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babunello [35]
4 years ago
13

Explain why many sustainable fishing and forestry practice depend on the actions of consumers like you.

Engineering
2 answers:
ololo11 [35]4 years ago
7 0

Answer:

Find explanation below.

Explanation:

Sustainable fishing and forestry practice are measures taken t ensure that the fishes and forests are preserved for future generations. Continuous lumbering and fishing can result to a depletion of these resources and in the worst scenario, their extinction. So it lies on both the fishers or lumberers and consumers to take preventive measures. As consumers of these resources we can help in solving these challenges by;

1.  Limiting our consumption of these resources. We can seek alternatives to them as the case may be.

2. Secondly we should take our time to learn about these natural life and the dangers they face. This would help us in choosing sources that adhere to conservative practices.

ehidna [41]4 years ago
6 0

Answer:

The reason why many sustainable fishing and forestry practice depend on the actions of consumers like me is:

  • <u>The offer is aimed at satisfying the demand for these products</u>.

Explanation:

Sustainable fishing and  forestry are based on allowing future generations to access the same products that we currently have, for this reason, a sustainable fishing practice can be waiting for fish to reproduce, in order to not to overexploit them and thus to extinguish them, however, as mentioned in the answer, the offer is aimed at satisfying the demand, therefore, <u>if consumers like me demand sea products even though they are not in the breeding season or without providing the time for the ecosystem to recover the extracted products, the supply will be forced to supply the demand despite the extinction of a species; The same applies to sustainable forestry, if they carry out reforestation work on wood products in proportions 10-1 (10 plants planted for each tree cut) or higher, but the consumer does not provide the prudent time to obtain wood products again, the indiscriminate felling in order to supply the product required by the consumer</u>.

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
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Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

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