Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. the clock hand has a length of 15.0 cm and a mass of 6.00 g. take the second hand to be a slender rod rotating with constant angular velocity about one end.

Explanation:

First of all, let's calculate the moment of inertia of the second hand. The moment of inertia of a slender rod rotating about one end is given by

where m is the mass of the rod and L is its length. For the second hand, we have

m = 6.00 g = 0.006 kg

L = 15.0 cm = 0.15 m

So, the moment of inertia is

Then, we have to calculate the angular speed of the second hand, which is given by:

where T is the period of the second hand, which is T=60 s. Substituting,

Now we can finally calculate the angular momentum of the second hand, which is equal to the product of the moment of inertia I and the angular speed :

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.