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asib1214

  • one year ago

Yelp!!!

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  1. rvc
    • one year ago
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    @Michele_Laino

  2. rvc
    • one year ago
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    what is your ans?

  3. Michele_Laino
    • one year ago
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    since we have three complete cycles within 4.8 meters, then the wavelength is: 4.8/3=1.6 meters

  4. rvc
    • one year ago
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    since we have 3 waves so divide 4.8 by 3 because one wave is equal to wavelength

  5. rvc
    • one year ago
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    lol sorry i did not see your post @Michele_Laino

  6. Michele_Laino
    • one year ago
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    no worries :) @rvc

  7. Michele_Laino
    • one year ago
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    the requested velocity, is: 4.8/6=...meters/seconds

  8. Michele_Laino
    • one year ago
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    period is T= 6 /3 =...second frequency f= 1/T =... Hz

  9. Michele_Laino
    • one year ago
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    if the velocity of our wave is unchanged, then we can apply this formula: \[\Large \lambda f = v\] from which we get: \[\Large \lambda = \frac{v}{f} = \frac{{0.8}}{2} = ...meters\]

  10. asib1214
    • one year ago
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    this is how i did it, i first took the inverse of the Ferquency.....F=1/T which agve me the period

  11. Michele_Laino
    • one year ago
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    how did you get 14.5 meters/second

  12. Michele_Laino
    • one year ago
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    v=traveled space/time interval= 4.8/6=0.8 m/second

  13. Michele_Laino
    • one year ago
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    period is T=6/3= 2 seconds

  14. Michele_Laino
    • one year ago
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    I think taht: since within 6 seconds we have three complete waves, then the period is:T=6/3= 2 seconds

  15. Michele_Laino
    • one year ago
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    so the corresponding frequency is: f=1/T= 1/2=0.5 Hertz

  16. asib1214
    • one year ago
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    when solving for the very first question wavelength.

  17. asib1214
    • one year ago
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    i mean second cuz time is given after the wavelength question :(

  18. asib1214
    • one year ago
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    so do we assume we don't know the time while solving for wavelength

  19. Michele_Laino
    • one year ago
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    as I wrote before, in the last part of the first question, we have to assume that the speed of our wave is the same as before, namely, it is 0.8 m/second

  20. asib1214
    • one year ago
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    @rvc do i figure out the amplitude? Please help!!!

  21. asib1214
    • one year ago
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    @Astrophysics

  22. asib1214
    • one year ago
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    how do i figure out the amplitude? Please help!!!

  23. asib1214
    • one year ago
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    the picture is at the top.

  24. Astrophysics
    • one year ago
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    |dw:1435727528455:dw| you can find the amplitude using \[Amplitude = \frac{ Distance }{ Frequency }\]

  25. Astrophysics
    • one year ago
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    They are related.

  26. Astrophysics
    • one year ago
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    No, the amplitude here is measured in decibels here. I think you're just confusing it because of the formula/ units but not really sure about what this actually means.

  27. Astrophysics
    • one year ago
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    \[\lambda = \frac{ v }{ f }\]

  28. asib1214
    • one year ago
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    can you quickly guide me what to do Please?!?!?!?!

  29. Astrophysics
    • one year ago
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    Slow down, ok so we have |dw:1435728513829:dw| so we have 3 complete cycles here, so our wavelength is 4.8/3

  30. Astrophysics
    • one year ago
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    \[y = A \sin \omega t\] you can use this A here is the amplitude the omega is angular frequency and t is the time period.

  31. Astrophysics
    • one year ago
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    \[\omega = f 2 \pi\]

  32. Astrophysics
    • one year ago
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    It's one in the same, math and physics are interrelated...

  33. Astrophysics
    • one year ago
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    Lol, no problem, it can get confusing I guess, but I think you just need a decent understanding of what everything means and you will get it, you can try khan academy they have some nice videos on such topics I'm sure

  34. rvc
    • one year ago
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    i m extremely sorry @asib1214 i was away from the laptop for a long time. Thanks @Astrophysics :) Sorry once again :)

  35. rvc
    • one year ago
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    Do you have the formula for amplitude?

