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4 years ago

Briefly describe the characteristics of each soil horizon from the top layer to the bottom layer

1 answer:

7 0

There is six horizen. 1. O Horizon - The top, organic layer of soil, 2. A Horizon - The layer called topsoil; 3. E Horizon - This layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand. 4. B Horizon - Also called the subsoil - this layer is beneath the E Horizon and above the C Horizon. 5. C Horizon - it's called regolith: the layer beneath the B Horizon and above the R Horizon. 6 R Horizon - this is last and the unweathered rock layer that is beneath all the other layers.

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A certain metal wire has a cross sectional area of 1 mm2 and is 1 m long. when it is hung from the ceiling and a 10 kg mass is h

kvasek [131]

From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm

5 0

3 years ago

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the book is

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the book is

F=44N

For more information on this visit

brainly.com/question/23379286

8 0

2 years ago

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0

3 years ago

A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the

olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

$f$ = frequency of the tone = 329.6 Hz

$T$ = Time period of the sound wave

we know that, Time period and frequency are related as

$T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s$

$v$ = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m$

$v$ = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m$

8 0

3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it

inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

$F_{b}=\rho_{air}\times V_{b}\times g$

Where, $\rho_{air}$ = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

$F_{b}=1.20\times321\times9.81$

$F_{b}=3778.8\ N$

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0

3 years ago