Download 2 Significant Figures File

Significant Figures
Uncertainty of Measurement
All measurements are assumed to be
approximate with the last digit estimated.
0
1
2
The length in
“cm” here is
written as:
1.45 cm
The last digit “5” is estimated as 1/2
of the interval between 4 and 5.
Estimated Measurements (Cont.)
Length = 1.45 cm
0
1
2
The last digit is estimated, but is significant. It
tells us the actual length is between 1.40 cm
and 1.50. It would not be possible to estimate
yet another digit, such as 1.453.
This measurement of length can be given in
three significant figures - the last is estimated.
Significant Figures
When writing numbers, zeros used ONLY to
help in locating the decimal point are NOT
significant—others are. See examples.
0.0062 cm
4.0500 cm
0.1061 cm
2 significant figures
5 significant figures
4 significant figures
50.0 cm
3 significant figures
50,600 cm
3 significant figures
Pacific-Atlantic Rule
0.0020 = (2)
70020 = (4)
15.800 = (5)
385 = (3)
95.0 = (3)
6,000 = (1)
Rule 1. When approximate numbers are
multiplied or divided, the number of
significant digits in the final answer is the
same as the number of significant digits in
the least accurate of the factors.
45 N
 6.97015 N/m2
Example 1: P 
(3.22 m)(2.005 m)
Least significant factor (45) has only two (2)
digits so only two are justified in the answer.
The appropriate way
to write the answer is:
P = 7.0 N/m2
Rule 2. When approximate numbers are added
or subtracted, the number of significant digits
should equal the smallest number of decimal
places of any term in the sum or difference.
Ex 2: 9.65 cm + 8.4 cm – 2.89 cm = 15.16 cm
Note that the least precise measure is 8.4 cm.
Thus, answer must be to nearest tenth of cm
even though it requires 3 significant digits.
The appropriate way
to write the answer is:
15.2 cm
Example 3. Find the area of a metal plate
that is 8.71 cm by 3.2 cm.
A = LW = (8.71 cm)(3.2 cm) = 27.872 cm2
Only 2 digits justified:
A = 28 cm2
Example 4. Find the perimeter of a metal
plate that is 8.71 cm long and 3.2 cm wide.
P = 8.71 cm + 3.2 cm + 8.71 cm + 3.2 cm
Ans. to tenth of cm:
P = 23.8 cm
Working with Numbers
Classroom work and
laboratory work should
be treated differently.
In class, the
uncertainties are
shown in the
numbers given.
Round to the
smallest number of
significant figures.
In lab, we know the
limitations of the
measurements. The
estimated digit is
either 0 or 1/2 of
the smallest unit of
the device .
Classroom Example: A car traveling
initially at 46.0 m/s undergoes constant
acceleration of 2.0 m/s2 for a time of 4.30
s. Find total displacement, given formula.
1 2
d  vi t  at
2
1
d  (46 m / s)( 4.3s)  (2.0m / s 2 )( 4.3s) 2
2
 197.8m  18.49m  216.29m
For class work, we assume all given info is
accurate to 2 significant figures.
d = 220 m
Lab Example 1: The length of a sheet of
metal is measured as 233.5 mm and the
width is 9.0 mm. Find the area.
(Multiplication Rule)
Note that the precision of each measure
is to the nearest tenth of a millimeter.
However, the length has four significant
digits and the width has only two.
How many significant digits are in the
product of length and width (area)?
Two (9.0 has least significant digits).
Lab Example 1 (Cont.): The length of a
sheet of metal is measured as 233.5 mm
and the width is 9.0 mm. Find area.
Area = LW = (233.5 mm)(9.0 mm)
Area = 2101.5 mm2
W = 9.0 mm
But we are entitled to
only two significant
digits. Therefore, the
answer becomes:
Area = 2100 mm2
L = 233.5 mm
Lab Example 2: Find the perimeter of
sheet of metal, with measured L = 233.5
mm and W = 9.0 mm. (Addition Rule)
P = 233.5 mm + 9.0 mm + 233.5 mm + 9.0 mm
P = 485 mm
Note:
is
Note: The
The answer
result has
determined
by the
more significant
digits
least
precise
measure.
than the
width
factor
(the
tenth
of a mm)
in this
case.
W = 9.0 mm
L = 233.5 mm
Perimeter = 485.0 mm