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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Its a derivation from momentum changes in a system of variable mass..my Question is that why a sign of velocity of big mass relative to small mass is changed in 2nd step..

  2. anonymous
    • one year ago
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    @mathmate

  3. anonymous
    • one year ago
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    @shamim

  4. anonymous
    • one year ago
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    @IrishBoy123

  5. IrishBoy123
    • one year ago
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    can hardly read it but they also switched U and V around to get relative velocity so i presume that it assists presentation

  6. anonymous
    • one year ago
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  7. mathmate
    • one year ago
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    Yeah, it's very hard to read. Can you write it out using the drawing tool (the "draw" button below)?

  8. anonymous
    • one year ago
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    No i am unable to use drawing tool because i am using tablet ..

  9. IrishBoy123
    • one year ago
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    clearly you are starting with \(\large \vec F = \frac{d}{dt}\vec p = \frac{d}{dt} (m \vec v)\) and running with that gives \(\large m \frac{d \vec v} {dt} +\frac{dm}{dt} (\vec v - \vec u)\) and then the switch to \(\large m \frac{d \vec v} {dt} -\frac{dm}{dt} (\vec u - \vec v)\) where \( (\vec u - \vec v)\ = \vec v_{rel} \) all looks very OM except we do not actually know what you are modelling! eg what is \(\vec u\) ?

  10. IrishBoy123
    • one year ago
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    "very OM" = very OK

  11. anonymous
    • one year ago
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    U is velocity of big mass where v is the velocity of ejected mass

  12. anonymous
    • one year ago
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    I think u r right tht signs doesnt mean.. this iw due to relativity..

  13. Michele_Laino
    • one year ago
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    since the subsequent identity, has been used: \[\Large {\mathbf{v}} - {\mathbf{u}} = - \left( {{\mathbf{u}} - {\mathbf{v}}} \right)\]

  14. anonymous
    • one year ago
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    Okay thanks to all for helping me.. :)

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