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crcosmin32
Pe: 2018-10-21T15:48:26+03:00 In: In:

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O minge cu densitatea ρ=600kg/m^3 este scufundata in apa la adancimea l=10 m si este lasata libera.
Neglijand frecarile in apa si in aer sa se calculeze timpul in care mingea ajunge la suprafata apei si inaltimea maxima la care se ridica deasupra apei. ( g=10m/s^2).

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3 raspunsuri

  1. baiatul122001

    F_{A}-G=m*a_{u}
    m_{H_{2}O}*g-m*g=m*a_{u}
    V\rho_{H_{2}O}g-V\rho g=V\rho a_{u}
    a_{u}=\frac{g(\rho _{H_{2}O}-\rho )}{\rho }=\frac{10(1000-600)}{600}=\frac{4000}{600}=6,66 \frac{m}{s^{2}}
    l=v_{o}t+\frac{a_{u}t^{2}}{2}<=>l=\frac{a_{u}t^{2}}{2}=>t^{2}=\frac{2l}{a_{u}}=>t=\sqrt{\frac{2l}{a_{u}}}=\sqrt{\frac{2*10}{6,66}}\approx \sqrt{3}=1,73 s
    v_{1}=v_{o}+a_{u}t<=>v_{1}=6,66*1,73=11,52 m/s

    h_{max}=\frac{v_{1}^{2}}{2a_{u}}=\frac{11,52^{2}}{2*6,66}=9,96 m

  2. Integrator
    maestru (V)

    baiatul122001 post_id=112308 time=1540303331 user_id=22243 wrote:
    F_{A}-G=m*a_{u}
    m_{H_{2}O}*g-m*g=m*a_{u}
    V\rho_{H_{2}O}g-V\rho g=V\rho a_{u}
    a_{u}=\frac{g(\rho _{H_{2}O}-\rho )}{\rho }=\frac{10(1000-600)}{600}=\frac{4000}{600}=6,66 \frac{m}{s^{2}}
    l=v_{o}t+\frac{a_{u}t^{2}}{2}<=>l=\frac{a_{u}t^{2}}{2}=>t^{2}=\frac{2l}{a_{u}}=>t=\sqrt{\frac{2l}{a_{u}}}=\sqrt{\frac{2*10}{6,66}}\approx \sqrt{3}=1,73 s
    v_{1}=v_{o}+a_{u}t<=>v_{1}=6,66*1,73=11,52 m/s

    h_{max}=\frac{v_{1}^{2}}{2a_{u}}=\frac{11,52^{2}}{2*6,66}=9,96 m


    Bună dimineața,

    Dacă nu făceați acele aproximări , atunci h_{max}=10 m.

    Toate cele bune,

    Integrator

  4. baiatul122001

    Buna ziua , aveti dreptate ! 😀 😀 😀

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