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wlad13 [49]
3 years ago
9

Question 6 (5 points)

Physics
1 answer:
myrzilka [38]3 years ago
4 0

Answer:

B. Two tuning forks that vibrate at the same frequency are near each other. One tuning fork is struck with a mallet so that it vibrates.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Resonance can be defined as a phenomenon in which an object is forced to vibrate as a result of the vibration of a second object at the same natural frequency of the first object. Thus, it's an increase in the amplitude of oscillation that occurs when the frequency of a periodically applied force is equal to the natural frequency of the system on which it's acting.

Hence, the scenario that best describe a condition in which resonance can occur is two tuning forks that vibrate at the same frequency are near each other. One tuning fork is struck with a mallet so that it vibrates.

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b. 460.8 m/s

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The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

f=\frac{v}{2L}

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

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v = 6,000 m/s is the speed of the wave

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Answer:

The maximun distance is  z_1 = z_2 = 0.0138m

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       The wavelength are  \lambda _ 1 = 540nm  (green) = 540 *10^{-9}m

                                           \lambda_2 = 450nm(blue) = 450 *10^{-9}m

        The distance of seperation of the two slit is d = 0.180mm = 0.180 *10^{-3}m

        The distance from the screen is D = 1.53m

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

           z = \frac{m \lambda D}{d}

   Where m is  the order of the fringe

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        z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}

             z_1=0.00459m_1 m

                 z_1= 4.6*10^{-3}m_1 m ----(1)

For the second  wavelength  we have              

        z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}

        z_2 = 0.003825m_2

        z_2 = 3.825 *10^{-3} m_2 m  ----(2)

From the question we are told that the two sides coincides with one another so

            zy_1 =z_2

         4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m

          \frac{m_1}{m_2}  = \frac{3.825 *10^{-3}}{4.6*10^{-3}}

Hence for this equation to be solved

       m_1 = 3

and  m_2 = 4

Substituting this into the  equation

                      z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}

      Hence z_1 = z_2 = 0.0138m

                       

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