Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t
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.
If the collision is inelastic, there is every possibility that the large body will drag the small stationary body along with it in the direction of the collision. Some amount of heat, light and sound energy will also be produced due to the kinetic energy of the large body. I hope the answer helps you.
The slope of a speed-time graph is the acceleration represented by the graph.
All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.
Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e 
.
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .
