Transcript Section 2.2 - Midland Park School District

Section 2.2
Scientific Notation and
Dimensional Analysis
Objectives
• Express numbers in scientific notation
• Use dimensional analysis to convert
between units.
Scientific Notation
• It is easier to work with very large or very
small numbers when they are in scientific
notation.
• Scientific notation expresses a number as
2 factors: a number between 1 and 10
times ten raised to a power or exponent.
Converting into
Scientific Notation
• The decimal must be moved to create a number
•
between 1 and 10
The number of places that the decimal is moved
are counted.
– If the decimal is moved to the LEFT, the number of
places is converted to a POSITIVE exponent.
– If the decimal is moved to the RIGHT, the number of
places is converted to a NEGATIVE exponent.
• 1, 625 g = 1.625 x 103 g
• 0.00283 m = 2.83 x 10-3 m
Practice Problems
• Express the following in scientific
notation
– 0.540 s
– 540,000 s
• Express the following in standard
notation
– 9.87 x 10-5 g
– 7.2 x 10-1 g
Doing Math in
Scientific Notation
• Adding & Subtracting
– the exponents must be the same
– add or subtract the first factor and keep
the power of ten the same
• Practice Problems
– 2.7 x 107 + 1.7 x 108
– 4.5 x 10-12 - 3.9 x 10-13
Doing Math in
Scientific Notation
• Multiplying & Dividing
– In multiplication, multiply the first factors
and add the exponents
– In division, divide the first factors and
subtract the exponents
• Practice Problems
– 3.5 x 10-3 x 9.6 x 1010
– 4.8 x 10-12/5.2 x 108
Dimensional Analysis
• Dimensional analysis is a method for
problem-solving that focuses on the units
used to describe matter.
• Dimensional analysis uses conversion
factors. A conversion factor is a ratio of
equivalent values that express the same
quantity in different units.
Conversion Factors
• For example, we know:
1 meter = 102 cm
• This expression relates equivalent values
in different units.
• If we express this equivalency as a ratio,
we will have a conversion factor.
• There are 2 possible ratios that can be
written:
1 meter or 102 cm
102 cm
1 meter
Practice Problems
Write the possible conversion factors for the
following equivalencies.
• 1 hr = 60 s
• 4 tbs = ¼ cup
• 1 liter = 103 milliliters
• 1 km = 103 m
Dimensional Analysis
• Problem: Convert 48 km to m.
1. Identify the known (given) unit & the
unknown unit.
km is known & m is unknown
2. Find an equivalent relationship between
the units & write the possible CFs
1 km = 103 m; the CFs: 1 km and 103 m
103 m
1 km
Dimensional Analysis
• Problem: Convert 48 km to m.
3. Multiply the given quantity & unit by the
conversion factor that has the given unit in
its denominator.
48 km x 103 m = 4.8 x 104 m
1 km
Dimensional Analysis
48 km x 103 m = 4.8 x 104 m
1 km
NOTE:
The given unit will cancel out when multiplied
by the conversion factor, giving you the unit
you need.
The conversion factor itself is equal to 1 so the
quantity does NOT change, but the units do!
Practice Problems
(Express answers in correct scientific notation.)
1. Convert 245 ms to s.
2. How many meters are in 5600 dm?
3. Convert 250 kg to g.
4. How many milliliters are in 34.8 L?
5. How many mm’s are in 0.067 m?
Multiple Conversion Factors
• In some cases you may need to do more
than one conversion to end up with the
proper unit.
• For example: Convert 0.0915 km to dm.
1. Known is km, unknown is dm.
2. Since I do not know an equivalency between
km & dm, I recall what I do know:
1 km = 103 m and 10 dm = 1 m
Multiple Conversion Factors
2. (cont.) 1 km = 103 m and 10 dm = 1 m
The CFs:
1 km , 103 m 10 dm , 1 m
103 m 1 km
1m
10 dm
3. a. First, multiply the given by the CF
involving the given unit.
0.0915 km x 103 m = 9.15 x 101 m
1 km
Multiple Conversion Factors
3. b. Then, multiply the answer to step 3a
by one of the other CFs. (Pick the CF in
which the unit cancels.)
9.15 x 101 m x 10 dm = 9.15 x 102 dm
1m
Note: The multiplication can be done in one
step with 2 CFs.
0.0915 km x 103 m x 10 dm = 9.15 x 102 dm
1 km
1m
Practice Problems
1. How many cups is 3.5 tsp? (3 tsp = 1
2.
3.
4.
5.
tbs. & 4 tbs. = ¼ cup)
Convert 300 min into days.
How many yds. is 89 in.? (1 yd = 3 ft. &
1 ft. = 12 in.)
What is the speed of 65 km/min in
m/min?
What is the speed in #4 in m/s?
More Practice Problems
1. What is 550 m/s in mi/hr? (1 km = 0.62
mi)
2. What is 790 m3 in km3?
3. The density of oxygen is 1.43 x 10-3
g/mL. Express this in kg/m3.
4. Convert 1000 ft2 into in2 and yd2.
Word Problems
• Word problems are solved in exactly the
same way as any other dimensional
analysis problem.
• The difference is this - you are NOT
directly given the known and unknown.
Rather, you must pick them out from a
descriptive paragraph.
• Also, you are often given one or more
conversion factors in the problem. You
must also be able to pick them out.
Word Problems
John goes to the local hardware store and finds that
they carry a large selection of widgets. The widgets he
needs are 8 for $5.25. How many widgets can he buy
for $20.00?
• As with all word problems, you should make a list and
identify the information that is given to you.
8 widgets = $5.25
$20.00
# of widgets
This is a conversion factor.
This is the KNOWN.
This is the UNKNOWN.
• Now that everything is identified, you solve the problem
as you would any other dimensional analysis problem.