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Step-by-Step Solution
Step 1: Identify the known quantities
• Length of the second hand (radius of circular motion), $R = 0.1\,\text{m}$.
• Time period of one revolution for the second hand, $T = 60\,\text{s}$.
Step 2: Calculate the angular speed
The angular speed $ \omega $ of the second hand is given by:
$$ \omega = \frac{2\pi}{T}. $$
Substituting $T = 60\,\text{s}$:
$$ \omega = \frac{2\pi}{60} \approx 0.105\,\text{rad}\,\text{s}^{-1}. $$
Step 3: Use the centripetal acceleration formula
For uniform circular motion, the centripetal acceleration $a$ at the tip of the hand is:
$$ a = \omega^2 \, R. $$
Substituting $\omega \approx 0.105\,\text{rad}\,\text{s}^{-1}$ and $R = 0.1\,\text{m}$:
$$ a = (0.105)^2 \times 0.1 \,\text{m}\,\text{s}^{-2} \approx 1.1 \times 10^{-3}\,\text{m}\,\text{s}^{-2}. $$
Step 4: Conclude the order of magnitude
Since the calculated acceleration is about $1.1 \times 10^{-3}\,\text{m}\,\text{s}^{-2}$, the average acceleration is of the order of $10^{-3}\,\text{m}\,\text{s}^{-2}$.
