When an electron jumps from high-to low-energy levels:

Physics

1 answer:

AysviL [449]3 years ago

5 0

Answer:

B. A wave of visible light is produced.

Explanation:

The visible light spectrum is in the centre of the Electromagnetic Spectrum, so any time an electron jumps from high to low levels of radiation, it has to cross through the visible spectrum.

I hope this is correct, and as always, I am joyous to assist anyone at any time.

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Which season is signaled by average lower temperature and indirect, angled sunlight?

anygoal [31]

When the earth in its translation movement has the earth's axis is further from the sun and the energy per unit area is minimal, we have the winter season

The translation is the movement of the earth around the Sun, this movement establishes the duration of the year, also the earth's axis is inclined about 23.5º in relation to the orbit.

A consequence of this inclination of the Earth's axis is that the solar rays reach the surface of the planet with different inclinations, when they arrive more perpendicularly the energy deposited by the surface unit is greater and therefore the temperature increases, this period is called Summer.

When the rays arrive with the greater inclination, the energy deposited per unit area is minimal, which is why the temperature decreases, this period is called Winter.

In the periods when the axis is almost vertical we have an average absorption of energy, these two periods are called Spring and Autumn.

In conclusion, the winter season occurs when the terrestrial axis is furthest from the sun and the energy per unit of area is minimal.

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Bezzdna [24]

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2 years ago

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The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;