Answer:
Electrons are located in specific orbit corresponding to discrete energy levels
Explanation:
In Bohr's model of the atom, electron orbit the nucleus in specific levels, each of them corresponding to a specific energy. The electrons cannot be located in the space between two levels: this means that only some values of energy are possible for the electrons, so the energy levels are quantized.
A confirmation of Bohr's model is found in the spectrum of emission of gases. In fact, when an electron jumps from a higher energy level to a lower energy level, it emits a photon whose energy is exactly equal to the difference in energy between the two levels: since the energy levels are discrete, this means that the emitted photons cannot have any value of wavelength, but also their wavelength will appear as a discrete spectrum. This is exactly what it is observed in the spectrum of emission of gases.
Answer:
Acceleration, 
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :




So, the acceleration on the rough ice
and negative sign shows deceleration.
I would say that strong language typically reveals strong feelings: strong language would reveal desperation, or very strong feelings: weak feelings would not demand it, and would be more likely accompanied by more controlled behavior.
When a mirror is rotated . . .
-- The incident ray doesn't turn. It's just the line from the source to the mirror.
It would be there, in the same place, even if there was no mirror.
-- The normal turns. It's the line perpendicular to the mirror, so it must turn
with the mirror.
-- Since the normal tuns and the incident ray doesn't, the angle between them
must change. And since the angle of the reflected ray is equal to the angle of
the incident ray, the reflected ray must also turn.