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Tju
3 months ago
13

Soil erosion is a process where the top soil layer is washed away because of the effects of water or wind. An agricultural area

loses its topsoil because of excess flooding. What three actions can be taken to stop the erosion and conserve the fertility of the land?
Plant vegetation so that their roots hold the soil firmly.
Build walls around the farm area.
Add natural fertilizers to the soil.
Cut down tall, strong trees that stand in the middle of the farm.
Increase the plowing of soil and grazing in the farm area.
(Can be more than one answer)
Biology
2 answers:
inysia [2.3K]3 months ago
7 0
<span>Grow plants to ensure their roots stabilize the soil</span>
inysia [2.3K]3 months ago
4 0

Solution: To combat soil erosion, plant vegetation to ensure the roots stabilize the soil.

Construct barriers around the agricultural land.

Incorporate natural fertilizers into the soil.

Soil erosion involves the detachment of the topsoil layer, which is then transported elsewhere due to agents such as wind and water. The uppermost soil layers contain a wealth of nutrients and organic matter, making them crucial for seed germination and plant development. Erosion eliminates these nutrient-rich layers, decreasing soil fertility and hindering plant growth.

Vegetation helps anchor the soil with their root systems, reducing erosion caused by wind and water. The construction of barriers around fields prevents floodwaters from infiltrating, therefore mitigating erosion from water. Adding natural fertilizers enriches the soil with nutrients that support plant growth, allowing roots to better secure the soil, which in turn helps to prevent erosion.

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In a sample of 2,400 people, 1,482 were found to have the dominant (taster) phenotype. Assuming that the population is in Hardy-
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Answer:

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Explanation:

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The phenotype for recessive tasters = 2400 – 1482 = 918.

Recessive phenotypes = 2400 – 1482 = 918.

The frequency of the recessive phenotype = q² = 918 / 2400 = 0.38.

The frequency of the recessive allele q = √0.38 = 0.62.

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The heterozygous frequency =  2pq = 2 × 0.38 × 0.62 = 0.47.

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