Question 1, chap 1, sect 1.
part 1 of 1
10 points
A newly discovered Jupiterlike planet has
an average radius 11
.
2 times that of the Earth
and a mass 313 times that of the Earth.
Calculate the ratio of new planet’s mass
density to the mass density of the Earth.
Correct answer: 0
.
222787 (tolerance
±
1 %).
Explanation:
Let :
R
np
= 11
.
2
R
E
and
M
np
= 313
M
E
.
The volume of a sphere of radius
R
is
V
=
4
3
π R
3
. The (average) density
ρ
of a body is
the ratio of the body’s mass to its volume,
ρ
=
M
V
.
Planets are spherical, so the (average)
density of a planet of a given mass
M
and a
given radius
R
is
ρ
=
M
4
3
π R
3
.
Comparing the newly discovered planet to the
Earth, we have
ρ
np
ρ
E
=
M
np
4
3
π R
3
np
M
E
4
3
π R
3
E
=
M
np
M
E
parenleftbigg
R
np
R
E
parenrightbigg
3
=
313
(11
.
2)
3
=
0
.
222787
.
Question 2, chap 1, sect 1.
part 1 of 1
10 points
A spherical balloon with radius
r
inches has
volume
4
3
πr
3
. Find a function that represents the amount
of air required to inflate the balloon from a
radius of
r
inches to a radius of
r
+ 5 inches.
1.
4
3
π
(
r
+
5)
3
2.
4
3
π
(
75
r
+
5
r
2
+
125
)
3.
4
3
πr
3
4.
4
3
π
(
25
r
+
15
r
2
+
125
)
5.
4
3
π
(
75
r
+
15
r
2
+
125
)
correct
Explanation:
Question 3, chap 1, sect 2.
part 1 of 1
10 points
What is the mass of water required to fill
a circular hot tub 4 m in diameter and 1
.
9 m
deep?
Correct answer: 23864 kg (tolerance
±
1 %).
Explanation:
Let :
d
= 4 m
,
l
= 1
.
9 m
,
and
ρ
= 1000 kg
/
m
3
.
The volume of the tub is
V
=(
πR
2
)
l
=
π
d
2
4
l
=
π
(4 m)
2
4
1
.
9 m
= 23
.
864 m
3
.
The mass is
m
=
ρ V
= (1000 kg
/
m
3
) (23
.
864 m
3
)
=
23864 kg
.
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Question 4, chap 1, sect 3.
part 1 of 1
10 points
Identify the equation below which is dimen
sionally
incorrect
.
A, x, y
and
r
have units of length.
k
here
has units of inverse length.
v
0
and
v
have units
of velocity.
a
and
g
have units of acceleration.
ω
has units of inverse time.
t
is time,
m
is
mass,
V
is volume,
ρ
is density, and
F
is force.
1.
F
=
m g
bracketleftbigg
1 +
v
r g
bracketrightbigg
correct
2.
t
=

v
0
+
radicalBig
v
2
0
+ 2
a x
a
+
v
2
a
2
t
3.
g
=
F
m
+
ρ
m
g V
4.
F
=
m ω
2
r
5.
v
=
radicalbig
2
g y
+
radicalbigg
r F
m
6.
y
=
A
cos(
k x

ω t
)
7.
v
=
±
ω
radicalbig
A
2

x
2
8.
v
0
=
radicalbig
v
2

2
a x
Explanation:
•
Check
y
=
A
cos(
k x

ω t
):
Since
k
is an inverse length,
k x
is dimension
less, and similarly,
ωt
is also dimensionless,
so
k x

ω t
is dimensionless, as an argument
of a trigonometry function must always be.
The value of a trigonometry function (
e.g.
cos(
k x

ω t
)) is also dimensionless.
So the
right hand side has the dimension of length.
[
y
] =
L
[
A
cos(
k x
)] =
L .
This equation is dimensionally correct.
•
Check
F
=
m ω
2
r
:
The dimension of force is [
F
] =
ML
T
2
.
The
dimension of right hand side is
bracketleftbig
m ω
2
r
bracketrightbig
=
M
parenleftbigg
1
T
parenrightbigg
2
L
=
M L
T
2
.
So it is also dimensionally correct.
•
Check
v
=
radicalbig
2
g y
+
radicalbigg
r F
m
:
Since
bracketleftbigg
F
m
bracketrightbigg
=
L
T
2
= [
g
], and [
y
] = [
r
] =
L
,
bracketleftBig
radicalbig
2
g y
bracketrightBig
=
bracketleftBigg
radicalbigg
r F
m
bracketrightBigg
=
radicalbigg
L
2
T
2
=
L
T
.
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 Fall '08
 Kaplunovsky
 Mass, Kilogram, Orders of magnitude, Helen –

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