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nath

  • 2 years ago

Consider a particle with initial velocity that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis. What is the x component Vx of V? What is the y component Vy of V?

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  1. Hunus
    • 2 years ago
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    |dw:1329692762479:dw| The x component Vx of V is the velocity vector V projected onto the x axis. We obtain this magnitude as follows \[v_{x}=12\cos(120)=-12\cos(60)=-12(\frac{1}{2})=-6\] The y component Vy of V is the velocity vector V projected onto the y axis. We obtain this magnitude as follows \[v_{y}=12\sin(120)=12\sin(60)=12\frac{\sqrt{3}}{2}=6\sqrt{3}=10.4\] |dw:1329693129338:dw|

  2. nath
    • 2 years ago
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    why does sin have a positive value?

  3. Hunus
    • 2 years ago
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    y has a positive value in the top two quadrants

  4. Hunus
    • 2 years ago
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    |dw:1329693437661:dw|

  5. nath
    • 2 years ago
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    so to figure out positive and negative i have to look at 12

  6. Hunus
    • 2 years ago
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    You have to look at the direction it's pointing in. For theta equal to 120 or '60 above the negative x axis' the vector is pointing in the negative x direction and in the positive y direction

  7. nath
    • 2 years ago
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    so what is u have something like this :

  8. nath
    • 2 years ago
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    srry i mean

  9. nath
    • 2 years ago
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    |dw:1329693963427:dw|

  10. nath
    • 2 years ago
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    now its moving clockwise

  11. Hunus
    • 2 years ago
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    The vector A you described is at an angle of (180+60) 240 degrees with respect to the positive x axis as you can see from your picture it is directed toward the negative x negative y direction, so both x and y will be negative.

  12. nath
    • 2 years ago
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    why do i have to add 180?

  13. Hunus
    • 2 years ago
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    |dw:1329694073557:dw| Because we're making the angle with respect to the positive x axis to make it simple to take the sine and cosine of the vector

  14. Hunus
    • 2 years ago
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    By adding 180 to 60 we're changing it from your original picture |dw:1329694180018:dw| and making it |dw:1329694218137:dw|

  15. nath
    • 2 years ago
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    or u can say the angle is -60

  16. Hunus
    • 2 years ago
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    If we're taking it to be with respect to the positive x axis |dw:1329694331261:dw| then it would be -120 degrees. -60 degrees looks like this |dw:1329694414386:dw|

  17. nath
    • 2 years ago
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    |dw:1329694584288:dw|

  18. nath
    • 2 years ago
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    thats how u show it

  19. nath
    • 2 years ago
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    cause its moving clockwise

  20. Hunus
    • 2 years ago
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    Yes.

  21. nath
    • 2 years ago
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    so in this question is the 60 degrees going clockwise

  22. Hunus
    • 2 years ago
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    It typically doesn't matter how you arrive at the angle, but it appears that the angle was rotated counter clockwise 60 degrees. The vector is at 240 degrees with respect to the positive x axis.

  23. nath
    • 2 years ago
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    so qhen do i know when to add 180 or minus 180

  24. Hunus
    • 2 years ago
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    Look at the positive x axis and see how many degrees it takes to get to your vector. This is only showing whether a vectors components are positive of negative.

  25. Hunus
    • 2 years ago
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    or negative*

  26. nath
    • 2 years ago
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    can u please give me an example of each type if itsnt too much trouble

  27. Hunus
    • 2 years ago
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    |dw:1329695075597:dw| Is vector M's x component positive or negative? What about its y component?

  28. nath
    • 2 years ago
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    the x compoinent is negative while the y component is positive

  29. Hunus
    • 2 years ago
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    \[v_x=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]so we know that the value of Vx is negative while the value of Vy is positive. From this we could say that \[v_{y}=|M| \sin(30)=\frac{|M|}{2}\] and \[v_{x}=-|M|\cos(30)=-|M|\frac{\sqrt{3}}{2}\] Where |M| is the magnitude of the vector M However we could also say that, to get to vector M from the positive x axis |dw:1329695464157:dw| we would have to go through an angle of 150 degrees. Thus would make our calculations \[v_y=|M|\sin(150)=\frac{|M|}{2}\] As you can see we arrive at the same answer which is why it doesn't matter which way you get to the angle.

  30. Hunus
    • 2 years ago
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    \[v_{x}=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]

  31. nath
    • 2 years ago
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    what does 150 degrees tellus ?

  32. nath
    • 2 years ago
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    are we saying the angle 150 is equal to 30 degrees

  33. Hunus
    • 2 years ago
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    150 degrees tells us what the angle is from the positive x axis. That's what we usually like to use as a reference point when using sines and cosines.

  34. Hunus
    • 2 years ago
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    What we are saying is that we only need the reference angle (how many degrees it is from the closest x axis) if we know if x is positive or negative or y is positive or negative

  35. nath
    • 2 years ago
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    so basically we are saying the angle is 150 degrees away from the orgin

  36. Hunus
    • 2 years ago
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    From the positive x axis

  37. nath
    • 2 years ago
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    what about the negative x axis

  38. Hunus
    • 2 years ago
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    150 degrees tells us how far we would have to rotate a vector from the positive x axis (0 degrees) to get to where the vector is.

  39. nath
    • 2 years ago
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    ohhh i get it

  40. Hunus
    • 2 years ago
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    Good :)

  41. nath
    • 2 years ago
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    and to find the negative value what would i have to do

  42. Hunus
    • 2 years ago
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    Well, the degrees from the positive x axis tells us whether it's positive or negative. For example, between 0 and 90 degrees, x and y are negative, between 90 and 180 degrees x is negative and y is positive, between 180 and 270 degrees both x and y are negative and from 270 to 360 degrees x is positive and y is negative. If you, for example, put in 6sin(340) you will get a negative value because y is negative in that quadrant. If you put in 12cos(320) you will get a positive value because x is positive in that quadrant.

  43. Hunus
    • 2 years ago
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    between 0 and 90 x and y are positive*** sorry

  44. nath
    • 2 years ago
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    so it doesnt matter whether u say 30 degreees or 150 degrees

  45. nath
    • 2 years ago
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    thanks . could u please help me with another question

  46. Hunus
    • 2 years ago
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    It matters, but only as far as the dirrection goes. This is because the MAGNITUDE of cos(30) is the same as the MAGNITUDE of cos(150), but one is negative and one is positive

  47. Hunus
    • 2 years ago
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    Sorry for the delay on that response

  48. nath
    • 2 years ago
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    and that has to do withe quadrant whether is positive or negative

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