Answer:
Explanation:
We can see from the question that
[tex]T 2\pi\sqrt{\frac{I}{mgh} }[/tex]
and [tex]I = \frac{ml^2}{3}[/tex]
[tex]m = 16[/tex]% of [tex]67kg[/tex]
=> [tex]m =10kg[/tex]
from the question [tex]l =48[/tex]% of [tex]1.83m[/tex]
Substituting this into the equation
[tex]I = \frac{10.72 * (0.8784)^2}{3}[/tex]
=> [tex]I = 2.7571 \ kg m^2[/tex]
[tex]h = 0.5 * L[/tex]
[tex]h = 0.5 * 0.8784[/tex]
[tex]h = 0.4392m[/tex]
From the equation above
[tex]T = 2 \pi \sqrt{\frac{2.7571}{10 .72 * 9.81 * 0.4392} }[/tex]
[tex]T = 1,534\ sec[/tex]