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    This is the mark scheme answer to a question I have....

    Acceleration = omega squared x radius = (omega squared x diameter ) / 2
    So.... 2pi squared x diameter x frequency squared = a


    My question is where does the division by 2 disappear ?
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    (Original post by HG1)
    This is the mark scheme answer to a question I have....

    Acceleration = omega squared x radius = (omega squared x diameter ) / 2
    So.... 2pi squared x diameter x frequency squared = a


    My question is where does the division by 2 disappear ?
    Im guessing: (4pi squared x frequency squared x diameter) / 2 <---- that equation comes from subbing omega=2πf into omega squared
    4 / 2, would get rid of the 2 on the bottom,
    hence: 2pi squared x diameter x frequency squared = a

    That's would I think anyways not 100% sure myself mate
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    (Original post by fisher96)
    Im guessing: (4pi squared x frequency squared x diameter) / 2 <---- that equation comes from subbing omega=2πf into omega squared
    4 / 2, would get rid of the 2 on the bottom,
    hence: 2pi squared x diameter x frequency squared = a

    That's would I think anyways not 100% sure myself mate

    Yeah that makes perfect sense mate ! Thanks a lot , I just wasn't seeing it like that for some reason lol
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    (Original post by HG1)
    Yeah that makes perfect sense mate ! Thanks a lot , I just wasn't seeing it like that for some reason lol
    Distance between the 8nc and 2nc charges is 60mm
    At a point between the charges, on the line joining them, the resultant electric field strength is zero. How far is this point from the + 8.0 nC charge?

    Would you mind helping me with this question ?
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    (Original post by HG1)
    Distance between the 8nc and 2nc charges is 60mm
    At a point between the charges, on the line joining them, the resultant electric field strength is zero. How far is this point from the + 8.0 nC charge?

    Would you mind helping me with this question ?
    Which year paper is that? I've worked it out but gotta make sure I got the right answer lol
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    Ok, I've found the question and my answer was right.
    There is a few ways to do this question, I'll list my method here:
    So it's asking for the distance from the 8nc charge. Where Electric field = 0, this will be nearer the +2nc charge
    Name:  question.png
Views: 76
Size:  27.6 KB
    Hope that image shows what I mean. Also we call the distance from the +8nc to this point "X" or whatever letter you want, and the distance therefore from the +2nc to that point is "60mm - X"
    Attachment 521285521289
    Hope that file up their also opens ^

    You basically make the electric field of one equal the other. Then using some maths you can get the answer. I've written out what I did step by step, hope it makes sense.
    Attached Images
     
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    (Original post by fisher96)
    Ok, I've found the question and my answer was right.
    There is a few ways to do this question, I'll list my method here:
    So it's asking for the distance from the 8nc charge. Where Electric field = 0, this will be nearer the +2nc charge
    Name:  question.png
Views: 76
Size:  27.6 KB
    Hope that image shows what I mean. Also we call the distance from the +8nc to this point "X" or whatever letter you want, and the distance therefore from the +2nc to that point is "60mm - X"
    Attachment 521285521289
    Hope that file up their also opens ^

    You basically make the electric field of one equal the other. Then using some maths you can get the answer. I've written out what I did step by step, hope it makes sense.
    Thank you so much much mate ! Such a great help I've Been trying to do this for ages lol
 
 
 
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