The dimensions of K in the equation W = frac{ | vedclass

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(A) Given the equation $W = \frac{1}{2}K{x^2}$,where $W$ is work and $x$ is displacement.Dimensional formula of work $W$ is $[M^1 L^2 T^{-2}]$.Dimensi

Vedclass · Accepted Answer

${M^1}{L^0}{T^{ - 2}}$

Vedclass · Answer

(A) Given the equation $W = \frac{1}{2}K{x^2}$,where $W$ is work and $x$ is displacement.Dimensional formula of work $W$ is $[M^1 L^2 T^{-2}]$.Dimensional formula of displacement $x$ is $[L^1]$.Rearranging the equation for $K$,we get $K = \frac{2W}{x^2}$.Substituting the dimensions: $[K] = \frac{[M^1 L^2 T^{-2}]}{[L^1]^2} = \frac{[M^1 L^2 T^{-2}]}{[L^2]} = [M^1 L^0 T^{-2}]$.Thus,the dimensions of $K$ are $[M^1 L^0 T^{-2}]$.

The dimensions of $K$ in the equation $W = \frac{1}{2}K{x^2}$ are:

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