Planting of crops to purify -lucern, shrubs, trees(possible job creation)

2 answers:

8 0

Plants serve as an effective means for purifying water. Examples include Lucerne and various bushes which easily absorb water and the pollutants contained within it. The process of using nature to cleanse the environment is not novel; the earth naturally employs both soil and plant life for self-purification. Water is constantly in motion, being consumed by plants, soil, and even clouds—essentially involved in a cycle. As water passes through the roots and foliage of plants, the liquid gets absorbed into their tissues. This natural process involves initial absorption and subsequent release from the leaves and roots. Plant species like Reed, Bur-Reed, Sedges, and Water Lilies are noteworthy examples. Sweet flag and water soldier can also recycle water through their root structure. Plants can also extract heavy metals, parasites, harmful microorganisms, and chemical pollutants from the water, though not perfectly. Once these contaminants are taken up by plants, questions arise regarding their disposal. Some of the waste will break down with the plant itself, but chemicals and heavy metals will necessitate special treatment upon the plant's death. An alternative approach for handling this plant waste involves using mushroom spores, which create a thread-like root network known as mycelium; this mycelium produces substances that break down many harmful contaminants, effectively using one plant to clean another.

kicyunya [3.2K]1 month ago

8 0

These plants were utilized to eliminate harmful substances from the environment. In today's world, such crops have evolved into a significant asset and can be a source of profit when managed properly, creating multiple job opportunities along the way. Hope this information is useful.

You might be interested in

Consider the uniform electric field E = (2.5 j + 3.5 k) × 103 N/C. (a) Calculate the electric flux through a circular area of ra

kicyunya [3294]

The desired electric flux Φ is calculated from the electric field E=(2.5• j + 3.5• k) ×10³ N/C and a circular path with a radius r=2.5m. The electric flux across a surface is represented as Φ=∮E•dA. Given the area A lies in the yz-plane, the normal orientation flows in the x-direction with A=πr² leading to dA=2πrdr •i. Thus, Φ evolves to Φ=∮(2.5j + 3.5k)×10³•(2πrdr i). Integrating from r=0 to r=2.5m and noting that components in different directions yield zero, results in Φ equaling 0 Nm²/C. Regarding the second part, when the area vector is at a 45° angle to the xy-plane, we redefine dA as (2πrCos45 i + 2πrSin45 j) dr, leading to the new flux calculation as Φ=10³∮ 5πrSin45 dr, integrating from 0 to 2.5m. With substitutions made, the result comes to Φ=34.71×10³ Nm²/C.

3 0

2 months ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

ValentinkaMS [3465]

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.

In this scenario, you can demonstrate it as follows: 

the initial velocity is Vo 
the launch angle is α 

the initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity becomes 
Vh = Vo×cos(α) 

the total flight duration is the period required to return to a height of 0 m, thus 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time =? 
a = gravitational acceleration = g (= -9.8 m/s²) 
therefore 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now let's examine the horizontal distance. 
r = v × t 
where 
r = horizontal range =? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
therefore 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

1 month ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

ValentinkaMS [3465]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is $\frac{4}{3} \pi (\frac{D}{2})^3$