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.
The Hooke's law is a principal of physics that states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance.
Answer: Fmax = 5.54*10^-12 N
Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.
The maximum force experienced by a charge of magnitude (q) is given as
Fmax = qvB
Where v = velocity of electron.
The velocity of the electron can be gotten by using the work energy theorem.
The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.
mv²/2 = qV.
Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.
By substituting the parameters, we have that
(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000
(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2
v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31
v² = 64.36*10^(-16)/9.11×10^-31
v² = 7.0647×10^15
v = √7.0647×10^15
v = 8.40×10^7 m/s
Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41
Fmax = 5.54*10^-12 N
