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lina2011 [118]
3 years ago
11

In complete sentences describe the energy flow between the Sun, the Earth, and space

Physics
1 answer:
agasfer [191]3 years ago
6 0
Answer: The energy from the sun passes through space in the form of invisible waves to the earth surface. It heats up the earth’s surface causing variation in climate.
Explanation:
The amount of incoming energy from the Sun decides the weather and climate of earth. If the energy that is incoming and outgoing on the earth, then climate is in equilibrium. The balance is depending on the scattering, absorption, reflection and transformation of energy.
The energy from sun passes through space and reaches the earth’s surface. On reaching surface, the solar energy warms the atmosphere releasing heat energy which gets transferred throughout the planets system by radiation, conduction and convection. Conduction happens in the atmosphere within first several millimeters close to the surface. This heated air expands as it is dense and rises causing transfer of heat to atmosphere through convection process. It results in formation of clouds.
The radiant energy from sun is transmitted via space in form of invisible waves. But much of the suns radiant energy, is transmitted back to atmosphere. The objects on earth like land, plants, animals absorb radiant energy as heat of which one third gets re-radiated back to atmosphere that is absorbed by carbon dioxide and water vapor. The atmosphere radiates heat energy back to earth increasing the earth temperature. This trapping of radiation is greenhouse effect.
The thermal energy obtained by convection currents are responsible for wind, cloud formation, and weather formation. The hydrosphere that comprises of 70% of earth’s surface absorbs solar energy.
On the basis of the above explanation is:
The energy from the sun passes through space in the form of invisible waves to the earth surface. It heats up the earth’s surface causing variation in climate.
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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
IgorLugansk [536]

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

3 0
3 years ago
Why do we calculate the "range" of the measurments?
KatRina [158]

help show if a straight "trend line" represents the experimental data

6 0
3 years ago
In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot
Fittoniya [83]

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

6 0
3 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
You are an evolutionary biologist studying a population of bats in the rain forest in Brazil. Most of the population possesses m
Oxana [17]

Answer:

Option (B)

Explanation:

In the stabilizing natural selection, the extreme traits from both the ends are eliminated by natural selection and natural selection favors the intermediate trait. So over time individuals having the intermediate traits are selected over the individuals having extreme traits.

So here the population of the bat which possesses moderate wing length is selected over the individual with extreme traits like individuals with short wings and long wings. As a result, the population of moderate length wing bats increased.

Therefore the correct answer is (B)- stabilizing natural selection.

3 0
3 years ago
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