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2 years ago

When assessing energy resources, it is helpful to use a measure called eroi, which is ________?

1 answer:

4 0

This is the ratio of energy returned to the amount invested. It is just like any other type of ROI value, in that it seeks to be maximized while investing the least amount of resource possible. Maximizing the ROI allows for a business to use its resources much more sparingly and wisely, in comparison to organizations that have to spend more on resources for a smaller gain.

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The state of health and functioning of the liver is often assessed with dye-tracer techniques. The dye used most frequently is b

elena-s [515]

Answer

1.Radioactive or chemical decay

2. X(t) = Xoe^-kt

lnY(t) = -t + lnYo

4. 6.07mg

Explanation:

Let the liver and and blood compartment be represented by the symbol L and B respectively

For the liver

Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,

∫dX/dt = -K1X

By rearrangement and integration;

∫dX/X = -K1t

ln X = -K1t + C

Since At t= 0, X = Xo then

C = lnXo

The equation becomes:

lnX = -K1t + lnXo

lnX - lnXo = -K1t

ln(X/Xo) = -K1t

X/Xo = e^-kt

X(t) = Xoe^-kt

X(t) = Xoe^-0.5t........(2)

Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t

B---------> B(t)

t=0 Yo 0

t=t. Y-b. b

dY/dt = -K12(Y-b)...........(3)

Let Y-b = Y(t)

∫dY/dt = -∫K12t

By rearrangement and integration;

∫dy/Y(t) = ∫-K12dt

lnY(t) = -K12t + C1

at t= 0, C1 = ln(Yo)

Therefore ln Y(t) = -K12t + lnYo

But K12 =1

ln Y(t) = -t + lnYo.............(4)

(3) The assumptions used here

is that of a decay for the liver . The amount remaining taking as the amount of a substance

(4) using the equation 2,

X(t)= Xo e^-K1t........(2)

For time t = 1hour, and an initial amount X = 10mg, K = 0.5

X(t) = 10× e^-0.5

X(t) = 10 × 0.607

X(t) = 6.07mg

(5) within the scope of information presented, I have no data to make this judgment.

3 0

2 years ago

Where has the population reached the carrying capacity in the Graph shown above? (need help quick!)

malfutka [58]

Answer:

The correct answer is option D. point D.

Explanation:

Carrying capacity is characterized as the maximum populace size that a domain can continue inconclusively. For most species, there are four factors that factor into figuring the carrying capacity of an individual population: food accessibility, water resources, habitat, and ecological conditions.

In a graph representation of population size in a time period increases until it reaches carrying capacity, carrying capacity is representing in the graph where the increase in curve becomes stable and makes a plateau.

Thus, the correct answer is option D. point D.

7 0

2 years ago

The problem of a limiting amino acid is most simply overcome by

Reptile [31]

Answer:

By combining vegetarian protein sources you can ensure that you are getting all 9 amino acids.

Explanation:

Giving up meat creates you to not get all your limited amino acids. This causes incomplete proteins and key amino acids from a diet. These amino limiting amino acids are: lysine, threonine, methionine, and tryptophan. Limiting amino acids are found in the shortest supply from incomplete proteins. Incomplete proteins are those found in plant food sources and geletin.

8 0

2 years ago

The picture shows a contractile vacuole of a unicellular freshwater organism. The contractile vacuole regulates the flow of wate

yuradex [85]

The concentration of water is greater outside the cell than inside the cell

8 0

2 years ago

Suppose a human blood cell containing a 0.9 percent solute concentration were put into a container of 0 percent solute solution

Vladimir [108]

Answer and explanation:

If a human blood cell with a 0.9% solute concentration were to be put into a container of 0% solute solution, the cell would get BIGGER.

The cell contains a more concentrated solution than the solution in the container. The difference in concentration would produce an osmotic gradient that would cause water from the container to get inside the cell to even the concentrations - this is going to make the cell much bigger because the entering water would bloat the cell.

In this example, the solution in the container is hypotonic in relation to the cell, while the solution inside the cell is hypertonic in relation to the solution in the container. This is why the water will be moving from outside of the cell to the inside of the cell.

6 0

2 years ago