What are some methods humans have used to control flooding and what has been done to worsen the effects of flooding?

Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will

djyliett [7]

Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

$P\cdot V = n \cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure.

$V$ - Volume.

$n$ - Molar quantity, in moles.

$R_{u}$ - Ideal gas constant.

$T$ - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

$P_{1}\cdot V_{1} = P_{2}\cdot V_{2}$ (2)

If we know that $\frac{V_{2}}{V_{1}} = \frac{1}{3}$ , then the resulting pressure of the system is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 3\cdot P_{1}$

The resulting pressure is 3 times the initial pressure.

4 0

3 years ago

When did agriculture first began?

timofeeve [1]

Answer:

Humans invented agriculture between 7,000 and 10,000 years ago, during the Neolithic era, or the New Stone Age. There were eight Neolithic crops: emmer wheat, einkorn wheat, peas, lentils, bitter vetch, hulled barley, chickpeas, and flax.

5 0

3 years ago

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An apple with a mass of 0.95 kilograms hangs from 3.0 meters above the ground. What is the potential energy of the apple

Vesnalui [34]

As Potential energy =mgh

m= 0.95kg

h=3 meter

g = 9.8 m/sec^2. ( acceleration due to gravity)

So P.E =(0.95)(9.8)(3)kgm^2/s^2

P.E =27.93 joules

3 0

3 years ago

Does plants have prokaryotic cells?

vova2212 [387]

No they have eukaryotic cells

7 0

3 years ago

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$