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iragen [17]
3 years ago
15

A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

illate in a vertical plane. Determine the period of oscillation of the pendulum if L = 5.95 m.
Physics
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

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Answer:

Part A:

The proton has a smaller wavelength than the electron.  

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Part B:

The proton has a smaller wavelength than the electron.

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Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

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\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

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Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

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For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

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Then, equation 2 can be used:

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\lambda = 1.29x10^{-13}m    

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