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2 years ago

6

A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea

sing at a rate of 9 inches per second. When this distance is 61 inches, how fast is the distance between the bottom of the ladder and the foot of the wall changing? The ladder is 152 inches long. (Do not include units in your answer, and round to the nearest hundredth.)

1 answer:

Kitty [74]2 years ago

3 0

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L² ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223 (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

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21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

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with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

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In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

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$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

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25)

The index of refraction of a material is

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26)

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The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

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The critical angle is

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So, the critical angle is larger for the glass-water interface.

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The complete question is shown on the first uploaded image

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