Question

Particle is moving with speed 10ms parallel to x axis from 0,2m fimd angular speed of the parricle about origin at t=0.2 sec

Answer 1

r = perpendicular distance of particle from the origin  = 2 meter

v = speed of the particle parallel to x-axis = 10 meter per second

w = angular speed of the particle moving parallel to x-axis about the origin

we know that angular speed , linear speed and perpendicular distance are related as

w = v/r

inserting the values mentioned above

w = 10 meter per second / 2 meter

w = 5 radian/sec

Answer 2

Answer:

2.5 rad/s

Distance travelled after 0.2 sec = 0.2 * 10 = 2             [as a = 0]

position of particle (2,2)

Radius after 0.2 sec = distance between origin and particle

                                   =  $\sqrt{2^{2} +2^{2} }$ = $2\sqrt{2}$

By conservation of angular momentum

$mv_{1} r_{1} =mv_{2} r_{2}$

$(10)(2)=(v_{2} )(2\sqrt{2} )$

$v_{2} =\frac{10}{\sqrt{2} }$

​ω = $\frac{v_{2} }{r_{2} }$

   = $\frac{10}{(\sqrt{2})(2\sqrt{2}) }$

   = $\frac{10}{4}$

   = 2.5 rad/s

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