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15_InstSolManual_PC

# 15_InstSolManual_PC - FLUID MOTION E XERCISES Section 15.1...

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Unformatted text preview: FLUID MOTION E XERCISES Section 15.1 Density and Pressure 16. The mass of molasses, which occupies a volume equal to the capacity of the jar, is 3 (1600 kg/m ) m V ρ ∆ = ∆ = × 3 3 (0.75 10 m ) 1.2 kg.- × = 17. I NTERPRET This problem is about the volume fraction of water that’s made up of the atomic nuclei. D EVELOP The average density of a mixture of two substances, with definite volume fractions, is tot 1 2 1 1 2 2 1 2 av 1 2 tot 1 2 1 2 1 2 1 2 m m m V V V V V V V V V V V V V ρ ρ ρ ρ ρ + + = = = = + + + + + where 1 2 V V V = + tot is the total volume. The density of water is approximately the average density 3 3 ( 10 kg/m ) av ρ = of the nuclei 17 3 1 ( 10 kg/m ) ρ = and empty space 2 ( 0), ρ = provided we neglect the mass of the atomic electrons. E VALUATE The volume fraction of nuclei in water is 3 14 av 1 17 1 10 10 10 tot V V ρ ρ- = = = A SSESS The result agrees with the fact that almost all the mass of an atom is concentrated in its nucleus which is made up of protons and neutrons that are much more massive than the electrons. 18. (a) The density of the compressed air is 3 3 / 8.8 kg 0.050 m 176 kg/m . m V ∆ ∆ = = = (b) The same mass of air, at density 3 1.2 kg/m would occupy a volume of 3 3 / 8.8 kg/(1.2 kg/m ) 7.33 m . V m ρ = = = ( Note: The volumes are small enough that any variation in the density of the air due to gravity may be ignored.) 19. I NTERPRET This problem is about expressing pressure in SI units, using suitable conversion factors. D EVELOP The pressure at a depth h in an incompressible fluid of uniform density ρ is given by Equation 15.3: p p gh ρ = + where p is the pressure at the surface of the liquid. By definition, one torr is the pressure that will support a column of mercury 1 mm high: 1 torr (1 mm). Hg g ρ ≡ E VALUATE From the definition above, we obtain 4 3 2 3 1 torr (1 mm) (1.36 10 kg/m )(9.81 m/s )(10 m) 133 Pa Hg g ρ- ≡ = × = Similarly, 1 in (or 25.4 mm) of Hg is 4 3 2 3 25.4 torr (25.4 mm) (1.36 10 kg/m )(9.81 m/s )(25.4 10 m) 3.39 kPa Hg g ρ- ≡ = × × = A SSESS One atmospheric pressure (1 atm) of 101.3 kPa supports a column of Hg 760 mm (or 29.92 in) high. So 1 torr is 1/760 of 1 atm. 20. From Equation 15.2, the pressure of 1 in. of water is 2 3 3 2 H O (10 kg/m )(9.81 m/s )(0.0254 m) 249 Pa. g h ρ ∆ = = 21. I NTERPRET This problem is about computing the weight of a column of air in the atmosphere. D EVELOP As shown in Equation 15.1, pressure measures the normal force per unit area exerted by a fluid, / . p F A = In this problem the fluid is the air. Since the atmospheric pressure supports the entire weight of the air column, we have air atm ....
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15_InstSolManual_PC - FLUID MOTION E XERCISES Section 15.1...

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