All categories

3 years ago

What are the major process of water cycle

Physics

2 answers:

Luba_88 [7]3 years ago

7 0

Answer:

The major processes of water cycle are evaporation,condensation,transpiration,precipitation and run off.....

Cerrena [4.2K]3 years ago

5 0

Answer:

the major process of water cycle are :

Evaporation
Condensation
Precipitation

hope it helps!

You might be interested in

Galaxy evolution is a very active area of research. Look for information on a future observatory/mission that will investigate g

Maslowich

The James Webb Space Telescope (JWST) is a space observatory developed among 20 countries, is being built and operated jointly by NASA, the European Space Agency and the Canadian Space Agency, to be the scientific successor of Hubble and Spitzer. The JWST will offer an unprecedented resolution and sensitivity, some of its main objectives is observe the formation of the first galaxies of the universe and study the formation and evolution of galaxies. The JWST will allow to observe wavelengths within the so-called near-infrared astronomy, but also the orange and red visible light. The importance of this spectrum range is due to the fact that the infrared light can penetrate dusty regions and reveal what is inside. It will be launched in March 2021, the dimensions of its main mirror are 20,197 m × 14,162 m (66.26 ft × 46.46 ft). The JWST will orbit the sun 1.5 million kilometers away from the Earth

3 0

4 years ago

Cations are formed by gaining protons losing protons gaining electrons losing electrons

krek1111 [17]

Cations are formed by losing electrons

Cations are basically positive ions

It's not lose or gain protons because atoms can't lose or gain protons. The protons don't move.
It's not gain electrons because electrons are negative. More electrons=more negative

If you can't remember cations are positive
CAT-ions are PAWS-itively charged

7 0

3 years ago

Explain the runaway refrigerator effect and the role it may have played in the evolution of Mars.

Fiesta28 [93]

Answer:

Explained

Explanation:

The runway refrigerator effect and the role it would have played in the evolution of Mars can be summarized as follows

The weak gravity of Mars does not allow it have a gaseous atmosphere over it. The thin atmosphere would have lead to lower temperature on mars.In in turn would have lead to freezing of gases thus lowering the temperature further. The thinner atmosphere and colder temperature would lead to loss of most water in the planet.

6 0

3 years ago

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its total mechanical energy(ME)PLEASE HELP

lord [1]

Answer:

Explanation:

ME=mvh

m=0.0780kg

v=4.84ms⁻¹

h=5.36m

ME=0.0780*4.84*5.36

ME=2.02joules

6 0

3 years ago

Read 2 more answers

An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus

frutty [35]

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

<u>Given:</u>

$u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

$v$ = final velocity of the electron = $3u$

$x$ = initial position of the electron from the proton = very distant = $\infty$

<u>Assume:</u>

$m$ = mass of an electron = $9.1\times10^{-31}\ kg$

$e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

$p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

$k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

$y$ = final position of the electron from the proton

$\Delta K$ = change in kinetic energy of the electron

$W$ = work done by the electrostatic force

$F$ = electrostatic force

$r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.

8 0

4 years ago