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amistre64 Group Title

We used vector to find wind speed. A balloon is timed as it rises inside, assume a *vertical component only The balloon is taken outside, and timed. The angle it makes is also recorded. Using the vectors, calculate the wind speed.

  • 2 years ago
  • 2 years ago

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  1. amistre64 Group Title
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    |dw:1348235064295:dw|

    • 2 years ago
  2. amistre64 Group Title
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    i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?

    • 2 years ago
  3. amistre64 Group Title
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    Teacher wanted us to use pythag thrm to determine the wind speed. But the data collected gave a sqrt(-#) which would not allow the pythag thrm to be used. I decided to take the more general Law of Cosines.

    • 2 years ago
  4. amistre64 Group Title
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    i did try it, and since the hypotenuse is smaller than the "leg", it results in a non-real value.

    • 2 years ago
  5. amistre64 Group Title
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    his assumption is that the wind never blows things downward i spose

    • 2 years ago
  6. amistre64 Group Title
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    speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s |dw:1348235842574:dw|

    • 2 years ago
  7. amistre64 Group Title
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    i followed per instructions; he noted on the lab report that the time of the outside balloon cannot be greater than the inside record

    • 2 years ago
  8. Algebraic! Group Title
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    exactly.

    • 2 years ago
  9. Algebraic! Group Title
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    |dw:1348235955911:dw|

    • 2 years ago
  10. Algebraic! Group Title
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    Vy is the same for both

    • 2 years ago
  11. amistre64 Group Title
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    Im wondering if windspeed is only a horizontal measure? or does directionality not matter?

    • 2 years ago
  12. Algebraic! Group Title
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    "assume a horizontal component only"

    • 2 years ago
  13. amistre64 Group Title
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    there was noting stated prior in the lab to indicate that assumption tho.

    • 2 years ago
  14. amistre64 Group Title
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    *nothing

    • 2 years ago
  15. amistre64 Group Title
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    the data recorded showed that it rose slower outside, and the note i got on the lab report afterwards was that the data cannot show that. But he also says that we cannot manipulate the data ... i get the feeling that im getting contradictory advice from him :/

    • 2 years ago
  16. amistre64 Group Title
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    i wasnt. but if we only assume the wind blows things from a horizontal, or upwards ... then the data would show it to rise faster. Rise is prolly a bad term for the outside part.

    • one year ago
  17. amistre64 Group Title
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    the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers

    • one year ago
  18. amistre64 Group Title
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    the lab was designed to get you working with vectors. The speed of the wind could have been calculated from the outside data alone if we were to assume a horizontal wind speed only. .97*sin(23)

    • one year ago
  19. UnkleRhaukus Group Title
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    did you really get 23°?

    • one year ago
  20. UnkleRhaukus Group Title
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    *measure

    • one year ago
  21. amistre64 Group Title
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    well, it was the best i could do with the little plastic protractor :)

    • one year ago
  22. UnkleRhaukus Group Title
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    if you were getting <11° you could use the small angle approximation

    • one year ago
  23. amistre64 Group Title
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    the teacher suggested the pythag thrm as a means of overcoming the inherent inaccuracies in using the protractors. There was no way to tell if it was "level" and if the person catching the balloon to read the thing hadnt moved it.

    • one year ago
  24. UnkleRhaukus Group Title
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    i really dont think i understand this experiment , what results did you get /

    • one year ago
  25. amistre64 Group Title
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    \[\sqrt{.97^2+1.05^2-2(.97)(1.05)~cos(23^o)}\]

    • one year ago
  26. amistre64 Group Title
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    the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.

    • one year ago
  27. UnkleRhaukus Group Title
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    was it a helium ballon ?

    • one year ago
  28. amistre64 Group Title
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    yes, we used 3 balloons filled with helium. those shiney things you get at a party supply place. I think one of them had Dora the Explorer on it :)

    • one year ago
  29. UnkleRhaukus Group Title
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    reminds me some of walter lewin 's experiments

    • one year ago
  30. UnkleRhaukus Group Title
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    what were you timing exactly ?

