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balandron [24]
3 years ago
11

The absorption line spectrum shows what we see when we look at a hot light source (such as a star or light bulb) directly behind

a cooler cloud of gas. Suppose instead that we are looking at the gas cloud but the light source is off to the side instead of directly behind it. In that case, the spectrum would __________.
Physics
1 answer:
Over [174]3 years ago
4 0

Answer:

The absorption line spectrum shows what we see when we look at a hot light source (such as a star or light bulb) directly behind a cooler cloud of gas. Suppose instead that we are looking at the gas cloud but the light source is off to the side instead of directly behind it. In that case, the spectrum would be an emission spectrum.

Explanation:

Kirchhoff’s laws establish that:

  • A solid, liquid or dense incandescent gas emits a continuous spectrum.  
  • A hot and diffuse gas produces bright spectral lines (emission lines).
  • A gas of lower temperature against a source of continuum spectrum, produces dark spectral lines (absorption lines) superposed in the continuum spectrum.

According with Kirchhoff's laws it is get an emission line spectrum¹ in the scenario at which the observer is looking directly at a gas cloud with the light source off to the side from the line of sight.

In this case the atoms, molecules or ions in the medium are excited by the radiation that comes from the light source. That is known as an electronic transition², an electron in the atom or ion will absorb the photon coming from the light source and pass to a higher energy level, the electron, upon returning to its base state will emit a photon or a series of photons.

Keys terms:

¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).

²Electronic transition: When an electron passes from one energy level to another, either for the emission or absorption of a photon.

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Fill in the table below with the properties of acids and bases.
Whitepunk [10]

Answer:

1) Acids have a sour taste , Bases have a bitter taste.

2) Acids turn blue litmus paper into red , Bases turn red litmus paper into blue.

3) Acids react with most metals to form Hydrogen gas but only a few base react with a few metals to form Hydrogen gas

4) Both will conduct electricity. Both acids & bases are good electrolytes. Strong acids & bases conduct more electricity than that of weak acids & bases

5 0
3 years ago
he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
Paraphin [41]

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂  __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69  = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

<u>R₂ = 16.92 Ω</u>

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

<u>R₁ = 50.77 Ω</u>

4 0
3 years ago
mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
valina [46]

Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

The two equations solved:

3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\

Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

5 0
3 years ago
David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2
dangina [55]
A. 441 m B: 46.0 m/s
5 0
2 years ago
An object accelerates to a velocity of 34m/s over a time 13 s the acceleration it experience was 15 m/s2 what was it’s intital v
miv72 [106K]
Vf =Vi + at
Vi =Vf - at
= 34 - 15•13
= - 161 m/s :/
7 0
3 years ago
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