Answer:
Deposition
Explanation:
- Deposition occurs when water slows or ceases moving, the wind dies or stops blowing, or glaciers melt. The deposited material can also be used to construct new landforms. Waves, for example, can dump sediment in places offshore, where it might accumulate to form sand dunes.
When the wind calms down or vegetation stops or slows the breeze, the sediment particles begin to fall. Water is another factor that may erode, move, or deposit sediment. Flowing water is a key erosive agent. Water transports dirt and rock fragments as it moves. Warm, wet air will not travel if wind systems are not present. Water will still evaporate, but it will not move, therefore everywhere else than a major body of water will dry up. Lakes may be fine since evaporating water will flow back into them, and the sea will be fine, but everywhere else will become extremely dry very rapidly. Wind is constantly blowing somewhere on the world at any given time. Winds are usually quiet near the middle of a high pressure system. Wind is the passage of air from a high pressure location to a low pressure area.... So essentially air is always moving. Weathering and erosion are caused by wind. Weathering is caused by wind blowing debris against cliffs and huge rocks. This wears down the rock, reducing it to sand and dust. Sand and dust are also eroded by wind. 2. Rocks are tough and durable, but they don't last forever. Weathering and erosion are processes that occur as a result of forces such as wind and water breaking down rocks. Weathering is the process through which rocks deteriorate. Weathering is caused by a variety of factors, including climate change.
Ice is only thing which is mightier than steel because it can breaks things which are made up of steels like ships but in the sunlight ice melts away it means it cowards away.
Answer:
10.5 m/s
Explanation:
For the first chestnut:
y₀ = 10 m
v₀ = 0 m/s
a = -9.8 m/s²
y = y₀ + v₀ t + ½ at²
y = 10 + (0) t + ½ (-9.8) t²
y = 10 − 4.9t²
When y = 7.5:
7.5 = 10 − 4.9t²
t = 5/7
When y = 0:
0 = 10 − 4.9t²
t = 10/7
For the second chestnut:
y₀ = 10 m
y = 0 m
a = -9.8 m/s²
t = 10/7 s − 5/7 s = 5/7 s
y = y₀ + v₀ t + ½ at²
0 = 10 + v₀ (5/7) + ½ (-9.8) (5/7)²
0 = 10 + 5/7 v₀ − 2.5
v₀ = -10.5
The second chestnut must be thrown downwards at 10.5 m/s.
You could pet the wrong bullets in the wrong gun
Answer:
<em>10.46m/s</em>
Explanation:
Given
Distance = 5.58m
Required
magnitude of the velocity'
Using the equation of motion expressed as
v² = u² + 2gS
v² = 0 +2(9.8)(5.58)
v² = 109.368
v= √109.368
v = 10.46m/s
<em>Hence the magnitude of the velocity of the of the chain is 10.46m/s</em>