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tekilochka [14]
3 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]3 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 object doesn't move you have done no work.
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A snowball that will be used to build a snowman is at the top of a only hill. If the
Darina [25.2K]

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

E=mgh

g is acceleration due to gravity

h is height of the hill

h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m

So, the height of the hill is 24.11 m.

5 0
3 years ago
g A 5.60-kilogram hoop starts from rest at a height 1.80 m above the base of an inclined plane and rolls down under the influenc
swat32

Answer:

4.24m/s

Explanation:

Potential energy at the top= kinetic energy at the button

But kinetic energy= sum of linear and rotational kinetic energy of the hoop

PE= mgh

KE= 1/2 mv^2

RE= 1/2 I ω^2

Where

m= mass of the hoop

v= linear velocity

g= acceleration due to gravity

h= height

I= moment of inertia

ω= angular velocity of the hoop.

But

I = m r^2 for hoop and ω = v/r

giving

m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2

and m's cancel

g h = v^2

Hence

v= √gh

v= √10×1.8

v= 4.24m/s

4 0
3 years ago
Read 2 more answers
An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e
zimovet [89]

Answer:

a) v = 6.25\times 10^{5} m/s

b) v = 1.73\times 10^{4} m/s

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is 4.55\times 10^5 m/s

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

a =\frac{q E}{m}

a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}

a = 2.58\times 10^{11} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

      = 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355

v = 6.25\times 10^{5} m/s

b) for proton

a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}

a = -1.41\times 10^{8} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

       = 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355

v = 1.73\times 10^{4} m/s

3 0
3 years ago
Which of the following forms of energy must an object have if it is moving at a constant velocity?
lbvjy [14]

Answer:

Kinetic energy

Explanation:

5 0
3 years ago
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