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2 years ago

Describe 3 ways that wind erosion moves sediment

1 answer:

8 0

Suspension, saltation, and surface creep are the three types of soil movement which occur during wind erosion. While soil can be blown away at virtually any height, the majority (over 93%) of soil movement takes place at or below one meter.

Suspension occurs when very fine dirt and dust particles are lifted into the wind. They can be thrown into the air through impact with other particles or by the wind itself. Once in the atmosphere, these particles can be carried very high and be transported over extremely long distances. Soil moved by suspension is the most spectacular and easiest to recognize of the three forms of movement.

Saltation - The major fraction of soil moved by the wind is through the process of saltation. In saltation, fine soil particles are lifted into the air by the wind and drift horizontally across the surface increasing in velocity as they go. Soil particles moved in this process of saltation can cause severe damage to the soil surface and vegetation. They travel approximately four times longer in distance than in height. When they strike the surface again they either rebound back into the air or knock other particles into the air.

Creep - The large particles which are too heavy to be lifted into the air are moved through a process called surface creep. In this process, the particles are rolled across the surface after coming into contact with the soil particles in saltation.

Hope this helps

Tried to be detailed ;)

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Janelle wants to buy some strings of decorative lights for her home. She is trying to decide between two strings of lights that

inysia [295]

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2 years ago

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A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

$PE= \frac{GMm}{d}$

The kinetic energy can be written as,

$KE= \frac{1}{2} mv^2$

Where,

$G = 6.67*10^{-11}m^3/kg\cdot s^2$ Gravitational Universal Constant

$m = 5.972*10^{24}kg$ Mass of Earth

$h = 56*10^6m$ Height

$r = 6.378*10^6m$ Radius of Earth

From the conservation of energy:

$\frac{1}{2} mv^2 = \frac{GMm}{d}$

Rearranging to find the velocity,

$v = \sqrt{\frac{2Gm}{d}} \rightarrow$ Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

$d = r+h$

$v = \sqrt{\frac{2Gm}{r+h}}$

Replacing the values we have that

$v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}$

$v = 3.6km/s$

Therefore the escape velocity is 3.6km/s

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2 years ago

Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What

likoan [24]

F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N

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2 years ago

A weightlifter lifts a 13.0-kg barbel from the ground an moves it a distance of 1.3 meters. What is the work se does on the barb

marta [7]

The work done on the barbell is -165.62 Nm.

Explanation:

Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.

In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.

Thus, $Force = - Mass * Acceleration due to gravity = - 13 * 9.8 =-127.4 N$

Now, the force is -127.4 N and the displacement is 1.3 m.

So, $Work done = F*d$

$Work done = -127.4* 1.3 = -165.62 Nm$

So, the work done on the barbell is -165.62 Nm.

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2 years ago

A cart moves along a track at a velocity of 3.5 cm/s. when a force is applied to the cart, its velocity increases to 8.2 cm/s. i

oksian1 [2.3K]

Acceleration, in physics, is the rate of change of velocity of an object with respect to time. <span> Anytime an object's velocity is changing, the object is said to be </span><span>accelerating. It can be calculated as follows:

acceleration = 8.2 - 3.5 / 1.5 = 3.1 m/s</span>²

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