Consdier the equation
`(d)/(dt)[intvecF.dvecs]=A[vecF.vecp]` Then dimension of A will be (where `vecF=` force, `dvecs=` small displacement, t=time and `vecp=` linear momentum).

Consider the equation d/(dt)(int vec(F). dvec(S))=A(vec(F).vec(p)) where vec(F) equiv force, vec(s) equiv displacement, t equiv time and vec(p) = momentum. The dimensional formila of A will be

Consider the equation d/(dt)(int vec(F). dvec(S))=A(vec(F).vec(p)) where vec(F) equiv force, vec(s) equiv displacement, t equiv time and vec(p) = momentum. The dimensional formula of A will be

Consider the equation d/(dt)(int vec(F). dvec(S))=A(vec(F).vec(p)) where vec(F) equiv force, vec(s) equiv displacement, t equiv time and vec(p) = momentum. The dimensional formula of A will be

Find the dimensions of a/b in the equation. F=asqrtx+bt^(2) where F is force, x is distance and t is time.

Force on a particle in one-dimensional motion is given by F=Av+Bt+(Cx)/(At+D) , where F= force, v= speed, t= time, x= position and A, B, C and D are constants. Dimension of C will be-

Force on a particle in one-dimensional motion is given by F=Av+Bt+(Cx)/(At+D) , where F= force, v= speed, t= time, x= position and A, B, C and D are constants. Dimension of C will be-

Force on a particle in one-dimensional motion is given by F=Av+Bt+(Cx)/(At+D) , where F= force, v= speed, t= time, x= position and A, B, C and D are constants. Dimension of C will be-

The distance x of a particle moving in one dimensions, under the action of a constant force is related to time t by the equation, t=sqrt(x)+3 , where x is in metres and t in seconds. Find the displacement of the particle when its velocity is zero.