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vodka [1.7K]
4 years ago
15

A sprinter accelerates from rest to 10.0 m/s in 1.35 s l. What is her acceleration?

Physics
1 answer:
ale4655 [162]4 years ago
8 0

Answer:

dsfsdfgfdsggfgsgsfsfgfsgffsfgfgfgf

Explanation:

sgdfggsfdsgfgsgsmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

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A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the
MArishka [77]

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

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