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 33 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 = mV , the above relationship becomes mm A A B B He He He B BH e 33 7000 0.18 kg m kg m ρρ =⇒ = = () = = 7000 7000 1260 Referring to Table 15.1, we find that the liquid is glycerine.
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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 m6 m m m 3 12 3 12 2 144 1 2 The mass of water in the swimming pool is mV == ( ) ρ 1000 kg m 144 m kg 33 144 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 mm m total gasoline water 0.050 kg 0.050 kg 0.100 kg =+ = + = The total volume is VV V total gasoline water m m gasoline gasoline water water ρρ = × 0.050 kg kg m 0.050 kg 1000 kg m m 33 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 V V total gasoline water water water gasoline gasoline avg water water gasoline gasoline water gasoline 333 3 3 cm kg / m kg / m cm kg / m = + + + = () + = 50 1000 680 100 840 Assess: The above average densities are between those of gasoline and water, and are reasonable.
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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: pp g d =+ = × + () ( ) × + × = × = 0 54 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 atm 32 88 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 mV water water water alcohol water alcohol alcohol water 3 3 33 790 kg m 1000 kg m 8.0 m m == = = () = ρρ ρ 632 . (b) The pressure at the bottom of the cubic tank is pp g d =+ 0 water : p + ( ) 1 013 10 6 32 1 194 10 5 13 5 .. . Pa 1000 kg m 9.80 m s Pa 32 where we have used the relation dV = water .
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15.7. Visualize: Solve: The pressure at the bottom of the vat is pp g d =+ = 0 ρ 1.3 atm . Substituting into this equation gives 1 013 10 1 3 1 013 10 1550 5 55 .. . . ×+ () = × ⇒= Pa 9.8 m s 2.0 m Pa kg m 2 3 ρρ The mass of the liquid in the vat is mV d == = = π 0.5 m kg m 0.5 m 2.0 m kg 3 22 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 ρρ == 900 1000 . Visualize: Solve: The pressure at the bottom of the oil layer is pp g d 10 1 =+ ρ oil , and the pressure at the bottom of the water layer is p p gd p gd gd p 21 20 1 2 2 5 5 1 013 10 118 10 + ⇒= × () + ( ) + ( ) water oil water 32 Pa 900 kg m 9.80 m s 0.50 m 1000 kg m 9.80 m s 1.20 m Pa . . Assess: A pressure of 1.18 × 10 5 Pa = 1.16 atm is reasonable.
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15.9.
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Chapter15 - 15.1. Solve: The density of the liquid is = m...

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