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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

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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