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3 years ago

During a flood, a stream overflows its banks, and water covers the adjacent ____.

Physics

1 answer:

nata0808 [166]3 years ago

6 0

Answer:

Flood Plain

Explanation:

The amount of water that circulates through a river, the flow, varies in time and space. These variations define the hydrological regime of a river. Temporary variations occur during or just after episodes of rains or thaws. Much of the water that falls in the catchment basin circulates underground, or feeds underground aquifers and takes much longer to feed the river flow and can reach it days, weeks or months after the rain generated by the runoff. The runoff that goes to the river is what increases its flow. In extreme cases, flooding can occur when the water supply is greater than the river's ability to evacuate it, overflowing and covering nearby flat areas or floodplain. In this distribution between the runoff water (or stream) that goes directly to the channel and water that infiltrates, feeds the aquifers and maintains the flow in the river in times without precipitation depends largely on the geomorphological integrity of the entire river system .

In natural dynamics, the river systems have their own space that has been modeled by the floodwaters and is made up of the channel, the banks and the plain or flood plain. Its dimensions have been defined by the main flood events that this river has attended. Floodplains are wide and flat areas built by the river in its floodwaters. They are flooded frequently and are covered by sediments and nutrients that fertilize the soil act as natural reservoirs, reducing the speed of the downstream current. They store floodwater and rainfall in aquifers (underground area).

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An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs

OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

$E=\dfrac{n^2\pi^2h^2}{8mL^2}$

For ground state, n = 1

$E_1=\dfrac{\pi^2h^2}{8mL^2}$ .............(1)

For first excited state, n = 2

$40=\dfrac{2^2\pi^2h^2}{8mL^2}$ .............(2)

Dividing equation (1) and (2), we get :

$\dfrac{E_1}{40}=\dfrac{1}{4}$

$E_1=10\ meV$

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0

3 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$