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 ;)
Answer:
The answer to your question is at the point where it is thrown.
Explanation:
Kinetic energy is the energy that possesses a body due to its motion. Its formula is
Ke = 1/2 mv²
Then, the kinetic energy is maximum when the velocity is the highest, and this is at the point where it is thrown, after this point, the velocity will be diminished and at the highest point will be equal to zero.
Answer:
At the point when an item is moved into the air toward a path other than straight up or down, the speed, quickening, and uprooting of the article don't all point a similar way. In circumstances like this, when taking care of the difficult we ought to apply the procedure of settling vectors into segments. At that point we apply the more straightforward one-dimensional types of the conditions for every segment. At long last, we can recombine the parts to decide the resultant.
Explanation:
At the point when an item is moved into the air toward a path other than straight up or down, the speed, quickening, and uprooting of the article don't all point a similar way. In circumstances like this, when taking care of the difficult we ought to apply the procedure of settling vectors into segments. At that point we apply the more straightforward one-dimensional types of the conditions for every segment. At long last, we can recombine the parts to decide the resultant.
Objects that are tossed or dispatched into the air and are dependent upon gravity are called projectiles. The way of a shot is a bend called parabola. In the event that an article has an underlying speed in some random time span, there will be level movement all through the trip of the shot.
Shot movement is free fall with an underlying speed. The underlying flat speed of a shot is equivalent to the level speed all through the shot's flight.
To discover the speed go a shot anytime during its flight, discover the vector aggregate of the parts of the speed now. We should utilize the Pythagorean hypothesis to discover the extent of the speed and the digression capacity to discover the course of the speed. On the off chance that an article has an underlying vertical part of speed and a level segment of speed, the item's movement ought to be settled into its segments, and afterward the sine and cosine capacities can be utilized to locate the vertical and even segments of the underlying speed. The speed of a shot dispatched at a point to the ground has both flat and vertical parts. The vertical movement is like that of an item that is hurled straight with an underlying speed.
Your equation is:
![2NaOH + H_{2}SO_{4} ==\ \textgreater \ Na_{2}SO_{4} + H_{2}O](https://tex.z-dn.net/?f=2NaOH%20%2B%20H_%7B2%7DSO_%7B4%7D%20%20%3D%3D%5C%20%5Ctextgreater%20%5C%20%20Na_%7B2%7DSO_%7B4%7D%20%2B%20H_%7B2%7DO)
An equation is balanced only if there are the same number of atoms of each element on both sides of the arrow - aka same number of atoms of each element in both reactants (left of the arrow) and products (right of the arrow).
It'll be easiest to tackle this by counting up the number of atoms of each element on the left and on the right and comparing those numbers. If there is a number in front of the entire compound, that means that number applies to all elements in the compound. If the number is a subscript (little number to the right of the element), that means that number only applies to the element that the subscript is attached to:
1) On the left, you have:
![2NaOH + H_{2}SO_{4}\\ \\ Na: \: 2 \: atoms\\ O: \: 2 + 4 = 6 \: atoms\\ H: \: 2 + 2 = 4 \: atoms\\ S: \: 1 \: atom](https://tex.z-dn.net/?f=2NaOH%20%2B%20H_%7B2%7DSO_%7B4%7D%5C%5C%0A%5C%5C%0ANa%3A%20%20%5C%3A%202%20%5C%3A%20atoms%5C%5C%0AO%3A%20%5C%3A%202%20%2B%204%20%3D%206%20%5C%3A%20atoms%5C%5C%0AH%3A%20%5C%3A%202%20%2B%202%20%3D%204%20%5C%3A%20atoms%5C%5C%0AS%3A%20%20%5C%3A%201%20%5C%3A%20atom)
2) On the right, you have:
![Na_{2}SO_{4} + H_{2}O\\ \\ Na: \: 2 \: atoms\\ O: \: 4 + 1 = 5 \: atoms\\ H: \: 2 \: atoms\\ S: \: 1 \: atom](https://tex.z-dn.net/?f=Na_%7B2%7DSO_%7B4%7D%20%2B%20H_%7B2%7DO%5C%5C%0A%5C%5C%20%0ANa%3A%20%5C%3A%202%20%5C%3A%20atoms%5C%5C%20%0AO%3A%20%5C%3A%204%20%2B%201%20%3D%205%20%5C%3A%20atoms%5C%5C%20%0AH%3A%20%5C%3A%202%20%5C%3A%20atoms%5C%5C%20%0AS%3A%20%5C%3A%201%20%5C%3A%20atom)
You can see that the number of oxygen and hydrogen atoms aren't equal on both the left (reactants) and the right (products), so the equation is unbalanced.
Your final answer is "T<span>he equation is
unbalanced because the number of hydrogen atoms and
oxygen is
not equal in the reactants and in the products."</span>