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Ostrovityanka [42]
3 years ago
12

These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz

one
Question 1 options:

smog


acid rain


chlorofluorocarbons (CFCs)


EPA
Physics
2 answers:
Grace [21]3 years ago
6 0
<span>The correct answer is: Chlorofluorocarbons (CFCs)

Explanation:
In simple words, CFCs are the chemical compounds made up of the following three elements:
1. Carbon
2. Chlorine
3. Fluorine

When CFCs compounds reach the upper atmosphere, the ultraviolet rays coming from the Sun break those compounds into individual elements, which then react with one of the oxygen atom of the ozone (O3) to form new compounds like chlorine monoxides. That way the ozone molecules get destroyed, resulting into the ozone depletion. Hence, the correct answer is Chlorofluorocarbons (CFCs).</span>
kicyunya [14]3 years ago
6 0

Answer:

The correct answer is option C,  

chlorofluorocarbons (CFCs

Explanation:

The chlorofluorocarbons (CFCs) were being used in air conditioners, aerosol spray cans, and industrial cleaning products since 1960s. These CFCs are inflammable, but once they are into atmosphere they move to the upper stratosphere and there they enter into a never ending chain reaction thereby releasing chlorine . The free chlorine react with ozone and destroy its molecule.

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The temperature on the Earth would increase.

The temperature of the stratosphere would decrease.
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Cardiorespiratory fitness improves the efficiency of the cardiovascular and the respiratory systems in delivering oxygen to the
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Your answer should be T
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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
If an object absorbs all colors but red, we see
julia-pushkina [17]
[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]

The answer is A, red.
"Remember, the color you see is light REFLECTING off the surface of that object. If all colors are absorbed in to the surface EXCEPT red, red must be reflected, and you'll see red." - Yahoo User @Chap
8 0
3 years ago
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a
nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi

Q = \frac{\pi^2 r^4}{2}}

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0
3 years ago
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