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Alisiya [41]
3 years ago
9

The process by which wind removes surface materials is called

Physics
2 answers:
Maurinko [17]3 years ago
4 0

Answer:

The process by which wind removes surface materials is called deflation.

34kurt3 years ago
3 0

Answer:

Moving water that flows over the land is called runoff. An area of wave-washed sediment along a coast is called a beach. Deflation is the process by which wind removes surface materials.

Explanation:

hope this helps. if it did, please mark brainliest :)

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ANSWER ASAP PLEASEEE Kendra wonders what would happen if the oceans did not exist on Earth. Based on her understanding of the oc
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that there would be more clouds.

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If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has
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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
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To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

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The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.
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<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION : lnPso2cl12=-kt+lnPso2cl1 initial partial pressure Pso2cl12 the rate constant k and the time t lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375) so lnPso2cl12=3.002 we take the base e antilog: lnPso2cl12=e3.002 Pso2cl12=20 torr we use the integrated first order rate lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002 we use the same rate constant and initial pressure k=4.5*10-2*s-1 Pso2cl12=375 Pso2cl12=1* so2cl12 Pso2cl12=37.5 torr subtract in Pso2cl12 grom both side lnPso2cl12- lnPso2cl12=-kt ln(x)-ln(y)=ln (x/y) ln (Pso2cl12/Pso2cl20)=-kt we get t -1/k*ln(Pso2cl12/Pso2cl20)=t t=51 s</span>
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