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JenniferSmart1

@Outkast3r09

  • one year ago
  • one year ago

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  1. Outkast3r09
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    i see boxes =P

    • one year ago
  2. JenniferSmart1
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    Mechanical energy vertical displacent energy

    • one year ago
  3. Outkast3r09
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    umm? ok question?

    • one year ago
  4. JenniferSmart1
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    I've been up since 5 am :'( can't think anymore. what happens to energy in this situation?

    • one year ago
  5. JenniferSmart1
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    now we have potential energy correct?

    • one year ago
  6. JenniferSmart1
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    V?

    • one year ago
  7. JenniferSmart1
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    mgh?

    • one year ago
  8. JenniferSmart1
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    E= U + K

    • one year ago
  9. JenniferSmart1
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    \[\frac 12 mv^2\]

    • one year ago
  10. JenniferSmart1
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    C=Q/V?

    • one year ago
  11. Outkast3r09
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    what do you mean?

    • one year ago
  12. Outkast3r09
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    what happens to the energy in terms of mechanics?

    • one year ago
  13. JenniferSmart1
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    yes

    • one year ago
  14. Outkast3r09
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    well lets see it starts with a potential energy and no kinetic

    • one year ago
  15. JenniferSmart1
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    we're only looking at it before and after it finished moving, so: no K I think

    • one year ago
  16. Outkast3r09
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    so what happened to the energy you are trying to ask?

    • one year ago
  17. Outkast3r09
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    since you end up with no kinetic or potential?

    • one year ago
  18. Outkast3r09
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    oh it's lifting the object

    • one year ago
  19. JenniferSmart1
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    yes

    • one year ago
  20. JenniferSmart1
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    sorry

    • one year ago
  21. Outkast3r09
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    the capacitor i'm assuming is using the energy to do work , and potential

    • one year ago
  22. JenniferSmart1
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    sounds right

    • one year ago
  23. JenniferSmart1
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    so C=Q/V voltage from where? battery? outlet?

    • one year ago
  24. JenniferSmart1
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    yes battery

    • one year ago
  25. Outkast3r09
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    where rae you getting it? lol outlet and battery are different lol

    • one year ago
  26. Outkast3r09
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    alright

    • one year ago
  27. JenniferSmart1
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    I guess I have to relate \[C=QV\] to \[U=\frac 12 CV^2\] to \[mgh\] what are we solving for? the displacement?

    • one year ago
  28. JenniferSmart1
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    we know V \[mgh=\frac 12 CV^2\] we know mg and V

    • one year ago
  29. JenniferSmart1
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    how can we get C? where is work? LOL

    • one year ago
  30. Outkast3r09
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    what about work?

    • one year ago
  31. JenniferSmart1
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    \[W=\int F\cdot dr\] or E dr

    • one year ago
  32. JenniferSmart1
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    http://en.wikipedia.org/wiki/Work_%28electrical%29

    • one year ago
  33. JenniferSmart1
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    http://en.wikipedia.org/wiki/Work_%28electrical%29 work and potential energy!!!!

    • one year ago
  34. Outkast3r09
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    would work be electrical in this case though since it's mechanical work?

    • one year ago
  35. JenniferSmart1
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    sure

    • one year ago
  36. Outkast3r09
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    is that yes or no lol

    • one year ago
  37. JenniferSmart1
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    yes?

    • one year ago
  38. JenniferSmart1
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    potential energy

    • one year ago
  39. Outkast3r09
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    So what is the actual question... i haven't actually gotten a question.. I've gotten 2 formulas and a procedure it sounds like lol

    • one year ago
  40. Outkast3r09
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    also dr = h

    • one year ago
  41. Outkast3r09
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    ;)

    • one year ago
  42. Outkast3r09
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    W=Fd=Fh \[mgh+Fh=

    • one year ago
  43. Outkast3r09
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    \[h(mg+F)\]?

    • one year ago
  44. JenniferSmart1
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    we dont have to integrate i'm assuming

    • one year ago
  45. Outkast3r09
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    no it's straight up

    • one year ago
  46. Outkast3r09
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    alright

    • one year ago
  47. Outkast3r09
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    now we need the work done

    • one year ago
  48. JenniferSmart1
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    ?

    • one year ago
  49. Outkast3r09
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    yes isn't that what it's asking lol

    • one year ago
  50. Outkast3r09
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    so you're lifting the mass meaning there is work being done = energy being used to lift it

    • one year ago
  51. JenniferSmart1
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    I'm tired dude.... ok so we used up E

    • one year ago
  52. JenniferSmart1
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    E= U+0

    • one year ago
  53. Outkast3r09
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    ok so E is actually your capacitor right?

    • one year ago
  54. JenniferSmart1
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    I hope so....let's say it is

    • one year ago
  55. Outkast3r09
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    Alright so E is equation 24-8 in page you showed me

    • one year ago
  56. Outkast3r09
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    since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]

    • one year ago
  57. Outkast3r09
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    \[mg(0)+\frac{1}{2}m(0^2)+U_c\]

    • one year ago
  58. Outkast3r09
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    so all energy is coming from the capacitor

    • one year ago
  59. Outkast3r09
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    now the other side you have \[mgh+\frac{1}{2}mv^2+W_{app}+W_d\]i'm assuming drag will be really low so it's not taken into account

    • one year ago
  60. JenniferSmart1
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    what's W app? and W d?

    • one year ago
  61. Outkast3r09
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    but since we're looking at it at rest in two places you have \[mgh+0+Fh\] \[U_c=mgh+Fh=(mg+F)h\]

    • one year ago
  62. Outkast3r09
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    this is what i'm thinking but i'm only like 50% certain lol

    • one year ago
  63. Outkast3r09
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    if it isn't that much don't worry about it

    • one year ago
  64. Outkast3r09
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    \[F=ma\] \[(mg+ma)h=(m(g+a))h\]

    • one year ago
  65. Outkast3r09
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    \[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol

    • one year ago
  66. Outkast3r09
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    W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

    • one year ago
  67. Outkast3r09
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    that's sort of a toughie question... we haven't gotten to capacitors and such yet

    • one year ago
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