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The vector -5.2 has a magnitude of 32 and points in the positive direction.?

  1. Feb 10, 2010 #1
    Find the x component of the vector A?

    Find the magnitude of the vector A?

    Please also explain your work too

    thanks in advance
     
  2. jcsd
  3. Feb 10, 2010 #2
    What do you mean by "the vector -5.2"?
     
  4. Feb 10, 2010 #3

    rl.bhat

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    Hi MrGoodyear812, welcome to the PF.
    You have not posted the actual problem. In the title, what is -5.2 and 32?
     
  5. Feb 10, 2010 #4
    OH! SORRY!

    it should be the vectore

    -5.2A
     
  6. Feb 10, 2010 #5

    rl.bhat

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    If you want the components of a vector, you most know the angle made by the vector with x-axis. What is that angle?
     
  7. Feb 10, 2010 #6
    Thats why this problem is killing me...its not enough info

    thats honestly everything the book gave us
     
  8. Feb 10, 2010 #7
    I don't get it, is -5.2A the name of the vector? That doesn't make any sense. Are -5 and 2 the components? Doesn't make sense that they would just ask for the x component. What the heck is the A for?

    Maybe if you could write the problem statement word for word it would help.
     
  9. Feb 10, 2010 #8
    5.2 is the value of hte vector is what im understanding....idk im so confused..........

    and that is word for word
     
  10. Feb 10, 2010 #9
    Just got this email from my prof

    Bx = 34m in this case --> Ax = (34m) / (-5.2) = ?? (a negative number), so vecA is pointing in the negative x-direction. The magnitude of A will just be |Ax| (since there's no y-component), which is positive.
     
  11. Feb 10, 2010 #10
    He just emailed me this:

    Bx = 34m in this case --> Ax = (34m) / (-5.2) = ?? (a negative number), so vecA is pointing in the negative x-direction. The magnitude of A will just be |Ax| (since there's no y-component), which is positive.
     
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