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Ruchi.

  • 2 years ago

question?

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  1. Ruchi.
    • 2 years ago
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    |dw:1339902704337:dw| FIND ACCELERATION?

  2. rebeccaskell94
    • 2 years ago
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    D: I don't know D: @tomas.A answer it! :D

  3. rebeccaskell94
    • 2 years ago
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    It's a her...don't be rude.

  4. Tomas.A
    • 2 years ago
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    Rebecca don't he so shy answer her

  5. Tomas.A
    • 2 years ago
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    that's what i said ^

  6. rebeccaskell94
    • 2 years ago
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    liezz

  7. Ruchi.
    • 2 years ago
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    are'nt you members geetting mah problem on right..? do help me pleas.!!

  8. rebeccaskell94
    • 2 years ago
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    lol um I don't know how to do this. I thought Tomas would be smart enough but I guess not ;D

  9. Tomas.A
    • 2 years ago
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    i forgot all the physics when i finished it

  10. rebeccaskell94
    • 2 years ago
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    lol someone else has arrived to help you!

  11. A.Avinash_Goutham
    • 2 years ago
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    it's easy........acc = net pulling force/eff mass

  12. Ruchi.
    • 2 years ago
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    i know that . if you can then explain it.

  13. A.Avinash_Goutham
    • 2 years ago
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    see the net pullin force = 5g-2g

  14. saifoo.khan
    • 2 years ago
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    |dw:1339903618208:dw|

  15. A.Avinash_Goutham
    • 2 years ago
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    not it's 7a i guess

  16. Ruchi.
    • 2 years ago
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    i get its answer 7/3 is its right

  17. A.Avinash_Goutham
    • 2 years ago
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    let's do it the long way.......... 5g-T=5a

  18. A.Avinash_Goutham
    • 2 years ago
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    T-t=(Wb)a t-2g=2a

  19. Ruchi.
    • 2 years ago
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    how is it?

  20. A.Avinash_Goutham
    • 2 years ago
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    simultaneous equations

  21. saifoo.khan
    • 2 years ago
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    a = 3/10

  22. A.Avinash_Goutham
    • 2 years ago
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    wait wat is the wait of B? nd i dont think it's 3/10

  23. Ruchi.
    • 2 years ago
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    its not given.

  24. A.Avinash_Goutham
    • 2 years ago
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    it's 3g/7........i guess

  25. A.Avinash_Goutham
    • 2 years ago
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    if T=t

  26. UnkleRhaukus
    • 2 years ago
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    Atwood machine

  27. Ruchi.
    • 2 years ago
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    we hav to use only 1st and 3rd equation?

  28. yakeyglee
    • 2 years ago
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    Use Newton's second law. \[\sum \vec F_i = m \vec a\]\[m_A g - m_Cg = (m_A+m_B+m_C)a\]

  29. ujjwal
    • 2 years ago
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    Agreed with @yakeyglee, And if there is friction between B and the surface on which it is kept, don't forget to subtract frictional force from the net force in your calculation.

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