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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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Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

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a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

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A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m
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Explanation:

Let the speeds of father and son are v_f\ and\ v_s. The kinetic energies of father and son are K_f\ and\ K_s. The mass of father and son are  m_f\ and\ m_s

(a) According to given conditions, K_f=\dfrac{1}{3}K_s

And m_s=\dfrac{1}{4}m_f

Kinetic energy of father is given by :

K_f=\dfrac{1}{2}m_fv_f^2.............(1)

Kinetic energy of son is given by :

K_s=\dfrac{1}{2}m_sv_s^2...........(2)

From equation (1), (2) we get :

\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}

v_s^2=4(v_f+1.5)^2

Using equation (3) in above equation, we get :

v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s

(b) Put the value of v_f in equation (3) as :

v_s=7.09\ m/s

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The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
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The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

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