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anonymous

  • one year ago

Please help me out. :)

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  1. anonymous
    • one year ago
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    ok

  2. anonymous
    • one year ago
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    Sure

  3. anonymous
    • one year ago
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    #208 only. :) thankyou.. ^_^

  4. anonymous
    • one year ago
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    Can you please type the question so it is easier to see

  5. anonymous
    • one year ago
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    oww.. okay.. sure. :) wait a minute..

  6. anonymous
    • one year ago
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    Can you see it clear?

  7. anonymous
    • one year ago
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    Yep thanks let me find the answer

  8. anonymous
    • one year ago
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    thankyou.. :) i'll try to answer it again. ^_^

  9. anonymous
    • one year ago
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    Hey there. We will have to use Pythagoras here, part of Mechanics exercises.

  10. anonymous
    • one year ago
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    ahh.. but how to illustrate the several forces?

  11. anonymous
    • one year ago
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    Relax, I will show you.

  12. anonymous
    • one year ago
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    For a, you will have diagrammatically, the following representative situation |dw:1434984303367:dw|

  13. anonymous
    • one year ago
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    Finding the hypotenuse will be the resultant force: \[DeltaP=\sqrt{(-200)^{2}+100^{2}}=223.7=224N\] For the angle, we can use Sine function, giving us\[\sin \beta=\frac{ 200 }{ 224 }\] \[\beta=\sin^{-1} \frac{ -200 }{ 224 }=-63.2\] Given that you do not want negative angles, we use the property 90-B=90-63.2=26.7Degrees. Can you use the same methodology for the other exercises?

  14. anonymous
    • one year ago
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    What do you mean?

  15. anonymous
    • one year ago
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    |dw:1434985341630:dw|

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spraguer (Moderator)
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