PHY303K-Spring07-Exam4-Moore

PHY303K-Spring07-Exam4-Moore - midterm 04 – SAENZ LORENZO...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 < − A , < x < d Using the principle of superposition, the re- sultant pulse is y = y 1 + y 2 = braceleftbigg 2 A , − d ≤ x < , < 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 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 ....
View Full Document

This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 16

PHY303K-Spring07-Exam4-Moore - midterm 04 – SAENZ LORENZO...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online