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YumYum247

  • one year ago

Please help me Astrophysics!!!! :"D

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  1. YumYum247
    • one year ago
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  2. YumYum247
    • one year ago
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    @sweetburger Please help me!!!! :)

  3. YumYum247
    • one year ago
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    Can anyone please help me find the resultant angle?!?!?!?

  4. YumYum247
    • one year ago
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    @MAEMAEHOCKEY final got your tagname :D

  5. anonymous
    • one year ago
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    give me a couple of mins

  6. YumYum247
    • one year ago
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    np, take your time :)

  7. anonymous
    • one year ago
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    i thought it would be simple

  8. YumYum247
    • one year ago
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    and i thought ......nvm. i'll just beg at someone's else's door. -_-

  9. anonymous
    • one year ago
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    i willl help u i am happy to

  10. YumYum247
    • one year ago
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    really?????? :D

  11. anonymous
    • one year ago
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    yaaa

  12. YumYum247
    • one year ago
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    yuss!!!

  13. anonymous
    • one year ago
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    ok i think i got it

  14. YumYum247
    • one year ago
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    Hey don't bother, i don't wanna make you go through stress just for a question...leave it, i'll just ask someone else...Chill...:"D

  15. anonymous
    • one year ago
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    what .......i am not stressed

  16. YumYum247
    • one year ago
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    i'm just saying if your here to be a chill bird, then be a chill bird, i ain't trapping you in a cage. :"D

  17. anonymous
    • one year ago
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    i am not

  18. YumYum247
    • one year ago
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    Aait then.... i just wanna let you know that i've already done 90% of the question, i just need to find the resultant angle. and Thanks!!!

  19. anonymous
    • one year ago
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    ok i think i got it if i am doing it right

  20. YumYum247
    • one year ago
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    please proceed....:D

  21. anonymous
    • one year ago
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    24 right ???

  22. mathmate
    • one year ago
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    |dw:1442883489245:dw|

  23. mathmate
    • one year ago
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    can you confirm if the 24 and 32 degrees as shown in the marked up diagram? |dw:1442883620661:dw|

  24. YumYum247
    • one year ago
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    hold up let me attach a picture of the actual question....don't go anywhere...promise me!!! O_O

  25. mathmate
    • one year ago
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    ok!

  26. YumYum247
    • one year ago
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  27. mathmate
    • one year ago
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    |dw:1442883835366:dw| Yes, this is a little better, with the force magnitudes. One suggestion I make for your future solutions is to draw the forces in the same orientation as the original. This way, there is a lesser chance of making mistakes.

  28. YumYum247
    • one year ago
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    ok :)

  29. mathmate
    • one year ago
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    |dw:1442883951868:dw| So R^2=15^2+12^2-2(15)(12)cos(124)=23.8812 as you had it.

  30. mathmate
    • one year ago
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    |dw:1442884257124:dw| sin(\(\theta\)/12=sin(124)/R so sin(\(\theta\)=12 sin(124)/23.8812=0.416582 \(\theta\)= asin(0.416582)=24.619 degrees (measured from the 15N force, or 24.619-24=0.619 degrees from the axis of the direction of the vehicle.

  31. mathmate
    • one year ago
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    correction: a little confusion for R. Should have written R^2=......., and R=sqrt(...)=23.8812 N

  32. mathmate
    • one year ago
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    @YumYum247 Still there? To find the angle, I used the sine rule, but forgot to indicate it.

  33. YumYum247
    • one year ago
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    yes i'm still here....!!!

  34. YumYum247
    • one year ago
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    0.61deg.....but that's not even a one whole degree tho :O

  35. YumYum247
    • one year ago
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    so you used... SineA/a = SineC/c SinA/15 = Sin124/24

  36. mathmate
    • one year ago
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    That's with respect to the axis of the vehicle. |dw:1442885722739:dw|

  37. mathmate
    • one year ago
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    Yes, but sin(theta)/12 = sin(124)/R. Sine rule sometimes does not give the correct angles when the unknown angle is near 90 degrees (angle or supplement). But here we know the unknown angle is acute, so there's no problem.

  38. YumYum247
    • one year ago
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    yah sorry i realized that.... :D

  39. YumYum247
    • one year ago
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    Again thank you infinite :"D

  40. mathmate
    • one year ago
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    You're welcome! :)

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