Answer:
a =1.61 × 10⁻⁶ m/s²
Explanation:
given,
length of minute hand = 0.53 m
length of hour hand = 0.26 m
the time taken by the minute hand to complete one revolution is T=3600 s
the angular frequency is
[tex]\omega =\dfrac{2\pi }{T}[/tex]
[tex]\omega =\dfrac{2\pi }{3600}[/tex]
[tex]\omega=0.001745 rad/sec[/tex]
the acceleration is
[tex]a = r \omega^2[/tex]
[tex]a = (0.53) \times (0.001745)^2[/tex]
a =1.61 × 10⁻⁶ m/s²