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4 years ago

Which explains how the Coriolis effect causes air circulation?

Physics

2 answers:

skelet666 [1.2K]4 years ago

6 0

Answer:

The answer to your question is A. Hope this helped

stira [4]4 years ago

3 0

The Coriolis effect causes air circulation when <span>Air from the equator moves toward the poles slower than the ground below it, so friction causes the air to veer east. In contrast, air from the poles moves toward the equator faster than the ground below it, so friction causes it to veer to the west. The answer is letter A</span>

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What is latent heat? Group of answer choices Energy released when water evaporates. Energy hidden in water vapor in the air. Ene

Andrews [41]

Answer:

Energy absorbed or hidden when water evaporates

Explanation:

The heat that is required to make a phase change is known as latent heat.

A phase change occurs when matter changes state. For example from solid to liquid, from liquid to gas, among others.

When changing from liquid to gas (for example when water evaporates), the heat necessary for this to happen is called latent heat of vaporization. The word latent means hidden, because a change in temperature is not perceived during the phase change, even when heat is being added, thus it is said that the heat is hidden or latent.

So the answer is:

Energy absorbed or hidden when water evaporates.

*Another type of latent heat is the latent heat of fusion, which is when a solid becomes liquid.

7 0

4 years ago

Which of the following has zeros that are not significant?

guapka [62]

10.0 J

if it was 10.01 it would be significant, but its 10.0 which makes it a whole number

4 0

3 years ago

A sheet of red paper will look black when illuminated with

vovikov84 [41]

It's cyan. I believe it's because it has green and blue in it.

6 0

3 years ago

Read 2 more answers

A coin is dropped from a height of 421 m. calculate the velocity of the coin after 3 s

Keith_Richards [23]

Answer:

29.4 m.s

Explanation:

Vf = vo + at v o = original velocity = 0 in this case

Vf = at

= 9.81 m/s^2 * 3 = 29.4 m/s

7 0

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

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago