De Kt Thuong Ki

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  • 7/24/2019 De Kt Thuong Ki

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    kim tra thng k 2015-2016

    Cu 2 (2d). Mt vt c t trn mt phng nghing hp vi mt phng nm ngang mt gc

    04 = . H s ma st ! " 0#0$. %&c '( )*+ ta+ gi, vt sa( th- ch vt t/ ch(+n ng. %*+

    g"1#2 m3s.

    H5i m(n trt h6t 7(8ng 9ng $0 m ! t: )&c ;

    Cu 4(3d). Eh h vt ch(+n ng nh hFnh vG# vt ?nm trn mt phng nghing. t m?" $ !g# m" I !g.Jc gi,a mt phng nghing vi mt phng ngang )KL"I0. H s ma st ca vt m?vi mt phng nghing )K

    ! " 0#. H8+ Dc Bnh ch(+n ng ca h sa( !hi th- vtch(+n ng# tCnh )/c cNng O=+> P5 7(a !hi )ng O=+vK rQng rRc# O=+ !hSng gi8n. %*+ g " 1#2 Tm3sU.

    Cu 1(3d).Eh mt ch*t im ch(+n ng vi phAng trFnh . . 4.r t i t j k = + + TmU.

    a. n tc tVi t"Is >

    . Jia tc t: 0s 6n Is>!. W(X V ch(+n ng>

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