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:
Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is
D) lower than the original pitch and decreasing as he falls.
Explanation:
As we know by the Doppler's effect of sound we have
so we will have
so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency
Here we know that the speed of the source is increasing with time as the source is falling under gravity
So we can say that the pitch of the sound will decrease with time
<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
Answer:
A uniform thin rod with an axis through the center
Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.
We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write
λ = m/l (orm) = λl
If we take the differential of each side of this equation, we find
d
m
=
d
(
λ
l
)
=
λ
(
d
l
)
since
λ
is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,
d
l
=
d
x
in this situation. We can therefore write
d
m
=
λ
(
d
x
)
, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain
I=∫r2dm=∫x2dm=∫x2λdx.
The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us
I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.
The frequency, f, of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz (Hz), since one hertz is equal to one wave per second.
