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Leona [35]
2 years ago
8

Match each item with the clean water regulation it describes. (2 points)

Physics
1 answer:
Leona [35]2 years ago
6 0

The correct match of each item to the clean water regulation it describes is as follows:

  • Regulates pollutants discharged into surface waters: Clean water act
  • Covers both surface and ground waters: Safe drinking water act
  • Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
  • Funds sewage treatment plants: Clean water act

<h3>What are the functions of clean water regulation?</h3>

Clean Water Act (CWA) is a regulatory body that establishes the basic structure for the regulation of pollutants discharge and maintenance of quality standards of the surface waters.

On the other hand, the Safe Drinking Water Act was founded to oversee the protection of the quality drinking water. The regulatory body is primarily concerned with potable water all waters, whether from above ground or underground sources.

Therefore, the correct match of each item to the clean water regulation it describes is as follows:

  • Regulates pollutants discharged into surface waters: Clean water act
  • Covers both surface and ground waters: Safe drinking water act
  • Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
  • Funds sewage treatment plants: Clean water act

Learn more about clean water regulation at: brainly.com/question/2142268

#SPJ1

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Answer:

Explanation:

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1. The apparatus used are:

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D is a microscope

2. The uses of the apparatus are:

A - produces the light required to so as to see clearly the movement of the particles.

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3 years ago
wite a paragraph explaining the difference between things that have matter and things that don't have matter.
mihalych1998 [28]

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6 0
3 years ago
A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

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Explanation:

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How far would a jet going 155 m/s travel in 9 s?
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Answer:

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