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Bad White [126]
4 years ago
9

The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the w

idth of the park. Find the dimensions of the park.
Physics
1 answer:
goldfiish [28.3K]4 years ago
8 0

Answer:

l= 4 mi   : width of the park

w= 1 mi  : length of the park

Explanation:

Formula to find the area of ​​the rectangle:

A= w*l       Formula(1)

Where,

A is the area of the  rectangle in mi²

w is the  width of the rectangle in mi

l is the  width of the rectangle in mi

Known data

A =  4 mi²

l = (w+3)mi    Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l  

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values ​​of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w  = 1 mi  :width of the park

We replace w  = 1 mi  in the equation (1) to calculate the length of the park:

l=  (w+3) mi

l= ( 1+3) mi

l= 4 mi

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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

       

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

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             \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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