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sweet-ann [11.9K]
2 years ago
9

1. What kind of light phenomena happen due to the refraction of light upon hitting the small crystals that make up cirrus or cir

rostratus clouds?
2. A light phenomena which are formed when light passes through clouds and strikes
18

the ice crystals in them?
3. A light phenomenon formed from the combination of several light properties like refraction, reflection, and dispersion?
4. Rainbow are created after a rain shower and . 5. What do you call a visible accumulations of tiny water droplets or ice crystals
in the Earth’s atmosphere?
6. Howmanycolorsinrainbow?
7. What kind of rainbows are formed when light strikes small raindrops with almost the same size?
8. What kind of rainbows are formed when two inner reflections are done instead of one; this causes the reversal of colors in a secondary rainbow in comparison to a primary rainbow?
9. What do you call as a reflective surface that is curved inward and away from the light source?
10. What kind of mirror are curved outward like the bottom
of a spoon?
Physics
1 answer:
Rama09 [41]2 years ago
8 0

1. The light phenomena happen due to the refraction of light upon hitting the small crystals that make up cirrus or cirrostratus clouds is called Sundogs.

2.A light phenomena which are formed when light passes through clouds and strikes the ice crystals in them is called Halos.

3.A light phenomenon formed from the combination of several light properties like refraction, reflection, and dispersion is Rainbow.

4.  Rainbow are created after a rain shower and Sun light.

5. A visible accumulations of tiny water droplets or ice crystals.

in the Earth’s atmosphere are called Clouds.

6. There are seven colors in a rainbow.

7. The kind of rainbows are formed when light strikes small raindrops with almost the same size are Supernumerary rainbows.

8.The kind of rainbows are formed when two inner reflections are done instead of one; this causes the reversal of colors in a secondary rainbow in comparison to a primary rainbow are Secondary rainbows.

9. A reflective surface that is curved inward and away from the light source is concave mirror.

10. The kind of mirror are curved outward like the bottom

of a spoon is convex mirror.

<h3>What is reflection?</h3>

When a light strikes on a shiny surface, it reflects back the light ray in the same medium. The phenomenon is called reflection and the ray is called the reflected ray.

Thus, all the questions are answered.

Learn more about reflection.

brainly.com/question/2254222

#SPJ1

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50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t
r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

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

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

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

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

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

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

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

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

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

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
Why does sound propagate faster in solid bodies than in liquids and faster in liquids than in air?
storchak [24]

Answer:

see below

Explanation:

this is because particles in solids are packed very closely together, thus , the particles collide with each other frequently and thus transfer of energy is faster. however, particles in liquid are closely packed but not as close as in solid so the particles do not collide as frequently. thus, transfer of energy slower than in solid. furthermore, the particles in gas are spaced far apart from each other, thus the particles don't collide with each other frequently, thus transfer of energy is very slow in gas.

hope you get it,

please mark

4 0
3 years ago
What is the relationship between mass and weight
sergiy2304 [10]
Mass is how heavy is it, weight is the size both are the same
5 0
3 years ago
Read 2 more answers
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is
scZoUnD [109]

Answer:4.34 miles

Explanation:

first  Elevation =19^{\circ}

After 1 minute Elevation changes to 59^{\circ}

Ditsance travelled in 1 minute =600\times \frac{1}{60}=10 mile

Now

tan59=\frac{H}{x}

H=xtan59

tan19=\frac{H}{10+x}

H=\left ( 10+x\right )tan19

Equating H

we get

1.319x=10tan19

x=2.61 miles

H=2.61\times tan59=4.34 miles

4 0
3 years ago
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