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BloomLocke367

  • one year ago

In the Coral Sea the East Australian Current flows 0.9m/s 22.0° West of South. A sea turtle riding the current, dude, swims due East at 2.58m/s. A fish desperate to find his son, Nemo, is moving across the turtles back at 0.25m/s at 35.0° South of East. a. What is the turtle's velocity? b. What is the fish's velocity? c. How many stripes does the fish have? ^^that is a legitimate question on my homework XD

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  1. BloomLocke367
    • one year ago
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    @sweetburger

  2. sweetburger
    • one year ago
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    might have to google a picture of nemo's dad for c lol

  3. BloomLocke367
    • one year ago
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    ye cx

  4. Jacob902
    • one year ago
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    c

  5. BloomLocke367
    • one year ago
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    I think c is a trick question XD he didn't specify which color stripes lol anyhoo, can you help me with the actual problems? cx

  6. sweetburger
    • one year ago
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    This is how i drew my graph may not be correct because currents confuse me sometimes. |dw:1443218314286:dw| this is kinda just for reference I think we just need to go ahead and find the x and y values for each vector and add them up.

  7. BloomLocke367
    • one year ago
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    how'd you draw the vectors?

  8. BloomLocke367
    • one year ago
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    wait

  9. BloomLocke367
    • one year ago
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    I see now lol

  10. BloomLocke367
    • one year ago
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    so now I break them up into components like normal?

  11. sweetburger
    • one year ago
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    so ill make my lists |dw:1443218569870:dw|

  12. sweetburger
    • one year ago
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    yes break into components

  13. sweetburger
    • one year ago
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    grabbing calc brb

  14. BloomLocke367
    • one year ago
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    so the first set is the current, the second is the turtle, and the last is the fish?

  15. sweetburger
    • one year ago
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    yes

  16. BloomLocke367
    • one year ago
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    wait.. why did you subtract 22 from 270?

  17. BloomLocke367
    • one year ago
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    wait.. nevermind cx I keep going back to the reference drawing and answering my own question XD

  18. BloomLocke367
    • one year ago
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    woe woe woe.. why is the second one 360?

  19. BloomLocke367
    • one year ago
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    (I checked the graph already too lol)

  20. sweetburger
    • one year ago
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    |dw:1443218818993:dw|

  21. sweetburger
    • one year ago
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    alright so considering its right on the x axis pointing to positive infinity you could say either cos(0) or cos(360)

  22. BloomLocke367
    • one year ago
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    why isn't it 180?..

  23. sweetburger
    • one year ago
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    and for the y values it would be the same value for sin(0) sin(360)

  24. sweetburger
    • one year ago
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    180 is on the x axis pointing towards negative infinity

  25. sweetburger
    • one year ago
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    all your x values turn negative and you y values stay 0 when the angle is 180

  26. BloomLocke367
    • one year ago
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    I don't know why this is so confusing for me :/ thanks for being patient.. but I don't understand

  27. sweetburger
    • one year ago
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    ill try to explain it

  28. BloomLocke367
    • one year ago
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    I'll be right back.. dinner is ready so I'm about to grab me some and I'll come back.

  29. sweetburger
    • one year ago
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    |dw:1443219370172:dw|

  30. sweetburger
    • one year ago
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    the angles i showed above will give you the same exact value lets use something from the question above .25m/s cos (335) = .227 if you say .25m/s cos(695) it also equals .227 the 695 came from (360+335)

  31. BloomLocke367
    • one year ago
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    back cx

  32. BloomLocke367
    • one year ago
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    oh okay I get it. the diagram helped XD

  33. BloomLocke367
    • one year ago
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    like how it is in the unit circle

  34. sweetburger
    • one year ago
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    alright so now to find the angle |dw:1443219938255:dw|

  35. sweetburger
    • one year ago
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    yes exactly!

  36. BloomLocke367
    • one year ago
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    270-22 is 248, right? XD sorry, did it in my head. I just want to be sure

  37. sweetburger
    • one year ago
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    |dw:1443220007377:dw|

  38. sweetburger
    • one year ago
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    yes u did the math right lol

  39. BloomLocke367
    • one year ago
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    woe you're going to fast O.O give me time to catch up lol I'm slow XD

  40. sweetburger
    • one year ago
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    my bad just let me know if you have any confusions

  41. sweetburger
    • one year ago
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    wait im not sure i even needed to solve what i solved. the question isnt asking for this stuff *facepalm*

  42. BloomLocke367
    • one year ago
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    O.O so what do I do?

  43. sweetburger
    • one year ago
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    i think this was a simpler 2 part problem in which we subtract the current vector from the speed vector the for the turtle to find the speed of the turtle

  44. BloomLocke367
    • one year ago
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    I have to go :O thanks for the help.. I may come back tomorrow to ask for more help XD

  45. sweetburger
    • one year ago
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    then using the found speed of the turlte combine the vector of the fish

  46. sweetburger
    • one year ago
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    alright im sorry :/

  47. BloomLocke367
    • one year ago
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    it's okay XD I'm going to the football game. Thank you! you still have helped me understand a lot. talk to ya later!!

  48. sweetburger
    • one year ago
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    alright take care

  49. BloomLocke367
    • one year ago
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    @sweetburger sorry I haven't been on. Can you help me now?

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