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3 years ago

Scientist often use ______________ which are programs that combine what is known about atmospheric circulation

Physics

2 answers:

Serggg [28]3 years ago

7 0

Answer:

General circulation model.

Explanation:

A general circulation model (GCM) is a type of climate model that employs a mathematical model of the general circulation of a planetary atmosphere or ocean. GCM uses the Navier–Stokes equations on a rotating sphere with thermodynamic terms for various energy sources (radiation, latent heat). These equations are the basic equations for computer programs used to simulate the Earth's atmosphere or oceans.

allsm [11]3 years ago

6 0

Answer:

The missing word is Computer Based Climate Models.

Explanation:

Computer-Based Climate Models are simply software which uses data collected about atmospheric conditions, air circulation, ocean circulation, interactions between weather elements and various climatic cycles to predict climate and weather.

These are often used by Meteorologists. Meteorologists are simply scientists who study and predict climate and weather.

Cheers!

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A 20-kilogram child is riding on a 10-kg sled over a frictionless icy surface at 8.0 meters per second. Calculate the kinetic en

Veseljchak [2.6K]

Answer:

K = 960 J

Explanation:

Given that,

Mass of a child = 20 kg

Mass of a sled = 10 kg

Speed of child on sled = 8 m/s

We need to find the kinetic energy of the sled with the child.

The total mass of child and the sled = 20 kg + 10 kg

= 30 kg

The formula for the kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 30\times (8)^2\\\\K=960\ J$

Hence, the kinetic energy of the sled with the child is 960 J.

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2 years ago

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3 years ago

Read 2 more answers

How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

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2 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

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Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

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3 years ago

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2 years ago