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3 years ago

What is the Ozone layer?

2 answers:

4 0

"<span>a layer in the earth's stratosphere at an altitude of about 6.2 miles (10 km) containing a high concentration of ozone, which absorbs most of the ultraviolet radiation reaching the earth from the sun."

Hope this helps!
</span>

djverab [1.8K]3 years ago

4 0

"The ozone layer is the layer in the upper atmosphere that contains a higher concentration of ozone than the rest of the atmosphere"

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On a beautiful night in Washington D.C., you see a mirror image of US Capitol and surrounding scenery. What causes this mirror i

Mekhanik [1.2K]

The correct response is D. This is because light is reflected of the building onto the water that is hitting the building.

3 0

3 years ago

Read 2 more answers

Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.

puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

5 0

3 years ago

Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity i

AlladinOne [14]

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

8 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

The point is at the edge of the disk and the component bodies are:

Murrr4er [49]

Answer:

a. A uniform disk of radius and mass .

Explanation:

The moment of inertia I of an object depends on a chosen axis and the mass of the object. Given the axis through the point, the inertia will be drawn from the uniform disc having a radius and the mass.

4 0

3 years ago