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9 Group Title

please show ur Work >>>> Dimension of a pressure ?

  • 2 years ago
  • 2 years ago

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  1. 9 Group Title
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    i need to see the steps of solving it plz :)

    • 2 years ago
  2. ash2326 Group Title
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    Do you know the relation between Pressure and force?

    • 2 years ago
  3. 9 Group Title
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    i know \[p = \frac{ F }{ a }\] So \[= \frac{ m a }{a }\]

    • 2 years ago
  4. ash2326 Group Title
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    \[P=ma / A\] \[a\ne A\] can you tell me the difference between a and A?

    • 2 years ago
  5. 9 Group Title
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    idk , u tell me :B

    • 2 years ago
  6. 9 Group Title
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    i just wanna know how we got L to the power -1 and T to the power -2

    • 2 years ago
  7. 9 Group Title
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    how we got the negative powers ?

    • 2 years ago
  8. ash2326 Group Title
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    a= acceleration= distance/ (time )^2= L/T^2=\(LT^{-2}\) A= area= \(L^2\)

    • 2 years ago
  9. 9 Group Title
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    oh yeah

    • 2 years ago
  10. ash2326 Group Title
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    Can you try now?

    • 2 years ago
  11. 9 Group Title
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    gemme a minute

    • 2 years ago
  12. ash2326 Group Title
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    sure :)

    • 2 years ago
  13. 9 Group Title
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    = \[\frac{ M L T ^{-2} }{L T ^{-2} }\]

    • 2 years ago
  14. 9 Group Title
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    is that right -0-

    • 2 years ago
  15. ash2326 Group Title
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    Why you have \(Lt^{-2}\) in the denominator?

    • 2 years ago
  16. 9 Group Title
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    because pressure = \[\frac{ m a }{a }\] and \[a = l T ^{-2}\]

    • 2 years ago
  17. 9 Group Title
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    u just said that >_>

    • 2 years ago
  18. ash2326 Group Title
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    Read my post again,

    • 2 years ago
  19. 9 Group Title
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    is p pressure Force over area ?

    • 2 years ago
  20. ash2326 Group Title
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    Yes :)

    • 2 years ago
  21. 9 Group Title
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    i thought its force over acceleration -.-

    • 2 years ago
  22. 9 Group Title
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    okie my bad

    • 2 years ago
  23. ash2326 Group Title
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    No problem, you got it now?

    • 2 years ago
  24. 9 Group Title
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    yeah -,- thanks again

    • 2 years ago
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