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In-s [12.5K]
2 years ago
11

Describe 3 ways that wind erosion moves sediment

Physics
1 answer:
Pepsi [2]2 years ago
8 0
Suspension, saltation, and surface creep are the three types of soil movement which occur during wind erosion. While soil can be blown away at virtually any height, the majority (over 93%) of soil movement takes place at or below one meter.

Suspension occurs when very fine dirt and dust particles are lifted into the wind. They can be thrown into the air through impact with other particles or by the wind itself. Once in the atmosphere, these particles can be carried very high and be transported over extremely long distances. Soil moved by suspension is the most spectacular and easiest to recognize of the three forms of movement.

Saltation - The major fraction of soil moved by the wind is through the process of saltation. In saltation, fine soil particles are lifted into the air by the wind and drift horizontally across the surface increasing in velocity as they go. Soil particles moved in this process of saltation can cause severe damage to the soil surface and vegetation. They travel approximately four times longer in distance than in height. When they strike the surface again they either rebound back into the air or knock other particles into the air.

Creep - The large particles which are too heavy to be lifted into the air are moved through a process called surface creep. In this process, the particles are rolled across the surface after coming into contact with the soil particles in saltation.

Hope this helps

Tried to be detailed ;)
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A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.
kakasveta [241]
The first law of thermodynamics says that the variation of internal energy of a system is given by:
\Delta U = Q + W
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system: 
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
\Delta U = -1275 J+855 J=-420 J
6 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go
jeka94
Heat source is the answer
7 0
2 years ago
Read 2 more answers
The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
snow_lady [41]
Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0
2 years ago
Read 2 more answers
What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?
Sindrei [870]

Answer:

9.12\cdot 10^{-8} m

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

E=13.6 eV= 2.18\cdot 10^{-18}J

The energy of the photon is related to the wavelength by

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

Solving the formula for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m

7 0
2 years ago
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