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korosh23

  • one year ago

Physics questions!

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  1. korosh23
    • one year ago
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  2. korosh23
    • one year ago
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    Please answer the orange questions which there is a question mark. Each question in a different page. So three medals.

  3. korosh23
    • one year ago
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    Ok #36 please! I have my answer. I will send it in a sec!

  4. korosh23
    • one year ago
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  5. Michele_Laino
    • one year ago
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    I got this: |dw:1433908300305:dw| so I can write this: \[\Large {m_1}{v_1} - {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v\] from which I got: v= +4 m/sec

  6. korosh23
    • one year ago
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    How !?

  7. Michele_Laino
    • one year ago
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    using my reference system, we can write: \[\Large \begin{gathered} {{\mathbf{v}}_{\mathbf{1}}} = \left( {{v_1},0} \right) \hfill \\ {{\mathbf{v}}_{\mathbf{2}}} = \left( { - {v_2},0} \right) \hfill \\ \end{gathered} \]

  8. korosh23
    • one year ago
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    Are you sure that is m1v1 - m2v2 or m1v1 + m2v2?

  9. Michele_Laino
    • one year ago
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    no, since what is conserved is the vector which represents the total momentum, actually our equation is: \[\Large {m_1}{{\mathbf{v}}_{\mathbf{1}}} + {m_2}{{\mathbf{v}}_{\mathbf{2}}} = \left( {{m_1} + {m_2}} \right){\mathbf{v}}\]

  10. korosh23
    • one year ago
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    Exactly, it is a closed system. Energy is conserved. Now lets plug in the numbers.

  11. Michele_Laino
    • one year ago
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    now we have to go from vectors to components, and as I wrote before, we can write: \[\begin{gathered} {{\mathbf{v}}_{\mathbf{1}}} = \left( {{v_1},0} \right) \hfill \\ {{\mathbf{v}}_{\mathbf{2}}} = \left( { - {v_2},0} \right) \hfill \\ \end{gathered} \]

  12. korosh23
    • one year ago
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    what is v2? Motor or truck?

  13. Michele_Laino
    • one year ago
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    v_2 is the motor

  14. Michele_Laino
    • one year ago
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    furthermore: \[\Large {v_1} = \left| {{{\mathbf{v}}_{\mathbf{1}}}} \right|,\quad {v_2} = \left| {{{\mathbf{v}}_{\mathbf{2}}}} \right|\]

  15. korosh23
    • one year ago
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    Wait a moment.

  16. korosh23
    • one year ago
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    In the question it says the truck is going west. West is the negative direction. Motorist is going opposite of west which is east! It is postive velocity. Truck should have -ve velocity.

  17. Michele_Laino
    • one year ago
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    ok! Then our correct drawing is: |dw:1433908946956:dw| and the previous equation will become: \[\Large \begin{gathered} - {m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v \hfill \\ {{\mathbf{v}}_{\mathbf{1}}} = \left( { - {v_1},0} \right) \hfill \\ {{\mathbf{v}}_{\mathbf{2}}} = \left( { + {v_2},0} \right) \hfill \\ \end{gathered} \]

  18. Michele_Laino
    • one year ago
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    and the answer is: v= -4 m/sec

  19. korosh23
    • one year ago
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    That means I was right? I got -4 m/s

  20. Michele_Laino
    • one year ago
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    yes!

  21. korosh23
    • one year ago
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    Ok great, thank you! I have to go, I will ask my other questions later.

  22. Michele_Laino
    • one year ago
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    ok!

  23. korosh23
    • one year ago
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    Just wondering are you here in openstudy tomorrow. On Thursday I have a physics exam. I was wondering if I have few quick question, Is it ok I ask you?

  24. Michele_Laino
    • one year ago
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    yes! I will stay in OpenStudy tomorrow, and I can help you

  25. korosh23
    • one year ago
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    At what time are you in openstudy, and when you leave?

  26. Michele_Laino
    • one year ago
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    I will be here in OpenStudy at 6:00 am (Italy Time zone)

  27. korosh23
    • one year ago
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    Ok I will manage mt time. Thank you, you are a very supportive tutor.

  28. Michele_Laino
    • one year ago
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    :)

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