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3 years ago

What are energy sources

Physics

1 answer:

ladessa [460]3 years ago

7 0

Answer:

Solar Energy. The primary source of energy is the sun.

Wind Energy. Wind power is becoming more and more common.

Geothermal Energy.

Hydrogen Energy

Tidal Energy.

Wave Energy.

Hydroelectric Energy.

Biomass Energy.

Explanation:

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Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul

timofeeve [1]

Answer:

a) $K_{max}$ = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

$K_{max}$ = h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

E = 2,486 10⁻¹⁸ J

Let's reduce to eV

E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

E = 15.5 eV

Now we can calculate the kinetic energy

$K_{max}$ = h c / f - fi

$K_{max}$ = 15.5 -13.6

$K_{max}$ = 1.9 eV

b) Extra energy = 10.2 eV

The total kinetic energy of electrons is

Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

ΔE = E photon - Total kinetic energy electron

ΔE = 15.5 - 12.1

ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

$E_{n}$ = 13.606 / n2

n = √ 13.606 / En

n = √ (13606 / 3.4)

n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

8 0

3 years ago

Compare wave A with wave B correctly in relation to amplitude.

olga nikolaevna [1]

We know for a fact that D and A can be eliminated from the answer choices because those are both false. we know that both B and C are both very accurate knowing that wave B does have a higher pitch than wave A but also wave A has a louder sound wave than wave B. so i'd say the answer is (C).

8 0

3 years ago

Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will

djyliett [7]

Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

$P\cdot V = n \cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure.

$V$ - Volume.

$n$ - Molar quantity, in moles.

$R_{u}$ - Ideal gas constant.

$T$ - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

$P_{1}\cdot V_{1} = P_{2}\cdot V_{2}$ (2)

If we know that $\frac{V_{2}}{V_{1}} = \frac{1}{3}$ , then the resulting pressure of the system is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 3\cdot P_{1}$

The resulting pressure is 3 times the initial pressure.

4 0

2 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

dem82 [27]

Angle (θ) = 60°
Force (F) = 20 N
Distance (s) = 200 m
Therefore, work done
= Fs Cos θ
= (20 × 200 × Cos 60°) J
= (20 × 200 × 1/2) J
= (20 × 100) J
= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Read 2 more answers

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?