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CH. 2 - MEASUREMENT. III. Unit Conversions. A. Dimensional Analysis. The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out. A. Dimensional Analysis. Steps: 1. Identify starting & ending units. 2. Line up conversion factors so units cancel.
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CH. 2 - MEASUREMENT III. Unit Conversions
A. Dimensional Analysis • The “Factor-Label” Method • Units, or “labels” are canceled, or “factored” out
A. Dimensional Analysis • Steps: 1. Identify starting & ending units. 2. Line up conversion factors so units cancel. 3. Multiply all top numbers & divide by each bottom number. 4. Check units & answer.
A. Dimensional Analysis • Lining up conversion factors: = 1 1 in = 2.54 cm 2.54 cm 2.54 cm 1 = 1 in = 2.54 cm 1 in 1 in
B. Conversion Factors • Conversion Factors to Know: • 1 in = 2.54 cm • 1 mile = 2,580 ft • 1 lb = 454 g • 1 kg = 2.2 lbs • 1 L = 1.06 qt
mega- kilo- 1 M(u) = k(u) = 103 (u) 106 (u) BASE UNIT deci- (u) = 1(u) = 100 101 d(u) centi- 1 (u) = 102 c(u) milli- 1 (u) = 103 m(u) micro- 1 (u) = 106 (u) nano- 1 (u) = 109 n(u) pico- 1 (u) = 1012 p(u) B. Conversion Factors Prefix Conversion Factor
cm in C. Practice Problems 1) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? 8.0 cm 1 in 2.54 cm = 3.2 in
2) A light bulb requires 0.032 amps to operate. What is this quantity in milliamps? A mA C. Practice Problems 0.032 A 103 mA 1 A = 32 mA
qt mL C. Practice Problems 3) How many milliliters are in 1.00 quart of milk? 1 L 1.057 qt 1000 mL 1 L 1.00 qt = 946 mL
cm pieces C. Practice Problems 4) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? 1 piece 1.5 cm 100 cm 1 m 1.3 m = 86 pieces
5) A road race is 805 dm long. What is the length of the race in kilometers? dm km C. Practice Problems 1 m 102dm 1000 km 1 m 805 dm = 8.05 x 10-2 km
cm yd C. Practice Problems 6) The Eagles needs 550 cm for a 1st down. How many yards is this? 1 ft 12 in 1 yd 3 ft 1 in 2.54 cm 550 cm = 6.0 yd
in3 L C. Practice Problems 7) How many liters of water would fill a container that measures 75.0 in3? 1 L 1000 cm3 (2.54 cm)3 (1 in)3 75.0 in3 = 1.23 L
lb cm3 C. Practice Problems 8) You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3. 1 cm3 19.3 g 1 kg 2.2 lb 1000 g 1 kg 1.5 lb = 35 cm3