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3 years ago

General Circulation Models (GCM) :_________

Physics

1 answer:

enot [183]3 years ago

3 0

Answer:

Explanation:

GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.

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Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?

MrMuchimi

T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,

Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.

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8 0

3 years ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

We know that

$X_l=\omega L=2\pi fL$

$2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10$

f = 2421.127 kHz

6 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

$h_{Hg-TOP}=675mmHg=0.675m$ the barometric reading at the top of the building

$h_{Hg-BOT}=695mmHg=0.695m$ the barometric reading at the bottom of the building

$\rho _{air}=1.18 kg/m^{3}$ density of air

$\rho _{Hg}=13600 kg/m^{3}$ density of mercury

$g=9.8/m^{2}$ gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building $P_{TOP}$ and the perssure at the bottom $P_{BOT}$ :

$P_{TOP}=\rho _{Hg} g h_{Hg-TOP}$ (1)

$P_{BOT}=\rho _{Hg} g h_{Hg-BOT}$ (2)

From (1): $P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa$ (3)

From (2): $P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa$ (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure $\Delta P$ :

$\Delta P=P_{BOT} - P_{TOP}$ (5)

$\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa$ (6)

On the other hand, we have a column of air with a cross-section area $A$ and the same height of the building, lets name it $h_{air}$ .

As pressure is defined as the force $F$ exerted on a specific area $A$ , we can write:

$\Delta P=\frac{F}{A}$ (7)

If we isolate $F$ we have:

$F= A \Delta P$ (8)

Also, the force gravity exerts on this column of air (its weight) is:

$F=m_{air} g$ (9)

Knowing the density of air is: $\rho_{air}=\frac{m_{air}}{V_{air}}$ (10)

where the volume of air can be written as: $V_{air}=(A)(h_{air})$ (11)

Substituting (1) in (10):

$\rho_{air}=\frac{m_{air}}{(A)(h_{air}}$ (12)

Isolating $m_{air}$ :

$m_{air}=(\rho_{air}) (A) (h_{air})$ (13)

Substituting (13) in (9):

$F=(\rho_{air}) (A) (h_{air}) (g)$ (14)

Matching (8) and (14)

$A \Delta P=(\rho_{air}) (A) (h_{air}) (g)$ (15)

Isolating $h_{air}$ :

$h_{air}=\frac{\Delta P}{g \rho_{air}}$ (16)

Substituting the known and calculated values:

$h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})}$ (17)

Finally:

$h_{air}=230.50 m$ This is the height of the building

8 0

3 years ago

A cube has sides that are each equal to 7 centimeters in length. What is the volume of the cube?

RUDIKE [14]

Answer:

Explanation:

7³ = 343

4 0

3 years ago

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How do fission nuclear reactions differ from fusion nuclear reactions?

Serhud [2]

The path of the raction occurs on the basis of mass of the nuclei involved in reaction.
In case of nuclear fusion, two or more nuclei having less mass fuse(combine, join) together to form a new nuclei(heavier mass but it is relatively stable). During fusion, matter is not conserved because some of the matter is converted into energy(light). This reaction evolves a huge amount of energy and there comes Einstein's famous Energy-mass equivalence formula E=mc^2! :D. The nuclear reaction occuring in stars(including our sun ) is "fusion".

Fission occurs with heavier nuclei such as that of Uranium-235. Which splits into smaller subatomic particles like gamma, neutrons and enormous amount of energy.
Both, Fission and Fusion releases enormous amount of energy and modern nuclear weapons works on the principle of nuclear fission.

6 0

3 years ago

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