Problem_Set_10_Solutions

# Problem_Set_10_Solutions - 20. To find the pressure at the...

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20. To find the pressure at the brain of the pilot, we note that the inward acceleration can be treated from the pilot’s reference frame as though it is an outward gravitational acceleration against which the heart must push the blood. Thus, with 4 ag = , we have 33 2 brain heart 1torr 120 torr (1.06 10 kg/m )(4 9.8 m/s )(0.30 m) 133 Pa 120 torr 94 torr 26 torr. pp a r ρ =− = −× × =

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38. If the alligator floats, by Archimedes’ principle the buoyancy force is equal to the alligator’s weight (see Eq. 14-17). Therefore, 22 HO () bg FFmg A h g ρ == = . If the mass is to increase by a small amount mm →= + Δ , then 2 bb FF A h h g + Δ . With 0.010 bbb FFF m g Δ= − = , the alligator sinks by 3 32 0.01 0.010(130 kg) 6.5 10 m 6.5 mm (998 kg/m )(0.20 m ) b F mg h Ag Ag ρρ Δ Δ= = = = × = .
48. Let ρ be the density of the cylinder (0.30 g/cm 3 or 300 kg/m 3 ) and ρ Fe be the density of the iron (7.9 g/cm 3 or 7900 kg/m 3 ). The volume of the cylinder is V c = (6 × 12) cm 3 = 72 cm 3 = 0.000072 m 3 , and that of the ball is denoted V b . The part of the cylinder that is submerged has volume V s = (4 × 12) cm 3 = 48 cm 3 = 0.000048 m 3 . Using the ideas of section 14-7, we write the equilibrium of forces as ρ gV c + ρ Fe gV b = ρ w gV s + ρ w gV b ¡ V b = 3.8 cm 3 where we have used ρ w = 998 kg/m 3 (for water, see Table 14-1). Using V b = 4 3 π r 3 we find r = 9.7 mm.

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54. (a) The equation of continuity provides (26 + 19 + 11) L/min = 56 L/min for the flow rate in the main (1.9 cm diameter) pipe. (b) Using v = R / A and A = π d 2 /4, we set up ratios: 2 56 2 26 56/ (1.9) / 4 1.0. 26/ (1.3) / 4 v v π =≈
62. (a) The volume of water (during 10 minutes) is () 2 3 11 15m s 10min 60s min 0.03m 6.4m . 4 Vv t A π §· == = ¨¸ ©¹ (b) The speed in the left section of pipe is 2 2 21 1 22 3.0cm 15m s 5.4m s. 5.0cm Ad vv v = = (c) Since 1 2 2 2 1 0 and , p vg h p h h h p p ρρ ++= ++ , which is the atmospheric pressure, 5 3 3 20 12 5 1.01 10 Pa 1.0 10 kg m 5.4m s 1.99 10 Pa 1.97atm. pp ρ ªº =+ = × + × ¬¼ = Thus, the gauge pressure is (1.97 atm – 1.00 atm) = 0.97 atm = 9.8 × 10 4 Pa.

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When the water level rises to height h 2 , just on the verge of flooding, 2 v , the speed of water in pipe M , is given by 2 12 2 2 1 ( ) 2 ( ) 13.86 m/s. 2 gh h v v ρρ −= ¡ =− = By continuity equation, the corresponding rainfall rate is 2 5 2 1 (0.030 m) (13.86 m/s) 2.177 10 m/s 7.8 cm/h. (30 m)(60 m) A vv A π §· == = × ¨¸ ©¹ 72. We use Bernoulli’s equation 22 11 1 2 2 2 p vg h p h ++= ++ .
78. To be as general as possible, we denote the ratio of body density to water density as f (so that f = ρ / w = 0.95 in this problem). Floating involves equilibrium of vertical forces acting on the body (Earth’s gravity pulls down and the buoyant force pushes up). Thus,

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## This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_10_Solutions - 20. To find the pressure at the...

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