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lawyer [7]
3 years ago
14

one layer of earth's atmosphere is called the stratosphere. At one point above earth's surface the stratosphere extends from an

altitude of 16 km to an altitude of 50 km. Write a compound inequality to show the altitudes that are within the range of the stratosphere. (please show work! i really need to know how to figure out these word problems. thanks!)
Physics
2 answers:
masya89 [10]3 years ago
8 0

According to all the given requirements 

stratosphere x starts at an altitude of 16 km

that means  

<span>16≤x

</span> and,as given ,stratosphere ends at an altitude of 30 km:
which means
<span>x≤30</span> combining both of them we get

<span>16≤x≤30</span> So the range of the stratosphere is from 16 to 30 km. 

Agata [3.3K]3 years ago
3 0

Answer:

The Stratosphere layer is the layer next to the troposphere and it extends from an elevation of about 10-15 km to 50-60 km above the surface of earth. In this layer, the temperature increases with the increasing height. This increase in temperature is because of the presence of the ozone. This ozone is also called good ozone as it protects the lives on earth from the harmful ultraviolet radiations emitted from the sun.

In order to show the elevation range of the stratosphere layer, in the form of compound inequalities, we take two equations from the given data-

=> X ≥ 16 -------- (equation 1), where X= thickness of the stratosphere layer.

=> X ≤ 50 -------- (equation 2)

Now, from equation 1 and equation 2, we have,

=> 16 ≤ X ≤ 50

So, in conclusion, we can say that the thickness of stratosphere ranges from 16 to 50 km.

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astra-53 [7]

Answer:

1. 40/20=2meters per second

2. 0

3.240/20=12 meters per second

6 0
3 years ago
When on object’s spectral lines are shifted from their rest wavelengths to longer wavelengths, we say that the object’s spectrum
neonofarm [45]

Answer:

Both statements are true.

Explanation:

When a celestial object (stars, galaxies) is moving away from an observer its spectral lines¹ will be shifted to the red part of the spectrum² (longer wavelength), in the other hand if the celestial body is moving toward the observer, the spectral lines will be shifted to the blue part of the spectrum (shorter wavelength). That is known as the Doppler shift.

This Doppler shift can be explained with the Doppler Effect³, which is defined for the case of light as:    

\frac{\Delta \lambda}{\lambda_{0}} = \frac{v}{c}    (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the rest wavelength, v is the velocity of the source and c is the speed of light.

This redshift in distant galaxies is a strong evidence for the expansion of the universe. Equation 1 also allows the measurement of radial velocity from celestial objects.  

Summary:  

Blueshift:

\lambda

Redshift:

\lambda >\lambda_{0}

¹Spectral lines: Determines the presence of particular elements in the photosphere of an star.

²Spectrum: Decomposition of light in its characteristic colors (wavelengths).

³Doppler Effect: Change in the frequency of a wave as a consequence of the movement from a source relative to an observer or vice versa.

4 0
4 years ago
What is the wavelength for this wave?
Alexus [3.1K]

Answer:

160

Explanation:

6 0
3 years ago
Read 2 more answers
Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,
NikAS [45]
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer:  1.3439 x 10⁵ (days²/AU³)

8 0
3 years ago
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solong [7]

Answer:

100Kg.m/s

Explanation:

From the question, we obtained the following information:

M= Mass = 25kg

V = Velocity = 4m/s

Momentum =?

Momentum = MV = 25x4= 100Kg.m/s

8 0
4 years ago
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