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3 years ago

7

Which soil horizon is located closest to the earth's crust?

2 answers:

Pachacha [2.7K]3 years ago

7 0

Answer: O horizon

Horizons refers to the distinct layers of soil lying parallel to the earth surface. Horizons develop as a result of soil formation. Soil forms as a result of weathering or rocks and addition of organic matter from the decomposition of plant and animal waste. Each horizon differs from the others on the basis of color, texture, type of particles present in the soil, type of minerals present and amount of organic matter present in the soil.

O horizon is the soil horizon that is located closest to the earth's crust. This horizon consist of undecayed or partially decayed animal and plant waste like shedded leaves, bark, animal skin and feces. As, the matter remains undecomposed, therefore, this horizon consists of low amount of organic matter and it is less fertile for plant growth.

Tom [10]3 years ago

4 0

The soil horizon that is located closest to the earth's crust is :
O horizon
On this area, lies a lot of rich nutrients and mineral

hope this helps

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This is a bond in which a single pair of electrons is shared between a pair of atoms.

Murrr4er [49]

Answer:single bond

Explanation:

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3 years ago

The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid

faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

$w=2250*2\pi *\frac{1}{60}\\ w=235.61$

$I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}$

Now using Inertia an w

$E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J$

average power= $\frac{11.4J}{0.230s}=50.2 W$

b).

power=t*w

P= $11.5465*0.25*235.61$

P=2.957 kW

4 0

3 years ago

Read 2 more answers

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

So

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

What makes a model not useful?

lina2011 [118]

Answer is b hope this helps

8 0

3 years ago

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

3 years ago