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Leya [2.2K]
3 years ago
11

The process by which natural forces move weathered rock and soil from one place to another is called

Engineering
2 answers:
melomori [17]3 years ago
6 0

Answer:

Erosion is the geological process in which earthen materials are worn away and transported by natural forces such as wind or water. A similar process, weathering, breaks down or dissolves rock, but does not involve movement.Erosion is the opposite of deposition, the geological process in which earthen materials are deposited, or built up, on a landform.

 

Most erosion is performed by liquid water, wind, or ice (usually in the form of a glacier). If the wind is dusty, or water or glacial ice is muddy, erosion is taking place. The brown color indicates that bits of rock and soil are suspended in the fluid (air or water) and being transported from one place to another. This transported material is called sediment.

Explanation:

<em>Physical erosion describes the process of rocks changing their physical properties without changing their basic chemical composition. Physical erosion often causes rocks to get smaller or smoother. Rocks eroded through physical erosion often form clastic sediments. Clastic sediments are composed of fragments of older rocks that have been transported from their place of origin.</em><em>Landslides and other forms of mass wasting are associated with physical weathering. These processes cause rocks to dislodge from hillsides and crumble as they tumble down a slope. </em>

<em>Landslides and other forms of mass wasting are associated with physical weathering. These processes cause rocks to dislodge from hillsides and crumble as they tumble down a slope.  </em>

<em>Plant growth can also contribute to physical erosion in a process called bioerosion. Plants break up earthen materials as they take root, and can create cracks and crevices in rocks they encounter.</em>

<h2><em>I</em><em> </em><em>HOPE</em><em> </em><em>I</em><em> </em><em>HELPED</em><em> </em><em>YOU</em></h2>
11Alexandr11 [23.1K]3 years ago
3 0

Answer:

D. erosion

Explanation:

I took the test and got it right

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1,334

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3 years ago
For an automotive spark-ignition engine, the combustion duration (from time of ignition through completion) is approximately one
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1.72

Explanation:

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3 years ago
A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

P_4=50\ kPa

h_4=0.7(2304.7)+340.5

    =1953.83 \ KJ/Kg

and,

P_3=15 \ mPa

h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

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h_f=340.54

or,

V_f=0.001030 \ m^3/Kg

then,

⇒ h_2=340.54+0.001030(15\times 10^{3}-50)

        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

         =\frac{2610.8-1953.83}{2610.8-1836.26}

         =84.818 (%)

The thermal efficiency of cycle will be:

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4 0
3 years ago
Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points
AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component  based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure (P_{in}) = 11 MPa

Outside pressure (P_{out}) = 1 MPa

Yield strength (\sigma_y) = 1000 MPa

Therefore:

\sigma_y =\frac{\sigma_y}{N}, Substituting values,

\sigma_y =\frac{\sigma_y}{N}=\frac{1000}{3.5}=285.714 MPa

The minimum required thickness for a steel pressure vessel (t) is given by the equation:

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}. Substituting values

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}=\frac{65*10^{-3} *(11-1)10^{6} }{285.714*10^{6} } =2.275*10^{-3} =2.275 mm

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

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Answer:

answer is A. EG,Et, and Ed

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