Physics HW9

# Physics HW9 - kim(jkk547 – HW09 – Tsoi –(58020 1 This...

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Unformatted text preview: kim (jkk547) – HW09 – Tsoi – (58020) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An incompressible, non-viscous liquid of den- sity ρ flows with speed v u into a pipe of diam- eter d u . The diameter of the pipe decreases to d y at its exit end. The elevation of the en- trance is h above the elevation of the exit end of the pipe. The pressure at the exit of the pipe is P y . P u v u d u P y v y d y h What is the velocity v y of the liquid flowing out of the exit end of the pipe? 1. v y = parenleftbigg A u d 2 y parenrightbigg v u 2. v y = parenleftbigg A u d y parenrightbigg v u 3. v y = parenleftbigg A y d u parenrightbigg v u 4. v y = parenleftbigg A y d 2 u parenrightbigg v u 5. v y = parenleftbigg d u d y parenrightbigg v u 6. v y = parenleftBigg d 2 y d 2 u parenrightBigg v u 7. v y = parenleftbigg d 2 u d 2 y parenrightbigg v u 8. v y = parenleftbigg d y d u parenrightbigg v u 9. None of these 002 (part 2 of 2) 10.0 points Note: P atm is atmospheric pressure. Applying Bernoulli’s principle, what is the pressure P u at the entrance end of the pipe? 1. P u = P y + ρ g h 2. P u = P y + ρ g h + 1 2 ρ ( v u 2- v y 2 ) 3. P u = P y- ρ g h 4. P u = P y- ρ g h + 1 2 ρ ( v u 2- v y 2 ) 5. P u = ρ g h + 1 2 ρ ( v u 2- v y 2 ) 6. P u = P y + ρ g h + 1 2 ρ ( v y 2- v u 2 ) 7. P u = P y- P atm- ρ g h + 1 2 ρ ( v u 2- v y 2 ) 8. P u = P y- ρ g h + 1 2 ρ ( v y 2- v u 2 ) 9. P u =- ρ g h + 1 2 ρ ( v y 2- v u 2 ) 10. P u = P y- P atm + ρ g h + 1 2 ρ ( v y 2- v u 2 ) 003 10.0 points10....
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Physics HW9 - kim(jkk547 – HW09 – Tsoi –(58020 1 This...

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