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446_PartUniversity Physics Solution

446_PartUniversity Physics Solution - 14-10 Chapter 14...

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14-10 Chapter 14 (b) The density of the liquid is 3 3 3 0.355 kg 1000 kg/m , 0.355 10 m = × and so the volume flow rate is 3 3 3 1.30 kg/s 1.30 10 m /s 1.30 L/s. 1000 kg/m = × = This result may also be obtained from ( )( ) 220 0.355 L 1.30 L/s. 60.0 s = (c) 3 3 1 4 2 1.30 10 m /s 6.50 m/s 2.00 10 m v × = = × . 2 1 /4 1.63 m/s v v = = . (d) ( ) ( ) 2 2 1 2 2 1 2 1 1 2 p p ρ v v ρ g y y = + + . ( ) 3 2 2 2 1 1 2 152 kPa (1000 kg/m ) [(1.63 m/s) (6.50 m/s) ] (9.80 m/s )( 1.35 m) p = + + . 1 119 kPa p = . E VALUATE : The increase in height and the increase in fluid speed at point 1 both cause the pressure at point 1 to be less than the pressure at point 2. 14.45. I DENTIFY : Apply Bernoulli°s equation to the two points. S ET U P : 1 2 y y = . 1 1 2 2 v A v A = . 2 1 2 A A = . E XECUTE : 2 2 1 1 1 1 1 2 2 2 2 2 p gy v p gy v ρ ρ ρ ρ + + = + + . 1 1 2 1 2 1 (2.50 m/s) 1.25 m/s 2 A A v v A A = = = . 2 2 4 3 2 2 4 1 1 2 1 1 2 2 2 ( ) 1.80 10 Pa (1000 kg/m )([2.50 m/s] [1.25 m/s] ) 2.03 10 Pa p p v v ρ = + = × + = × E VALUATE : The gauge pressure is higher at the second point because the water speed is less there. 14.46. I DENTIFY and S ET U P : Let point 1 be where 1 4.00 cm r = and point 2 be where 2 2.00 cm. r = The volume flow rate vA has the value 3 7200 cm /s at all points in the pipe. Apply Eq.(14.10) to find the fluid speed at points 1 and 2 and then use Bernoulli°s equation for these two points to find 2 . p E XECUTE : 2 3 1 1 1 1 7200 cm , v A v r π = = so 1 1.43 m/s v = 2 3 2 2 2 2 7200 cm , v A v r π = = so 2 5.73 m/s v = 2 2 1 1 1 1 1 2 2 2 2 2 p gy v p gy v ρ ρ ρ ρ + + = + + 1 2 y y = and 5 1 2.40 10 Pa, p = × so 2 2 5 1 2 1 1 2 2 ( ) 2.25 10 Pa p p v v ρ = + = × E VALUATE : Where the area decreases the speed increases and the pressure decreases. 14.47. I DENTIFY : F pA = , where A is the cross-sectional area presented by a hemisphere. The force bb F that the body builder must apply must equal in magnitude the net force on each hemisphere due to the air inside and outside the sphere. S ET U P : 2 4 D A π = . E XECUTE : (a) ( ) 2 bb 0 . 4 D F p p π = (b) The force on each hemisphere due to the atmosphere is 2 2 5 (5.00 10 m) (1.013 10 Pa/atm)(0.975 atm) 776 π × × = Ν . The bodybuilder must exert this force on each hemisphere to pull them apart. E VALUATE : The force is about 170 lbs, feasible only for a very strong person. The force required is proportional to the square of the diameter of the hemispheres. 14.48. I DENTIFY : Apply 0 p p gh ρ = + and 0 ( ) p V V B Δ Δ = − ,where B is the bulk modulus. S ET U P : Seawater has density 3 3 1.03 10 kg/m ρ = × . The bulk modulus of water is 9 2.2 10 Pa B = × . 5 air 1.01 10 Pa p = × . E XECUTE : (a) 5 3 3 2 3 8 0 air 1.01 10 Pa (1.03 10 kg/m )(9.80 m/s )(10.92 10 m) 1.10 10 Pa p p gh ρ = + = × + × × = × (b) At the surface 3 1.00 m of seawater has mass 3 1.03 10 kg × . At a depth of 10.92 km the change in volume is 8 3 3 0 9 ( ) (1.10 10 Pa)(1.00 m ) 0.050 m 2.2 10 Pa p V V B Δ × Δ = − = − = − × . The volume of this mass of water at this depth therefore is 3 0 0.950 m V V V = + Δ = . 3 3 3 3 1.03 10 kg 1.08 10 kg/m 0.950 m m V ρ × = = = × . The density is 5% larger than at the surface.
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