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3 years ago

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

Physics

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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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$

$= 12.56*10^{-6} - 8.37*10^{-6}$

$= 4.2 *10^ {-6}T$

Since $B_{da} > B_{bc}$ then the direction of the net charge would be into the page

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erica [24]

Answer:

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geniusboy [140]

From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560

let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
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