Chapter 1 Problem Solutions (Easy)
1.7
The desired number of atoms is the length of the line divided by the diameter of
one atom, or
(1 cm)
=
(0.1 nm)
=
(10
!
2
m)
=
(10
!
10
m)
=
10
8
. (See Table 11 for SI
prefixes.)
1.10
(a)
(35.0 mi/h)(1609 m/mi)(1 h
=
3600 s)
=
15.6 m/s.
(b)
(35.0 mi/h)(5280 ft/mi)(1 h
=
3600 s)
=
51.3 ft/s.
1.12
The number of seconds in a year is approximately
(365.24 d)(86,400 s/d)
=
3.1557
!
10
7
s
(see Appendix C), which differs from the
mnemonic
!
"
10
7
s
in the third decimal place. The percent difference, to
two significant figures, is
100(3.1557
!
3.1416)
=
3.1416
=
0.45%.
1.13
1 m
3
=
(10
2
cm)
3
=
10
6
cm
3
.
1.20
With reference to Appendix C,
550
!
10
9
bbl
y
"
#
$
%
&
’
42
gal
bbl
"
#
$
%
&
’
1 y
3.156
!
10
7
s
"
#
$
%
&
’
=
7.32
!
10
5
gal
s
"
#
$
%
&
’
3.786 L
gal
"
#
$
%
&
’
=
2.77
!
10
6
L
s
1.25
We can first solve for
a
=
2
x
=
t
2
to find its dimensions (which are usually denoted
by square brackets). Thus
[
a
]
=
[
x
]
=
[
t
2
]
=
LT
!
2
.
1.35
Substitute the given values into the equation for the temperature
T
=
S
4
!
"
#
$
%
&
’
1
=
4
=
1.4
(
10
3
kg
)
s
*
3
4
(
5.7
(
10
*
8
kg
)
s
*
3
)
K
*
4
"
#
$
%
&
’
1
=
4
=
2.8
(
10
2
K
The result is expressed in scientific notation with two significant figures.
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 Spring '07
 Hicks
 Human Body, SI prefix, Litre

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