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Section 2.2

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.

hadley-rutledge
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Section 2.2

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  1. Section 2.2 Scientific Notation and Dimensional Analysis

  2. Objectives • Express numbers in scientific notation • Use dimensional analysis to convert between units.

  3. 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.

  4. 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 Converting intoScientific Notation

  5. 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

  6. Doing Math inScientific 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

  7. Doing Math inScientific 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

  8. 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.

  9. 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

  10. 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

  11. Dimensional Analysis • Problem: Convert 48 km to m. • Identify the known (given) unit & the unknown unit. km is known & m is unknown • 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

  12. Dimensional Analysis • Problem: Convert 48 km to m. • 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

  13. 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!

  14. Practice Problems(Express answers in correct scientific notation.) • Convert 245 ms to s. • How many meters are in 5600 dm? • Convert 250 kg to g. • How many milliliters are in 34.8 L? • How many mm’s are in 0.067 m?

  15. 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. • Known is km, unknown is dm. • 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

  16. Multiple Conversion Factors • (cont.) 1 km = 103 m and 10 dm = 1 m The CFs: 1 km , 103 m10 dm , 1 m 103 m 1 km 1 m 10 dm • 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

  17. Multiple Conversion Factors • 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 1 m 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 1 m

  18. Practice Problems • How many cups is 3.5 tsp? (3 tsp = 1 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?

  19. More Practice Problems • What is 550 m/s in mi/hr? (1 km = 0.62 mi) • What is 790 m3 in km3? • The density of oxygen is 1.43 x 10-3 g/mL. Express this in kg/m3. • Convert 1000 ft2 into in2 and yd2.

  20. 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.

  21. 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 This is a conversion factor. $20.00 This is the KNOWN. # of widgets This is the UNKNOWN. • Now that everything is identified, you solve the problem as you would any other dimensional analysis problem.

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