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(D) The magnitude of the change in velocity $\Delta \vec{v}$ for a particle moving with constant speed $v$ through an angle $	heta$ is given by the f

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$10$

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(D) The magnitude of the change in velocity $\Delta \vec{v}$ for a particle moving with constant speed $v$ through an angle $	heta$ is given by the formula: $\Delta v = 2v \sin\left(\frac{	heta}{2}ight)$ Given: Speed $v = 10\,m/s$ Angle $	heta = 60^{\circ}$ Substituting the values: $\Delta v = 2 	imes 10 	imes \sin\left(\frac{60^{\circ}}{2}ight)$ $\Delta v = 20 	imes \sin(30^{\circ})$ Since $\sin(30^{\circ}) = 0.5$: $\Delta v = 20 	imes 0.5 = 10\,m/s$ Therefore,the magnitude of the change in velocity is $10\,m/s$.

$A$ particle is moving along a circular path with a constant speed of $10\,m/s.$ What is the magnitude of the change in velocity of the particle,when it moves through an angle of $60^{\circ}$ around the centre of the circle? .......... $m/s$

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