admin November 20, 2020 Measurement, Numerical Problems 2 Comments Question 7: Show that (a) KE = ½ mv^{2} (b) PE_{g} = mgh Solution If the dimensions on both sides of the equations are same, the equations will be equal. Since the dimensions of energy are, J = N m ⇒ J = kg ms^{-2}, therefore, dimensions of energy = [M][LT^{-2}] = [MLT^{-2}] …. (A) In order the given equations to be correct, each equation must have these dimensions. Dimensions of KE Comparing with the dimensions of energy in (A), we see both dimensions are same. Therefore, KE=1/2mv^{2} Dimensions of PE Again compare with equation (A), we see the dimensions are same. Therefore, PE_{g} = mgh. MCQs List of Questions Previous Question Tagged (a) KE = ½ mv2, (b) PEg = mgh, Derive dimensions of kinetic energy., Derive dimensions of potential energy, Question 7: Show that, Solved numerical problem 7 on measurement for first year physics new course. Post navigation Previous Previous post: Dimensions of gravitational constant G … msaNext Next post: Twin Paradox 2 Comments Pingback:Dimensions of gravitational constant G … msa – msa Pingback:Numerical Problems on Measurement … msa – msa Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment Name * Email * Website Save my name, email, and website in this browser for the next time I comment.

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