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# ch18 - CHAPTER 18 FLUID MOTION density of the compressed...

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CHAPTER 18 FLUID MOTION Section 18-1: Describing Fluids: Density and Pressure Problem 1. The density of molasses is 1600 kg / m 3 . Find the mass of the molasses in a 0.75-liter jar. Solution The mass of molasses, which occupies a volume equal to the capacity of the jar, is Δ m = ρ Δ V = (1600 kg / m 3 )(0 . 75 × 10 - 3 m 3 ) = 1 . 2 kg . Problem 2. Salad dressing is made from one part (by volume) of vinegar (density 1 . 0 g / cm 3 ) to three parts olive oil (density 0 . 92 g / cm 3 ) . What is the average density of the dressing? Solution The average density of the mixture is its total mass, Δ m = Δ m vin + Δ m oil , divided by its volume Δ V, which is 1 4 vinegar and 3 4 oil. Therefore, ρ av = m vin + Δ m oil ) / Δ V = ( ρ vin 1 4 Δ V + ρ oil 3 4 Δ V ) / Δ V = [ 1 4 (1 . 0) + 3 4 (0 . 92)](g / cm 3 ) = 0 . 94 g / cm 3 . Problem 3. The density of atomic nuclei is about 10 17 kg / m 3 , while the density of water is 10 3 kg / m 3 . Roughly what fraction of the volume of water is not empty space? Solution The average density of a mixture of two substances, with definite volume fractions, is ρ av = ρ 1 ( V 1 /V ) + ρ 2 ( V 2 /V ) , where V 1 + V 2 = V is the total volume. (Try this formula in the preceding problem.) The density of water is approximately the average density ( ρ av = 10 3 kg / m 3 ) of the nuclei ( ρ 1 = 10 17 kg / m 3 ) and empty space ( ρ 2 = 0) provided we neglect the mass of the atomic electrons, so the volume fraction of nuclei in water is ( V 1 /V ) = ρ av 1 = 10 3 / 10 17 = 10 - 14 . Problem 4. Compressed air with mass 8.8 kg is stored in a gas cylinder with a volume of 0 . 050 m 3 . (a) What is the density of the compressed air? (b) How large a volume would the same gas occupy at typical atmospheric density of 1 . 2 kg / m 3 ? Solution (a) The density of the compressed air is Δ m/ Δ V = 8 . 8 kg / 0 . 050 m 3 = 167 kg / m 3 . (b) The same mass of air, at density 1 . 2 kg / m 3 would occupy a volume of V = m/ρ = 8 . 8 kg / (1 . 2 kg / m 3 ) = 7 . 33 m 3 . (Note: The volumes are small enough that any variation in the density of the air due to gravity may be ignored.) Problem 5. A plant hangs from a 3.2-cm diameter suction cup affixed to a smooth horizontal surface (Fig. 18-42). What is the maximum weight that can be suspended (a) at sea level and (b) in Denver, where atmospheric pressure is about 0.80 atm? Suction cup figure 18-42 Problem 5. Solution (a) The force exerted on the suction cup by the atmosphere is F = PA = P atm ( πd 2 / 4) = (1 . 013 × 10 5 Pa) π (0 . 016 m) 2 = 81 . 5 N (perfect vacuum inside cup assumed). This is equal to the maximum weight. (b) At Denver, P = 0 . 8 P atm , so the maximum weight is 80% of that in part (a), or 65.2 N (a slight variation in g with altitude is neglected). Problem 6. The pressure unit torr is defined as the pressure that will support a column of mercury 1 mm high. Meteorologists often give barometric pressure in inches of mercury , defined analogously. Express

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2 CHAPTER 18 each of these units in SI. The density of mercury is 1 . 36 × 10 4 kg / m 3 .
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ch18 - CHAPTER 18 FLUID MOTION density of the compressed...

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