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8090 [49]
3 years ago
5

. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass

will be​
Physics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

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Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
koban [17]

Answer:

A

   N  = 1340.86 \ slits  / cm

B

    \theta  = 15.7^o

Explanation:

From the question we are told that  

      The wavelength is  \lambda  =  650 \  nm  =  650  *10^{-9} \  m  

        The angle of  first bright fringe is  \theta  =  5^o  

        The order of the fringe considered is  n  =1

Generally the condition for constructive interference is  

       dsin (\theta ) = n * \lambda

=>    d =  \frac{1 *  650 *10^{-9 }}{ sin(5)}

=>    d = 7.458 *10^{-6} \  m

Converting to cm

           d = 7.458 *10^{-6} \  m = 7.458 *10^{-6}  * 100 =  0.0007458 \  cm

Generally the number of grating pre centimeter is  mathematically represented as

           N  =  \frac{1}{d}

=>         N  =  \frac{1}{0.0007458}

=>         N  = 1340.86 \ slits  / cm

Considering question B  

   From the question we are told that

     The first wavelength is  \lambda_1 =  650 \ nm  =  650 *10^{-9} \  m

     The second wavelength is  \lambda_2 = 429 \  m   =   420 *10^{-9 } \  m

      The order of the fringe is  n  =  2

       The grating is  N =  5000 \  slits / cm

Generally the slit width is mathematically represented as

              d =  \frac{1}{N  }

=>          d =  \frac{1}{ 5000  }

=>          d =   0.0002 \  c m  =  2.0 *10^{-6} \ m

Generally the condition for constructive interference for the first ray is mathematically represented as

         d sin(\theta_1) =  n *  \lambda_1

=>      \theta_1 = sin^{-1} [\frac{ 2 *  \lambda }{d}]

=>       \theta_1 = sin^{-1} [\frac{ 2 *   650 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_1 = 40.5 ^o

Generally the condition for constructive interference for the second ray is mathematically represented as

         d sin(\theta_2) =  n *  \lambda_2

=>      \theta_2 = sin^{-1} [\frac{ 2 *  \lambda_1 }{d}]

=>       \theta_2 = sin^{-1} [\frac{ 2 *   420 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_2 = 24.8  ^o

Generally the angular separation is mathematically represented as

            \theta  =  \theta_1 - \theta_1

=>          \theta  = 42.5^o -  24.8^o

=>          \theta  = 15.7^o

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Explanation:

It's correct

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The answer would be Ca. Calcium
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