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4 years ago

How much of the earths water is usable as a freshwater resource

Physics

2 answers:

o-na [289]4 years ago

8 0

Only 3% of the earth's water supply is considered as fresh water.

Water covers in excess of 70 percent of the world's surface. It controls the Earth's atmosphere.

The ceaseless development of water in and around the earth is known as the hydrologic cycle or water cycle. According to the U.S. Land Survey,70 percent of the earth is water, almost 97 percent is in the seas. Sea water is excessively salty for drinking, assembling, and cultivating. The new water accessible for us to utilize is around three percent of the world's water supply. Three-fourths of the three percent new water is inaccessible in light of the fact that it is in icecaps and different ice sheets. Of all the water on Earth, in excess of 99 percent of Earth's water is unusable by people and numerous other living things!

Grace [21]4 years ago

7 0

Only about 0.3% of our freshwater , which is found in the surface of water lakes rivers and swamps.

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A student walks the hallway for 25 m stops to talk and continues down the hallway another 10 m what is the distance and displace

brilliants [131]

Answer:

Distance = displacement = 35m

Explanation:

The distance of the student is how far he has gone.

Distance = 25m + 10m

Distance = 35m

Displacement is the distance specified in specific direction. Since the student walk in the sane direction, thence the displacement is also 35m

3 0

3 years ago

Are temperature and thermal energy the same thing justify your answer

Sever21 [200]

Temperature is the measurement of the average energy of the particles in a solid, liquid or gas and thermal energy is the total energy in a set amount of solid, liquid or gas. Therefore, the temperature and thermal energy is not the same thing. They are both about the particle theory, which is a theory that all particles of solid, liquid or gas are always in motion. But the difference between the two is that temperature is the "measurement" of the particles in a solid, liquid or gas and the thermal energy is the total energy in a set amount of solid, liquid or gas.

6 0

4 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )