Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>FALSE. </em>The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. <em>FALSE. </em>The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. <em>FALSE. </em>During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>TRUE. </em>Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. <em>TRUE. </em>The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. <em>TRUE. </em>In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
<span>The answer is 6 kg the mass of the second
object. By using Inversely proportional formula it means that (14 kg) (3 m/s</span>²<span>)
= M (7 m/s</span>²<span>). Where M is the mass of the second object. For the
Newton’s second law of motion formula which is: Force = mass x acceleration, we
have:</span>
<span>F = (14 kg) (3 m/s</span>²<span>) = 42 N</span>
Therefore:
<span>42 N = M (7 m/s</span>²)
<span>M = (42 N) / (7 m/s</span>²<span>)</span>
M = 6 kg mass of the second object
Answer:
kenitec energy
Explanation:
because kinetic and mass have same
Answer:
The kinetic energy of the arrow is equaled to the potential energy of the stretching of the bow, which in this case is 50 J.
Explanation:
Potential energy converts to kinetic as soon as it begins to move.
Answer:
4.9 minutes
Explanation:
Given; T(t) = Ce^-kt + Ts
Now;
T(t) = 190 degrees Fahrenheit
Ts = 60 degrees
To obtain C;
190 = Ce^0 + 60
190 - 60 = C
C = 130
Hence, to find k when t=11
172 = 130 e^-11k + 60
172 -60/130 = e^-k
e^-k = 0.86
ln(e^-k) = ln( 0.86)
-k = -0.15
k = 0.15
Hence at 122 degrees, t is;
T(t) = Ce^-kt + Ts
122 = 130e^-0.15t + 60
122 - 60/130 = e^-0.15t
0.477 = e^-0.15t
ln (e^-0.15t) = ln (0.477)
-0.15t = -0.74
t = 0.74/0.15
t = 4.9 minutes
