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YumYum247

  • one year ago

Yelp me please!!!!

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  1. YumYum247
    • one year ago
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  2. YumYum247
    • one year ago
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    I've already done the question but can someone please check it for me!!! ?!?!?! d_p

  3. YumYum247
    • one year ago
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    @Astrophysics Please help me.....!!! :")

  4. YumYum247
    • one year ago
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  5. YumYum247
    • one year ago
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    can you zoo in on it?!?!?!?!

  6. YumYum247
    • one year ago
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    @johnweldon1993 Please help me!!!

  7. johnweldon1993
    • one year ago
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    Just 1 sec finishing up a previous post :)

  8. johnweldon1993
    • one year ago
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    Okay back sorry about that...let me take a look

  9. YumYum247
    • one year ago
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    can you see my answer key?!?!?!?!?!?!

  10. johnweldon1993
    • one year ago
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    It didnt load at first for some reason but yeah I got it

  11. johnweldon1993
    • one year ago
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    So everything looks perfect...did you just want confirmation or did this turn out incorrect?

  12. YumYum247
    • one year ago
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    no just for confirmation but can you help me with net force?!?!?!?! ?

  13. YumYum247
    • one year ago
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    i know that the net force is 24N but at what angle tho?!?!?!

  14. johnweldon1993
    • one year ago
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    Oh okay... well to make your life SO much easier I'm going to show you a quicker easier way to do this question And it will answer the angle question as well :)

  15. YumYum247
    • one year ago
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    Thank you!!! :")

  16. YumYum247
    • one year ago
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    i think i'm looking for the resultant angle between "C" and "A"....i think...

  17. johnweldon1993
    • one year ago
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    Net force of these 2 forces |dw:1442876451291:dw| If we figure out the components of these vectors...we can add them and find the resultant that way...since we know \(\large \text{Resultant} = \sqrt{F_x^2 + F_y^2}\) So...we know that the Force and the COSINE of the angle, give us our 'x' components for the vectors... And the Force and the SINE of the angle give us the 'y' components *Let the 12N force be V1 and 15N be V2 and let us break them into components \[\large V_1 = <12cos(32), 12sin(32)>\] \[\large V_2 = <15cos(24), 15sin(24)>\] Summing the components of the 2 Give us \[\large \text{Resultant} = <12cos(32) + 15cos(24), 12sin(32) + 15sin(24)> \\ = <23.87976, 0.25798>\] And the magnitude of the resultant is our Resultant vector \[\large V = \sqrt{23.87976^2 + 0.25798^2} = 23.88N\]

  18. YumYum247
    • one year ago
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    yes i get about the same amount of force.....

  19. johnweldon1993
    • one year ago
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    Now I'm not sure if you've seen the whole breaking vectors into components...but it is very easy as you can see...and quick! And 1 good thing about it is....you see how we found our resultant to be \[\large V = <23.87976, 0.25798>\] Well...those are the COMPONENTS of the vector SO, a neat little trick is...to find the angle this vector is pointing...we simply take the 'y' components....divide it by the 'x' component...and take the inverse tangent of that answer So \[\large tan^{-1}\theta = \frac{y}{x}\] \[\large \theta = tan^{-1} (\frac{0.25798}{23.87976}) = ??\]

  20. YumYum247
    • one year ago
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    but the thing is that all the angles are given.....so what are we looking for?!?!?!?!

  21. johnweldon1993
    • one year ago
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    |dw:1442877340365:dw|

  22. johnweldon1993
    • one year ago
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    We are to find that "single force" and that "SOME angle"

  23. johnweldon1993
    • one year ago
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    We already found the force...now we need to find what angle it is being applied at

  24. YumYum247
    • one year ago
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    with respect to what?

  25. johnweldon1993
    • one year ago
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    With respect to the horizontal

  26. YumYum247
    • one year ago
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    ok so am i using the same varialbes i used previously to find hte net force....A and C?

  27. johnweldon1993
    • one year ago
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    Nope those are out of the question now because we replaced those 2 forces with this single force We are just using the components of the vector we found like I wrote above

  28. YumYum247
    • one year ago
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    ok

  29. johnweldon1993
    • one year ago
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    I fell like this isnt making sense... So we have 2 forces acting on something By adding those 2...we found a "resultant" force that we can replace those 2 forces with i.e....it wouldnt matter if we used those 2 forces acting at those different angles....or used this 1 new vector acting at some angle....the box would feel the same exact force

  30. johnweldon1993
    • one year ago
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    So instead of having 2 forces of 12N acting at 32 degrees above the horizontal....and 15N acting at 24 degrees below the horizontal We instead us just 1 force of 23.88N acting at SOME angle with respect to the horizontal

  31. johnweldon1993
    • one year ago
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    And that SOME angle....is solved using that formula I wrote above \[\large \text{Angle} = \tan^{-1}(\frac{0.25798}{23.87976}) = ?\]

  32. YumYum247
    • one year ago
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    ok thank you very much.....it's too noisey here....:"D i can't focus.

  33. YumYum247
    • one year ago
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    but i think i got it.....!!! :D

  34. YumYum247
    • one year ago
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    i'm getting 0.61 deg....that's not even a one complete degree....

  35. YumYum247
    • one year ago
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    i don't want to use the component method.... x comp and y comp, please just tell me from what i've done, how do i go about find the angle....Please help :")

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