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seropon [69]
3 years ago
5

When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho

fer lines are taken into account and useful for determining elements within a star?
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

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A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rat
Inessa05 [86]

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

4 0
3 years ago
Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocit
ser-zykov [4K]
The reaction time of Boris is t(r), so before that, Boris will not have jumped. Thus, H(b)(t) = 0
The vertical displacement will simply be
D(t) = H(a)(t)
7 0
3 years ago
Different between pulling and pushing force​
yuradex [85]

Answer:

Push and pull are both forces, but the difference is in their direction at which it is applied. If the force applied in the direction of motion of the particle then we call it push. If that force applied in the direction OPPOSITE to the motion of the particle then it it termed as pull

4 0
3 years ago
When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________
Nana76 [90]

Answer: Oblique impact

Explanation:

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be oblique impact.

On the other hand, when the directions of motion of the two colliding particles are moving along a line of impact, then it's refered to as central impact.

7 0
3 years ago
A rescue team is searching for Andrew, a geologist who was stranded while conducting research in the mountains of Colorado. The
Sindrei [870]

Answer:

I = 9.82 10⁻⁷ W / m²

Explanation:

The intensity of the sound wave is the energy of the wave between the order per unit area of ​​the same

      I = P / A = E / T A

the energy is calculated by integrating the mechanical energy in a period, where the mass is changed by the density and ‘s’ is the amplitude of the sound wave

     I = ½ ρ v (w s)²

   

     I = ½ 1.35 328 (2π 530 2.00 10⁻⁸)²

     I = 221.4 (4.435 10⁻⁹)

     I = 9.82 10⁻⁷ W / m²

4 0
3 years ago
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