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3 years ago

As air rises, why do clouds form?

Physics

1 answer:

MissTica3 years ago

3 0

Answer:

C. the air cools to its dew-point temperature and condenses into liquid water

Explanation:

Clouds are formed when the air rises heated by terrestrial irradiation. As it warms, the air rises and rises until it reaches its dew point, which will be when the water vapor condenses on (dew) or ice crystals. These droplets, which have a spherical shape and measure between 0.004 and 0.1 mm, are in continuous motion when they are suspended in the air and when they are subjected to updrafts, so they collide with each other and group together. Depending on the atmospheric conditions, an increase in their thickness may occur so as to cause them to precipitate.

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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 ( $\frac{1}{3}V(R)$ )

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which force field can accelerate an electron, but never change its speed? a) electric field b) magnetic field c) both of these d

Drupady [299]

Option a; Electric field can accelerate an electron, but never change its speed

An electric field (also known as an E-field) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It can also refer to the physical field of a charged particle system. Electric fields are created by electric charges and time-varying electric currents. Electric and magnetic fields are both aspects of the electromagnetic field, one of nature's four fundamental interactions (also known as forces). Electric fields are significant in many areas of physics and are used in electrical technology. In atomic physics and chemistry, for example, the electric field is the attractive force that holds the atomic nucleus and electrons together in atoms. It is also the driving force behind chemical bonds between atoms.

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