oldhomewk 01 – YOO, HEE – Due: Jan 17 2008, 4:00 am
1
Question 1, chap 1, sect 5.
part 1 of 1
10 points
A newly discovered giant planet has an av
erage radius 17 times that of the Earth and a
mass 181 times that of the Earth.
Calculate the ratio of the new planet’s den
sity to the Earth’s density.
Correct answer: 0
.
036841 (tolerance
±
1 %).
Explanation:
Let :
R
n
= 17
R
E
and
m
n
= 181
m
E
.
Density is the ratio of mass to volume,
ρ
=
m
V
.
A spherical planet of average radius
R
has volume
4
3
π R
3
and hence density
ρ
=
m
4
3
π R
3
.
For two planets of respective radii
R
1
and
R
2
and masses
m
1
and
m
2
we have
ρ
1
ρ
2
=
m
1
4
3
π R
3
1
m
2
4
3
π R
3
2
=
parenleftbigg
m
1
m
2
parenrightbigg
parenleftbigg
R
1
R
2
parenrightbigg
3
=
181
(17)
3
=
0
.
036841
.
Question 2, chap 1, sect 5.
part 1 of 1
10 points
A sphere of metal has a radius of 5
.
7 cm
and a density of 6
.
81 g
/
cm
3
.
What is the mass of the sphere?
Correct answer: 5282
.
75 g (tolerance
±
1 %).
Explanation:
Let :
r
= 5
.
7 cm
and
ρ
= 6
.
81 g
/
cm
3
.
Density is mass per unit volume, so
ρ
=
m
V
m
=
ρ V
=
ρ
4
3
π r
3
=
(
6
.
81 g
/
cm
3
)
4
3
π
(5
.
7 cm)
3
=
5282
.
75 g
.
Question 3, chap 1, sect 99.
part 1 of 1
10 points
A cylinder, 20 cm long and 8 cm in radius,
is made of two different metals bonded end
toend to make a single bar. The densities are
5 g
/
cm
3
and 6
.
1 g
/
cm
3
.
20 cm
8 cm
What length of the lighter metal is needed
if the total mass is 21565 g?
Correct answer: 13
.
4041
cm (tolerance
±
1
%).
Explanation:
Let :
ℓ
= 20 cm
,
r
= 8 cm
,
ρ
1
= 5 g
/
cm
3
,
ρ
2
= 6
.
1 g
/
cm
3
,
and
m
= 21565 g
.
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oldhomewk 01 – YOO, HEE – Due: Jan 17 2008, 4:00 am
2
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 .
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 Spring '08
 Turner
 Physics, Mass, Correct Answer

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