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BARSIC [14]
3 years ago
5

A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick

. the indices of refraction of the glass for red and violet light are 1.641 and 1.667, respectively. calculate the dispersion angle (i.e., the difference between the directions of red and violet light within the glass block). 0.0168 deg submit answer incorrect. tries 1/16 previous tries how wide is the beam when it reaches the bottom of the block, as measured along the bottom surface of the block? submit answer tries 0/32 when the rainbow-colored beam emerges from the bottom of the block, the exiting rays of light are once again parallel. calculate the distance between the exiting red and violet rays (i.e., the thickness of the rainbow).
Physics
1 answer:
DerKrebs [107]3 years ago
7 0
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
 So the distance between the red and violet rays is 0.07038 cm.
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Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

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A camera, a drill, and a hand-squeeze flashlight use either a generator or a motor as a component. Identify which component each
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3 years ago
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A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
BigorU [14]

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  f=  -75 \ cm =  - 0.75 \ m

b

  P  =  -1.33 \ diopters

Explanation:

From the question we are told that

    The  image distance is  d_i =  -75 cm

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  d_o =  \infty

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              \frac{1}{f}  =  \frac{1}{d_i}  -   \frac{1}{d_o}

substituting values

             \frac{1}{f}  =  \frac{1}{-75}  -   \frac{1}{\infty}

=>         f=  -75 \ cm =  - 0.75 \ m

Generally the power of the corrective lens is  mathematically represented as

        P  =  \frac{1}{f}

substituting values

       P  =  \frac{1}{-0.75}

        P  =  -1.33 \ diopters

7 0
3 years ago
A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a
Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

V_c = l^3

V_c = 6^3 = 216cm^3

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

V_d = \pi r^2*l

V_d = \pi (\frac{2}{2})^2*6

V_d = 6\pi cm^3

In this way the net volume would be

\Delta V = V_c-V_d

\Delta V = 216cm^3-6\pi cm^3

\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

F= mg

m = \frac{F}{g}

m = \frac{6.6}{9.8}

m = 0.673kg

PART A) From the relation of density as a unit of mass and volume we have to

\rho = \frac{m}{V}

\rho = \frac{0.673}{197.15*10^{-6}}

\rho = 3413.64kg/m^3

PART B) To find the weight of the cube then we apply the ratio of

W = mg

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4 years ago
he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s
alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

F=\mu mg

where

\mu is the coefficient of friction

g=9.8 m/s^2 is the acceleration of gravity

So the work done by it is (in magnitude)

W=Fd=\mu mg d

where

d = 2.0 m is the distance covered

Therefore,

\frac{1}{2}mv^2 = \mu mg d

And solving for \mu,

\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245

8 0
3 years ago
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