PHY303K-Spring07-Exam4-Moore

# PHY303K-Spring07-Exam4-Moore - midterm 04 SAENZ LORENZO Due...

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midterm 04 – SAENZ, LORENZO – Due: May 2 2007, 11:00 pm 1 Question 1 part 1 of 1 10 points The figure shows two wave pulses that are approaching each other. P Q Which of the following best shows the shape of the resultant pulse when the centers of the pulses, points P and Q , coincide? 1. correct 2. 3. 4. 5. Explanation: Notice that the two pulses have the same width and amplitude. Choosing the point P (the same as point Q when the two pulses coincide) as the origin, the two pulses can be described as: P : y 1 = A , d x d Q : y 2 = braceleftBigg A , d x < 0 A , 0 < x < d Using the principle of superposition, the re- sultant pulse is y = y 1 + y 2 = braceleftbigg 2 A , d x < 0 0 , 0 < x < d bracerightbigg . P Q P + Q Question 2 part 1 of 1 10 points The small piston of a hydraulic lift has a cross-sectional area of 1 . 9 cm 2 and the large piston has an area of 29 cm 2 , as in the figure below. 44 kN F 1 . 9 cm 2 area 29 cm 2 What force F must be applied to the small piston to raise a load of 44 kN? 1. 2515 . 07 N 2. 2594 . 12 N 3. 2676 . 92 N 4. 2774 . 75 N 5. 2882 . 76 N correct 6. 2975 N 7. 3075 N 8. 3170 . 67 N

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midterm 04 – SAENZ, LORENZO – Due: May 2 2007, 11:00 pm 2 9. 3278 . 87 N 10. 3384 . 62 N Explanation: Let : A 1 = 1 . 9 cm 2 , A 2 = 29 cm 2 , W = 44 kN , and F = F . According to Pascal’s law, the pressure ex- erted on A 1 must be equal to the one exerted on A 2 . The pressure P 1 = F A 1 must be equal to the pressure P 2 = W A 2 due to the load. F A 1 = W A 2 , so F = A 1 A 2 W = (1 . 9 cm 2 ) (29 cm 2 ) (44000 N) = 2882 . 76 N . Question 3 part 1 of 1 10 points Assume: The buoyant force of the air is negligible. Given: The density of the oil is 867 kg / m 3 . The density of the water salt mixture is 1163 kg / m 3 . A rectangular parallelepiped block of uni- form density floats in a container which con- tains oil and salt water as shown. The block sticks up above the oil by a distance 30 cm. The oil thickness is 40 cm. The block’s depth in the salt water is 20 cm. The horizontal area of the block is 0 . 03 m 2 . 40 cm 20 cm 30 cm M Calculate the mass of the block. 1. 13 . 96 kg 2. 14 . 409 kg 3. 14 . 862 kg 4. 15 . 344 kg 5. 15 . 836 kg 6. 16 . 328 kg 7. 16 . 857 kg 8. 17 . 382 kg correct 9. 17 . 922 kg 10. 18 . 504 kg Explanation: Let : d 1 = 30 cm = 0 . 3 m , d 2 = 40 cm = 0 . 4 m , d 3 = 20 cm = 0 . 2 m , and A = 0 . 03 m 2 , The mass of the salt water displaced is M 3 = ρ w d 3 A = (1163 kg / m 3 ) (0 . 2 m) (0 . 03 m 2 ) = 6 . 978 kg . The mass of the oil displaced is M 2 = ρ o d 2 A = (867 kg / m 3 ) (0 . 4 m) (0 . 03 m 2 ) = 10 . 404 kg . Since the weight of the block M block g is equal to the weight of the displaced liquid, we have M block g = [ M 2 + M 3 ] g
midterm 04 – SAENZ, LORENZO – Due: May 2 2007, 11:00 pm 3 M block = M 2 + M 3 = (10 . 404 kg) + (6 . 978 kg) = 17 . 382 kg . Question 4 part 1 of 1 10 points Note: P atm = 101300 Pa / atm. The viscos- ity of the fluid is negligible and the fluid is incompressible.. A liquid of density 1390 kg / m 3 flows with speed 1 . 4 m / s into a pipe of diameter 0 . 12 m. The diameter of the pipe decreases to 0 . 05 m at its exit end. The exit end of the pipe is 7 . 81 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 . 1 atm.

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