Pickering, Tracey – Homework 1 – Due: Jan 23 2006, noon – Inst: Drummond
1
This
printout
should
have
9
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
This problem shows how dimensional analysis
helps us check our work and sometimes even
help us find a formula.
A rope has a cross section
A
= 11 m
2
and
density
ρ
= 1950 kg
/
m
3
. The “linear” density
of the rope
μ
, is defined to be the mass per
unit length, in the form
μ
=
ρ
x
A
y
.
Based on dimensional analysis,
find the
powers
x
and
y
.
1.
x
= 1
, y
=

1
2.
x
=

1
, y
= 1
3.
x
=

2
, y
= 1
4.
x
=

1
, y
=

1
5.
x
=

2
, y
=

1
6.
x
= 1
, y
= 1
correct
7.
x
=

1
, y
= 2
8.
x
= 1
, y
= 2
9.
x
=

2
, y
= 2
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
=
M L

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

3
x
=

1
2
y
=

1 + 3 = 2
y
= 1
Thus the answer is (
x, y
) = (1
,
1).
002
(part 1 of 1) 10 points
The
modern
standard
of
length
is
1
m
and
the
speed
of
light
is
approximately
2
.
99792
×
10
8
m
/
s
.
Find the time Δ
t
for light to cover 1 m at
the given speed.
1.
Δ
t
≈
3
×
10

6
s
2.
Δ
t
≈
3
.
3
×
10

8
s
3.
Δ
t
≈
3
×
10

9
s
4.
Δ
t
≈
3
.
3
×
10

10
s
5.
Δ
t
≈
3
.
3
×
10

7
s
6.
Δ
t
≈
3
.
3
×
10

6
s
7.
Δ
t
≈
3
×
10

8
s
8.
Δ
t
≈
3
×
10

10
s
9.
Δ
t
≈
3
×
10

7
s
10.
Δ
t
≈
3
.
3
×
10

9
s
correct
Explanation:
Since distance is speed times velocity, we
have
speed =
distance
time
.
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 Fall '08
 Turner
 Mole, Mass, Work, International System of Units, Kilogram, Orders of magnitude, Atomic mass unit

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