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Final Exam Questions

  1. May 8, 2006 #1
    Okay, I am back for one more time. Last thread of the semster. Thanks for everyone's help for the other 3.


    Question 6

    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]


    How would you go about finding this problem

    I know torque = FrSinTheta
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 8, 2006 #2

    berkeman

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    Draw the diagram showing the force on the valve stem due to gravity. What is the angle between the force on the valve stem and the radius from the axle to the valve stem? Do you see why the answer to question 5 also involves this angle?
     
  4. May 8, 2006 #3
    If you know the equation for torque (a vector quantity), then all you have to do is find the magnitude.
     
  5. May 8, 2006 #4
    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

    Question 2, I dont get why it is true

    V ave for A = -20 m/s
    V ave For B = 10 m/s
    V ave For C = 40 m/s

    (10+-20+40)/(3) does not equal 0
     
    Last edited by a moderator: May 2, 2017
  6. May 8, 2006 #5
    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

    Question 8

    Change in Y is 1

    So what i was thinking of doing is

    Actually i have no idea

    I think the formula i should be using is V^2 = V0^2 + 2a(change in X)

    A = 9.8
    V0 = 12sin65
    so i found V to be 11.74

    Then i just used

    V = V0 + at
    which gave me .084 sec

    Any help, thanks
     
    Last edited by a moderator: May 2, 2017
  7. May 8, 2006 #6
    O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

    Question 21
     
    Last edited by a moderator: May 2, 2017
  8. May 8, 2006 #7

    Doc Al

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    Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned $1/hour. Are you sure that your average earning rate is (101)/2 = $50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

    The right way to find average velocity is displacement/time.
     
    Last edited by a moderator: May 2, 2017
  9. May 8, 2006 #8
    I want to thank you for all your help, I aced all the exams, just this final and i am done. Thanks
     
  10. May 8, 2006 #9

    Doc Al

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    Hint: What's the horizontal velocity and acceleration?
     
    Last edited by a moderator: May 2, 2017
  11. May 8, 2006 #10
    Horizontal Velocity = 12 Cos 65 = 5.0714

    Acc = -9.8

    What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that
     
  12. May 8, 2006 #11

    Doc Al

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    Here's one way to look at it. To get across as fast as possible, just swim straight across (with respect to the water, which will drag you downstream, but who cares?). How long does it take the water to drag the swimmer the given distance downstream? Since he swims across the river in that same time, what must be his swimming speed (with respect to the water)?
     
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  13. May 8, 2006 #12

    Doc Al

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    Right (in m/s).

    Wrong! (What's the horizontal acceleration?)

    What you do with it is ignore it. You don't need it. Focus on the horizontal component of the motion. You have the speed and the distance.
     
  14. May 8, 2006 #13
    am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed
     
  15. May 8, 2006 #14

    Doc Al

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    You can certainly use vectors, but that's more work than needed. I don't see a 45-45-90 triangle.
     
  16. May 8, 2006 #15
    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06 [Broken]

    Question 9

    So i got 1.97 Sec for 8

    so ithinkin of using X = X0 + V0T + .5 a T^2

    But that doesnt work
     
    Last edited by a moderator: May 2, 2017
  17. May 9, 2006 #16

    Doc Al

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    OK.

    Why not? It works for me. (Be sure you are using vertical components and proper signs.)
     
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  18. May 9, 2006 #17

    Okay so let's see if i got what you are saying

    Vave = X/T

    1.8 = 55/ T

    T = 30.555 sec

    V ave = 75 / 30.5555

    V ave = 2.5

    Is this the correct way?
     
    Last edited by a moderator: May 2, 2017
  19. May 9, 2006 #18

    Doc Al

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    Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)
     
  20. May 9, 2006 #19
    okay question 5 and 6

    http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06 [Broken]


    Was question 5 basically you know it or you don't type of thing?

    And free body digram would be

    Stem Valve Pointing down = mg = .245

    T = F r sin theta
     
    Last edited by a moderator: May 2, 2017
  21. May 9, 2006 #20

    So like (1.8^2) + (2.5^2) = X^2
     
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