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NeTakaya
2 years ago
12

will mark brainlest 1) Now that you have studied the facts about the Severn River, you are ready to make a decision. What questi

ons would you want answered if you were to determine whether to build a tidal power plant at the Severn River?
Chemistry
1 answer:
chubhunter [2.5K]2 years ago
5 0

Answer:

The questions I would like to answer are:

Will the productivity of the river severn offset the cost of installing the plant?

Is it possible to build a dam on site?

Is it possible to establish submerged turbines in this location?

How would human life and nature be impacted by the plant?

Explanation:

Tidal power plants are signals that use tidal energy to generate electricity. The installation of this type of plant is quite expensive and involves the establishment of dams on site to take advantage of the differences between high and low tides and the establishment of submerged turbines to take advantage of sea currents.

To decide whether I would decide to build a tidal plant on the Severn River, I would need to evaluate a few things to make sure that the project is viable. For that I would ask myself some questions like:

Will the productivity of the river severn offset the cost of installing the plant? (If the installation costs are greater than the capacity of this river to promote high generation of electricity, it would be better to abandon the project)

Is it possible to build a dam on site? Is it possible to establish submerged turbines in this location? (It is necessary to consider the physical and chemical characteristics of the place to know if it is possible to install these two equipment and in a way that they work efficiently)

How would human life and nature be impacted by the plant? (Projects like this include major environmental and social impacts. These impacts must be analyzed and ways to mitigate them must be considered and included in the project.)

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Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

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dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

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integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

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\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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