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Unformatted text preview: CHAPTER 18 FLUID MOTION Section 181: Describing Fluids: Density and Pressure Problem 1. The density of molasses is 1600 kg / m 3 . Find the mass of the molasses in a 0.75liter jar. Solution The mass of molasses, which occupies a volume equal to the capacity of the jar, is m = 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, m = m vin + m oil , divided by its volume V, which is 1 4 vinegar and 3 4 oil. Therefore, av = ( m vin + m oil ) / V = ( vin 1 4 V + oil 3 4 V ) / 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 av = 1 ( V 1 /V ) + 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 ( av = 10 3 kg / m 3 ) of the nuclei ( 1 = 10 17 kg / m 3 ) and empty space ( 2 = 0) provided we neglect the mass of the atomic electrons, so the volume fraction of nuclei in water is ( V 1 /V ) = 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 m/ V = 8 . 8 kg / . 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/ = 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 3.2cm diameter suction cup affixed to a smooth horizontal surface (Fig. 1842). 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&amp;#13; cup figure 1842 Problem 5. Solution (a) The force exerted on the suction cup by the atmosphere is F = PA = P atm ( d 2 / 4) = (1 . 013 10 5 Pa) (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)....
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This note was uploaded on 08/28/2009 for the course PHYS 2330 taught by Professor Wilke during the Spring '08 term at Rensselaer Polytechnic Institute.
 Spring '08
 WILKE

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