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lawyer [7]
2 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]2 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]2 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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Answer:

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8 0
2 years ago
Read 2 more answers
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0
3 years ago
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