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IsTim

  • 4 years ago

A marble rolls off a table with a velocity of 1.93m/s[horizontally]. The tabletop is 76.5cm above the floor. If air resistance is negligible, determine the velocity at impact

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  1. IsTim
    • 4 years ago
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    Apparently I misinterpreted my question. So now I got a problem.

  2. TuringTest
    • 4 years ago
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    is it 4.3

  3. IsTim
    • 4 years ago
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    I took this as a regular acceleration question, and used \[V _{2}^{2}=v _{1^{2}}+2ad\]

  4. TuringTest
    • 4 years ago
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    good...

  5. anonymous
    • 4 years ago
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    how is it rolled off

  6. IsTim
    • 4 years ago
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    Yeah, its 4.33.How?

  7. IsTim
    • 4 years ago
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    No other info given. Previous questions tho.

  8. IsTim
    • 4 years ago
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    Time is 0.396, and displacement is .763m.

  9. TuringTest
    • 4 years ago
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    solve what you have for V2 V1 is zero the number you get is the y-component of the velocity

  10. TuringTest
    • 4 years ago
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    the x-component does not change

  11. IsTim
    • 4 years ago
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    Im suppose to break it up into x and y components?

  12. IsTim
    • 4 years ago
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    I got 5.799m/s (down)

  13. TuringTest
    • 4 years ago
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    they are already broken up\[V_x=1.93\]because it rolls of horizontally, and that component is constant because no force acts in that direction. The vertical component is\[V_y=\sqrt{2gh}\]and the total velocities magnitude is\[|V|=\sqrt{V_x^2+V_y^2}\]

  14. TuringTest
    • 4 years ago
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    rolls off*

  15. anonymous
    • 4 years ago
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    reaction force?

  16. IsTim
    • 4 years ago
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    How did you get such formulas? Both 2?

  17. TuringTest
    • 4 years ago
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    like I said, the horizontal component is constant because there is no force acting on the object in the x-direction, so the horizontal velocity is the 1.93m/s it rolled of the table with

  18. anonymous
    • 4 years ago
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    can angular rotation come into play?

  19. IsTim
    • 4 years ago
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    I'm not that advd yet.

  20. anonymous
    • 4 years ago
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    okies

  21. TuringTest
    • 4 years ago
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    the y-component is what your formula leads to for the up-down portion of the motion if you recognize that the initial downward velocity is zero\[\large v_y^2=v_{yi}^2+2gh=(0)^2+2gh\to v_y=\sqrt{2gh}\]

  22. TuringTest
    • 4 years ago
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    @mth none of that stuff is necessary so why use it?

  23. anonymous
    • 4 years ago
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    if air resistance negligable gravity?

  24. IsTim
    • 4 years ago
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    Gravity applies.

  25. TuringTest
    • 4 years ago
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    gravity is in my formulation

  26. anonymous
    • 4 years ago
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    on the rolling of the marble

  27. IsTim
    • 4 years ago
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    So I rearrange V22=v12+2ad

  28. IsTim
    • 4 years ago
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    Apparently I can't copy equations.

  29. anonymous
    • 4 years ago
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    i was asking

  30. TuringTest
    • 4 years ago
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    V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h leads to Vy2=2gh

  31. anonymous
    • 4 years ago
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    so potential energy does not play in part

  32. IsTim
    • 4 years ago
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    im doing potential energy later. i hope.

  33. TuringTest
    • 4 years ago
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    V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h *sorry got shifted around

  34. IsTim
    • 4 years ago
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    So my result is 5.79m/s. What's my next course of action.

  35. TuringTest
    • 4 years ago
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    where did you get that number?

  36. anonymous
    • 4 years ago
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    sorry thats all i can do :( my math sucks i like theoretics much better

  37. IsTim
    • 4 years ago
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    (1.93^3+2(9.8)(0.763))^1/2

  38. IsTim
    • 4 years ago
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    IT's alright mth.

  39. IsTim
    • 4 years ago
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    I re-did it, and got 4.32.

  40. TuringTest
    • 4 years ago
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    I keep trying to tell you that the equations are separate Vx=1.93 Vy=(2*9.890.763)^1/2 you have Vx plugged in where the initial speed goes, which is zero.

  41. IsTim
    • 4 years ago
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    Oh. Shoot.

  42. TuringTest
    • 4 years ago
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    if you wanted both those things in the same equation it would have to be a vector formula, and I don't think you've done parametric stuff yet.

  43. IsTim
    • 4 years ago
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    this time i got 3.87

  44. TuringTest
    • 4 years ago
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    \[V_x=1.93\]\[V_y=\sqrt{2gh}\]\[|V|=\sqrt{V_x^2+V_y^2}=\sqrt{1.93^2+2(9.8)(0.765)}\approx4.33\]I'm going to sleep here so I gotta wrap it up...

  45. IsTim
    • 4 years ago
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    Pythagorean Theorem. Got it.

  46. TuringTest
    • 4 years ago
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    |dw:1327908694189:dw|Vy changes independently from Vx. That is a handy thing to know. Good luck!

  47. IsTim
    • 4 years ago
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    I got 4.32.

  48. TuringTest
    • 4 years ago
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    4.326 ?

  49. IsTim
    • 4 years ago
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    Ok. Thank yo uvery much.

  50. IsTim
    • 4 years ago
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    4.3218

  51. IsTim
    • 4 years ago
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    (3.867^2+1.93^2)^1/2

  52. TuringTest
    • 4 years ago
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    \[\sqrt{1.93^2+2(9.8)(0.765)}\]is what I have in my calculator

  53. IsTim
    • 4 years ago
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    Oops. Instead of .765, I used .763.

  54. TuringTest
    • 4 years ago
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    Cheers :D

  55. IsTim
    • 4 years ago
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    Yup. Thanks, if you ever decide to re-visit.

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