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GuDViN [60]
3 years ago
13

If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelength (as measured in water)

would there be constructive interference in the light that reflects back downward
Physics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

      λ_n =  λ / n

where  λ_n is the wavelength in the material and  λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

       2t = (m + ½)  λ_n

       2nt = (m + ½)  λ

let's apply this equation to our case

            λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

        n = n_oil / n_water = 1.5 / 1.33

        n = 1,128

       

let's calculate

        λ = 2 1,128 t / (m + ½)

        λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

          t = 0.001 mm = 1 10⁻⁶ m

the formula remains

        λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values ​​of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

     m + ½ = 2,256 10⁻⁶ / λ

     m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

     m = 2,256 10-6 / 400 10⁻⁹ - ½

     m = 5.64 - 0.5

     m = 5.14

     m = 5

red light  λ = 700 10⁻⁹m

      m = 2,256 1-6 / 700 10⁻⁹ - ½

      m = 3.22 - 0.5

      m = 2.72

      m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

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A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coeff
natulia [17]

Answer:

a

H  =212.6 \  J

b

v  =  7.647  \  m/s

Explanation:

From the question we are told that

   The child's weight is  W_c  =  287 \ N

    The length of the sliding surface of the playground is  L =  7.20 \  m

    The coefficient of friction is  \mu =  0.120

      The angle is \theta = 31.0 ^o

      The initial  speed is  u =  0.559 \  m/s

Generally the normal force acting on the child is mathematically represented as

=>    N  =  mg  *  cos \theta

Note  m *  g  =  W_c

Generally the frictional force between the slide and the child is    

         F_f  =  \mu *  mg  *  cos \theta

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      F =m* g sin(\theta) - F_f

Here  F is the resultant force and it is represented as  F =  ma

=>   ma =   m* g sin(31.0)  - \mu *  mg  *  cos (31.0)

=>   a =  g sin(31.0)-  \mu *  g  *  cos (31.0)

=>  a =    9.8 *  sin(31.0) - 0.120 *  9.8  *  cos (31.0)

=>a =  4.039 \ m/s^2

So

   F_f  =  0.120  * 287  *  cos (31.0)

=> F_f  = 29.52 \  N

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     H  =  F_f  * L

=>  H  = 29.52 *  7.2

=>  H  =212.6 \  J

Generally from kinematic equation we have that

    v^2  =  u^2  +  2as

=>  v^2  =  0.559^2  +  2 * 4.039 * 7.2

=>  v  =  \sqrt{0.559^2  +  2 * 4.039 * 7.2}

=>  v  =  7.647  \  m/s

   

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3 years ago
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
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Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

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3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 2.07 s at the same height as the point of release.
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(a) The ball will accelerate at a speed of 9.81 m/s² while it is in flight.

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<h3 /><h3>What is speed?</h3>

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The information provided in the issue is;

u is the fall's beginning speed in m/sec;

h is the fall's distance

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b. When the ball reaches its highest point, its velocity will be zero.

c. The ball's starting velocity is;

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0=u-gt

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d. The formula is used to determine the highest height it can reach.

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As a result, the ball's greatest height, beginning velocity, and acceleration are all 9.81 meters per second, zero meters per second, 20.30 meters per second, and 21 meters accordingly.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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Normal force is the force exerted when an object is on an surface. So an example could be a pile of books on top of a table.

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