Question

1)Dyne X second stands for the unit of
a) force b) Momentum
c)energy d)power

2)A man walks 4 km towards north in one hour and 3 km towards east in next hour .His average velocity is .plzz give Answer with solution
a)5 km/hr b)7 km/hr
c)2.5 km/hr d)3.5 km/hr

3)A car traveling in a straight line moves with uniform velocity v1 over a distance x and another car moves with a uniform velocity v2 over a further distance y.If x=y,then the average velocity v is given by
a)v1+v2/2 b)1/v=1/v1+1/v2
c)root of v1 v2 c)1/v=1/v1 -1/v2

Answer 1

1.  Dyne - unit of force in CGS system.  it is equal to 10⁻⁵ of a Newton.

2.  4 km towards North in one hour: 
     and then 3 km towards East in next one hour.
       Final displacement of the person =  √4²+3²  = 5 km from the point of start.
   The velocity =  total displacement / total time = 5 km /2 hours = 2.5 km/hr

   if you want the average speed then it is = total distance / total time
                      = 7 km /2 hours = 3.5 kmhr
 
3.   We could directly say that

   The car travels a uniform velocity of v1 over a distance x  and a uniform velocity v2 over distance y.  let the times of travel be t1 and t2 respectively.  We assume that  the car travels in the same direction throughout.

        average velocity = ( v1 * t1 + v2 * t2 ) / (t1 + t2)
                                = ( v1 * x/v1 + v2 * y/v2 ) / (x/v1 + y/v2)
                                =  (x + y) (v1 v2) / (v1 y + v2 x)
           if x = y, then,
                        v = 2 x v1 v2 / (v1 + v2) x
                           = 2 / [ 1/v1 + 1/v2 ]
             or,    2/v = 1/v1 + 1/v2

Answer 2

1) Dyne×sec= momentum

Expl:

momentum= m × v

=f/a × v (where m=f/a)

=dyne/(m/s^2)× m/s

=dyne-sec