5.
5
Explanation:
H
2
O (water) has two hydrogen atoms and
one oxygen atom for a total of 3 atoms.
004
(part 2 of 2) 10.0 points
How many elements are in a water molecule?
1.
3
2.
5
3.
2
correct
4.
1
5.
4
Explanation:
A water molecule is composed of hydrogen
and oxygen.
005
(part 1 of 2) 10.0 points
A perfectly spherical piece of metal is found
at the bottom of a wishing well. The mass of
the object is 1
.
15 kg and the radius is 0
.
19 m.
What is its density?
Correct answer: 40
.
0266 kg
/
m
3
.
Explanation:
Let :
m
= 1
.
15 kg
and
R
= 0
.
19 m
.
The volume is
V
=
4
3
π R
3
=
4
3
π
(0
.
19 m)
3
= 0
.
0287309 m
3
,
so the density is
ρ
=
m
V
=
1
.
15 kg
0
.
0287309 m
3
=
40
.
0266 kg
/
m
3
.
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schroeder (cs39963) – Resource I: Units, Dimensions, Conversions – Balasubramanya – (1401310) 2
006
(part 2 of 2) 10.0 points
What would be its weight if it had the same
volume and were made of pure gold?
The
density of pure gold is 19300 kg
/
m
3
.
Correct answer: 5434
.
16 N.
Explanation:
W
=
m g
= (
ρ
′
V
)
g
= (19300 kg
/
m
3
) (0
.
0287309 m
3
)
×
(9
.
8 m
/
s
2
)
=
5434
.
16 N
.
007
10.0 points
One cubic meter (1.0 m
3
) of aluminum has a
mass of 2700 kg, and a cubic meter of iron has
a mass of 7860 kg.
Find the radius of a solid aluminum sphere
that has the same mass as a solid iron sphere
of radius 4
.
44 cm.
Correct answer: 6
.
33971 cm.
Explanation:
Let :
m
Al
= 2700 kg
,
m
Fe
= 7860 kg
,
and
r
Fe
= 4
.
44 cm
.
Density is
ρ
=
m
V
.
Since the masses are the
same,
ρ
Al
V
Al
=
ρ
Fe
V
Fe
ρ
Al
parenleftbigg
4
3
π r
3
Al
parenrightbigg
=
ρ
Fe
parenleftbigg
4
3
π r
3
Fe
parenrightbigg
parenleftbigg
r
Al
r
Fe
parenrightbigg
3
=
ρ
Fe
ρ
Al
r
Al
=
r
Fe
3
radicalbigg
ρ
Fe
ρ
Al
= (4
.
44 cm)
3
radicalBigg
7860 kg
2700 kg
=
6
.
33971 cm
.
008
10.0 points
In
1990
there
were
approximately
7
×
10
6
people living in New York City. New
York
City
has
an
area
of
approximately
8
.
31
×
10
2
square kilometers.
What was the population density of New
York City in 1990?
Correct answer: 8423
.
59 people
/
km
2
.
Explanation:
Let :
N
= 7
×
10
6
people
and
A
= 8
.
31
×
10
2
.
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 Fall '08
 STAFF
 Physics, Correct Answer, Orders of magnitude, Schroeder

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