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fiasKO [112]
3 years ago
9

Explain the human impact on local lakes and ponds, such as Rough River, Nolin, or Caneyville Watershed. Answer in 6-8 sentences

while using ecosystem vocabulary.
Physics
2 answers:
Finger [1]3 years ago
3 0

Answer:the human impact had some enivornmental effects.

Explanation:

For Climate it has no impact.

Air Quality, it had minor and temporary impacts from construction activities.

While for the Topography,

Geology and Soils Minor, localized

impacts within project

footprint where observed and there was a potential major impacts within the footprint of the lake.

Surface Water Hydrology, it has minor short term adverse impact and a major, long term adverse impacts to Rough River hydrology between the dam and new outlet works Increased risk of major, long

term changes to hydrology upstream of dam. For the Groundwater there was no significant adverse impact. While the Vegetation had minor adverse impact Minor adverse impact.

Aquatic Resources had long-term beneficial impact.

zheka24 [161]3 years ago
3 0

Answer:

Explanation:

The having to have no climate impact this would happen

The quality of air , it has to have the minor and temporary impacts from activities that is been constructed

In the case of the topography

The geology and minor soils, are localized.

The impacts within project footprint that was observed and it was gotten that a great potential impacts around the lake footprint.

Surface Water Hydrology.

The surface water hydrology is said to have a little short term dangerous effect and a great long term dangerous effect on the rough river hydrology that is the dam and outlet which has new great work Increased risk.

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
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Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

<h3>Step B:</h3>

<em>Using the same method as above:</em>

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

<em>Finally we could calculate the total work done and heat added as follows:</em>

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
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\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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