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

The motion of a particle moving along x-axis is represented by the equation (dv)/(dt)=6-3v , where v is in m/s and t is in second. If the particle is at rest at t = 0 , then

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:

Two solution of H_(2)SO_(4) of molarities x and y are mixed in the ratio of V_(1) mL : V_(2) mL to form a solution of molarity M_(1) . If they are mixed in the ratio of V_(2) mL : V_(1) mL , they form a solution of molarity M_(2) . Given V_(1)//V_(2) = (x)/(y) gt 1 and (M_(1))/(M_(2)) = (5)/(4) , then x : y is

For this reaction N_(2(g))+3H_(2(g))to2NH_(3(g)) The relation between (d(NH_(3)))/(dt) and -(d(H_(2)))/(dt) is :-

Solution of (dy)/(dx)+2x y=y is (a) ( b ) (c) y=c (d) e^( e ) (f) x-( g ) x^((( h )2( i ))( j ) (k))( l ) (m) (n) (b) ( o ) (p) y=c( q ) e^( r ) (s) (t) x^((( u )2( v ))( w )-x (x))( y ) (z) (aa) (c) ( d ) (e) y=c( f ) e^(( g ) x (h))( i ) (j) (k) (d) ( l ) (m) y=c (n) e^( o ) (p)-( q ) x^((( r )2( s ))( t ) (u))( v ) (w) (x)

Equation of motion of a body is (dv)/(dt) = -4v + 8, where v is the velocity in ms^-1 and t is the time in second. Initial velocity of the particle was zero. Then,

For the reaction B+2D to 3T, -(d[B])/(dt)=k[B][D]^(2) . The expression for -(d[D])/(dt) will be

The function u=e^x sin x ; v=e^x cos x satisfy the equation a. v(d u)/(dx)-u(d v)/(dx)=u^2+v^2 b. (d^2u)/(dx^2)=2v c. (d^2v)/(dx^2)=-2u d. (d u)/(dx)+(d v)/(dx)=2v

To establish an instantaneous displacement current of I ampere in the space between the plates of a parallel plate capacitor of (1)/(2) farad, the value of (d V)/(dt) is

The oxidation electrode potential E, of a 0.1 M solution of M^(+) ions (E_(RP)^(o) = - 2.36 V) is :