    • one year ago
  31. amistre64 Group Title
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    i held the balloon on the ground, released it, and timed how long it took it to tighten the string it was attached to.

    • one year ago
  32. amistre64 Group Title
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    so we were calculating the speed along a vector path

    • one year ago
  33. amistre64 Group Title
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    using both vectors (inside and outside) we would have been able to calculate the speed (magnitude) of the resultant vector - which in turn would have been the wind speed

    • one year ago
  34. UnkleRhaukus Group Title
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    so your vectors are ; \(\{ (1.05,0°) , (0.97, 23°) \} \) ?

    • one year ago
  35. amistre64 Group Title
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    yes

    • one year ago
  36. amistre64 Group Title
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    those are more like the polar coordinants of the vectors

    • one year ago
  37. UnkleRhaukus Group Title
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    so converting to cartesian co-ordinates \[\{\quad(0,1.05)\quad,\quad\left(0.97 \cos(23°), 0.97\sin (23°)\right)\quad\}\]

    • one year ago
  38. UnkleRhaukus Group Title
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    |dw:1348240487279:dw|

    • one year ago
  39. amistre64 Group Title
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    the theta is in the wrong spot

    • one year ago
  40. amistre64 Group Title
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    spose to be 23 degrees west of north

    • one year ago
  41. UnkleRhaukus Group Title
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    ah ,

    • one year ago
  42. UnkleRhaukus Group Title
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    so theta is 90-23 ?

    • one year ago
  43. amistre64 Group Title
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    in your setup, yes

    • one year ago
  44. UnkleRhaukus Group Title
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    you can find the resultant right?

    • one year ago
  45. amistre64 Group Title
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    spose we just use this setup, and focus on the vectors |dw:1348240769620:dw| the resultant can be determined by the law of cosines

    • one year ago
  46. UnkleRhaukus Group Title
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    i thought you were ment to use pythag. ?

    • one year ago
  47. amistre64 Group Title
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    he said we could use whatever method we wanted to find the resultant, and when I used the law of cosines, he said I was wrong for using it.

    • one year ago
  48. UnkleRhaukus Group Title
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    do you get the same answer both ways /

    • one year ago
  49. amistre64 Group Title
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    which leads me to wonder, is windspeed only a horizontal measurement? if so, then the use of 2 vectors is a bad setup

    • one year ago
  50. amistre64 Group Title
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    no, you dont get the same answer both ways

    • one year ago
  51. amistre64 Group Title
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    |dw:1348241029039:dw| this would have been the resultant using pythag thrm; but my data does not present itself as a right triangle

    • one year ago
  52. amistre64 Group Title
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    there was no "answer" to compare to. So the answer itself is irrelevant. It was the process that I used to find it that he said was wrong. and I dont know why

    • one year ago
  53. amistre64 Group Title
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    how can i use the pythag thrm with the data collected? i cant.

    • one year ago
  54. amistre64 Group Title
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    \[0.97^2-1.05^2=-0.1616\] \[\sqrt{-0.1616}=nonReal~\#\]

    • one year ago
  55. amistre64 Group Title
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    lol, i just noticed a typo in the original post :) assume a *vertical* component only

    • one year ago
  56. amistre64 Group Title
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    im not sure what to make of demitris reply ...

    • one year ago
  57. amistre64 Group Title
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    the data wasnt "given". I went out and collected the data by performing the experiment

    • one year ago
  58. amistre64 Group Title
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    • one year ago
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  59. amistre64 Group Title
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    the 23 im using of course is from a faulty memory.

    • one year ago
  60. amistre64 Group Title
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    but not by use of pythag

    • one year ago
  61. amistre64 Group Title
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    you cant have a hypotenuse shorter than a leg ...

    • one year ago
  62. experimentX Group Title
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    does wind only have horizontal component?

    • one year ago
  63. amistre64 Group Title
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    experiment: i dont know :/

    • one year ago
  64. amistre64 Group Title
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    the lab just wanted us to determine the resultant vector using the data we collected.

    • one year ago
  65. experimentX Group Title
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    well .. maybe the wind is blowing from |dw:1348242000106:dw|

    • one year ago
  66. amistre64 Group Title
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    well, magnitude of the resultant that is

    • one year ago
  67. amistre64 Group Title
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    i agree, if the wind has a downward component, than the pythag will not be useful

    • one year ago
  68. amistre64 Group Title
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    nothing stated in the lab said to find a horizontal component ....

    • one year ago
  69. experimentX Group Title
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    still even we consider wind, this is not just Pythagoras problem ...

    • one year ago
  70. experimentX Group Title
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    is it @amistre64 ?

    • one year ago
  71. amistre64 Group Title
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    college, yes

    • one year ago
  72. experimentX Group Title
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    any info given on acceleration?

    • one year ago
  73. amistre64 Group Title
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    no acceleration given. just direction and speed to define the vectors with

    • one year ago
  74. amistre64 Group Title
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    it was noted when i got the lab report handed back to me, that the time outside needed to be quicker than the time inside. But my data did not reflect that.

    • one year ago
  75. Algebraic! Group Title
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    of course it does...

    • one year ago
  76. experimentX Group Title
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    |dw:1348242332765:dw|

    • one year ago
  77. amistre64 Group Title
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    pfft, typoed that

    • one year ago
  78. amistre64 Group Title
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    the speed recorded outside needed to be quicker than the speed inside

    • one year ago
  79. amistre64 Group Title
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    so, shorter time

    • one year ago
  80. Algebraic! Group Title
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    same diff.

    • one year ago
  81. amistre64 Group Title
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    the pythag thrm was suggests as a means to circumvent the inaccuracy of degree measurement. But that only works out if the data collected does not have the wind pushing down on the balloon outside.

    • one year ago
  82. experimentX Group Title
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    |dw:1348242525949:dw| we can calculate the acceleration on balloon due to wind. \[{d^2x \over dt^2} = -k {dx \over dt} \]

    • one year ago
  83. amistre64 Group Title
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    thanks for reading it, but my results were calculated already using the cosine rule. I am wondering if it was appropriate to use it in the first place.

    • one year ago
  84. amistre64 Group Title
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    the acceleration seems to be a bit outside the scope that the lab was intended for

    • one year ago
  85. amistre64 Group Title
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    yes, becasue I use the cosine rule, and included it in the doc that a wrote up

    • one year ago
  86. experimentX Group Title
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    not sure ... if this is mathematics |dw:1348243032602:dw| you can simply get that component using cosine rule.

    • one year ago
  87. amistre64 Group Title
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    experiment, I agree :)

    • one year ago
  88. experimentX Group Title
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    lol ... anyway this was getting interesting. :)

    • one year ago
  89. amistre64 Group Title
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    we got a new printer in the office, and this ine allows scans to flash drives :) this is the instruction for the lab, just in case you can spot a place that I may have missed.

    • one year ago
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  90. amistre64 Group Title
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    this is just a physics 1 class, very basic stuff so far

    • one year ago
  91. experimentX Group Title
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    seems plausible. I've been having problems with physics quite a lot lately.

    • one year ago
  92. amistre64 Group Title
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    here in florida, temp outside is usually higher than inside :)

    • one year ago
  93. amistre64 Group Title
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    thats a good notice on the outside only part. the balloons were already a part of the lab in class; i had no control on prepping them. the inside part would have been an instruction from the teacher, since its not explicitly stated anywhere in the handout. My only idea as to why we would do the inside part would be to better establish the vertical part of the experiment, but it seems to have added more trouble than it was worth :) Thanks

    • one year ago
  94. amistre64 Group Title
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    thats what i had in mind at the start i prolly could have included that thought in the "results" section of the lab report.

    • one year ago
  95. hipster Group Title
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    @demitris, please enable private messaging so i may message you about a post that you'd tagged me in a few minutes ago

    • one year ago
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