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monroe17

The displacement vectors A and B both have magnitudes of 3.30 m. The direction of vector A is 16.6 degrees. Find the magnitude of |A+B|, |A-B|, |B-A|, and |A-2B| graphically. PLEASE HELP ;/

  • one year ago
  • one year ago

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  1. Outkast3r09
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    each vector has a y and x component

    • one year ago
  2. monroe17
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    i and j :) ?

    • one year ago
  3. Outkast3r09
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    well you're given a magnitude

    • one year ago
  4. Outkast3r09
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    |dw:1359604762762:dw|

    • one year ago
  5. Outkast3r09
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    you there?

    • one year ago
  6. Outkast3r09
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    |dw:1359605259789:dw|

    • one year ago
  7. monroe17
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    yeah, im here.. trying to understand

    • one year ago
  8. Outkast3r09
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    |dw:1359605406061:dw| these need direction due to vectors .. it says A direction is 16.6 degrees sooooo

    • one year ago
  9. Outkast3r09
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    |dw:1359605476238:dw|

    • one year ago
  10. Outkast3r09
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    so in order to make sense of each vector you need to break them down into right triangles that show components of x and y i'll start with A|dw:1359605592468:dw|

    • one year ago
  11. Outkast3r09
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    now using trig functions you can find the components correct ? can you do that?

    • one year ago
  12. monroe17
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    using sohcahtoa right?

    • one year ago
  13. monroe17
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    A_x=sin(16.6)*A?

    • one year ago
  14. Outkast3r09
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    yes

    • one year ago
  15. monroe17
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    i got 3.16 as side A_y..

    • one year ago
  16. Outkast3r09
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    so now we have \[A_x=Acos(16.6)\] \[A_y=Asin(16.6)\]

    • one year ago
  17. Outkast3r09
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    you had them switched around =]

    • one year ago
  18. monroe17
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    and 0.95 as a_y.. A=3.30 right?

    • one year ago
  19. Outkast3r09
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    yep sounds correct

    • one year ago
  20. Outkast3r09
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    atleast according to my diagram lol

    • one year ago
  21. Outkast3r09
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    alright so the B is easy since there is only one component to it... which is the \[B_y\]

    • one year ago
  22. monroe17
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    B=3.30 but theres no theta?

    • one year ago
  23. Outkast3r09
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    yes but it's a simple line on the y axis so it only has one component an x is what i meant =/

    • one year ago
  24. Outkast3r09
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    so you have \[B_x=3.30\]

    • one year ago
  25. Outkast3r09
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    |dw:1359606759899:dw|

    • one year ago
  26. Outkast3r09
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    now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components

    • one year ago
  27. monroe17
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    "like components" ? 8.1?

    • one year ago
  28. monroe17
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    so B is 3.30 and for A do I add a_x and a_y and A?

    • one year ago
  29. Outkast3r09
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    well lets do it like my book says \[R_x=A_x+B_x\] \[R_y=A_y+B_y\]

    • one year ago
  30. Outkast3r09
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    \[|A+B|=\sqrt{R_x^2+R_y^2}\]

    • one year ago
  31. monroe17
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    R_x=4.25 R_y= 3.16

    • one year ago
  32. Outkast3r09
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    |dw:1359607102751:dw|

    • one year ago
  33. monroe17
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    5.3 m?

    • one year ago
  34. Outkast3r09
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    \[B_x=3.30\] \[A_x=3.30cos(16.6)\] \[A_y=3.30sin(16.6)\]

    • one year ago
  35. monroe17
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    6.6 m?

    • one year ago
  36. Outkast3r09
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    \[R_x=6.46\] \[R_y=.943\] \[|R|=\sqrt{6.46^2+.943^2}\]

    • one year ago
  37. Outkast3r09
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    i'm getting 6.5

    • one year ago
  38. Outkast3r09
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    6.53 to be more exact

    • one year ago
  39. monroe17
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    yeah i got that too now. i rounded wrong earlier.

    • one year ago
  40. monroe17
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    so for A-B would i just subtract?

    • one year ago
  41. Outkast3r09
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    yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|

    • one year ago
  42. Outkast3r09
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    a-b is the diagonal from left to right if i remember right so you have to take A and move it infront of b like this

    • one year ago
  43. Outkast3r09
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    |dw:1359607791579:dw|

    • one year ago
  44. monroe17
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    oh my lordy... i give up... ;/ my grade is good enough.. 96.3% on this assignment. I'm done with this question.. i still have to do a lab report and study for calc 2... thanks for the help though ;/

    • one year ago
  45. monroe17
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    I'll go back later when i have more imte tomorrow and try to figure it out lol.

    • one year ago
  46. Outkast3r09
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    i looked at my book you have to face the vectors in certain ways so a-b says to face the vecotrs inward to eachother \

    • one year ago
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