For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity.
Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g))
The equation reduces to 3.85 m = 6 * L so L = 0.642 m
chrsclrk · 7 years ago
Be because it’s the answer
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
Coastal areas near water bodies like sea and oceans
I would say 1000c as my answer
