# hw23a - aldawsari(ma28683 – HW23 – Betancourt –(16856...

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Unformatted text preview: aldawsari (ma28683) – HW23 – Betancourt – (16856) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 18.1-18.5 001 (part 1 of 2) 10.0 points An air bubble originating from a under-water diver has a radius of 6 mm at some depth h . When the bubble reaches the surface of the water, it has a radius of 7 . 9 mm. Assuming the temperature of the air in the bubble remains constant, determine the depth h of the diver. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 13 . 2185 m. Explanation: Let : ρ = 1000 kg / m 3 , g = 9 . 8 m / s 2 , P atm = 1 . 01 × 10 5 Pa , R 1 = 7 . 9 mm = 0 . 0079 m , and R 2 = 6 mm = 0 . 006 m . From the ideal gas law P V = nRT = constant , so P 1 V 1 = P 2 V 2 P 1 P 2 = V 2 V 1 P atm P atm + ρg h = 4 3 π R 3 2 4 3 π R 3 1 = R 3 2 R 3 1 R 3 1 P atm = R 3 2 ( P atm + ρg h ) . h = P atm ( R 3 1- R 3 2 ) ρg R 3 2 = 1 . 01 × 10 5 Pa (1000 kg / m 3 ) (9 . 8 m / s 2 ) × (0 . 0079 m) 3- (0 . 006 m) 3 (0 . 006 m) 3 = 13 . 2185 m . 002 (part 2 of 2) 10.0 points Determine the absolute pressure at this depth. Correct answer: 230 . 541 kPa. Explanation: The absolute pressure is P = P atm + ρg h = bracketleftBig 1 . 01 × 10 5 Pa + ( 1000 kg / m 3 )( 9 . 8 m / s 2 ) × (13 . 2185 m) bracketrightBig × 1 kPa 1000 Pa = 230 . 541 kPa ....
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## This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

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hw23a - aldawsari(ma28683 – HW23 – Betancourt –(16856...

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