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3 years ago

Comparing Wave A (black) to Wave B (green), Wave A has a

Physics

2 answers:

Gennadij [26K]3 years ago

7 0

I think it's C, longer wave length.

RoseWind [281]3 years ago

4 0

Yes the answer is c longer wavelength

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An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

R = Rydberg constant, $R=1.097\times 10^7\ m^{-1}$

$\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

Solving above equation we get the value of final n is,

$n_f=2.04$

$n_f=2$

So, it will relax in the n = 2. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

What is mean by combination reaction ?

mrs_skeptik [129]

$\underline{\purple{\large \sf Combination \: reaction :-}}$

Those reaction in which two or more substances combine to form a one new substance are called Combination reaction

In this reaction, We can add :

Two or more elements can combine to form a compound.
Two or more compounds can combine to from a one new compound.
An element and a compound can combine to form a new compound.

$\underline{\green{\large \sf For\: example :}}$

$\sf 2H_{2} + O_{2} \: \underrightarrow{Combination} \: 2H_{2}o$

In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.

3 0

3 years ago

The most important part of physical fitness is. Strength. B muscular endurance c flexibility. D. Cardiovascular

solniwko [45]

The answer is D. Cardiovascular

7 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

3 years ago

What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored

Musya8 [376]

The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.

<h3>What is the potential difference of the circuit?</h3>

The potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

where;

C is capacitance of the capacitor
V is the potential difference

For a parallel circuit the voltage in the circuit is always the same.

The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

2U = CV²

V = √2U/C

V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)

V = 12 V

The equivalent capacitance of C1 and C2 is calculated as follows;

1/C = 1/C₁ + 1/C₂

1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)

1/C = 1,173,611.11

C = 1/1,173,611.11

C = 8.52 x 10⁻⁷ C

The total capacitance of the circuit is calculated as follows;

Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C

Ct = 3.35 x 10⁻⁶ C

The total energy of the circuit is calculated as follows;

U = ¹/₂CtV²

U = ¹/₂(3.35 x 10⁻⁶ )(12)²

U = 2.41 x 10⁻⁴ J

Learn more about energy stored in a capacitor here: brainly.com/question/14811408

#SPJ1

7 0

1 year ago