446_PartUniversity Physics Solution

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

This preview shows pages 1–3. Sign up to view the full content.

14-10 Chapter 14 (b) The density of the liquid is 3 33 0.355 kg 1000 kg/m , 0.355 10 m = × and so the volume flow rate is 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) 1 42 1.30 10 m /s 6.50 m/s 2.00 10 m v × == × . 21 /4 1.63 m/s vv . (d) () 22 12 2 1 1 2 pp ρ ρ gy y =+ − + . ( ) 32 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 = . EVALUATE: 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. IDENTIFY: Apply Bernoulli&s equation to the two points. SET UP: yy = . 11 2 2 vA vA = . 2 AA = . EXECUTE: 111 2 2 2 pg y v y v ρ ρρ ++ = . (2.50 m/s) 1.25 m/s 2 ⎛⎞ ⎛ ⎞ = ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ . 4 3 2 2 4 ( ) 1.80 10 Pa (1000 kg/m )([2.50 m/s] [1.25 m/s] ) 2.03 10 Pa − = × + = × EVALUATE: The gauge pressure is higher at the second point because the water speed is less there. 14.46. IDENTIFY and SET UP: 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 EXECUTE: 23 1 1 7200 cm , vA v r π so 1 1.43 m/s v = 7200 cm , so 2 5.73 m/s v = y v y v = = and 5 1 2.40 10 Pa, p so 5 1 2 ( ) 2.25 10 Pa EVALUATE: Where the area decreases the speed increases and the pressure decreases. 14.47. IDENTIFY: Fp A = , 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. SET UP: 2 4 D A π = . EXECUTE: (a) 2 bb 0 . 4 D p π =− (b) The force on each hemisphere due to the atmosphere is 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. EVALUATE: 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. IDENTIFY: Apply 0 p h and 0 p V V B Δ Δ= ,where B is the bulk modulus. SET UP: Seawater has density 1.03 10 kg/m . The bulk modulus of water is 9 2.2 10 Pa B . 5 air 1.01 10 Pa p . EXECUTE: (a) 53 3 2 3 8 0a i r (1.03 10 kg/m )(9.80 m/s )(10.92 10 m) 1.10 10 Pa pp g h + × × = × (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 83 3 0 9 ( ) (1.10 10 Pa)(1.00 m ) 0.050 m pV V B Δ× = = × . The volume of this mass of water at this depth therefore is 3 0 0.950 m VV V . 3 3 1.08 10 kg/m 0.950 m m V × . The density is 5% larger than at the surface. EVALUATE: For water B is small and a very large increase in pressure corresponds to a small fractional change in volume.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Fluid Mechanics 14-11 14.49. IDENTIFY: In part (a), the force is the weight of the water. In part (b), the pressure due to the water at a depth h is gh ρ . Fp A = and mV = . SET UP: The density of water is 33 1.00 10 kg/m × .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

### Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online