Chapter 1
Page 1
CHAPTER 1 - Introduction, Measurement, Estimating
1. (
a
) Assuming one significant figure, we have
10 billion yr = 10
×
10
9
yr =
1
×
10
10
yr
.
(
b
) (1
×
10
10
yr)(3
×
10
7
s/yr) =
3
×
10
17
s
.
2. (
a
)
4 significant figures
.
(
b
) Because the zero is not needed for placement, we have
4 significant figures
.
(
c
)
3 significant figures
.
(
d
) Because the zeros are for placement only, we have
1 significant figure
.
(
e
) Because the zeros are for placement only, we have
2 significant figures
.
(
f
)
4 significant figures
.
(
g
)
2, 3, or 4 significant figures
, depending on the significance of the zeros.
3. (
a
) 1,156 =
1.156
×
10
3
.
(
b
) 21.8 =
2.18
×
10
1
.
(
c
) 0.0068 =
6.8
×
10
–3
.
(
d
) 27.635 =
2.7635
×
10
1
.
(
e
) 0.219 =
2.19
×
10
–1
.
(
f
)
22 =
2.2
×
10
1
.
4. (
a
) 8.69
×
10
4
=
86,900
.
(
b
) 7.1
×
10
3
=
7,100
.
(
c
) 6.6
×
10
–1
=
0.66
.
(
d
) 8.76
×
10
2
=
876
.
(
e
) 8.62
×
10
–5
=
0.000 086 2
.
5.
% uncertainty = [(0.25 m)/(3.26 m)] 100 =
7.7%
.
Because the uncertainty has 2 significant figures, the % uncertainty has 2 significant figures.
6.
We assume an uncertainty of 1 in the last place, i. e.,
0.01, so we have
% uncertainty = [(0.01 m)/(1.28 m)] 100 =
0.8%
.
Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.
7.
We assume an uncertainty of 0.2 s.
(
a
) % uncertainty = [(0.2 s)/(5 s)] 100 =
4%
.
Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.
(
b
) % uncertainty = [(0.2 s)/(50 s)] 100 =
0.4%
.
Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.
(
c
) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 =
0.07%
.
Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.
8.
For multiplication, the number of significant figures in the result is the least number from the multipliers;
in this case 2 from the second value.
(2.079
×
10
2
m)(0.072
×
10
–1
) = 0.15
×
10
1
m =
1.5 m
.
9.
To add, we make all of the exponents the same:
9
.
2
×
10
3
s + 8.3
×
10
4
s + 0.008
×
10
6
s = 0.92
×
10
4
s + 8.3
×
10
4
s + 0.8
×
10
4
s
= 10.02
×
10
4
s =
1.0
×
10
5
s
.
Because we are adding, the location of the uncertain figure for the result is the one furthest to the left.
In
this case, it is fixed by the third value.