The energy E of a particle at position x at time t is given by E=a/(t(b+x^(2)) Where a and b are constants. The dimensional formula of a is

The velocity v of a particle at time t is given by v=at+b/(t+c) , where a, b and c are constants. The dimensions of a, b, c are respectively :-

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=

If the velocity of a particle "v" as a function of time "t" is given by the equation,v=a(1-e^(-bt) ,where a and b are constants,then the dimension of the quantity a^2*b^3 will be

A particle of mass m is located in a region where its potential energy [U(x)] depends on the position x as potential Energy [U(x)]=(1)/(x^2)-(b)/(x) here a and b are positive constants… (i) Write dimensional formula of a and b (ii) If the time perios of oscillation which is calculated from above formula is stated by a student as T=4piasqrt((ma)/(b^2)) , Check whether his answer is dimensionally correct.

Let gravitational potential energy U is given by U=Asqrtx/(B+x^2) , where x is distance. The dimensional formula of A/B is

The displacement x of a particle at time t moving along a straight line path is given by x^(2) = at^(2) + 2bt + c where a, b and c are constants. The acceleration of the particle varies as

The relation between time t and distance x is t = ax^(2)+ bx where a and b are constants. The acceleration is