Chapter 15 - 15.1. Solve: The density of the liquid is !#...

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15.1. Solve: The density of the liquid is 3 3 3 3 0.240 kg 0.240 kg 960 kg m 250 mL 250 10 10 m m V ! "" # # # # $$ Assess: The liquid’s density is near that of water 3 (1000 kg/m ) 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, AB . VV ! Using the definition of mass density , mV " ! the above relationship becomes # $ # $ #$ 33 A B He He B He A B He B 7000 7000 7000 0.18 kg m 1260 kg m m m m m "" ! % ! % ! ! ! Referring to Table 15.1, we find that the liquid is glycerine.
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15.3. Model: The density of water is 3 1000 kg/m . Visualize: Solve: Volume of water in the swimming pool is ! " 3 1 2 6 m 12 m 3 m 6 m 12 m 2 m 144 m V # $ $ % $ $ # The mass of water in the swimming pool is ! "! " 3 3 5 1000 kg m 144 m 1.44 10 kg mV & # # # $
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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 total gasoline water 0.050 kg 0.050 kg 0.100 kg m m m ! " ! " ! The total volume is total gasoline water V V V !" gasoline water gasoline water m m ## 43 33 0.050 kg 0.050 kg 1.24 10 m 680 kg m 1000 kg m $ ! " ! % 23 total avg total 0.100 kg 8.1 10 kg m m V # $ & ! ! ! % % (b) The average density is calculated as follows: ’ (’ ( total gasoline water water water gasoline gasoline 3 3 3 water water gasoline gasoline avg 3 water gasoline 50 cm 1000 kg/m 680 kg/m 8.4 10 kg/m 100 cm m m m V V VV ! " ! " " " & ! ! ! % " 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 3 1030 kg/m . Solve: The pressure below sea level can be found from Equation 15.6 as follows: ! " ! "! " 5 3 2 4 0 5 8 8 3 1.013 10 Pa 1030 kg m 9.80 m s 1.1 10 m 1.013 10 Pa 1.1103 10 Pa 1.1113 10 Pa 1.10 10 atm p p gd # $ % $ & % & $ & % & $ & $ & where we have used the conversion 5 1 atm 1.013 10 Pa. $& Assess: The pressure deep in the ocean is very large.
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15.6. Model: The density of water is 3 1000 kg/m and the density of ethyl alcohol is 3 790 kg/m . Solve: (a) The volume of water that has the same mass as 3 8.0 m of ethyl alcohol is !" 3 33 water alcohol alcohol alcohol water 3 water water water 790 kg/m 8.0 m 6.3 m 1000 kg/m m m V V # $% & & & & & ’( )* (b) The pressure at the bottom of the cubic tank is 0 water : p p gd &+ ! " ! " 13 5 3 2 5 1.013 10 Pa 1000 kg/m 9.80 m s 6.3 1.19 10 Pa p & , + & , where we have used the relation water . dV &
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15.7. Visualize: Solve: The pressure at the bottom of the vat is 0 1.3 atm. p p gd ! " # " Substituting into this equation gives $% $ % $ %$ % 5 2 5 3 1.013 10 Pa 9.8 m s 2.0 m 1.3 1.013 10 Pa 1550.5 kg m !! & # " & " The mass of the liquid in the vat is $ % $ % $ % 22 33 0.50 m 1550.5 kg m 0.50 m 2.0 m 2.4 10 kg m V d !( ( " " " " &
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15.8. Model: 33 oil water The density of oil 900 kg m and the density of water 1000 kg m . !! "" Visualize: Solve: The pressure at the bottom of the oil layer is 1 0 oil 1 , p p gd ! "# and the pressure at the bottom of the water layer is $% 2 1 water 2 0 oil 1 water 2 5 3 2 3 2 5 2 1.013 10 Pa 900 kg m 9.80 m s 0.50 m 1000 kg m 9.80 m s 1.20 m 1.18 10 Pa p p gd p gd gd p " # " # # & " # # " Assess: A pressure of 5 1.16 atm ’" is reasonable.
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15.9. Model: The density of seawater 2 seawater 1030 kg m . ! " Visualize: Solve: The pressure outside the submarine’s window is out 0 seawater , p p gd "# where d is the maximum safe depth for the window to withstand a force F . This force is out in , F A p p "$ where A is the area of the window.
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Chapter 15 - 15.1. Solve: The density of the liquid is !#...

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