ch13 - CHAPTER 13 Fluids 1 A copper cylinder is 6 cm long...

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CHAPTER 13 Fluids 1* · A copper cylinder is 6 cm long and has a radius of 2 cm. Find its mass. Find the volume, then m = r V V = p r 2 h = 75.4 × 10 - 6 m 3 ; m = 8.93 × 10 3 V = 0.673 kg 2 · Find the mass of a lead sphere of radius 2 cm. Find the volume, then m = r V V = 4 p r 3 /3 = 3.35 × 10 - 5 m 3 ; m = 11.3 × 10 3 V = 0.379 kg 3 · Find the mass of air in a room 4 m × 5 m × 4 m. m = r V m = (1.293 × 80) kg = 103 kg 4 · A solid oak door is 200 cm high, 75 cm wide, and 4 cm thick. How much does it weigh? w = mg = r Vg ; oak is dense wood, take r = 0.9 × 10 3 w = (0.9 × 10 3 × 2 × 0.75 × 0.04 × 9.81) N = 530 N 5* ·· A 60-mL flask is filled with mercury at 0 o C (Figure 13-22). When the temperature rises to 80 o C, 1.47 g of mercury spills out of the flask. Assuming that the volume of the flask is constant, find the density of mercury at 80 o C if its density at 0 o C is 13,645 kg/m 3 . 1. Write r in terms of r 0 , V , and m 2. Evaluate r m = r 0 V ; m = m - m = r V ; r = r 0 - m / V r = 13,621 kg/m 3 6 · If the gauge pressure is doubled, the absolute pressure will be ( a ) halved. ( b ) doubled. ( c ) unchanged. ( d ) squared. ( e ) Not enough information is given to determine the effect. ( e ) 7 · Barometer readings are commonly given in inches of mercury. Find the pressure in inches of mercury equal to 101 kPa. See Equ. 13-9 101 kPa = 29.8 inHg 8 · The pressure on the surface of a lake is atmospheric pressure P at = 101 kPa. ( a ) At what depth is the pressure twice atmospheric pressure? ( b ) If the pressure at the top of a deep pool of mercury is P at , at what depth is the
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Chapter 13 Fluids pressure 2 P at ? ( a ) P = P at + r gh = 2 P at ; h = P at / r g ( b ) h Hg = P at / r Hg g h = [1.01 × 10 5 /(1 × 10 3 × 9.81)] m = 10.3 m h Hg = 10.3/13.6 m = 75.7 cm 9* · Find ( a ) the absolute pressure and ( b ) the gauge pressure at the bottom of a swimming pool of depth 5.0 m. ( a ) P = P at + r gh ( b ) P gauge = P - P at P = (1.01 × 10 5 + 9.81 × 10 3 × 5) = 1.5 × 10 5 N/m 2 = 1.5 atm P gauge = 0.5 atm 10 · When a woman in high heels takes a step, she momentarily places her entire weight on one heel of her shoe, which has a radius of 0.4 cm. If her mass is 56 kg, what is the pressure exerted on the floor by her heel? P = mg / A = mg / p r 2 P = [56 × 9.81/( p × 16 × 10 - 6 )] N/m 2 = 1.09 × 10 7 N/m 2 » 100 atm 11 · A hydraulic lift is used to raise an automobile of mass 1500 kg. The radius of the shaft of the lift is 8 cm and that of the piston is 1 cm. How much force must be applied to the piston to raise the automobile? Use Equ. 13-3 F = [1500 × 9.81 × (1/8) 2 ] N = 230 N 12 · Blood flows into the aorta through a circular opening of radius 0.9 cm. If the blood pressure is 120 torr, how much force must be exerted by the heart? F
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This note was uploaded on 01/23/2011 for the course PHYS 122 taught by Professor Cramer during the Winter '08 term at University of Washington.

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ch13 - CHAPTER 13 Fluids 1 A copper cylinder is 6 cm long...

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