The length of the second's hand in a watch is | vedclass

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(D) The length of the second's hand is $r = 1 \, cm$. The time period $T$ of the second's hand is $60 \, s$.The angular velocity is $\omega = \frac{2\

Vedclass · Accepted Answer

$\frac{\pi\sqrt{2}}{30} \, cm/s$

Vedclass · Answer

(D) The length of the second's hand is $r = 1 \, cm$. The time period $T$ of the second's hand is $60 \, s$.The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = \frac{\pi}{30} \, rad/s$.The linear velocity of the tip is $v = r\omega = 1 	imes \frac{\pi}{30} = \frac{\pi}{30} \, cm/s$.In $15 \, s$,the second's hand rotates by an angle $	heta = \omega t = \frac{\pi}{30} 	imes 15 = \frac{\pi}{2} = 90^\circ$.The change in velocity $\Delta v$ is given by $\Delta v = |\vec{v}_f - \vec{v}_i| = 2v \sin(	heta/2)$.Substituting the values: $\Delta v = 2 	imes \left(\frac{\pi}{30}ight) \sin(90^\circ/2) = 2 	imes \frac{\pi}{30} 	imes \sin(45^\circ)$.$\Delta v = 2 	imes \frac{\pi}{30} 	imes \frac{1}{\sqrt{2}} = \frac{\pi\sqrt{2}}{30} \, cm/s$.

The length of the second's hand in a watch is $1 \, cm$. The change in velocity of its tip in $15 \, seconds$ is:

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