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mayankdevnani

  • 2 years ago

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) \[2800 m/s^2\]

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

  2. tictac4456
    • 2 years ago
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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.

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

  4. tictac4456
    • 2 years ago
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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

  5. tictac4456
    • 2 years ago
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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

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

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

  8. mayankdevnani
    • 2 years ago
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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

  9. tictac4456
    • 2 years ago
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    yup

  10. mayankdevnani
    • 2 years ago
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    @tictac4456

  11. tictac4456
    • 2 years ago
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    thats correct

  12. mayankdevnani
    • 2 years ago
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    thnx..

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

  14. mayankdevnani
    • 2 years ago
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    @amistre64

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

  16. tictac4456
    • 2 years ago
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    the answer is A

  17. mayankdevnani
    • 2 years ago
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    howz that??

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

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

  20. mayankdevnani
    • 2 years ago
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    ohkkk!!

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

  22. mayankdevnani
    • 2 years ago
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    thnx...

  23. asnaseer
    • 2 years ago
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    yw :)

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

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

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

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

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

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

  30. tictac4456
    • 2 years ago
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    but never mind

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

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

  33. mayankdevnani
    • 2 years ago
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    k

  34. tictac4456
    • 2 years ago
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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

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

  36. tictac4456
    • 2 years ago
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    thnkz lol

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

  38. mayankdevnani
    • 2 years ago
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    right!!!

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

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

  41. tictac4456
    • 2 years ago
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    what grade

  42. mayankdevnani
    • 2 years ago
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    9 class

  43. mayankdevnani
    • 2 years ago
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    ok @tictac4456

  44. tictac4456
    • 2 years ago
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    me2 lol

  45. tictac4456
    • 2 years ago
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    what skool?

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