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anonymous

  • 4 years ago

The magnitudes of the vectors shown are I U I = 8 and I V I= 3. The vector V is vertical. Graphically determine the magnitude of the vector U - 2V.

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  1. anonymous
    • 4 years ago
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    |dw:1327371116945:dw|

  2. anonymous
    • 4 years ago
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    How to solve this problem ? Guys Thanks

  3. anonymous
    • 4 years ago
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    is your question this

  4. amistre64
    • 4 years ago
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    |dw:1327371324663:dw|

  5. amistre64
    • 4 years ago
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    im assuming v is up too

  6. amistre64
    • 4 years ago
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    |dw:1327371386093:dw|

  7. amistre64
    • 4 years ago
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    law of cosines ar any rate

  8. amistre64
    • 4 years ago
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    c^2 = 8^2 + 6^2 -2.8.6 cos(45)

  9. anonymous
    • 4 years ago
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    |dw:1327371436761:dw|

  10. amistre64
    • 4 years ago
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    then change it to the up direction and we get 45+90 as the angle istead of 45

  11. anonymous
    • 4 years ago
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    How ?

  12. amistre64
    • 4 years ago
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    |dw:1327371556791:dw|

  13. amistre64
    • 4 years ago
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    if we know 2 sides and the angles between them the law of cosines from trig gives us the third side

  14. amistre64
    • 4 years ago
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    45+90 = 135 so that just the only thing that chagnes from before

  15. amistre64
    • 4 years ago
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    c^2 = 8^2 + 6^2 -2.8.6 cos(135)

  16. amistre64
    • 4 years ago
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    -v is just the opposite direction; and -2 is stretched twice as far; 3+3 = 6

  17. anonymous
    • 4 years ago
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    in the book is given only the answer and the answer is that in the book I U - 2V I = 5.7

  18. amistre64
    • 4 years ago
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    that may be; what does the law of C give us?

  19. amistre64
    • 4 years ago
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    the wolf say that gives us 12.9

  20. amistre64
    • 4 years ago
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    5.66 if v is pointing up

  21. anonymous
    • 4 years ago
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    there is no rule in the question :(

  22. amistre64
    • 4 years ago
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    so either you meant the v is pointing up; and we get 5.66 or the book is wrong

  23. amistre64
    • 4 years ago
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    |dw:1327371929000:dw|

  24. anonymous
    • 4 years ago
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    if it's up

  25. anonymous
    • 4 years ago
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    |dw:1327382757289:dw|

  26. amistre64
    • 4 years ago
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    vectors are defined by length and direction

  27. anonymous
    • 4 years ago
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    How did you get 5.66?

  28. amistre64
    • 4 years ago
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    if v pointed up then -2v points down and we get u-2v = 5.7

  29. amistre64
    • 4 years ago
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    law of cosines; thats been posted

  30. amistre64
    • 4 years ago
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    i cant recall the dot version of it tho

  31. anonymous
    • 4 years ago
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    can you do the step of it please ?

  32. anonymous
    • 4 years ago
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    R^2=6^2+8^2-2*6*8*cos135 R=12.95

  33. amistre64
    • 4 years ago
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    we know 2 sides, u and 2v u = 8, 2v = 6 |u-2v|^2 = u^2 + v^2 -2uv cos(t)

  34. amistre64
    • 4 years ago
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    since t = 45 as we originally thought; just plug it in :) and we get 5.66

  35. amistre64
    • 4 years ago
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    i missed a few typos in there but hopefully you get the picture :)

  36. anonymous
    • 4 years ago
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    i got 32 :(

  37. anonymous
    • 4 years ago
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    i followed your steps but I didn't get the same answer

  38. amistre64
    • 4 years ago
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    or .... u = <8,8> v = <0,3> u = <8,8> -2v = <0,-6> ---------- <8,2> sqrt(8^2+2^2) = sqrt(68) ..... ugh, that aint 5.7 either is it

  39. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%288%5E2+%2B+6%5E2+-2.8.6+cos%2845%29%29

  40. amistre64
    • 4 years ago
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    my us off; forgot how to get a 45 :)

  41. anonymous
    • 4 years ago
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    Oh i see now Thank you so much that was very helpful :)

  42. amistre64
    • 4 years ago
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    sqrt(2) = 8 1 = c c = 4sqrt(2) thas prolly better

  43. amistre64
    • 4 years ago
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    u = <4.2,4.2> v = < 0 , 3 > u = <4.2,4.2> -2v = < 0 , -6> ---------- <4.2, 4.2-6> sqrt(32+36+32-24.2) hopefullt i did that right; assume .2 means sqrt(2)

  44. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%28%284sqrt%282%29%29%5E2%2B%284sqrt%282%29-6%29%5E2%29 yeah, when I actually do it right it works out the same :)

  45. anonymous
    • 4 years ago
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    so we didn't use cosine in this problem

  46. amistre64
    • 4 years ago
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    we didnt have to, but it was the first idea i had until this one popped up

  47. amistre64
    • 4 years ago
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    there more than one way to skin this problem

  48. anonymous
    • 4 years ago
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    but we have 8 in U not 4

  49. amistre64
    • 4 years ago
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    |dw:1327372925034:dw|

  50. amistre64
    • 4 years ago
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    the length of the vector and its angle lets me know how to construct its component parts when I do it right

  51. anonymous
    • 4 years ago
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    is V up or down in your question

  52. amistre64
    • 4 years ago
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    the answer key would suggest its up to begin with otherwise we cant get the 5.7

  53. anonymous
    • 4 years ago
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    that1s make sense

  54. amistre64
    • 4 years ago
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    we still had to use some trig, or maybe this is geometry to get the results we wanted tho.

  55. anonymous
    • 4 years ago
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    Thank you Amister64

  56. amistre64
    • 4 years ago
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    youre welcome, it was worth the practice :)

  57. anonymous
    • 4 years ago
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    |dw:1327384535229:dw|

  58. anonymous
    • 4 years ago
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    Thank You Cinar :)

  59. anonymous
    • 4 years ago
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    yw

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