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qwelly [4]
3 years ago
13

a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

erlocity of 50m/s by a person standing at the foot of the building.find the time after which the two stones meet.​
Physics
1 answer:
solniwko [45]3 years ago
6 0

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

                            x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

                S - x = U tₓ - ½ gtₓ²  (∵ the motion of the stone is in opposite direction)

Adding these two equations,

                      x + (S - x) = U tₓ

                               S = U tₓ

                               200 = 50 tₓ

∴                                  tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s   

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A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What
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Answer:

The  wavelength is \lambda_ 1 =  574.2 nm

Explanation:

From the question we are told that  

      The  thickness is t =  198 nm  =  198 *10^{-9 }\ m

      The refractive  index of the non-reflective coating is  n_m  =  1.45  

       The  refractive  index of glass is n_g  = 1.50

       

Generally the condition for  destructive  interference is mathematically represented as

            2 *  n_m *  t  *  cos (\theta) =  n  *  \lambda

Where \thata \theta is the angle of refraction which is  0° when the light is strongly transmitted

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        so  

             \lambda = \frac{2 *  n *  t  *  cos (\theta )}{n}

at the point n =  1  

           \lambda _1 = \frac{2 *  1.45  *  198*10^{-9}  *  cos (0 )}{1}

           \lambda_1  = 574.2 *10^{-9}

          \lambda_1  = 574.2 nm

at  n =2  

         \lambda _2  =  \frac{\lambda _1 }{2}

         \lambda _2  =  \frac{574.2*10^{-9} }{2}

         \lambda _2  =  2.87 1 *10^{-9} \ m

         \lambda _2  =  287. 1  nm

Now we know that the wavelength range of visible light is  between

           390 \ nm \to  700 \ nm

   So the wavelength of visible light that is been transmitted is  

          \lambda_ 1 =  574.2 nm

           

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