The eroded rock and soil materials that are transported downstream by a river are called its load. A river transports, or carries, its load in three different ways: in solution, in suspension, and in its bed load.
Mineral matter that has been dissolved from bedrock is carried in solution. Common minerals carried in solution by rivers include dissolved calcium, magnesium, and bicarbonate. Most of a river’s solution load comes from groundwater seeping into the river. Before it reaches the stream,thegroundwaterhastraveledthroughfracturesinthebedrock, chemically eroding rock along the way.
When river water looks muddy, it is carrying rock material in suspension. Suspended material includes clay, silt, and fine sand. Although these suspended materials are heavier than water, the turbulence of the stream flow stirs them up and keeps them from sinking. Turbulence includes swirls and eddies that form in water as a result of friction between the stream and its channel. The faster a stream flows, the more turbulent and muddy it becomes. A rough or irregular channel also increases turbulence.
A river may also transport rock materials in its bed load. The bed load consists of sand, pebbles, and boulders that are too heavy to be carried in suspension. These heavier materials are moved along the streambed, especially during floods. Boulders and pebbles roll or slide along the river bed. Large sand grains are pushed along the bottom in a series of jumps and bounces.
The relative amounts of a river’s load that are carried in solution, in suspension, and in the bed load depend on the nature of the river, the climate, the type of bedrock, and the season of the year. As a general rule, most of the load carried by the world’s streams and rivers is carried in suspension. The size of a river’s suspended load increases with human land use. Road and building construction and removal of vegetation make it easier for rain to wash sediment into streams and rivers.
Answer:
0.872<em>m/s</em>
Explanation:
Tangential velocity is given by the formula,
In the question given,
radius= 25meters
time= 180secs
pie= 3.14
number of laps= 1
The magnitude of tangential velocity equals;
<em>v </em>= 157<em>m</em>/180<em>secs</em>
Therefore, the magnitude of the tangential velocity
=0.872<em>m/secs</em>
Answer:
The velocity of the Mr. miles is 17.14 m/s.
Explanation:
It is given that,
Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m
We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,
g is the acceleration due to gravity
v = 17.14 m/s
So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.
In this problem,
Applied force(F) = 10 N
The object’s mass (m) is 5 kg.
Having said that,
An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.
i.e., Mass + Acceleration = Force (a)
F= m×a
Therefore,
A= F÷m
A= (10÷5) m/sec²
A= 2 m/sec²
Consequently, the object’s acceleration,
A=2 m/sec²
Concept of force and acceleration:
This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.
It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.
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Answer:
4NaBr + 2CaF2 ---------- 4 NaF + 2CaBrz
Explanation:
