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Lana71 [14]
3 years ago
6

A wave with a frequency of 14Hz has a wavelength of 3 meters. At what speed will this wave travel​

Physics
1 answer:
topjm [15]3 years ago
6 0

Answer:

42 m/s

Explanation:

by definition of the velocity(speed) of a wave,

The velocity of light, v, is the product of its wavelength, λ , and its frequency, f.

V= fλ

frequency - number of occurances in a unit time

(check the graph)

velocity=frequency*wave.length\\=14Hz*3m\\=42 ms^{-1}

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
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Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

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Explanation:

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