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frez [133]
3 years ago
10

Electric power plants are the main contributors of what? Oxygen pollution, Sulfur dioxide pollution, thermal pollution, mercury

poisoning
Physics
2 answers:
PIT_PIT [208]3 years ago
7 0

Answer:

sulfur dioxide pollution

Explanation:

Sulfur dioxide is a colorless and extremely polluting gas, resulting from the burning of fossil fuels.

The impurities of fossil fuels allow this gas to be released into the atmosphere, whenever this type of fuel is burned. The activities that most promote the burning of fossil fuels are the burning of diesel oil in vehicles and the burning of coal and oil in power plants. With that, we can say that sulfur dioxide pollution is the main impact that power plants have.

When it reaches the atmosphere, sulfur dioxide reacts with oxygen generating sulfur trioxide, which reacts with moisture in the air and forms sulfuric acid.

zloy xaker [14]3 years ago
6 0
The correct answer is sulfur dioxide pollution

Hope this helped :)
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A man can jog 10 miles in 90 minutes. What’s his speed in mph?
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Answer:

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Explanation:

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what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

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