X-ray formation
Energy difference in the first ionization of the same group of elements
Differential linear emission spectra of different elements
Explanation:
The formation of x-ray, energy difference in the first ionization of the same group of elements and differential linear emission spectra of different elements can be justified by the Bohr model of the atom.
- According to Neil Bohr, in an atom, the extranuclear part consists of electrons in specific spherical orbits around the nucleus.
- Bohr's model explained the discrete lines in atomic spectrum of different elements.
- Due to excitation of electrons, some rays are given off. Some of these are energetic like x-rays.
- The energy difference between the first ionization of the same group of elements is a proof that electrons are in specific spherical orbits.
- The higher the energy the level, the ease at which electrons can be removed and the lower the ionization energy.
Learn more:
Bohr model of the atom brainly.com/question/4986277
#learnwithBrainly
Answer: tu dia es INCORRECTO
Explanation:
Answer:
During the voyage Charles Darwin explored the Galapagos islands and noticed the same species have different adaptations in places. ... Charles noticed that each species has the same ancestor but they evolve to adapt over time so they can live longer.
Explanation:
Answer:
The increase in the gravitational potential energy is 29.93 joules.
Explanation:
Given that,
Mass of the box, m = 2.35 kg
It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m
We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :


U = 29.93 Joules
So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.