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3 years ago

How are the layers of the atmosphere separated

Physics

2 answers:

Volgvan3 years ago

8 0

Answer:

it depends on temperature^^

Explanation:

hope this helped^^

Makovka662 [10]3 years ago

4 0

Answer:

The atmosphere can be divided into layers based on its temperature, as shown in the figure below. These layers are the troposphere, the stratosphere, the mesosphere and the thermosphere. A further region, beginning about 500 km above the Earth's surface, is called the exosphere.

Explanation:

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

7 0

3 years ago

If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off

alexira [117]

If the box is a distance 1.81 m from the rear of the truck when the truck starts, ... Force of Friction = mu_s * Normal Force( M * G) ... The box starts moving! ... Now that the box is moving, the bed of the truck pulls at it with 17.4 ... out how long it will take the box to reach the back of the truck. ... T^2 = 2 * 1.81 / .64

4 0

3 years ago

Which of the following is true of greenhouse gases?

Sedaia [141]

Answer:

A. They trap energy in the atmosphere.

Explanation:

Greenhouse gases are defined as gases that absorb and emit radiation within the infrared range. These gases allow the sun’s rays to pass through the ozone layer and warm the earth, but prevent this warmth from escaping atmosphere.

Greenhouse gases cause the greenhouse effect on planets. Examples of greenhouse gases in Earth's atmosphere are water vapor, carbon dioxide, methane, ozone etc.

4 0

2 years ago

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of

Simora [160]

Answer:

Approximately $1.5\; \rm kg$ . (Assuming that this table is level, and that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ .

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$ .

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$ .

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$ .

Pressure is the ratio between normal force and the area over which this force is applied.

$\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}$ .

Equivalently:

$(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})$ .

In this question, $\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}$ .

It was found that $(\text{contact Area}) = 0.075\; \rm m^{2}$ (assuming that the entire side of this book is in contact with the table.

Hence:

$\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}$ .

If that the table is level, this normal force would be equal to the weight of this book:

$\text{weight} = 15\; \rm N$ .

Assuming that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ . The mass of this book would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}$ .

4 0

2 years ago

Arjun told to his friend that " Q can form a virtual image larger than the object by refraction." What is "Q"?

nadezda [96]

Q is a concave mirror.

Explanation:

The image formed by a concave mirror is observed to be virtual, erect and larger than the object.

7 0

2 years ago