1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bond [772]
3 years ago
9

The light which is produced in the core of the sun passes through layers of comparatively cold gas clouds on the surface of the

sun. These clouds absorb some of the sun's light producing dark lines in the spectrum of the sun and sky. Two of the lines, called Fraunhofer C and I) lines after their discoverer, are found at about 656 nm and 587 nm, respectively. Looking at your atomic spectroscopy data, which elements that you examined would absorb at these wavelengths
Physics
1 answer:
Norma-Jean [14]3 years ago
3 0

Answer:

the two lines observed correspond to Hydrogen and Helium

Explanation:

One of the great principles of atomic physics is that the emission and absorption lines of atoms are the same, in addition, each different element has a series of characteristic lines with which we can identify it.

Bohr's atomic model allows to fully explain the spectrum of lines and absorption of Hydrogen, in this case the line

            λ = 656 nm

corresponds to a hydrogen trasition of the Balmer series from state n = 3 to state n = 2

The other line identified λ = 587 nm does not correspond to any hydrogen line and neither to heavy element lines, so this line when it was discovered was assigned to a new element, which was called Helium.  This line is explained this element was assigned an atomic number (charge) of +2 and a mass of + 4 u, specifically it is a transition from level n = 3 to level n = 2

therefore the two lines observed correspond to Hydrogen and Helium

You might be interested in
A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op
Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

F cos \theta

where

F is the magnitude of the pull

\theta=30^{\circ} is the angle

- Backward: the force of friction, which is

F_f = 30 N

So, the equation of motion is

F cos \theta - F_f = ma

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

F cos \theta - F_f = 0

Therefore, we can find the pulling force:

F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N

(b) 43.9 N

In this case, the acceleration is

a=0.40 m/s^2

So, the equation of motion in this case is

F cos \theta - F_f = ma

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

\theta=30^{\circ}

F_f = 30 N

We find the new magnitude of F:

F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N

6 0
3 years ago
(II) A 12.0-cm-radius thin ring carries a uniformly distributed charge. A small 7.5-g sphere with a charge of is placed exactly
Bond [772]

Answer:

v = 29.q m/s

Explanation:

From the question, the charge is uniformly distributed and therefore, each point in the x-axis is equidistant to every point of the ring.

we can compute the potential at a point (x, 0, 0) without resorting to integration:

15 µC/[4πεo√(x² + R²)]

The potential at the center of the ring is: (15 µC)/{(4πεo) x (12.0 cm)} = 1.123 x 10^(6) V.

However, 2m from the center of the ring, the potential gives:

(15 µC)/{(4πεo x√((2m)² + (12.0 cm)²} = 6.73 x 10^(4) V.

The potential difference is: 6.73 x 10^(4) - 1.123 x 10^(6) V = -1.06 x 10^(6) V.

The change in potential ΔU = qΔV = (3 µC)(-1.06 x 10^(6) V) = -3.17 J.

In conservation of energy, the kinetic energy change ΔK = -ΔU = 3.17 J. Since the particle starts from rest, the final KE = ΔK:

KE = 3.17J = (1/2)mv²

(Multiply both sides by2/m

3.17 x (2/m) = v²

(2 x 3.17 J)/(7.5 g) = v²

v² = 840 m²/s²

Take square root of both sides to get;

v = 29.1 m/s

7 0
3 years ago
Here is something fun to learn about space. I have a link that takes you out of this world to see the spacecraft JUNO, that is c
guapka [62]

Answer:

wow thanks for telling me this website is really cool!

8 0
3 years ago
Read 2 more answers
If the moon moved closer to the earth, how would the gravitational force between them change?.
hichkok12 [17]

If the moon moved closer to the earth, The gravitational force between them will increase.

This can be understood by Newton's law of universal gravitation. This is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

F_{gravitation} = G\frac{M_{1} M_{2}}{r^{2} }

If the moon moved closer to earth, the distance between them will be reduced, and as the force of gravitation is inversely proportional to the distance between the masses, the gravitation force increases.

Learn more about gravitation force here

brainly.com/question/12528243

#SPJ4

7 0
2 years ago
A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0
4 years ago
Other questions:
  • Instead of using his book to hold down fluttering papers, Sam looks for a paperweight. What type of problem-solving obstacle is
    9·1 answer
  • Determine the mechanical energy of this object: 1-kg ball rolls on the ground at 2 m/s
    10·1 answer
  • A man stands on the roof of a building of height 13.0and throws a rock with a velocity of magnitude 30.0 at an angle of 34.9 abo
    8·1 answer
  • when a body receives a stimulus, the nervous system is called to release adrenaline. under what circumstances can this happen?
    12·2 answers
  • Question 11(Multiple Choice Worth 3 points)
    12·1 answer
  • A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the rms value of the cur
    7·1 answer
  • The width, length, and height of a large, custom-made shipping crate are 1.12 m, 1.25 m, and 0.83 m, respectively. The volume of
    15·1 answer
  • What is the acceleration of a 50kg object with 100 newtons force applied to it
    7·2 answers
  • What happens to light when its wavelength changes?
    13·1 answer
  • When do things move faster? Day or night?​
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!