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svetoff [14.1K]
4 years ago
9

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edg

e of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?
Physics
1 answer:
uysha [10]4 years ago
4 0

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and  laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

i=arctan(2.2/.90)

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2=  refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

1\times sin(67.8) = 1.33\times sin(r)

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

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The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

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