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JenniferSmart1

  • 2 years ago

@Outkast3r09

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  1. Outkast3r09
    • 2 years ago
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    i see boxes =P

  2. JenniferSmart1
    • 2 years ago
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    Mechanical energy vertical displacent energy

  3. Outkast3r09
    • 2 years ago
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    umm? ok question?

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

  5. JenniferSmart1
    • 2 years ago
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    now we have potential energy correct?

  6. JenniferSmart1
    • 2 years ago
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    V?

  7. JenniferSmart1
    • 2 years ago
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    mgh?

  8. JenniferSmart1
    • 2 years ago
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    E= U + K

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

  10. JenniferSmart1
    • 2 years ago
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    C=Q/V?

  11. Outkast3r09
    • 2 years ago
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    what do you mean?

  12. Outkast3r09
    • 2 years ago
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    what happens to the energy in terms of mechanics?

  13. JenniferSmart1
    • 2 years ago
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    yes

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

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

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

  17. Outkast3r09
    • 2 years ago
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    since you end up with no kinetic or potential?

  18. Outkast3r09
    • 2 years ago
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    oh it's lifting the object

  19. JenniferSmart1
    • 2 years ago
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    yes

  20. JenniferSmart1
    • 2 years ago
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    sorry

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

  22. JenniferSmart1
    • 2 years ago
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    sounds right

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

  24. JenniferSmart1
    • 2 years ago
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    yes battery

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

  26. Outkast3r09
    • 2 years ago
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    alright

  27. JenniferSmart1
    • 2 years ago
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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?

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

  29. JenniferSmart1
    • 2 years ago
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    how can we get C? where is work? LOL

  30. Outkast3r09
    • 2 years ago
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    what about work?

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

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

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

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

  35. JenniferSmart1
    • 2 years ago
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    sure

  36. Outkast3r09
    • 2 years ago
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    is that yes or no lol

  37. JenniferSmart1
    • 2 years ago
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    yes?

  38. JenniferSmart1
    • 2 years ago
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    potential energy

  39. Outkast3r09
    • 2 years ago
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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

  40. Outkast3r09
    • 2 years ago
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    also dr = h

  41. Outkast3r09
    • 2 years ago
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    ;)

  42. Outkast3r09
    • 2 years ago
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    W=Fd=Fh \[mgh+Fh=

  43. Outkast3r09
    • 2 years ago
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    \[h(mg+F)\]?

  44. JenniferSmart1
    • 2 years ago
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    we dont have to integrate i'm assuming

  45. Outkast3r09
    • 2 years ago
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    no it's straight up

  46. Outkast3r09
    • 2 years ago
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    alright

  47. Outkast3r09
    • 2 years ago
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    now we need the work done

  48. JenniferSmart1
    • 2 years ago
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    ?

  49. Outkast3r09
    • 2 years ago
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    yes isn't that what it's asking lol

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

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

  52. JenniferSmart1
    • 2 years ago
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    E= U+0

  53. Outkast3r09
    • 2 years ago
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    ok so E is actually your capacitor right?

  54. JenniferSmart1
    • 2 years ago
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    I hope so....let's say it is

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

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

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

  58. Outkast3r09
    • 2 years ago
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    so all energy is coming from the capacitor

  59. Outkast3r09
    • 2 years ago
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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

  60. JenniferSmart1
    • 2 years ago
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    what's W app? and W d?

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

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

  63. Outkast3r09
    • 2 years ago
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    if it isn't that much don't worry about it

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

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

  66. Outkast3r09
    • 2 years ago
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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

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

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