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3 years ago

Which statement applies to the electron at state 2?

Physics

2 answers:

Inga [223]3 years ago

7 0

Answer:

d) It possesses greater energy than the ground state.

Explanation:

As we know that total energy associated with the electron is given by the equation

$E = -13.6 \frac{z^2}{n^2}$

total potential energy is given as

$U = -27.2 \frac{z^2}{n^2}$

total kinetic energy of the system is given as

$K = 13.6 \frac{z^2}{n^2}$

so in terms of total energy when electron moves to higher energy state the total energy will also increase.

So if it jump from this higher energy state to lower state it must radiate the energy

so correct answer would be

d) It possesses greater energy than the ground state.

iren [92.7K]3 years ago

6 0

The statement that applies to the electron at state 2 is it possesses greater energy than the ground state. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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What could be used as another word for electrical potential

tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

Explanation:

Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points

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A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

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(d) शून्य

Ede4ka [16]

The answer is a 20 or 140 u choose because u subtract or add these numbers and sorry I can’t speak sri lanka or Indian but I know how to read them

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Phantasy [73]

Answer:

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dezoksy [38]

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