  36. rvc
    • one year ago
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    so use the formula :)

  37. rvc
    • one year ago
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    The maximum displacement of the particle from the mean position is the amplitude.

  38. rvc
    • one year ago
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    :/

  39. rvc
    • one year ago
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    v=n lambda

  40. rvc
    • one year ago
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    :)

  41. rvc
    • one year ago
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    idk :(

  42. rvc
    • one year ago
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    @welshfella please help us

  43. welshfella
    • one year ago
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    i'll have to look later if you still need it gotta go right now

  44. welshfella
    • one year ago
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    there are 3 complete waves in the diagram so one wavelength is 4.8 / 3 m

  45. welshfella
    • one year ago
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    well the amplitude is the distance from the equilibrium point to the crest of the wave. I'm not sure if we have enough information ..

  46. welshfella
    • one year ago
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    i'll have to check that out.

  47. welshfella
    • one year ago
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    A = D/F where D is distance travelled by the wave / frequency F

  48. welshfella
    • one year ago
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    I'm a bit rusty with this stuff.

  49. welshfella
    • one year ago
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    yes i see hmmm

  50. welshfella
    • one year ago
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    1.6 m is correct for the wavelength i'm sure of that

  51. welshfella
    • one year ago
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    yea sorry my recall of this stuff is really hazy. one website i visited gave the formula A = D / F so i guess its must be right

  52. welshfella
    • one year ago
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    D is distance and F = freqency in cycles per second

  53. welshfella
    • one year ago
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    yea sorry i cant be of more help. If i have the time i'll have to revise this stuff - I did it in Physics years back!!

  54. Michele_Laino
    • one year ago
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    hint: in order to find the amplitude, you have to measure the distances D and H as below:

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  55. Michele_Laino
    • one year ago
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    you have to measures the distances H adn D on your drawing, using a scale ruler, then you have to solve this proportion: \[\Large D:4.8 = H:x\] where x is the requested amplitude

  56. Michele_Laino
    • one year ago
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    so we have: \[\Large x = \frac{{4.8 \times H}}{D} = ...\]

  57. Michele_Laino
    • one year ago
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    what is the distance D? Please measure it with a scale ruler

  58. Michele_Laino
    • one year ago
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    how many cm is D? how many cm is H?

  59. Michele_Laino
    • one year ago
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    ok! and how many cm is D?

  60. Michele_Laino
    • one year ago
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    ok! then if we apply our formula, we get: \[\Large x = \frac{{4.8 \times H}}{D} = \frac{{4.8 \times 1}}{{10.5}} = ...meters\]

  61. Michele_Laino
    • one year ago
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    what is x?

  62. Michele_Laino
    • one year ago
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    That's right! better is x=0.457 meters

  63. Michele_Laino
    • one year ago
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    it is a proportion

  64. Michele_Laino
    • one year ago
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    we have finished, since by means of that proportion, our amplitude, is 0.457 meters

  65. Michele_Laino
    • one year ago
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    yes! I am sure.

  66. Michele_Laino
    • one year ago
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    yes! since our procedure is correct!

  67. Michele_Laino
    • one year ago
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    yes! that's right!

  68. Michele_Laino
    • one year ago
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    thanks!

  69. rvc
    • one year ago
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    Thanks @Michele_Laino

  70. Michele_Laino
    • one year ago
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    :) @rvc

  71. asib1214
    • one year ago
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    @Astrophysics A tuning fork with a frequency of 420Hz emits sound with a wavelength of 0.82m in air. If the temperature of the air increases, what will happen to the wavelength and why?

  72. Astrophysics
    • one year ago
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    http://www.sengpielaudio.com/calculator-speedsound.htm

  73. asib1214
    • one year ago
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    i understand that the rectangular wave on the left has a positive displacement because it's above the equilibrium and the triangle trough has a negative displacement, the thing that i don't understand here is that, Why Triangle trough DOESN'T have any displacement?!?!?!?!

