When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?
2 answers:
When the electrons step down to lower energy level, emission line spectrum produced.
Further Explanation:
The spectrum of emission of an element is spectrum of wavelength of radiation emitted during the transition of electron from a higher energy level to lower energy level. The energy of photon emitted in this process is equal to difference between the two levels. There is possibility of different types of transition for an atom and energy difference is specific for each atom.
When electrons are excited in atom, extra energy forces electrons to higher energy orbital. When electron goes to lower state, energy releases in form of photon. This emitted photon form the emission line spectrum.
Emission is process in which higher energy state of the particle convert into a lower state through the emission of photon, which produces light. The frequency of emitted light is function of transitional energy.
Since energy of system conserved, difference in energy between two states equal to energy carried by photon. The energy state of transitions results in emission of large frequency range.
Learn more:
1. Broadcast wavelength of station: brainly.com/question/9527365
2. Stress under the tension: brainly.com/question/12985068
3. Average translational kinetic energy: brainly.com/question/9078768
Answer Details:
Grade: High School
Subject: Physics
Chapter: Modern physics
Keywords:
Electrons, step, lower, energy, levels, thin, cloud, hot, gas, emission, line, spectrum, ground, state, higher, radiation, transition, energy, light and photon.
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Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
By definition, the speed of an object is given by:
Where,
dr/dt: derived from the position with respect to time
Therefore, speed has units of length over units of time.
Thus, speed is a derived quantity, since it depends on the value of two other quantities.
Answer:
a derived quantity is:
C. Speed