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zysi [14]
3 years ago
8

Factors that influence Wellness

Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

Help, Fine, and Better.

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Science Tolliver
Nookie1986 [14]

Answer:

Objects; waves.

Explanation:

Waves interact with objects and other waves. Thus, waves are used on objects such as mobile phones and can be transformed from one form to another.

There are various types of waves in our physical environment such as gamma rays, x-rays, ultraviolet waves, radio waves etc.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.

Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data. Some examples of communication technologies that uses radio waves are radio set, mobile phones, television etc.

6 0
3 years ago
Use the drop-down menus to complete the
bagirrra123 [75]

Answer:

Cocoa mix is the: Solute

Water is the: Solvent

The solution has reached: Saturation

Explanation:

5 0
4 years ago
The ________________ is the region around the Sun, extending more than one million kilometers from its surface, where the temper
erma4kov [3.2K]
Corona is the right answer
8 0
3 years ago
Read 2 more answers
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
During summer in the northern hemisphere _____.
Kipish [7]
For part of our orbit the northern half of Earth is tilted toward the Sun. This is summer in the northern hemisphere; there are longer periods of daylight, the Sun is higher in the sky, and the Sun's rays strike the surface more directly, giving us warmer temperatures
7 0
3 years ago
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