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

the solution of (dy)/(dt)+(k)/(m)y=-g is

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

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

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

The solution of (d v)/(dt)+k/m v=-g is (a) ( b ) (c) v=c (d) e^(( e ) (f) (g) k/( h )g( i ) (j) t (k))( l )-( m )(( n ) mg)/( o ) k (p) (q) (r) (s) (b) ( t ) (u) v=c-( v )(( w ) mg)/( x ) k (y) (z) (aa) e^(( b b ) (cc) (dd) k/( e e ) m (ff) (gg) t (hh))( i i ) (jj) (kk) (c) ( d ) (e) v (f) e^(( g ) (h) (i) k/( j ) m (k) (l) t (m))( n )=c-( o )(( p ) mg)/( q ) k (r) (s) (t) (u) (d) ( v ) (w) v (x) e^(( y ) (z) (aa) k/( b b ) m (cc) (dd) t (ee))( f f )=c-( g g )(( h h ) mg)/( i i ) k (jj) (kk) (ll) (mm)

The velocity v of mass m of a rocket at time t is given by the equation : m (dv)/(dt)+V (dm)/(dt)=0 , where 'V' is the constant velocity of emission. If the rocket starts from rest when t=0 with mass m, prove that : v= V log ((m_(0))/(m)) .

A object falls from rest through a resistive medium. The equation of its motion is (dv)/(dt) = alpha -beta v. Then

Find the particular solution of v(dv)/(dx)=n^(2)x , given v=u when x=a

For the reaction 3A(g)overset(k)toB(g)+C(g) k is 10^(-4)L//mol.min . If [A] = 0.5M then the value of -(d[A])/(dt) (in ms^(-1) is:

For the reaction 3A(g)overset(k)toB(g)+C(g) k is 10^(-4)L//mol.min . If [A] = 0.5M then the value of -(d[A])/(dt) (in ms^(-1) is: