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Thread: NWS **THE OFFCIAL IA RECORD LENGTH THREAD** STARTED BY THE IA OLD MAN!!

  1. #3601
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    if u compress the spring there will be a force that developes and the potential energy will increase then when released the kinetic energy will increase.

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  2. #3602
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    if we know how much the spring is compressed then we can find the potential energy and restoring force. This would be difficult to calculate using kinematics b/c the restoring force will constantly change

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  3. #3603
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    So we need a relationship between the acceleration of the spring and the position(amount of compression)

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  4. #3604
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    it is possible to solve using f=ma=-(k*delta S)=m(ds/dt)

    but its very difficult

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  5. #3605
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    if the spring has negligible mass then the alternate method would be...


    conservation of momentum(P)

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  6. #3606
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    Quote Originally Posted by Sinfix_15 View Post
    You travel with so much luggage that it wont fit in a wagon? you dating a kardashian?

  7. #3607
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    The third method is through the following:

    Energy conservation

    J is the impulse=delta P

    F(in the S direction)=ma(in the S direction)=m(dv/dt)=m(dv/ds)(ds/dt)

    which changes variables to position, so..

    F=mv(dv/ds)

    integrate from initial S to final S, which are velocities of F with respect to position

    =integral mv*dv

    =delta K

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  8. #3608
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    Quote Originally Posted by phatboislim
    OOOooo0o0o thats decieving, I'm in class don't distract me!

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  9. #3609
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    dont be posting this educational BULLCRAP on this thread...i'm not in school muthafuccer lol



    Quote Originally Posted by Sinfix_15 View Post
    You travel with so much luggage that it wont fit in a wagon? you dating a kardashian?

  10. #3610
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    delta K= - integral k(s-s(@eq))ds = -integral k(delta S)d(delta S) = -k((delta S)^2)/2 evaluated from initial delta s to final

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  11. #3611
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    Quote Originally Posted by phatboislim
    dont be posting this educational BULLCRAP on this thread...i'm not in school muthafuccer lol
    well I am and you'll feel my pain

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  12. #3612
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    weaksos



    Quote Originally Posted by Sinfix_15 View Post
    You travel with so much luggage that it wont fit in a wagon? you dating a kardashian?

  13. #3613
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    K(final)+.5k(delta S final)^2=K(initial)+.5k(delta S initial)^2

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  14. #3614
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    lol@weaksos

    Time for the PRS(personal response System) question of the day!

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  15. #3615
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    Question:

    I am

    (1) Here in body & spirit
    (2) Here body but @ Spring break in spirit
    (3) I am operating someone else's PRS who is on spring break

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  16. #3616
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    5 people chose 3

    I chose 1

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  17. #3617
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    U(potential energy)=1/2 k(delta S)^2

    zero of PE delta S=0, s=s(@eq)

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  18. #3618
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    U becomes larger as you travel farther from the equilibrium position

    Stick and Slip ex.:

    you have a force pulling on a mass connected to another mass via a spring.

    As you pull the block to the right, the tension in the spring is increasing, when it reaches the max, then the block will move.

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  19. #3619
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    such that f max=mu(static)*N=kx

    mu(static) is the coefficient of static friction

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  20. #3620
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    Brett (One of the true OG's, No really... ask anyone)
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    www.facebook.com/brett.lowenthal1

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  21. #3621
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    Conservation of energy:

    two objects with masses m1 and m2

    and there is a spring between them

    they collide and compress the spring to a maximum extent and then the particles move away from each other at different velocities

    If I know the V1i and V2i (which are in opp directions)

    how fast are they moving during this collision?

    after the spring reaches eq pos. how fast are the particles moving?

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  22. #3622
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    so long as the spring is mass-less then u can calculate the speed before the collision

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  23. #3623
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    The max compression can be calculated by first finding the sum of the potential energy

    K1+K2+0(the elastic potential engergy before collision)

    DO NOT set this = to .5k(delta S)^2

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  24. #3624
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    this is wrong based on energy conservation

    at the collision the two objects become one and move together, so the potential energy before and after compare as follows:

    Pi=Pf
    m1(v1)i+m2(v2)i=(m1+m2)V

    so...

    K1+K2+0=.5k(delta S max)^2+.5m1*V^2+.5m2*V^2

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  25. #3625
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    KB with the whoring....jesus...and what the hell...i dont know what u are talking about. good thing u have the 30 sec rule lol.
    The Carbon Fibered R6

  26. #3626
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    alrigh /physics lesson on to the library!

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  27. #3627
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    im up in this bishh!!!!! .......1

  28. #3628
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    3628
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  29. #3629
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    ^^11,260..... my # is higher
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  30. #3630
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    whew, at the library now

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  31. #3631
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    Im fueling all those stalkers out there with all this location info im giving out

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  32. #3632
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    i have ur IP addy....
    The Carbon Fibered R6

  33. #3633
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    alright checking notebook for notebook check @ noon

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  34. #3634
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    Quote Originally Posted by TRYMY4.0
    i have ur IP addy....
    hook that up to your GPS and stalk like a pro!

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  35. #3635
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    *inserts flash drive*

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  36. #3636
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    8gigs! Fry's FTMFW!

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  37. #3637
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    *loading autocad*

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  38. #3638
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    It has to "re-install" every time you logon to a school computer which is teh ghey

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  39. #3639
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    *logs into Tsquare*

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  40. #3640
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    nm I have to check my email for my corrected DWFs, *switches over to email*

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