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Nimfa-mama [501]
3 years ago
11

Why does the winds move? What makes it move,? ​

Physics
2 answers:
juin [17]3 years ago
7 0

<em>The wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.</em>

<u><em>Why does the wind move?</em></u>

<em>he wind blows because of differences in air pressure. ... Instead, the wind blows anti-clockwise around the low pressure area in the Northern Hemisphere and clockwise in the Southern Hemisphere. This is the effect of the earth's rotation, which produces a force, called Coriolis, that deflects the wind from its path.</em>

<em />

<u><em>Why does air move to one place or another?</em></u>

<u><em /></u>

<em>Air moves because the sun generates heat on the earth, this causes hot air, hot air rises creating wind, the air which is made up of 02 C02 and many other gases is carried by wind. air moves from place to place due to the different temperature and pressure conditions on the earth</em>

lana66690 [7]3 years ago
5 0

Explanation:

The wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.

<h3>HOPE THIS HELP YOU !!! :)))</h3>
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den301095 [7]

Answer:

C

Explanation:

6 0
2 years ago
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How can you tell there is only one atom of oxygen in H2O?
kow [346]

Answer:

For H2O, there is one atom of oxygen and two atoms of hydrogen. A molecule can be made of only one type of atom. In its stable molecular form, oxygen exists as two atoms and is written O2. to distinguish it from an atom of oxygen O, or ozone, a molecule of three oxygen atoms, O3.

Explanation:

4 0
3 years ago
What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?
tigry1 [53]

(a)

Electronic configuration is given as follows:

[Kr]4d^{3}

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is [Kr]4d^{5}5s^{1}.

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be Mo^{3+}.

(b) The given electronic configuration is [Kr]5s^{2}4d^{2}.

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0
3 years ago
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

Q = 4.63 \times 10^6 J

3 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
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