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Alona [7]
2 years ago
8

The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the poo

l appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level

(b) The pool is filled halfway with water.
______m below ground level
Physics
1 answer:
Yuri [45]2 years ago
4 0

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

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tresset_1 [31]

The definition of the celestial bodies allows us to find that the correct answer for a body that is captured and is in planetary orbit is:

  • Moon

Asteroids are small rocky bodies that rotate around the Sun, when this body enters the atmosphere of a planet and reaches the surface it is called meteoroids.

A meteorite is a fragment of meteoroid, which has been divided in space or the atmosphere during the entrance to the planet, in general they are smaller

A meteor is the atmospheric phenomenon that occurs when the pattern meteorite or meteoroid enters, that is, it does not correspond to a celestial body.

An asteroid satellite or Moon is a celestial object that revolves captures and around another asteroid, this concept can be extended to an asteroid revolving captures and around a planet

A satellite is a celestial body that orbits a planet, its origin is varied and could be formed during the formation of the planet itself, or by capturing a nearby body during the initial formation of the solar system.

Let's examine the different answers

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True. A body captured by a planet is generally called the Moon.

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False. A meteoroid is a body that enters the atmosphere of the plant and reaches its surface.

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In conclusion, using the definition of celestial bodies we can find that the correct answer for a body that is captured and is in planetary orbit is:

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4 0
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List down all the jovian planets in order of increasing distance from the sun
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Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

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2. Venus  - 108 million km

3. Earth  - 150 million km

4. Mars  - 228 million km

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kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

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

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

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v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

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disa [49]

Just divide the both, you will get the answer!

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Explanation:

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