Question

Calculate the velocity of the tip of second's hand of a watch of length 1.5 cm

Answer 1

Answer:

L=1.5 cm=0.015 m

ASSUMPTION: THE CLOCK IS CIRULAR

$total \: \: distance \: covered = 2\pi \: l \\ = 2 \times 3.14 \times 0.015 = 0.0942 \: m \\ now \\ total \: time \: taken = 60 \: sec \\ so \\ velocity = \frac{totl \: disance \: covered \: }{total \: ime \: taken} = </p><p> \frac{0.0942}{60} =</p><p> 1.57 \: mm \: {sec}^{ - 1}$