The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`.

The potential energy of a particle varies with distance x from a fixed origin as V= (Asqrt(X))/(X + B) where A and B are constants . The dimension of AB are

The potential energy of a particle depends on its x-coordinates as U=(Asqrt(x))/(x^2 +B) , where A and B are dimensional constants. What will be the dimensional formula for A/B?

If x times momentum is work, then the dimensional formula of x is

The potential energy of a particle varies with distance 'x' as U=(Ax^(1/2))/(x^(2)+B),where A and B are constants.The dimensional formula for A times B is [M^(a)L^(b)T^(c)] then a=

The potential energy (U) of a particle can be expressed in certain case as U=(A^(2))/(2mr^(2))-(BMm)/(r) Where m and M are mass and r is distance. Find the dimensional formulae for constants.

The potential energy of a particle in a force field is: U = (A)/(r^(2)) - (B)/(r ) ,. Where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle is

The potential energy of a particle varies with x according to the reltiaon U(x) =x^(2)-4x . The point x =2 is a point of