Patel, Kinal – Homework 1 – Due: Sep 4 2007, 7:00 pm – Inst: D Weathers
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
The mass of the planet Saturn is 5
.
64
×
10
26
kg
and its radius is 6
×
10
7
m.
Calculate its density.
Correct answer: 623
.
357 kg
/
m
3
.
Explanation:
Let :
M
= 5
.
64
×
10
26
kg
and
R
= 6
×
10
7
m
.
Assuming Saturn to be a sphere (neglecting
the rings), we can use the formula for the
volume of a sphere of radius
R
:
V
=
4
3
π R
3
.
The density of Saturn is
ρ
=
M
V
=
M
4
3
π R
3
=
5
.
64
×
10
26
kg
4
3
π
(6
×
10
7
m)
3
=
623
.
357 kg
/
m
3
.
keywords:
002
(part 1 of 2) 10 points
This problem shows how dimensional analysis
helps us check and sometimes even find a
formula.
A rope has a cross section
A
= 11
.
4 m
2
and
density
ρ
= 2010 kg
/
m
3
. The “linear” density
of the rope
μ
, defined to be the mass per unit
length, can be written in the form
μ
=
ρ
x
A
y
.
Based on dimensional analysis, determine
the powers
x
and
y
by choosing an expression
below.
1.
μ
=
ρ
A
2.
μ
=
A
ρ
3.
μ
=
1
ρ A
2
4.
μ
=
A
ρ
2
5.
μ
=
1
ρ A
6.
μ
=
A
2
ρ
2
7.
μ
=
ρ
A
2
8.
μ
=
ρ A
2
9.
μ
=
ρ A
correct
10.
μ
=
A
2
ρ
Explanation:
Kilogram (kg): a unit of mass (
M
).
Meter (m): a unit of length (
L
).
[
x
] means ”the units of
x
”.
The units of both sides of any equation must
be the same for the equation to make sense.
The units of the left hand side (LHS) are
given as
[
μ
] =
M
L
=
ML

1
,
and the right hand side has
[
ρ
x
A
y
] =
M
L
3
¶
x
×
(
L
2
)
y
=
M
x
L

3
x
L
2
y
=
M
x
L
2
y

3
x
,
thus
M
+1
L

1
=
M
x
L
2
y

3
x
.
The powers of the units of mass and length
need to be the same as for the LHS above, so
x
= 1
2
y

3
x
=

1
.
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 Fall '07
 Weathers
 Acceleration, Mass, Work, Velocity, Correct Answer, patel

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