  74. Michele_Laino
    • one year ago
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    1) we have a destructive interference when 2 signals overlap each other and they cancel out each other, namely during that overlapping we have no signal, for example the dark fringes in a interference pattern 2) we have a constructive interference, when 2 signals overlap each other and they reinforce each other, so during that overlapping we have a signal whose intensity is greater than each of both overlapped signal. For example the bright fringes in a interference pattern

  75. asib1214
    • one year ago
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    i understand that when two waves interfere with each other, they form a constructive wave, and when two waves with opposite displacement destruct each other!!!

  76. Michele_Laino
    • one year ago
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    a destructive interference occurs, when from the overlapping between 2 signals, the resultant signal has a little intensity, less than the intensities of each signal

  77. asib1214
    • one year ago
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    When these weird shapes confuse me, like when a triangle interferes with a rectangle :( like how do you figure out if they are going to form a constructive or destructive waves :(

  78. Michele_Laino
    • one year ago
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    since the phase shift between the rectangle signal and the triangle signal is 180 degree, namely those signals interfere, each other, with opposite phases

  79. Michele_Laino
    • one year ago
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    you have to subtract the y-coordinates of the triangle signal, from the corresponding y-coordinate of the rectangle signal

  80. asib1214
    • one year ago
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    oh so since the triangle is flipped to 180 degress, but still how do i know the resultant displacement or what shape will wave form during the collision??!!?1

  81. Michele_Laino
    • one year ago
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    as I wrote before, you can get that shape if you subtract the y-coordinates of the triangle signal, from the corresponding y-coordinate of the rectangle signal

  82. asib1214
    • one year ago
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    can you please draw a quick drawing?!?!!?!?? :(

  83. asib1214
    • one year ago
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    please!?!?!?!?!?!!?!?

  84. Michele_Laino
    • one year ago
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    |dw:1436132812859:dw|

  85. Michele_Laino
    • one year ago
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    |dw:1436132987610:dw|

  86. asib1214
    • one year ago
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    i'll do the next two question after this example, please check them for me, I Love You!!! :")

  87. Michele_Laino
    • one year ago
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    then we have: |dw:1436133042674:dw|

  88. asib1214
    • one year ago
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    OMG i'm crying here, Thank you so much, please stick around, i'll do the next two questions on my own. Please check them for me!!!! Thank you!!!!

  89. asib1214
    • one year ago
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    wait one quick question here. how do you know the amplitude is 5, can you just assume and make up your own numbers to figureout the resultant displacement.

  90. asib1214
    • one year ago
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    i nother words, how do yo know what shape is bigger or smaller :(

  91. Michele_Laino
    • one year ago
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    the amplitude A is given by this computation: \[A = \frac{{\max value - \min value}}{2}\]

  92. asib1214
    • one year ago
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    OMG i'm going to ask you a very sily question but how did you come up with 5?!?!?

  93. Michele_Laino
    • one year ago
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    I chose those value arbitrarily

  94. Michele_Laino
    • one year ago
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    values*

  95. asib1214
    • one year ago
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    because the picture didn't give any numbers, so when i'm solving questions on my own, how do i assume the amplitude is 5 and 8 :(

  96. asib1214
    • one year ago
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    oh okay, let me try the next two questions and i'll get back to you.....i hope you'll be here!!! :")

  97. Michele_Laino
    • one year ago
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    no, you have to measure your numbers using a scale ruler

  98. asib1214
    • one year ago
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    ok so in what units???? cm?

  99. Michele_Laino
    • one year ago
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    yes!

  100. asib1214
    • one year ago
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    ok Thanks you Soooooooooooooooooooooooooooooooooooooooooooooooo MCH!!!! :")

  101. asib1214
    • one year ago
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    Love you forever!!! :")

  102. Michele_Laino
    • one year ago
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    Thanks!

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