15 1 Solve The density of the liquid is m V 0 240 kg 250 mL 0 240 kg 3 3 10 m 3 250 10 960 kg m 3 Assess The liquid s density is near that of water 1000 kg m and is a reasonable number 3 15 2 Solve The volume of the helium gas in container A is equal to the volume of the liquid in container B V That is A Using the definition of mass density the above relationship becomes m V V B m m B A B A m He He 7000 B m He B 7000 He 7000 0 18 kg m 1260 kg m 3 3 Referring to Table 15 1 we find that the liquid is glycerine 15 3 Model The density of water is Visualize 1000 kg m 3 Solve Volume of water in the swimming pool is V 6 m 12 m 3 m 6 m 12 m 2 m 144 m 3 1 2 The mass of water in the swimming pool is 1000 kg m 144 m V m 3 3 1 44 10 kg 5 15 4 Model The densities of gasoline and water are given in Table 15 1 Solve a The total mass is m total m gasoline m water 0 050 kg 0 050 kg 0 100 kg The total volume is V total V V gasoline water m gasoline water gasoline water m 0 050 kg 0 050 kg 680 kg m 1000 kg m 3 3 1 24 10 m 4 3 avg m V total total 0 100 kg 1 24 10 m 4 3 2 8 1 10 kg m 3 b The average density is calculated as follows avg V water water V water V gasoline gasoline V gasoline 50 cm 1000 kg m 680 kg m 3 3 100 cm 3 2 8 4 10 kg m 3 m total m gasoline water V water water V gasoline gasoline m 3 Assess The above average densities are between those of gasoline and water and are reasonable 15 5 Model The density of sea water is Solve The pressure below sea level can be found from Equation 15 6 as follows 1030 kg m 3 p p 0 gd 1 013 10 Pa 5 1 013 10 Pa 1 1103 10 Pa 1 1113 10 Pa 1 10 10 atm 5 3 3 1030 kg m 9 80 m s 8 8 2 1 1 10 m 4 where we have used the conversion Assess The pressure deep in the ocean is very large 1 atm 1 013 10 Pa 5 15 6 Model The density of water is 1000 kg m and the density of ethyl alcohol is 790 kg m 3 3 Solve a The volume of water that has the same mass as 8 0 m of ethyl alcohol is 3 V water water alcohol water water m m V alcohol alcohol water 3 790 kg m 1000 kg m 3 8 0 m 3 6 3 m 3 b The pressure at the bottom of the cubic tank is p gd water p 1 013 10 Pa 5 3 1000 kg m 9 80 m s 2 6 3 1 3 1 19 10 Pa 5 p 0 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 0 gd 1 3 atm 1 013 10 Pa 5 9 8 m s 2 2 0 m 1 3 1 013 10 Pa 5 1550 5 kg m 3 Substituting into this equation gives The mass of the liquid in the vat is m V 0 50 m 2 d 1550 5 kg m 3 0 50 m 2 0 m 2 4 10 kg 3 2 15 8 Model Visualize The density of oil oil 900 kg m and the density of water 3 3 1000 kg m water Solve The pressure at the bottom of the oil layer is water layer is p 1 p 0 gd 1 oil and the pressure at the bottom of the p 2 1 013 10 Pa 5 p 2 3 900 kg m 9 80 m s p 1 water gd 2 p 0 2 0 50 m gd 1 gd oil 3 1000 kg m 9 80 m s water 2 2 1 20 m 1 18 10 Pa 5 Assess A pressure of 1 18 10 Pa 1 16 atm is reasonable 5 15 9 Model The density of seawater Visualize seawater 1030 kg m 2 Solve The pressure outside the submarine s window is gd where d is the maximum safe depth for the window to withstand a force F This force is where A is the area of the window p out p 0 F A p seawater in p out With p in p 0 we simplify the pressure equation to p out p 0 seawater gd d F A F A seawater g d 1 0 10 N 6 0 10 m 1030 kg m 9 8 m s 2 2 2 3 2 km Assess A force of 1 0 10 N corresponds to a pressure of 6 F A 6 1 0 10 N 2 0 10 m 314 atm A depth of 3 km is therefore reasonable 15 10 Visualize We assume that the seal is at a radius of 5 cm Outside the seal atmospheric pressure presses on both sides of the cover and the forces cancel Thus only the 10 cm diameter opening inside the seal is relevant not the 20 cm diameter of the cover Solve Within the 10 cm diameter area where the pressures differ F to left p atmos A F to right p A gas where A r 2 is the area of the opening The difference between the forces is p atmos p gas A 101 300 Pa 20 000 Pa 7 85 10 m 3 2 0 64 kN 2 3 7 85 10 m to right F F to left Normally the rubber seal exerts a 0 64 kN force to the right to balance the air pressure force To pull the cover off an external force must pull to the right with a force 0 64 kN 15 11 Model The density of water is Visualize Please refer to Figure 15 17 Solve From the figure and the equation for hydrostatic pressure we have 3 1000 kg m p 0 gh p atmos Using p 0 0 atm and 5 p atmos 1 013 10 Pa 1000 kg m 9 8 m s we get 0 Pa 3 2 h 1 013 10 Pa h 10 3 m 5 Assess This large value of h is due to water having a much smaller density than mercury 15 12 Model Assume that the oil is incompressible Its density is Visualize Please refer to Figure 15 19 Because the liquid is incompressible the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the …
View Full Document
Unlocking...