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anonymous

  • 4 years ago

a stone of mass 0.5kg tied with a string of length 1m is moving in a horizontal circle with speed 4m/s.the tension in the string is...

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  1. anonymous
    • 4 years ago
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    |dw:1327670211788:dw| for vertical equilibrium

  2. AravindG
    • 4 years ago
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    hey salini

  3. AravindG
    • 4 years ago
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    heloo

  4. anonymous
    • 4 years ago
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    endha answer ariyo aravind?

  5. AravindG
    • 4 years ago
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    maybe

  6. AravindG
    • 4 years ago
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    tht answer to my qn was wnderful i was looking for tht

  7. anonymous
    • 4 years ago
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    endha?

  8. AravindG
    • 4 years ago
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    ?

  9. AravindG
    • 4 years ago
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    u gud in malayalam??

  10. anonymous
    • 4 years ago
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    ene kurichu endha vijaricha ney?

  11. AravindG
    • 4 years ago
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    enaal nallathu in this qn enikku thonunu tension is provided by the centripetal force

  12. AravindG
    • 4 years ago
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    helo

  13. anonymous
    • 4 years ago
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    annu pakshe e questionil angle kurichu onum thannela aa diagram yaane varachadhannu

  14. AravindG
    • 4 years ago
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    y do u need angle for mv^2/r

  15. anonymous
    • 4 years ago
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    |dw:1327671459182:dw|alla imagine chei if there was no angle then there wud be no force in the vertical direction to balance mg

  16. AravindG
    • 4 years ago
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    http://www.twiddla.com/735087

  17. JamesJ
    • 4 years ago
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    The centripetal force, \( F_c \) required to keep the object in uniform circular motion is \[ F_c = \frac{mv^2}{r} \] |dw:1327674336876:dw| F_c is the force pointing towards the center of rotation. Now, where is that F_c provided from? It is provided but the tension, T, in the string. It is the component in F_c in that direction. The other piece of T is the vertical component and you know that must be equal to the weight of the mass, mg. If the string makes an angle theta to the vertical, we must have therefore that \[ T\cos\theta = mg \ \ \ \ -- (1) \] \[ T \sin\theta = F_c = \frac{mv^2}{r} \ \ \ \ -- (2) \] Now you know the length of the string, it is 1 m. And you know the radius of the rotation, it is \( r \). Hence another expression involving \( \theta \) is \[ \sin\theta = \frac{r}{1} = r \ \ \ \ -- (3) \] You now have enough equations to be able to solve for \( T \).

  18. anonymous
    • 4 years ago
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    always ina horizontal circle there is some angle made with vertical or horizontal right?

  19. JamesJ
    • 4 years ago
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    By the string, yes.

  20. AravindG
    • 4 years ago
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    :P

  21. AravindG
    • 4 years ago
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    hey

  22. AravindG
    • 4 years ago
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    u thr?

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