chap 15 - FLUID MOTION 15 V2 V1 mtot m1 + m2 1V1 + 2V2 = 1...

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15.1 FLUID MOTION EXERCISES Section 15.1 Density and Pressure 16. The mass of molasses, which occupies a volume equal to the capacity of the jar, is ∆∆ mV == × ρ (/ ) 1600 3 kg m (. ) . . 075 10 12 33 ×= mk g 17. INTERPRET This problem is about the volume fraction of water that’s made up of the atomic nuclei. DEVELOP The average density of a mixture of two substances, with definite volume fractions, is ρρ av tot tot + + = + + = m V mm VV V 11 2 2 1 1 V V V 2 2 + + + where V tot =+ is the total volume. The density of water is approximately the average density ) av = 10 kg m of the nuclei ) 1 17 3 10 = kg m and empty space () , 2 0 = provided we neglect the mass of the atomic electrons. EVALUATE The volume fraction of nuclei in water is V V tot 1 1 3 17 14 10 10 10 = av ASSESS 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 /. . / . = 8 8 0 050 176 kg m kg m (b) The same mass of air, at density12 3 ./ kg m would occupy a volume of Vm = /.( . / ) . . 88 733 kg/ kg m m ( Note: The volumes are small enough that any variation in the density of the air due to gravity may be ignored.) 19. INTERPRET This problem is about expressing pressure in SI units, using suitable conversion factors. DEVELOP The pressure at a depth h in an incompressible fluid of uniform density is given by Equation 15.3: pp g h 0 where p 0 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: torr Hg g . EVALUATE From the definition above, we obtain 1 3 6 1 0 9 8 1 43 2 torr kg m m s ≡= × Hg g ( . / ) ( ) 10 133 3 = mP a Similarly, 1 in (or 25.4 mm) of Hg is 254 136 10 981 .( . ) ( . / ) ( . torr kg m × Hg g ms m kP a /) ) . 23 25 4 10 3 39 ASSESS 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 HO kg m m s m 2 10 9 81 0 0254 249 2 gh ∆= = ( / )( . / )( . ) Pa. 21. INTERPRET This problem is about computing the weight of a column of air in the atmosphere. DEVELOP As shown in Equation 15.1, pressure measures the normal force per unit area exerted by a fluid, pF A = /.In this problem the fluid is the air. Since the atmospheric pressure supports the entire weight of the air column, we have Fm gpA g air atm . 15
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15.2 Chapter 15 EVALUATE The weight of the air is Fm gpA g == = = air atm kPa m . kN (. ) ( ) 101 3 1 101 3 2 ASSESS This is the force that’s pushing down on the 1-m 2 cross-sectional area. It is enormous! 22. The force exerted on the ground is the elephant’s weight, so the average pressure under its foot is PF A / () ( . / ) / ( . ) . 4300 9 8 0 15 596 22 kg m s m kPa π = 23. INTERPRET This problem is about finding the force must be exerted on the paper clip to produce a certain pressure.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at Berkeley.

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chap 15 - FLUID MOTION 15 V2 V1 mtot m1 + m2 1V1 + 2V2 = 1...

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