All categories

4 years ago

Which factors improve soil fertility? Select the three correct answers. A. Humus B. Bedrock C. Horizon D. Nitrogen E. Manure

Physics

2 answers:

Katyanochek1 [597]4 years ago

5 0

Correct answer choices are :

A) Humus

E) Manure

D) Nitrogen

Explanation:

Soil fertility leads to the capability of a soil to support agricultural plant growth, i.e. to produce plant environment and occur in continued and steady yields of high quality. A fertile soil will include all the important nutrients for basic plant nourishment, as well as other nutrients required in smaller amounts.

alisha [4.7K]4 years ago

4 0

<h3>Answer;</h3>

Humus
Manure
Nitrogen

<h3>Explanation;</h3>

Soil fertility is a characteristic of soil that refers to the capability of soil to maintain and provide a suitable habitat for growth of agricultural crops, thus resulting to continuous productivity.
The fertility of soil may be influenced or enhanced by various factors which includes, the soil structure, organic matter, soil pH, and water retention among other factors.
Humus, nitrogen, and manure are among the factors that boosts or improves soil fertility, and thus enhances growth and sustainability of agricultural crops by the soil.

You might be interested in

A Plane crashes into the side of a mountain. If the crash took place over 9

Julli [10]

Given parameters:

Duration of crash = 9min

Velocity before crash = 2000 yards per second

Unknown:

Acceleration of the plane = ?

Solution:

Acceleration is the change of velocity with time.

Acceleration = $\frac{V - U}{t}$

where v is the final velocity

u is the initial velocity

t is the time taken

We need to convert the time into seconds from min;

60s = 1 min

So, 9min = 60 x 9 = 540s

So input the parameters and solve;

Acceleration = $\frac{2000 - 0 }{540}$ = 3.7yards/s²

The acceleration of the plane is 3.7yards/s²

6 0

3 years ago

Why is magnesium the limiting reactant in this experiment

Keith_Richards [23]

Answer:

Explanation:

Magnesium is being oxidized by the oxygen in the air to magnesium oxide. This is a highly exothermic combustion reaction, giving off intense heat and light. The reaction of the combustion of magnesium in oxygen is given below:

2Mg(s) + O2(g) → 2MgO(s)

The stoichiometric factor is 2 moles of magnesium are burned for every 1 mole of oxygen (2mol Mg/1mol O2). If the magnesium strip weighs 1 gram, then there is 0.04 mol of magnesium (1 gram divided by 24.3 grams/mol Mg) available in the reaction. The amount of oxygen required to completely react with the magnesium strip is:

0.04 mol Mg x (1 mol O2 / 2 mol Mg) = 0.02 mol O2 x 16 g/mol O2 = 0.32 gram O2.

The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.

4 0

3 years ago

A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the

evablogger [386]

Answer:

Explanation:

hope this helps!

7 0

3 years ago

HELP ASAP!! WILL TRY TO GIVE BRAINLIEST

VikaD [51]

Answer:

Facilitated diffusion and active transport both utilize proteins to transport substances across membranes. Differences between active transport and facilitated diffusion 1. Active transport requires an input of energy, usually ATP, while facilitated transport does not.

5 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago