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Sholpan [36]
2 years ago
7

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

ical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.
Physics
1 answer:
Dennis_Churaev [7]2 years ago
6 0

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle \theta_{a},  since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are n_{red} = 2.410 for red light and n_{blue} = 2.450 for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has n_{air} = 1.000.

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

\theta_{crit} = 24.09^{0}

Explanation:

Only the blue refracted ray is related to the critical angle in this question

n_{air} = 1.000

n_{blue} = 2.450

The relationship between the critical angle(\theta_{crit}), n_{air} and n_{blue} can be given as sin \theta_{crit} = \frac{n_{air} }{n_{blue} }

sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\  \theta_{crit} = 24.09^{0}

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Svetach [21]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Force and Power.

Since, according to the Newton's law,

Force = mass * Acceleration.

hence, here

Force = 142 N, accelration = 22.75 m/s2

hence, mass = 142/22.75

===> Mass = 6.24 Kg

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8 0
2 years ago
If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v
dexar [7]
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
5 0
2 years ago
A truck pulled a car of 2,350 kg a distance of 25 meters. If the car accelerates from 3 m/s to 6 m/s, whats the average force ex
faust18 [17]

Answer:

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Explanation:

4 0
2 years ago
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo
irina1246 [14]

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0
2 years ago
A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally a
just olya [345]

Explanation:

The given data is as follows.

  Mass of the ornament (m_{1}) = 0.9 kg

  Length of the wire (l) = 1.5 m

 Mass of missile (m_{2}) = 0.4 kg

 Initial speed of missile (u_{2}) = 12 m/s

         r = 1.5 m

According to the law of conservation of momentum,

                   p_{i} = p_{f}  

     m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

          m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

         0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v

              0 + 4.8 = 1.3v

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Now, the centrifugal force produced is calculated as follows.

            F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}

                       = (0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}

                       = 11.80 N

Hence, tension in the wire is calculated as follows.

              T = F_{c} + (m_{1} + m_{2})g

                 = 11.80 N + (0.9 + 0.4) \times 9.8

                 = 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

4 0
2 years ago
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