Chapter15 - 15.1 Solve The density of the liquid is = m...

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15.1. Solve: The density of the liquid is ρ = = = × × = m V 0.120 kg 100 mL 0.120 kg 100 m 1200 kg m 3 3 10 10 3 3 Assess: The liquid’s density is more than that of water (1000 kg/m 3 ) and is a reasonable number.

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15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, V A = V B . Using the definition of mass density ρ = m V , the above relationship becomes m m m m A A B B He He He B B He 3 3 7000 0.18 kg m kg m ρ ρ ρ ρ ρ ρ = = = ( ) = ( ) ( ) = 7000 7000 1260 Referring to Table 15.1, we find that the liquid is glycerine.
15.3. Model: The density of water is 1000 kg/m 3 . Visualize: Solve: Volume of water in the swimming pool is V = × × × × ( ) = 6 m m m 6 m m m m 3 12 3 12 2 144 1 2 The mass of water in the swimming pool is m V = = ( )( ) = × ρ 1000 kg m 144 m kg 3 3 1 44 10 5 .

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15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve: (a) The total mass is m m m total gasoline water 0.050 kg 0.050 kg 0.100 kg = + = + = The total volume is V V V total gasoline water = + = + m m gasoline gasoline water water ρ ρ = + = × 0.050 kg kg m 0.050 kg 1000 kg m m 3 3 3 680 1 235 10 4 . = = × = ρ avg total total 3 3 0.100 kg m 810 kg m m V 1 235 10 4 . (b) The average density is calculated as follows: m m m V V V V V V total gasoline water water water gasoline gasoline avg water water gasoline gasoline water gasoline 3 3 3 3 3 cm kg / m kg / m cm kg / m = + = + = + + = ( ) + ( ) = ρ ρ ρ ρ ρ 50 1000 680 100 840 Assess: The above average densities are between those of gasoline and water, and are reasonable.
15.5. Model: The density of sea water is 1030 kg/m 3 . Solve: The pressure below sea level can be found from Equation 15.6 as follows: p p gd = + = × + ( )( ) × ( ) = × + × = × = 0 5 4 5 1 013 10 1 1 10 1 013 10 1 1097 ρ . . . Pa 1030 kg m 9.80 m s m Pa .1103 10 Pa 1.1113 10 Pa atm 3 2 8 8 where we have used the conversion 1 1 013 10 5 atm Pa = × . . Assess: The pressure deep in the ocean is very large.

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15.6. Model: The density of water is 1,000 kg/m 3 and the density of ethyl alcohol is 790 kg/m 3 . Solve: (a) The volume of water that has the same mass as 8.0 m 3 of ethyl alcohol is V m m V water water water alcohol water alcohol alcohol water 3 3 3 3 790 kg m 1000 kg m 8.0 m m = = = = ( ) = ρ ρ ρ ρ 6 32 . (b) The pressure at the bottom of the cubic tank is p p gd = + 0 ρ water : p = × + ( )( ) ( ) = × 1 013 10 6 32 1 194 10 5 1 3 5 . . . Pa 1000 kg m 9.80 m s Pa 3 2 where we have used the relation d V = ( ) water 1 3 .
15.7. Visualize: Solve: The pressure at the bottom of the vat is p p gd = + = 0 ρ 1.3 atm . Substituting into this equation gives 1 013 10 1 3 1 013 10 1550 5 5 5 . . . . × + ( ) ( ) = ( ) × ( ) = Pa 9.8 m s 2.0 m Pa kg m 2 3 ρ ρ The mass of the liquid in the vat is m V d = = ( ) = ( ) ( ) ( ) = ρ ρπ π 0.5 m kg m 0.5 m 2.0 m kg 3 2 2 1550 5 2440 .

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15.8. Model: The density of oil kg m and the density of water kg m oil 3 water 3 ρ ρ = = 900 1000 . Visualize: Solve: The pressure at the bottom of the oil layer is p p gd 1 0 1 = + ρ oil , and the pressure at the bottom of the water layer is p p gd p gd gd p 2 1 2 0 1 2 2 5 5 1 013 10 1 18 10 = + = + + = × ( ) + ( )( ) ( ) + ( )( ) ( ) = × ρ ρ ρ water oil water 3 2 3 2 Pa 900 kg m 9.80 m s 0.50 m 1000 kg m 9.80 m s 1.20 m Pa . .
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