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Temka [501]
3 years ago
15

Some material consisting of a collection of microscopic systems is kept at a high temperature, so that all excited states are po

pulated and can participate in emission of photons. A photon detector capable of detecting photon energies from infrared through ultraviolet observes photons emitted with energies of 0.3 eV, 0.5 eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV. These are the only photon energy observed.
a) Draw and label a possible energy-level diagram for one of the microscopic systems, which has 4 bound states. On the diagram, indicate the transitions corresponding to the emitted photons. Explain briefly.
b) The material is now cooled down to a very low temperature, and the photon detector stops detecting photon emissions. Next a beam of light with a continuous range of energies from infrared thorugh ultraviolet shines on the material, and the photon detector observes the beam of light after it passes through the material. What photon energies in this beam of light are observed to be significantly reduced in intensity ("dark absorption lines")? Explain briefly.
Physics
1 answer:
Lelechka [254]3 years ago
6 0

Answer:

The responses to this question can be defined as follows:

Explanation:

During energy exchange E=hv, electrodes spring through one orbit to another

Please find the image file in the attachment.

Its absorption layer comprises 0.3 eV, 0.5 eV., 0.8 eV, 2.0 eV, 2.5 eV again, as light passes via material at low temperature those lines absorbed in the strata called absorption stratum.

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As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the
In-s [12.5K]

Answer:

It is conserved

Explanation:

Converted to heat energy due to the friction caused by the box rubbing on the floor

4 0
3 years ago
All living organisms reproduce by a live functioning baby leaving the mother's body.​
algol13

Answer:

there is no question

Explanation:

4 0
3 years ago
Find the volume of the tire with dimensions
Vinvika [58]

The volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

<h3>What is volume?</h3>

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

<h3>Volume of the tire</h3>

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

Radius of the tire = 0.5 x 26" = 13"

Volume of the tire = Area x thickness

Volume of the tire = πr² x h

where;

  • r is the radius of the tire
  • h is the thickness of the tube

Volume of the tire = π(13)² x (2.125)

Volume of the tire =  1,128.2 cubic inch

Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

Learn more about Volume of tire here: brainly.com/question/1972490

#SPJ1

7 0
2 years ago
Choose the 200 kg refrigerator. set the applied force to 400 n (to the right). be sure friction is turned off. what is the net f
Firlakuza [10]

The net force acting on the refrigerator is 400 N to the right.

<h3></h3><h3>FURTHER EXPLANATION</h3>

The net force or resultant force is the sum of all the forces acting on a body or an object in x and y axes.

  • Forces along the y-axis The forces that usually act on an object vertically (in the y-axis) are: gravitational force which is a downward force and the normal force which is an upward (perpendicular) force exerted by a surface on an object resting above it that keeps the object from falling.
  • Forces along the x-axis These include the force or forces applied to cause a left or right motion of an object along the horizontal plane (called the Applied Force) and the force that opposes the motion or friction.

In this problem the forces acting on the x and y - axes can be determined:

Along the x-axis:

  • gravitational force = -1960 N
  • normal force = +1960 N
  • Net force = -1960 N + 1960 N = 0

The gravitational force is the weight of the object obtained by multiplying the mass of the object (in kg) with the acceleration due to gravity, 9.8 m/s^2. It is given a negative (-) sign to indicate that it is a downward force.

Since the object is not falling through the surface, it can be assumed that the gravitational force and normal force are balanced. Hence, the size of the normal force is the same as the gravitational force but with the opposite direction indicated by the + sign for an upward force.

The forces along the x-axis are balanced  (i.e. net force is zero) so the object neither moves upward or downward.

Along the y-axis

  • applied force = +400 N
  • friction = 0
  • Net force = +400 N + 0 = +400 N

The applied force is +400 N. It is given a + sign to indicate that its direction is to the right.

The friction, as mentioned in the problem, is set to zero or "turned off".

The net force along the y-axis is +400. The forces are unbalanced so the object will move to the right as force is applied to it.

<h3>LEARN MORE</h3>
  • Balanced Forces brainly.com/question/760473
  • Unbalanced Forces brainly.com/question/4289010
  • Free body Diagrams brainly.com/question/12345810

Keywords: net force, resultant force

5 0
4 years ago
Read 2 more answers
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0
3 years ago
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