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svp [43]
2 years ago
8

Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba divers we

ar masks to allow them to form images properly underwater. However, this does affect the perception of distance, as you will calculate. Consider a flat piece of plastic (index of refraction npnpn_p) with water (index of refraction nwnwn_w) on one side and air (index of refraction nanan_a) on the other. If light is to move from the water into the air, it will be refracted twice: once at the water/plastic interface and once at the plastic/air interface.
Part A

If the light strikes the plastic (from the water) at an angle θw , at what angle θa does it emerge from the plastic (into the air)?

Express your answer in terms of nw , np , na , and θw . Remember that the inverse sine of a number x should be entered as asin(x) in the answer box.

Part B

What is the distance to the object in terms of θw and l ?

Express your answer in terms of θw and l .

Part C

If the distance to the object is more than about 0.4 m , then you can use the small-angle approximation tan(θ)≈θ . What is the formula for the distance D to the object, if you make use of this approximation?

Express your answer in terms of θw and l .

Part D

Now use the expression found in Part C for the distance between your eyes and the object at point O, and find the ratio of the apparent distance to the real distance, d/D . Remember that the apparent distance is the distance calculated by your eyes using the angle θa instead of the angle θw . Since we are dealing with small angles, you may use the approximations sin(x)≈x and asin(x)≈x .

Express your answer in terms of nw and na .
Physics
1 answer:
grin007 [14]2 years ago
8 0

Answer:

Check the explanation

Explanation:

A) There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface).

To find out how they relate, draw a picture with the path the light follows in the plastic and the normal to both surfaces.

Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normal are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law,   θa = asin[(nw/na)*sin(θw)]

B) D = l/tan(θw)

C) D = l/θw

D) d/D = na/nw

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When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t
Vlada [557]

A.

Gases do not have a definite volume.

Explanation:

This observation suggests that the gases do not have a definite volume. Gases moves randomly and will expand to take up the shape of containers they are introduced into.

  • The particles of a gas are very far apart.
  • They are held together by very weak attractive forces
  • Gases are random and they frequently collide with one another and the walls of the containers they are put into.
  • They have no fixed volumes

learn more:

Kinetic theory of matter brainly.com/question/12362857

#learnwithBrainly

3 0
2 years ago
A hoop is rolling (without slipping) on a horizontal surface so it has two types of kinetic energy: translational kinetic energy
LenaWriter [7]

Answer:

\dfrac{T}{K}=1

Explanation:

The translational kinetic energy of the hoop is given by :

K=\dfrac{1}{2}Mv^2..................(1)

M is the mass of the hoop

v is the velocity of the hoop

The rotational kinetic energy of the hoop is given by :

T=\dfrac{1}{2}I\omega^2

Since, I=MR^2

\omega=\dfrac{v}{R}

T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2..............(2)

From equation (1) and (2) :

\dfrac{T}{K}=1

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 1.

8 0
2 years ago
Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
BARSIC [14]
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
4 0
2 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
2 years ago
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