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Likurg_2 [28]
3 years ago
6

How did the appearance of the smallest creatures, such as the microbes and bacteria, in Earth's early history impact the geosphe

re?
A)
formation of rocks


B)
enriching the soil


C)
decreased moisture


D)
caused land erosion
Physics
2 answers:
shepuryov [24]3 years ago
8 0

B) Enriching the soil.

Hope this helps.

EleoNora [17]3 years ago
6 0

I believe the answer is B) enriching the soil. Bacteria converts energy in would organic matter into forms useful to the rest of the organization in the soil food web.

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The answer is a because tjhey do move at a molecular movemn
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erastovalidia [21]

Answer:11.5m

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Help would be greatly appreciated:) thank you! a pendulum clock is brought to mars. How does the bob move on Mars as compared to
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It runs slower <span>as gravity is lower so acceleration due to gravity is smaller</span>
8 0
3 years ago
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co
lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0
3 years ago
One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

Given:

temperature at the hotter end, T_H=100^{\circ}C

temperature at the cooler end, T_C=0^{\circ}C

length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

mass of ice melted at zero degree Celsius, m=8.7\times 10^{-3}\,kg

time taken for the melting of ice, t=15\times60=900\,s

thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

\dot{Q}=rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

\dot{Q}=3.2287\,W

Now using eq,(1)

3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}

k=205.4606\,W.m^{-1}.K^{-1}

8 0
3 years ago
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