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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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Answer: a all the outer planets are large

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3 years ago
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A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.
gavmur [86]

Answer:

(a) E = -1.02 \times 10^5~N/C

(b) E = -9.7 \times 10^4~N/C

Explanation:

(a) The electric field for a point charge is given by the following formula:

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}

Hence, dE is now:

dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}

Therefore, vertical component of dE now becomes

dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z =  \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C

(b) We will have a similar approach, but a simpler integral.

dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi

E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

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3 years ago
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How would you describe the layers of the earth
pychu [463]
The inner core is solid, while the outer core is liquid.
8 0
4 years ago
Plz help!!! Will give brainsiest!!!!
Zanzabum

Answer:

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Explanation:

The reaction is a neutralization reaction since an acid, aqeous H2SO4 reacts completely with an appropriate amount of alkali, aqueous KOH to produce salt, aqueous K2SO4 and liquid water, H2O only.

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Alkali + Acid → Salt + Water.

During this reaction, 2 moles of KOH neutralize 1 mole of H2SO4 to yield 1 mole of K2SO4 and 2 moles of H2O.

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