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Murrr4er [49]
2 years ago
8

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

s a mass of 2.40 kg, and the pivot is located 0.380 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Physics
1 answer:
zlopas [31]2 years ago
4 0

Answer:

The  moment of inertia is  I =1.0697 \ kg m^2

Explanation:

From the question we are told that

    The  frequency is  f  =  0.460 \ Hz

    The  mass of the pendulum is  m  =  2.40  \ kg

    The  location of the pivot from the center is d  =  0.380 \ m

     

Generally the period of the simple harmonic motion is mathematically represented as

        T   = 2 \pi  *  \sqrt{  \frac{I}{ m  *  g *  d  } }

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=>    I =  [ \frac{T}{2 \pi } ]^2 *  m*  g * d

But the period of this simple harmonic motion can also be represented mathematically as

        T  =  \frac{1}{f}

substituting values

      T  =  \frac{1}{0.460}

      T  =  2.174 \ s

So

      I =  [ \frac{2.174}{2 * 3.142 } ]^2 *   2.40*  9.8 * 0.380

      I =1.0697 \ kg m^2

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Explanation:

Given

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blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

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T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

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