Show that force can be expressed as the product of mass and acceleration. [Hint `F=(dp/dt)-(d/dt)mv- m(dv/dt)-ma`]

Assertion: A given force applied in turn to a number of different masses may cause the same rate of change in momentum in each but not the same acceleration to all. Reason: F=(dp)/(dt) and a=F/m

The rate of the reaction may be expressed by the following different ways . (1)/(2) (d[A])/(dt)=-(1)/(3)(d[B])/(dt)=(-d[C])/(dt) reaction is

A block os mass m moves on as horizontal circle against the wall of a cylindrical room of radius R. The floor of the room onwhich the block moves is smoth but the friction coefficient between the wall and the block is mu . The block is given an initial speed v_0 . As a function of the speed v write a. the normal force by the wall on the block. b. thefrictional force by the wall and c. the tangential acceleration of the block. d. Integrate the tangential acceleration ((dv)/(dt)=v(dv)/(ds)) to obtain the speed of the block after one revoluton.

The rate of a reaction is expressed in different ways as follows +1/2 (d[C ])/(dt)=-(1)/(5) -(d[D])/( dt)=+(1)/(3 ) [d[A ])/(dt) =-(d[B])/(dt) the reaction is ________.

The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bBrarr cC+dD is given as: rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt) or expression can be written as : rate =K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d) . At equilibrium, rate = 0 . The constants K, K_(1), K_(2) are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. For a gaseous reaction, the rate is expressed in terms of (dP)/(dt) in place of (dC)/(dt) or (dn)/(dt) where C is concentration, n is number of moles and 'P' is pressure of reactant. The three are related as:

IF t=v^2/2 , then (-(df)/(dt)) is equal to , (where f is acceleration)

The solution of (dv)/(dt)+(k)/(m)v=-g is