homework 01 – HELD, ETHAN – Due: Jan 23 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
A piece of pipe has an outer radius, an
inner radius, and length as shown in the figure
below.
36 cm
4
.
5 cm
2
.
5 cm
The density
is 7
.
8 g
/
cm
3
.
What is the mass of this pipe?
Correct answer:
12
.
3502
kg (tolerance
±
1
%).
Explanation:
Let :
r
1
= 4
.
5 cm
,
r
2
= 2
.
5 cm
,
ℓ
= 36 cm
,
and
ρ
= 7
.
8 g
/
cm
3
.
Basic Concepts:
The volume of the pipe will be the cross
sectional area times the length.
Solution:
V
= (
π r
2
1

π r
2
2
)
ℓ
=
π
[
r
2
1

r
2
2
]
ℓ
=
π
[(4
.
5 cm)
2

(2
.
5 cm)
2
] (36 cm)
= 1583
.
36 cm
3
.
Thus the density is
ρ
=
m
V
so
m
=
ρ V
=
ρ π
[
r
2
1

r
2
2
]
ℓ
= (7
.
8 g
/
cm
3
)
π
[(4
.
5 cm)
2

(2
.
5 cm)
2
] (36 cm)
= 12350
.
2 g = 12
.
3502 kg
.
Question 2
part 1 of 1
10 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 3
.
4 cm.
Correct answer: 4
.
85473
cm (tolerance
±
1
%).
Explanation:
Let :
m
Al
= 2700 kg
,
m
Fe
= 7860 kg
,
and
r
Fe
= 3
.
4 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
parenleftbigg
ρ
Fe
ρ
Al
parenrightbigg
1
3
= (3
.
4 cm)
parenleftbigg
7860 kg
2700 kg
parenrightbigg
1
3
=
4
.
85473 cm
.
Question 3
part 1 of 1
10 points
A cylinder, 17 cm long and 3 cm in radius,
is made of two different metals bonded end
toend to make a single bar. The densities are
4
.
5 g
/
cm
3
and 6
.
1 g
/
cm
3
.
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homework 01 – HELD, ETHAN – Due: Jan 23 2007, 4:00 am
2
17 cm
3 cm
What length of the lighter metal is needed
if the total mass is 2470 g?
Correct answer: 10
.
2135
cm (tolerance
±
1
%).
Explanation:
Let :
ℓ
= 17 cm
,
r
= 3 cm
,
ρ
1
= 4
.
5 g
/
cm
3
,
ρ
2
= 6
.
1 g
/
cm
3
,
and
m
= 2470 g
.
Volume of a bar of radius
r
and length
ℓ
is
V
=
π r
2
ℓ
and its density is
ρ
=
m
V
=
m
π r
2
ℓ
so that
m
=
ρ π r
2
ℓ
ℓ
x
ℓ

x
r
Let
x
be the length of the lighter metal;
then
ℓ

x
is the length of the heavier metal.
Thus,
m
=
m
1
+
m
2
=
ρ
1
π r
2
x
+
ρ
2
π r
2
(
ℓ

x
)
=
ρ
1
π r
2
x
+
ρ
2
π r
2
ℓ

ρ
2
π r
2
x .
Therefore
m

ρ
2
π r
2
ℓ
=
ρ
1
π r
2
x

ρ
2
π r
2
x
and
x π r
2
(
ρ
1

ρ
2
) =
m

ρ
2
π r
2
ℓ .
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 Fall '08
 Turner
 Work, Correct Answer, Orders of magnitude, ETHAN

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