Magnetic pole reversal worksheet

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

NISA [10]

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.

5 0

3 years ago

The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center

jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0

3 years ago

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c

Romashka [77]

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

6 0

3 years ago

10. John does 176 J of work lifting himself a distance of 0.40 m. How

ch4aika [34]

The answer is 440N

4 0

3 years ago

Which component of the eye contains the visual receptors

juin [17]

'Retina' of the eye contains the visual receptors.

7 0

3 years ago