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4 years ago

Most weather occurs in which atmospheric layers?

Physics

1 answer:

Slav-nsk [51]4 years ago

5 0

Answer:

Troposphere

Explanation:

The troposphere is the lowest layer of the Earth's atmosphere. The earth's atmosphere has the following layers:

- Troposphere - Stratosphere - Mesosphere -Thermosphere - Exosphere

The troposphere lies 10 km above the land surface and most weather phenomena and conditions (e. g. clouds, turbulence) occur there.

Most of the water vapor in the atmosphere is in the troposphere (about 99%) and most of the atmosphere's mass is contained in the troposphere (75%)

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An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Anvisha [2.4K]

Answer:

denisity = 52.33 g/c $m^{3}$

Explanation:

Density:

$d = \frac{m}{v}$

We have that m = 785 and that v = 15 c $m^{3}$ .

$d = \frac{785}{15}$

d = 52.33 $m^{3}$

4 0

3 years ago

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.

galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

$\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}$

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}$

Here, k is the Coulomb's constant whose value is

$k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the magnitude of charge will be

$\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}$

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

6 0

2 years ago

Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w

blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

Corona
Transition Region
Chromosphere
Photosphere
Convection Zone
Radiative Zone
Core

A brief description of them:

Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0

4 years ago

The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With

tangare [24]

Answer:

0.7 mJ

Explanation:

Identify the unknown:

The work required to turn the dial from 180° to 0°

List the Knowns:

Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F

Voltage of the battery: V = 130 V

Set Up the Problem:

Energy stored in a capacitor:

U_c=1/2*V^2*C

=1/2*Q^2/C

When the dial is set at 180°:

U_c=1/2*(130)^2*350*10^-12=10^-4

Q=√2*U_c*C=4*10^-7

When the dial is set at 0°:

U_c=1/2*(4*10^-7)^2/100*10^-12

=8*10^-4 J

Solve the Problem:

ΔU_c=7*10^-4 J

=0.7 mJ

note:

there maybe error in calculation but method is correct

4 0

3 years ago

Read 2 more answers

What is the mass of 3 m3 of a substance having density 1200 kg/m3

adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0

3 years ago