State the principle of homogeneity. test the dimensional homogeneity of equations— s = ut +1/2 at2
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♦ Units and Dimensions ♦
• Principle of Homogeneity : Dimensions of expressions on either sides of a dimensional equation are same
• Homogeneity Test :
[tex] s = [ M^0L^1T^0] \ ; \\ \\ ut + ( 1/2 )at^2 = [M^0L^1T^{-1}] [M^0L^0T^1] + [M^0L^1T^{-2}][M^0L^0T^{2}] \\ \ \ \ \ \ \ = [M^0L^1T^{-1+1}]+ [M^0L^1T^{-1+1}] \\ = [M^0L^1T^0][/tex]
Since we find that the Dimensional Formulas for both sides are equivalent, the homogeneity of equation is satisfied and hence, the equation is dimensionally correct
_____________________________________________________________
_____________________________________________________________
For further doubts concerning Dimensions and Homogeneity :
○ https://www.sciencetopia.net/physics/dimensional-equation-formula
_____________________________________________________________
Hope that helps
♦ Units and Dimensions ♦
• Principle of Homogeneity : Dimensions of expressions on either sides of a dimensional equation are same
• Homogeneity Test :
[tex] s = [ M^0L^1T^0] \ ; \\ \\ ut + ( 1/2 )at^2 = [M^0L^1T^{-1}] [M^0L^0T^1] + [M^0L^1T^{-2}][M^0L^0T^{2}] \\ \ \ \ \ \ \ = [M^0L^1T^{-1+1}]+ [M^0L^1T^{-1+1}] \\ = [M^0L^1T^0][/tex]
Since we find that the Dimensional Formulas for both sides are equivalent, the homogeneity of equation is satisfied and hence, the equation is dimensionally correct
_____________________________________________________________
_____________________________________________________________
For further doubts concerning Dimensions and Homogeneity :
○ https://www.sciencetopia.net/physics/dimensional-equation-formula
_____________________________________________________________
Hope that helps
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