All categories

3 years ago

What would cause soil horizons to become mixed?

Physics

1 answer:

AfilCa [17]3 years ago

7 0

Answer:

The process of soil formation generally involves the downward movement of clay, water, and dissolved ions, and a common result of that is the development of chemically and texturally different layers known as soil horizons. The typically developed soil horizons, as illustrated in Figure 5.16, are: O — the layer of organic matter A — the layer of partially decayed organic matter mixed with mineral material E— the eluviated (leached) layer from which some of the clay and iron

Explanation:

i'm built different

You might be interested in

What important event takes place during the interphase of the cell cycle?

yan [13]

The dna is multiplied obviously

5 0

3 years ago

Read 2 more answers

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

3 years ago

How is newton's first law used in basketball?

arlik [135]

Newton's first law of motion states that an object at rest tends to stay at rest, while an object in motion tends to stay in motion unless an external force acts upon it. This law appears in basketball when the player is shooting the ball. When the player is holding the ball, the ball is at rest but when a player shoots the ball, they use force to throw the ball in the hoop.

4 0

3 years ago

A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0

ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

6 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago