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mayankdevnani Group Title

A ball is dropped on to the floor from a height of 10 m. It reboundes to a height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact? a) \[1400 m/s^2\] b) \[2100 m/s^2\] c) \[700 m/s^2\] d) \[2800 m/s^2\]

  • 2 years ago
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  1. mayankdevnani Group Title
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    @amistre64 @hba @asnaseer @AccessDenied @jhonyy9 @jazy @ganeshie8

    • 2 years ago
  2. tictac4456 Group Title
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    The acceleration of a body is defined as the change of its velocity during an interval of time divided by the interval of time. Expressed algebraically, a = (Vo - Vf)/t where a = the acceleration, Vo = the initial velocity, Vf = the final velocity and t = the interval of time.

    • 2 years ago
  3. tictac4456 Group Title
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    From the basic a = (Vo - Vf) we derive the first as ......................................t ..............................Vf = Vo + at (the acceleration is assumed constant)

    • 2 years ago
  4. tictac4456 Group Title
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    The second defines the relationship between the distance traveled by a body exposed to a constant acceleration. The distance traveled over an interval of time derives from the product of the average velocity during the time interval and the time interval. The average velocity is (Vo + Vf)/2 and the time interval "t" which yields d = (Vo + Vf)t/2. Substituting the Vf derived above, d = [Vo + (Vo + at)]t/2 resulting in ...............................d = Vo(t) + a(t^2) (quite often, s is used in place of d) ......................................................2

    • 2 years ago
  5. tictac4456 Group Title
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    The third combines the first two by eliminating the time interval "t". Therefore, a(d) = (Vo - Vf)(Vo + Vf)t = (1/2)(Vf - Vo)(Vf + Vo) resulting in .....................................t.............2 ..............................Vf^2 = Vo^2 + 2ad

    • 2 years ago
  6. mayankdevnani Group Title
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    thnx.... @tictac4456

    • 2 years ago
  7. tictac4456 Group Title
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    no problem do u go to connexus

    • 2 years ago
  8. mayankdevnani Group Title
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    is it right?? While falling down, initial velocity = 0, final velocity = v, acceleration due to gravity = 9.8 m/s 2 , height = 10 m The velocity of the ball with which it hits the ground can be found using, v 2 = u 2 + 2as => v 2 = 0 + 2×9.8×10 => v = 14 m/s (downward) While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s2 [this time the velocity and acceleration are oppositely directed so we have the negative sign] , height = 2.5 m The velocity of the ball with which it leaves the ground can be found using, v 2 = u 2 + 2as => 0 = u 2 - 2×9.8×2.5 => u = 7 m/s (upward) So, Change in velocity = 7 m/s (upward) – 14 m/s (downward) => Δv = 7 m/s + 14 m/s [both in upward direction] => Δv = 21 m/s So, acceleration is, a = Δv/Δt = 21/0.01 = 2100 m/s 2

    • 2 years ago
  9. tictac4456 Group Title
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    yup

    • 2 years ago
  10. mayankdevnani Group Title
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    @tictac4456

    • 2 years ago
  11. tictac4456 Group Title
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    thats correct

    • 2 years ago
  12. mayankdevnani Group Title
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    thnx..

    • 2 years ago
  13. mayankdevnani Group Title
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    @satellite73 help me:)

    • 2 years ago
  14. mayankdevnani Group Title
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    @amistre64

    • 2 years ago
  15. tictac4456 Group Title
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    hey ur work is correct but the answer is wrong

    • 2 years ago
  16. tictac4456 Group Title
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    the answer is A

    • 2 years ago
  17. mayankdevnani Group Title
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    howz that??

    • 2 years ago
  18. tictac4456 Group Title
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    oh im srry never mind i didnt see the seconds ur answer is correct

    • 2 years ago
  19. tictac4456 Group Title
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    im 100% sure its right

    • 2 years ago
  20. mayankdevnani Group Title
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    ohkkk!!

    • 2 years ago
  21. asnaseer Group Title
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    I agree - your work and answer is correct

    • 2 years ago
  22. mayankdevnani Group Title
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    thnx...

    • 2 years ago
  23. asnaseer Group Title
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    yw :)

    • 2 years ago
  24. tictac4456 Group Title
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    do ya take entrepuener class

    • 2 years ago
  25. mayankdevnani Group Title
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    what do you mean??

    • 2 years ago
  26. tictac4456 Group Title
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    other than physics do u take entrepuener class

    • 2 years ago
  27. tictac4456 Group Title
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    its like a business class

    • 2 years ago
  28. mayankdevnani Group Title
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    no!! never!!! why??

    • 2 years ago
  29. tictac4456 Group Title
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    haha i kinda needed help

    • 2 years ago
  30. tictac4456 Group Title
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    but never mind

    • 2 years ago
  31. mayankdevnani Group Title
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    plz you can!!!

    • 2 years ago
  32. tictac4456 Group Title
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    ok lol but if u dont knw it u dont have to help k

    • 2 years ago
  33. mayankdevnani Group Title
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    k

    • 2 years ago
  34. tictac4456 Group Title
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    what is gross profit? A)the money left over after the cost of making a product or providing a service B) the cost of labor to produce a product C) the difference between a product's price and the revenue of the company D) a negative net profit, in which company experiences a loss

    • 2 years ago
  35. mayankdevnani Group Title
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    (finance) the net sales minus the cost of goods and services sold.

    • 2 years ago
  36. tictac4456 Group Title
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    thnkz lol

    • 2 years ago
  37. tictac4456 Group Title
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    i gave u a medal right

    • 2 years ago
  38. mayankdevnani Group Title
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    right!!!

    • 2 years ago
  39. tictac4456 Group Title
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    cool so did u graduate? or u still in high skool

    • 2 years ago
  40. mayankdevnani Group Title
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    high school!!!

    • 2 years ago
  41. tictac4456 Group Title
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    what grade

    • 2 years ago
  42. mayankdevnani Group Title
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    9 class

    • 2 years ago
  43. mayankdevnani Group Title
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    ok @tictac4456

    • 2 years ago
  44. tictac4456 Group Title
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    me2 lol

    • 2 years ago
  45. tictac4456 Group Title
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    what skool?

    • 2 years ago
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