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Sveta_85
2 days ago
7

What is another metaphor (aside from the fact that a piano can only produce certain notes) that could help someone understand en

ergy quantization at the atomic level? Explain your answer.
The text it is referring to is:
Because protons and electrons in an atom carry opposite charges, they are attracted to each other and thus maintain the boundaries of the atom as a singular closed system. This means that an atom’s energy is quantized, or fixed to a certain amount. Just like a piano can only produce a certain set of notes that correspond to its keys, an atom of a certain element can only contain a certain fixed amount of energy.
Physics
1 answer:
Ostrovityanka [942]2 days ago
4 0

Response:

A suitable metaphor that illustrates energy quantization at the atomic level is the number of teams in different phases of the Football World Cup. Here’s how energy corresponds to teams at various levels:

Phase 1 Group Stage

Teams at the group level = 32, Energy assigned per team = 1

Knockout stages

Round of 16

Teams in the round of 16 = 16, Energy per team = 2

Semi-final

Teams in the semi-final = 8, Energy per team = 4

Quarter-final

Teams in the quarter-final = 4, Energy for each team = 8

Final

Teams that win the World Cup = 1, Energy of the champion = 32

In each phase, teams have defined energy levels which cannot take intermediate values, similar to how particles within atoms have only distinct quantized energy levels.

Clarification:

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A positive charge moves in the direction of an electric field. Which of the following statements are true?
kicyunya [1025]

Answer:

The potential energy tied to the charge diminishes.

The electric field performs negative work on the charge.

Explanation:

3 0
9 days ago
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To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Softa [913]

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
12 days ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte
Yuliya22 [1153]

We will use the equations of rotational kinematics,

\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2             (A)

\omega^2= \omega^2_{0} +2\alpha\theta                                     (B)                                          

Here, \theta and \theta _{0} denote the final and initial angular displacements, respectively, whereas \omega and \omega_{0} represent final and initial angular velocities, and \alpha is the angular acceleration.

We are provided with \alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad and t = 4.20 \ s.

By substituting these values into equation (A), we have

62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s

Now, using equation (B),

\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0
9 days ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Softa [913]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Reservoir pressure = 10 atm

T_1 = Reservoir temperature = 300 K

P_2 = Exit pressure = 1 atm

T_2 = Exit temperature

R_s = Specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

Assuming isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

Flow density at exit measures 2.2721 kg/m³

4 0
11 days ago
Which pair of graphs represent the same motion of an object
Ostrovityanka [942]
The correct choice is C
3 0
15 days ago
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