PHY160-Solution-Ch1.1-F07

# PHY160-Solution-Ch1.1-F07 - 2 INTRODUCTION AND MATHEMATICAL...

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2 INTRODUCTION AND MATHEMATICAL CONCEPTS CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS PROBLEMS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION We use the fact that 0.200 g = 1 carat and that, under the conditions stated, 1000 g has a weight of 2.205 lb to construct the two conversion factors: (0.200 g)/(1 carat) = 1 and (2.205 lb)/(1000 g) = 1 . Then, 3106 carats ( 29 0.200 g 1 carat 2.205 lb 1000 g 1.37 lb = 3. REASONING a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km = 1 mi (see the page facing the inside of the front cover of the text). b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must convert miles to meters and hours to seconds. This is accomplished by using the conversions 1 mi = 1609 m and 1 h = 3600. SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we find the speed of the bicyclists is ( 29 mi mi Speed = 34.0 1 34.0 h = 1.609km h 1 mi km 54.7 h = b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, the speed of the bicyclists is ( 29 ( 29 mi mi Speed = 34.0 1 1 34.0 h = h 1609m 1 mi 1 h m 15.2 3600s s = _____________________________________________________________________________ _ 5. REASONING When converting between units, we write down the units explicitly in the calculations and treat them like any algebraic quantity. We construct the appropriate conversion factor (equal to unity) so that the final result has the desired units. SOLUTION a. Since 1.0 × 10 3 grams = 1.0 kilogram, it follows that the appropriate conversion factor is (1.0 × 10 3 g)/(1.0 kg) = 1 . Therefore,
Chapter 1 Problems 3 6 5 10 kg - ( 29 3 1.0 10 g 1.0 kg 1 3 5 10 g - = b. Since 1.0 × 10 3 milligrams = 1.0 gram, 3 5 10 g - ( 29 3 1.0 10 mg 1.0 g 1 5 mg = c. Since 1.0 × 10 6 micrograms = 1.0 gram, 3 5 10 g - ( 29 6 1.0 10 g 1.0 g μ 3 5 10 g = ____________________________________________________________________________________________ 7. REASONING AND SOLUTION a. F = [M][L]/[T] 2 ; ma

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PHY160-Solution-Ch1.1-F07 - 2 INTRODUCTION AND MATHEMATICAL...

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