DOING PHYSICS
E
XERCISES
Section 1.2 Measurements and Units
10.
SI prefixes are explained in Table 1.1.
(a)
3
3
6
9
10
MW
10
(10
W)
10
W.
=
=
(b)
and
(c)
9
6
10
W
10
kW
1 GW.
=
=
11.
I
NTERPRET
We interpret this as a problem involving the comparison of the diameter of two objects (a hydrogen
and a proton) expressed in different units.
D
EVELOP
Before any comparison can be made, the quantity of interest must first be expressed in the same units.
From Table 1.1, we see that a nanometer is
9
10
m,

and a femtometer (fm) is
15
10
m.

E
VALUATE
Using the conversion factors, the diameter of a hydrogen atom is
10
H
0.1 nm
10
m,
d

=
=
and the
diameter of a proton is
15
p
1 fm
10
m.
d

=
=
Therefore, the ratio of the diameters of a hydrogen atom and a proton
(its nucleus) is
10
5
H
15
p
10
m
10
10
m
d
d


=
=
A
SSESS
The hydrogen atom is about 100,000 times larger than its nucleus.
12.
The current definition of the meter is such that the speed of light in vacuum is exactly 299,792,458 m/s. The
distance traveled is the speed multiplied by the travel time,
9
1 ns
10
s

=
in this case. Thus
9
(299,792,458 m/s)
10
s
0.299792458 m,
d

=
×
=
or approximately 30 cm.
13.
I
NTERPRET
We interpret this as a problem involving expressing the period of cesium radiation in different units.
D
EVELOP
By definition,
1 s
9,192,631,770
=
periods of a cesium atomic clock. In addition, we know that
9
1 ns
10
s.

=
E
VALUATE
One period of cesium radiation is
10
1 s
1.087827757
10
s
0.1087827757 ns
9,192,631,770

=
×
=
A
SSESS
Since one nanosecond corresponds to about 9 periods of the cesium radiation, each period is about
1
9
of a
nanosecond. Note that there exists an alternative definition based on the frequency of the cesium133 hyperfine
transition, which is the reciprocal of the period.
14.
The prefix
eka
equals
18
10
,
so
18
3
16
14 Eg
14
10
(10
kg)
1.4
10
kg.

=
×
=
×
15.
I
NTERPRET
We interpret this as a problem involving expressing 1cm line as multiples of the diameter of hydrogen
atoms.
D
EVELOP
We first express the quantities of interest (diameter of a hydrogen atom and1cm line) in the same units.
Since a nanometer is
9
10
m

(Table 1.1), we see that
10
H
0.1 nm
10
m.
d

=
=
In addition,
2
1cm
10
m.

=
E
VALUATE
The desired number of atoms is the length of the line divided by the diameter of one atom:
2
8
10
10
m
10
10
m
N


=
=
A
SSESS
If 1 cm corresponds to
8
10
hydrogen atoms, then each atom would correspond to
8
10
10
cm
10


=
m
=
0.1 nm.
1.1
1
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1.2
Chapter 1
16.
s
(8.1 cm)(1.4)
11.3 cm.
r
θ
=
=
=
(
Note
: Radians are pure numbers with no units; for this reason, they are the
simplest angular measure mathematically.)
17.
I
NTERPRET
This is a problem that involves the definition of an angle subtended by a circular arc.
D
EVELOP
The angle in radians is the circular arc length
s
divided by the radius
R
, or
/ .
sR
θ
=
E
VALUATE
Using the equation above, we find the angle to be
2.1 km
0.62 rad
3.4 km
s
R
θ
=
=
=
Using the conversion factor
1
rad
180 ,
π
=
°
the result can be expressed as
180
0.62 rad
(0.62 rad)
35
rad
θ
π
°
=
=
≈
°
A
SSESS
Since a complete turn corresponds to
360 , 35
°
°
would be roughly 1/10 of a circle. The circumference of
a circle of radius
R
=
3.4 km is
2
(3.4 km)
21.4 km.
C
π
=
=
Therefore, we expect the jetliner to fly approximately
1/10 of
C
, or 2.1 km, in complete agreement with the problem statement.
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 Spring '09
 Prof.Kim
 Physics, Orders of magnitude, conversion factor, different units, DEVELOPWe

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