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Acceleration = rω^2 = r *[2π/T] ^2
r = 1.3 cm since this is a wrist watch.
T= 60s Secnd hand completes one revolution in 60 second.
a = 1.3 *[2π/ 60] ^2 = 0.0143 cm/s^2 = 1.43 m/s^2
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Hey!

I am assuming that you mean 1.3 cm-long? If not, the units are easily changeable, but I am going to go with 1.3 cm.

To start, this is a uniform circular motion problem, where the hand moves with constant velocity.
Since it is the second-hand of a wrist watch, we know the period of revolution is 60 minutes, or (60 min)*(60 seconds) = 3600 seconds

Now, before we can find the angular acceleration of the hand, we need to find the velocity at which the hand moves, which is expressed as:
T = (2πr)/v
Where T = period of revolution, r = radius of circle, v = velocity at which object moves about the circle.
So plugging in what we know:
(3600 s) = ((2π)(.013 m))/v
Multiplying both sides by velocity:
(3600 s)v = (2π)(.013 m)
Then dividing both sides by 3600 seconds:
velocity = ((2π)(.013 m))/(3600 s)
velocity = .0000269 m/s or 2.69*10^(-5) m/s

Now that we have our velocity, we can solve for the acceleration using the equation:
a = (v^2)/r
Plugging in what we know:
a = (.0000269 m/s)/(.013 m)
a = 5.57*10^(-8) m/s^2

Hope this helps!