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amid [387]
3 years ago
7

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

ace, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nmnm (in air). The index of refraction of water is 1.33.
a. The index of refraction of the oil is 1.20. What is the minimum thickness ttt of the oil slick at that spot?
b. Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
c. Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver?
Physics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

(a) 313nm

(b) 250nm

(c) 301nm

Explanation:

This problem involves the concept of thin-film interference. The scientist observes constructive interference. In this case m = 1.

(a) To find the thickness of the oil at that spot we need to first calculate the wavelength in the oil.

       λ = λo/n = 750/ 1.20 = 625nm

λ = wavelength in the material of interest into which light is entering (the oil)

λo = wavelength in the material from which light is coming in or entering.

n = refractive index of the material into which the light is entering.

Light travels from air into the oil. So the refractive index used above is that of the oil and not air.

The formula relating the thickness t to the wavelength is 2t = mλ

m = 1 so,

    2t = λ = 625nm

     t = 625/2 = 313nm

(b) Supposing the refractive index of the water is now n = 1.50

The wavelength in the oil λ = 750/1.50 = 500nm

 The thickness t = λ/2 = 500/2 = 250nm

(c) If the oil has a thickness of 200nm then the wavelength λ = 2t = 2 ×200nm = 400nm

The wavelength in the oil is 400nm. In order to find the wavelength in water we set this wavelength to λo = 400nm. So the wavelength in water seen by the diver is given by the formula

       λ = λo/n

n for water is 1.33

      λ = 400/1.33 = 300.8nm ≈ 301nm.

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3 years ago
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During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal
weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

  • The  Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º
  • The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

  • Vectors are decomposed into a coordinate system
  • The components are added
  • The resulting vector is constructed

 Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

              v = \frac{\Delta d}{t}

              Δd = v t

Where  v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

          t₁ = 2 min = 120 s

          Δd₁ = v₁ t₁

          Δd₁ = 600 120

          Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s    

  time t₂ = 1 min = 60 s

          Δd₂ = v₂ t₂

          Δd₂ = 400 60

          Δd₂ = 24 103 m

   

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

          cos (180 -35) = \frac{x_1}{\Delta d_1}

          sin 145 = \frac{y_1}{\Delta d1}

          x₁ = Δd₁ cos 125

          y₁ = Δd₁ sin 125

          x₁ = 72 103 are 145 = -58.98 103 m

          y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

          cos (90+ 30) = \frac{x_2}{\Delta d_2}

          sin (120) = \frac{y_2}{\Delta d_2}

          x₂ = Δd₂ cos 120

          y₂ = Δd₂ sin 120

          x₂ = 24 103 cos 120 = -12 10³ m

           y₂ = 24 103 sin 120 = 20,78 10³ m

             

The component of the resultant vector are

              Rₓ = x₁ + x₂

              R_y = y₁ + y₂

              Rx = - (58.98 + 12) 10³ = -70.98 10³ m

              Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

             R = \sqrt{R_x^2 +R_y^2}

             R = \sqrt{70.98^2 + 62.08^2}   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{62.08}{70.98}

             θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

            θ = 180 - θ'

            θ = 180 -41.2

            θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

             v = \frac{\Delta R}{t_3}  

             v = \frac{94.30}{120}  \ 10^3

              v = 785.9 m / s

in the opposite direction, that is, the angle must be

               41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

  • To find the initial Barry trajectory is 94.30 10³ m with n angles of  138.8º
  • The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here:  brainly.com/question/15074838

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What are some of the forces affecting rock climbers?​
alexdok [17]

Answer:

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Explanation:

Gravitational force is obviously one of the biggest obstacles in climbing. You are essentially going against this very strong force to pull your body mass up the beautiful terrain. Gravity is defined as the force of attraction between all masses in the universe, gravity is what allows the sport of climbing.

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Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you
Temka [501]

Answer:

   f = 6.37 Hz,       T = 0.157 s

Explanation:

The expression you have is

       y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

      y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

         w = 2π f

         f = w / 2π

from the equation w = 40 rad / s

        f = 40 / 2π

        f = 6.37 Hz

frequency and period are related

       f = 1 / T

       T = 1 / f

       T = 1 / 6.37

       T = 0.157 s

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