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2 years ago

What is the relationship between the movement of the atmosphere and surface currents

Physics

1 answer:

maxonik [38]2 years ago

3 0

Answer:

Waves and surface currents are wind-generated. The atmosphere pushes the ocean surface along. But as the ocean moves, heat is redistributed and eventually tossed back into the atmosphere. The exchange of heat influences the pressure field in the atmosphere, which modifies the wind field.

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Special relativity can be used to study an object in which frame of reference? A. A frame of reference that has no change in vel

Georgia [21]

Answer:

gangster

Explanation:

sorrynsjajshdnqmanshsha

4 0

2 years ago

Read 2 more answers

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

2 years ago

The part of an atom that is mostly empty space is the

Ghella [55]

<h3><u>Answer;</u></h3>

Electron cloud

<h3><u>Explanation;</u></h3>

<em><u>An atom is the smallest particle of an element that can take part in a chemical reaction. Atom is made up of two parts ; that is the nucleus and the electron cloud. The nucleus contain subatomic particles; protons and neutrons, while the electron cloud contains the electrons.</u></em>
<em><u>The electron cloud is the largest part of the atom and is mostly an empty space. Most of an atom is a cloud of electrons surrounding a space called the nucleus with tiny protons and neutrons.</u></em>

5 0

3 years ago

Read 2 more answers

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

2 years ago

What does the Pauli exclusion principle state? . . A.Electrons will be added to an atom at the lowest possible level or sublevel

BARSIC [14]

The Pauli exclusion principle state that : D. Two electrons occupy the same orbital only if they have opposite spins

This happen because he stated that in an atom or molecule, two electrons CANNOT have same four electronic quantum numbers

hope this helps

5 0

2 years ago