{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test4 - midterm 04 BOBBITT JAMES Due May 2 2007 11:00 pm...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
midterm 04 – BOBBITT, JAMES – Due: May 2 2007, 11:00 pm 1 Version number encoded for clicker entry: V1:1, V2:1, V3:2, V4:3, V5:1. Question 1, chap 18, sect 4. part 1 of 1 10 points The small piston of a hydraulic lift has a cross-sectional area of 4 . 2 cm 2 and the large piston has an area of 30 cm 2 , as in the figure below. 39 kN F 4 . 2 cm 2 area 30 cm 2 What force F must be applied to the small piston to raise a load of 39 kN? 1. 5061 . 22 N 2. 5279 . 49 N 3. 5460 N correct 4. 5647 . 06 N 5. 5865 N 6. 6057 . 45 N 7. 6249 . 25 N 8. 6446 . 51 N 9. 6658 . 33 N 10. 6875 N Explanation: Let : A 1 = 4 . 2 cm 2 , A 2 = 30 cm 2 , W = 39 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 = (4 . 2 cm 2 ) (30 cm 2 ) (39000 N) = 5460 N . Question 2, chap 17, sect 3. part 1 of 1 10 points This picture shows the displacements S of the air molecules in a traveling sound wave as a function of distance, x . + A −A S 0 λ/ 2 λ 3 λ/ 2 2 λ Sound Wave Which of the following tubes, open at the left end and open at the right end, is closest to the right length so as to resonate at its fundamental frequency when placed in this sound wave? 1. = 1 2 λ correct 2. = 1 4 λ 3. = 3 8 λ 4. = 1 8 λ 5. = 3 4 λ 6. = 3 2 λ
Image of page 1

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

View Full Document Right Arrow Icon
midterm 04 – BOBBITT, JAMES – Due: May 2 2007, 11:00 pm 2 7. = λ 8. = 2 λ Explanation: The tube is open at the left end requiring an antinode in the wave form. The tube is open at the right end requiring an antinode in the wave form. The displaced between the antinode on the left and the antinode on the right is shown in the figure below. This correspondance can be represented by the distance between = 1 4 λ and = 3 4 λ as shown. + A −A S 0 λ/ 2 λ 3 λ/ 2 2 λ Sound Wave So the length of a tube to produce the fundamental resonance should be = 1 2 λ . Second of four versions. Question 3, chap 16, sect 2. part 1 of 1 10 points A harmonic wave y = A sin[ k x ω t φ ] , where A = 1 meter, k has units of m 1 , ω has units of s 1 , and φ has units of radians, is plotted in the diagram below. +1 1 A (meters) x (meters) 6 12 18 At the time t = 0 Which wave function corresponds best to the diagram? 1. y = A sin bracketleftbigg parenleftbigg 2 π 15 m parenrightbigg x ω t parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 2. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x ω t parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 3. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x ω t parenleftbigg 4 π 3 parenrightbiggbracketrightbigg 4. y = A sin bracketleftbigg parenleftbigg 2 π 9 m parenrightbigg x ω t parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 5. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x ω t parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 6. y = A sin bracketleftbigg parenleftbigg 2 π 9 m parenrightbigg x ω t parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 7. y = A sin bracketleftbigg parenleftbigg 2 π 15 m parenrightbigg x
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern