anonymous
  • anonymous
test dimensionally if the equation v^2 = u^2 +2ax may be correct?
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
you need to square the dimensions used for velocity you need to aggregate the dimensions used for acceleration and displacement presuming you are using metric, you only need to think about metres & seconds, so it should be fairly straightforward for you to do.
anonymous
  • anonymous
v in dimension=L/T=u s=L a=L/T^2 L^2/T^2 = L^2/T^2 + 2 (L/T^2)(L) 0=2(L^2/T2) 0=0 hence equation is correct
IrishBoy123
  • IrishBoy123
bingo!!!

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anonymous
  • anonymous
Checking the equation: v^2 = [LT^-1]^2 = [L^2T^-2] u^2 = [LT^-1]^2 = [L^2T^-2] 2ax = [LT^-2][L] = [L^2T^-2] Since all the physical quantities have same dimensions, Therefore the equation is dimensionally correct.
anonymous
  • anonymous
shivi can you please explain what you did about 2ax
anonymous
  • anonymous
yea sure! Since 2 is a pure number, therefore it is dimensionless. a i.e. acceleration has dimension [LT^-2] x i.e. distance has dimension [L] So. [2ax] = [LT^-2] * [L] = [L^2T^-2] hope it helps! :)
anonymous
  • anonymous
thanks shivi...

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