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3 years ago

Wind erosion _____.

Physics

2 answers:

DerKrebs [107]3 years ago

6 0

Answer: A- removes fertile topsoil

A wind erosion is a natural phenomena in which the surface and subsurface layers of soil is removed and displaced by the action of wind. It occurs in flat, bare areas, having dry and sandy soil or any place where the soil is loose, dry and fine in texture.

The wind erosion removes the superficial layers of soil that are fertile, hence reduces the soil fertility of soil.

iren2701 [21]3 years ago

5 0

<span>
A- removes fertile topsoil</span>

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Give a graph of velocity v, time does a horizontal line represent ?

riadik2000 [5.3K]

The graph represents an object moving at constant velocity

Explanation:

A graph of velocity (v) versus time (t) of an object is generally used to give information about the motion of an object. In fact, from this graph, it is possible to infer several pieces of information:

The slope of the graph corresponds to the acceleration of the object. In fact, the acceleration of an object is defined as the rate of change of the velocity: $a=\frac{\Delta v}{\Delta t}$ , which also corresponds to the slope of the graph
The area under the curve between two times $t_1, t_2$ corresponds to the distance travelled by the object in that time interval.

In this problem, we have an object whose velocity-time graph corresponds to a horizontal line. We said that the slope of the graph corresponds to the acceleration of the object: since the slope of an horizontal line is zero, this means that the acceleration of this object is zero, and therefore the object is travelling at constant velocity.

Learn more about velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

8 0

3 years ago

An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia

sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = $\frac{1}{2}CV^{2}$

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = $\frac{1}{2}C_{1} V^{2}$ = $\frac{1}{2}\frac{C}{2} V^{2}$

U₁ = U/2

$\frac{U_{1} }{U} = \frac{1}{2}$

7 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1

Nesterboy [21]

Answer:

a) $a_c=3.41x10^{-5} \frac{m}{s^2}$

b) $a_c=3.34x10^{-5}\frac{m}{s^2}$

If we analyze the two values obtained for the centripetal acceleration we see that are similar. Based on this we can say that the centripetal force would be similar to the gravitational force between the Earth and Moon.

Explanation:

1) Notation and important concepts

Centripetal acceleration is defined as "The acceleration experienced while in uniform circular motion. It always points toward the center of rotation and is perpendicular to the linear velocity."

Angular frequency is defined as "(ω), or radial frequency, measures angular displacement per unit time and the units are usually degrees (or radians) per second. "

T = 27.3 d represent the time required by the Earth around an specific point given in the problem

G= universal constant $6.673x10^{-11}\frac{Nm^2}{kg^2}$

M= represent the mass of the Moon=7.35x10^{22}kg

2) Part a

$a=\frac{GM}{r^2}$ (1)

The Earth-Moon Distance Is $3.84x10^8 km$ (average Value) and both rotates at a point located about 4700 km from the center of Earth, so the radius for this case would be the difference between these two values

$r=(3.84x10^8 km)-(4.7x10^6)=3.793x10^8 m$

Since we have the radius now we can replace into equation (1)

$a=\frac{(6.673x10^{-11}\frac{Nm^2}{kg^2})(7.35x10^22 kg)}{(3.793x10^8 m)^2}=3.41x10^{-5} \frac{m}{s^2}$

And the acceleration due to the Moon's gravity would be $3.41x10^{-5} \frac{m}{s^2}$ at the point required.

3)Part b

For this case we can find the centripetal acceleration from this formula:

$a_c =r \omega^2$ (2)

But on this case we don't have the angular frequency so we can find it with this formula

$\omega =\frac{2\pi}{T}$ (3)

But since the period is on days we need to convert that into seconds

$27.3dx\frac{86400s}{1d}=2358720sec$

Replacing the value into equation (3) we got:

$\omega =\frac{2\pi}{2358720}=2.664x10^{-6}\frac{rad}{s}$

Now we can find the centripetal acceleration with the equation (2), the new radius on this case since our reference is the Earth and the point is located 4700km=4700000m from the center of Earth then the new value for the radius would be r=4700000m

$a_c =r \omega^2 =(4700000m)(2.664x10^{-6}\frac{rad}{s})^2=3.34x10^{-5}\frac{m}{s^2}$

5 0

3 years ago

What is the wavelength of a soundwave moving at 340 m/s with a frequency of 265 Hz?

lyudmila [28]

Wave speed = frequency * wavelength
Rearrange so it's equal to wavelength. Do this by diving both sides by frequency to leave you with:
Wave speed / frequency = wavelength
340 / 265 = 1.2830 m

4 0

3 years ago

Mirage is due to...

Irina18 [472]

<span>Mirage is due to (A). unequal heating of different parts of the atmosphere,
because that produces unequal density of different parts of the atmosphere,
which bends the light.
</span>

3 0

4 years ago