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March 26, 2017

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The position of a 2.75×10^5 N training helicopter under test is given by r =(0.020m/s 3 )t^3 i ^ +(2.2m/s)tj ^ −(0.060m/s 2 )t^2 k ^. find the net force of the helicopter at t=5.0s

  • physics - ,

    r =(0.020m/s³)t ³ i ^ +(2.2m/s)tj ^ −(0.060m/s² )t²k ^.
    v= dr/dt =(3•0.020m/s³ ) •t²• i ^ +(2.2m/s) • j ^ −2• (0.060m/s² )t•k ^.
    a=d²r/dt²=(6•0.020m/s ³) •t i ^ -2• (0.060m/s² )•k ^.
    F=ma =2.75•10⁵•{(6•0.020m/s ³) •t i ^ -2• (0.060m/s² )•k ^.}
    t=5 s
    F₅=2.75•10⁵•{(6•0.020m/s ³) •5• i ^ -2• (0.060m/s² )•k ^.}=
    =165000 i ^ - 33000•k ^.

  • physics - ,

    the actual answer for two sig figs are 1.7x10^4*i^-3.3x10^3*k^1

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