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anonymous

  • 5 years ago

hey, sorry about going off subject, i would like to ask a question on physics. Since there isn't anyone in that section i am helpless. here is my question: A cylinder of radius 20cm and mass 100g rolls down an inclined plane of height 60cm without slipping. the linear velocity of the cylinder at the bottom is?? (g=980cm/s/s)

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  1. anonymous
    • 5 years ago
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    mgh = KE

  2. anonymous
    • 5 years ago
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    1/2 mv^2+1/2 I w^2= Mgh

  3. anonymous
    • 5 years ago
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    i did that but i m not gettin the answer.

  4. anonymous
    • 5 years ago
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    KE = KE(r) + KE(t) 1/2(mr^2/2)w^2 + 1/2 x m x v^2

  5. anonymous
    • 5 years ago
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    rw=v w=v/r substitute

  6. anonymous
    • 5 years ago
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    3/4 x mv^2 = mgh

  7. anonymous
    • 5 years ago
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    i went through all that buddy. can you solve ang post your anser.

  8. anonymous
    • 5 years ago
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    *and *answer

  9. anonymous
    • 5 years ago
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    v=2.8 m/s

  10. anonymous
    • 5 years ago
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    ow rats! i made a calculation mistake! neway thnx buddy..

  11. anonymous
    • 5 years ago
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    no prob

  12. anonymous
    • 5 years ago
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    wat is the radius of curveture if a particle moves in a trajectory given by y=f(x) at any point(x,y)

  13. anonymous
    • 5 years ago
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    parameterize the eqn in terms of time....r =v^2/a(radial)

  14. anonymous
    • 5 years ago
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    it is given as \[R=[1+(dy/dx)^{2}\left[ \right]^{3/2}\]

  15. anonymous
    • 5 years ago
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    divied by d^2y\dx^2

  16. anonymous
    • 5 years ago
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    if u know the answer then y do u ask..i dont memorize formulae..its wiser to go through the basic approach...

  17. anonymous
    • 5 years ago
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    i want to kno how it came

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