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tekilochka [14]
2 years ago
11

How do the processes of weathering, erosion, and deposition affect the ecoregions of Texas? Give a specific example of how each

process affects a particular ecoregion.
Physics
1 answer:
NISA [10]2 years ago
5 0

Answer:

Weathering, erosion, and deposition from the terrestrial surface topography and soil characteristics. These processes, for example, have formed a variety of landforms in Texas like beaches, plateaus, mountains, and canyons as well as soil types like fertile soil, clay-rich soil, and sandy soil. The combination of topography, soil, and climatic conditions in an area defines the types of habitats that the area can support this is crucial to ecoregion classification. Ten separate ecoregions occur in Texas including 1) East Texas Pineywoods, 2) Gulf Coast Prairies and Marshes, 3) Oak Woods and Prairies, 4) Blackland Prairie, 5) cross timbers and prairies (6) Rolling Plains, (7) High Plains, (8) TransPecos, (9) South Texas Plains, (Brush Country), and (10) Edwards Plateau. Such ecoregions are named for the major types of habitats topographical features (e.g. Edwards Plateau) present in their areas. The weathering, erosion, and deposition of each of these ecoregions have an important influence.

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If the value of the electric field in an electromagnetic wave were doubled then
Norma-Jean [14]

Answer:

Electromagnetic wave are waves formed as a result of the oscillatory activities involving the electric and the magnetic field.

However in an Electromagnetic wave, the electric field and magnetic field carry equal amounts of energy and the magnitude of the electric field is directly proportional to the magnitude of the magnetic field. This direct proportionality gives rise to the speed of light being the constant between the two fields.

When the electric field is doubled then an equal action is to be set for the magnetic field so it doesn’t deviate from its main functions and characteristics.

4 0
3 years ago
20 POINTS!
CaHeK987 [17]
Potential energy is the answer
6 0
3 years ago
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

3 0
3 years ago
How you use physics in your<br> everyday life.
stepan [7]

Answer:

Explanation:

Physics gets involved in your daily life right from you wake up in the morning. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can't see, but hear or experience. Physics studies the origin, propagation, and properties of sound

5 0
3 years ago
a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
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