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anonymous

  • 5 years ago

I have a doubt which needs to be urgently solved cause i got a test tommorow!!!!!!!!The question is in the replies cause i cannot write a diagram or attach a diagram here!!!!! Its physics i know its a wrong section to ask at but i have no choice!!!!!!!

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  1. anonymous
    • 5 years ago
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    I want to know how can we find the vertical component of the force vector here for normal reaction using \[N _{a}\cos \Theta\] rather than\[N_{a}\sin \Theta\]

  2. watchmath
    • 5 years ago
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    It depends on where your \(\Theta\) is

  3. anonymous
    • 5 years ago
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    hang on i am scanning the diagram

  4. anonymous
    • 5 years ago
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    here is the diagram pls answer quickly!!!!!!!!!!

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  5. anonymous
    • 5 years ago
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    this fairly easy

  6. anonymous
    • 5 years ago
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    so answer it pls!!!!!!!!!!!!!

  7. anonymous
    • 5 years ago
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    well , just consider the angle between theta1 and the vector mg

  8. anonymous
    • 5 years ago
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    so then!!!!!!!!!!

  9. anonymous
    • 5 years ago
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    this will be 90 - theta

  10. anonymous
    • 5 years ago
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    then do trig in that triangle

  11. anonymous
    • 5 years ago
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    fine!!! further ahead!!!!!!!

  12. anonymous
    • 5 years ago
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    yeh.. im answering it , its just a bit slow lol

  13. anonymous
    • 5 years ago
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    k sorry!!!!!!

  14. anonymous
    • 5 years ago
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    i need to know why is the vertical component of say Na NaCos Theta and not Na sin Theta

  15. anonymous
    • 5 years ago
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    yeh , its basically because sin(90-x) = cos(x)

  16. anonymous
    • 5 years ago
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    lol, thats wrong cos(90-x) = sin(x)

  17. anonymous
    • 5 years ago
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    but is that the triangle we can use to do this cause i don't think so!!!!!!! and now i am confused with the two comments u made

  18. anonymous
    • 5 years ago
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    can any1 make a quick video and attach it i know i am asking way too much but pls do if possible!!!!!!!!

  19. anonymous
    • 5 years ago
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    ill draw a picture on paint , hopefully it will help lol

  20. anonymous
    • 5 years ago
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    k hopefully!!!!!!!!!!!!

  21. anonymous
    • 5 years ago
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  22. anonymous
    • 5 years ago
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    ill do another one

  23. anonymous
    • 5 years ago
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  24. anonymous
    • 5 years ago
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    second picture is just redrawing the top triangle in the previous picture, and marking in what you know

  25. anonymous
    • 5 years ago
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    now in the second triganle you apply trig sin( 90-x) = Na / Fy

  26. anonymous
    • 5 years ago
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    but sin(90-x) = cos(x)

  27. anonymous
    • 5 years ago
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    so Fy = Na / cos(x)

  28. anonymous
    • 5 years ago
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    wait dont think thats right

  29. anonymous
    • 5 years ago
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    thanks for ur effort but cud u use a the line tool to make the diagram cause this is really strange i am not able to make sense

  30. anonymous
    • 5 years ago
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    to be honest I dont really know what you are asking , what is the actual question ?

  31. anonymous
    • 5 years ago
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    and if you look at it , the second triangle , the one that has two sides ( one being the y axis and the other being the plane along the track

  32. anonymous
    • 5 years ago
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    that is a right angled trangle , and the angle between the track and the y axis is just 90-theta

  33. anonymous
    • 5 years ago
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    then you re draw the triangle and apply basic trig ratios and simply sin(90-x) to cos(x)

  34. anonymous
    • 5 years ago
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    i am trying to relate it to the concept of vectors which we can represent on the x axis and the y axis!!!!!

  35. anonymous
    • 5 years ago
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