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nignag [31]
3 years ago
6

An example of sunlight passing through clouds during sunset results in what is often called the cloud's silver lining. Pastel sh

ades of blue, pink, purple, and green can be observed at times of cloud cover. What is the cause of these patterns in the sky?
Physics
2 answers:
Furkat [3]3 years ago
8 0
Water in the atmosphere and within the clouds themselves cause light waves to refract and reflect, creating an assortment of colors. The patterns occur when light is absorbed by water droplets within the clouds. The amount of absorption that occurs depends on the temperature of the air.  The patterns occur when light is reflected from water droplets within the clouds. The amount of reflection that occurs depends on the wavelength of the light, and shorter wavelengths are reflected at a greater angle than longer ones. 

Hope this helps!!


IRINA_888 [86]3 years ago
6 0

Answer:

Answer is D

Explanation:

The patterns occur when light is diffracted from water droplets within the clouds. The amount of diffraction that occurs depends on the wavelength of the light, and shorter wavelengths are diffracted at a greater angle than longer ones.

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How much water
nevsk [136]

Answer:

The mass is  m  =  3.75 \  kg

Explanation:

From the question we are told that

     The initial temperature is  T_1 =  15^oC

      The final  temperature is  T_2 =  100 ^o C  \ (boiling point )

Generally the maximum heat produced by  1  Liter   of  natural gas is  9000 \ cal

 So the amount of heat produced by 100 L is  

              E = 9000 * 100

=>           E = 9000 00 \  cal

Generally given that the efficiency is  \eta =  0.35

Then actual heat received by the water is

         H  =  0.35 *  E

=> H  =  0.35 *   9000 00

=> H  =  315000 \  cal

Converting to kcal  

=> H  =  315000 \  cal = \frac{315000}{1000} =  315 \ kcal

Generally the specific heat of water is  

       c_w  =  1 kcal/ kg \cdot ^oC

Generally the heat received by the water is mathematically represented as

       H  =  m *  c_w  *  (T_2 - T_1)

=>     315  =  m *  1   *  ( 100  - 15 )

=>     m  =  3.75 \  kg

     

6 0
3 years ago
HELP PLEASE ASAP!!!<br> I CAN'T FAIL PHYSICS<br> xoxo thank you!
Allushta [10]
They are malleable and lustrous, and can conduct both electricity and thermal heat
5 0
2 years ago
engineers who design battery operated devices suck as sell phones and MP3 players try to make them as efficient as possible. An
ICE Princess25 [194]
If the 5,500 J of sound and light is the ONLY useful output
from the phone, then the phone's efficiency is

                   (5,500J / 10,000J)  =  0.55  =  55% .

But the test engineer forgot one little minor almost insignificant detail.
As a test engineer myself, I'd say that he needs to turn in his laptop
and soldering iron, and think about changing his career to a job for
which he may be better suited, like 8 hours a day in a highway toll-booth.  

What about that little radio transmitter and receiver inside the phone,
that maintain digital RF communication with the nearest cell tower ?
Without that microscopic radio transceiver ... plus 30 or 40 apps
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8 0
3 years ago
Which of the following is a force or motion that
faltersainse [42]

Answer: this answer is D. That's what I think it is

Explanation:

7 0
2 years ago
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
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