Moment of inertia for one rod is expressed as:
<span>
I(1)= I(end) + md^2 = (1\12)mL^2+m(L/2√2)^2=(5/24)mL^2
</span><span>
Therefore, for two rods:
</span><span>
I2 = 2I1 = (5/12)mL^2
</span><span>
For the moment of inertia at the pivot point,
</span><span>
I=I2+2md^2=(5/12)mL^2+2(m(L/2√2)^2)=(2/3)mL^2
</span><span>
Substituting the equations above to the equation for frequency:
</span><span>
f=(1/2π)√(2mgd/I)=(1/4π)√(6g/√2L)</span>
Answer:
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Explanation:rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
<span>a. address all components of fitness
</span>According to the exercise principle of balance, a workout should address all components of fitness.
<span>NOT:
</span>b. focus on balance and coordination
<span>c. target specific areas for improvement </span>
<span>d. vary exercise activities over time</span>
Answer:
-A.
Explanation:
: Hope it's Help:
[correct me if I'm not correct]
Answer:
x = A sin ω t describes the displacement of the particle
v = A ω cos ω t
a = -A ω^2 sin ω t
a (max) = -A ω^2 is the max acceleration (- can be ignored here)
ω = (K/ m)^1/2 for SHM
F = - K x^2 restoring force of spring
K = 4.34 / .0745^2 = 782 N / m
ω = (782 / .297)^1/2 = 51.3 / sec
a (max) = .0745 * 782 / .297 = 196 m / s^2
