5. Consider a particle in a two-dimensional, rigid, square box with side a. (a) Find the time independent wave function φ(x,y)describing an arbitrary energy eigenstate. (b)What are the energy eigenvalues and the quantum numbers for the three lowest eigenstates? Draw the energy level diagram

Extra problem: The basis of the l = 1 Hilbert space in the (L², L_z) representation are Y₁₁(θ, φ), Y₁₀(θ, φ), Y_1-1(θ, φ).

(1) Find the matrix expressions of L_x, L_y and L_z. Hint: make use of L₊ and L_-.

(2) Suppose the system is in a normalized state Ψ = c₁Y₁₁ +c₂Y₁₀. Find the possible values and corresponding possibilities when measuring L_z, L², and L_x, respectively.

Suppose a sled passenger with total mass 51 kg is pushed 26 m across the snow (μk = 0.20) at constant velocity by a force directed 37° below the horizontal. (a) Calculate the work of the applied force (in J).

(a)

Calculate the work of the applied force (in J).

(b)

Calculate the work of friction (in J).

(c)

Calculate the total work (in J).

I have the answers for this problem from a key, the answers are 3060, -3060, and 0 for a, b, and c respectively.

I have no clue how to get those answers.

Use well labeled diagrams to explain the following two phenomena: the first is that at sunset (or sunrise) the sun appears to be much redder than at mid-day, the second involves the pattern of polarization of the sky when the sun is about to set. Where is the sky most polarized? What orientation is this polarization and why?

Heat transfer

in a test in a double tube heat exchanger the following data are obtained
For hot fluid
Flow = 11.6 gal / sec
Outlet temperature = 30.1 ° C
Inlet temperature = 32 ° C
For cold fluid
Flow 11gal / min
Outlet temperature 25.1 ° C
Inlet temperature 24.2 ° C

For the fluid consider
k = 0.49 w / mK
cp = 3729.95 J / kgK
Prandtl number = 14.29
Density = 1035.02kg / m3

Get:
1) the interior and exterior convection coefficient
2) The heat transferred and the average log temperature
3) Area

Explain your procedure as widely as possible.

You're the operator of a 1.70

Discuss the big bang theory of the universe:

Why do we believe the Universe had a beginning?

What is the evidence that the Universe is expanding?

What is dark energy and how is that related to what the Universe looks like and what will happen to it in the future.

What is inflation?

How does Einstein’s theory of General Relativity fit into this?

Is this changing our view of our place in the cosmos as drastically as the Heliocentric theory changed?

What is the multiverse theory and how does it fit into religion?

What is Zener Diode?

Calculate the amount of energy required to escape from the surface of the following bodies, relative to that required to escape from the surface of Earth.

(a)    Uranus
                     energy to escape Earth

(b)    Mars
                             energy to escape Earth

Explain how a p - n junction works like a solar cell.

Please Summarize the follwing article (this is for a presentation)

"When you release a yo-yo, gravity acts on its center of mass to pull the yo-yo downward. Because the string of the yo-yo is wrapped around the yo-yo's axle, and because one end of the string is attached to your finger, the yo-yo is forced to rotate as it drops. If the yo-yo could not rotate, it would not drop.

Just as any object falling in a gravitational field, the rate of drop increases with time and so, necessarily, does the rotation rate of the yo-yo. The rate of drop and the rotation rate are greatest when the bottom is reached and the string is completely unwound. The spinning yo-yo contains angular momentum (or rotational kinetic energy) derived from the gravitataion potential energy through which the yo-yo has dropped.

Usually, the string is tied loosely around the axle so that the yo-yo can continue to spin at the bottom. Because the full length of the string has been paid out, the yo-yo can drop no further and, consequently, the rotation rate cannot increase further. If left in this condition, the friction between the axle and the string will eventually dissipate the energy of rotation or, equivalently, the angular momentum of the yo-yo and the yo-yo will come to rest.

However, a momentary tug on the string causes the friction between the string and the axle briefly to increase so that the axle no longer slips within the string. When the axle thus stops slipping, the angular momentum of the spinning yo-yo is sufficient to cause the string to wind around the axle. This, of necessity, causes the yo-yo to begin to 'climb' back up the string. After the first one or two rotations, the string can no longer slip, so the process of climbing up the string continues beyond the momentary application of the tug.

As the yo-yo continues to climb back up the string, the angular momentum (or kinetic energy of rotation) of the yo-yo is converted back into gravitational potential corresponding to the increasing height of the center of mass of the yo-yo. For this reason, the yo-yo's angular momentum and, hence, its rotation rate, steadily decreases as the yo-yo rises. This is, of course, the reverse of the process when the yo-yo was dropped.

If not for frictional losses, the yo-yo would climb all the way back up the string to your hand just as its rotational rate decreases to zero. But, due to friction, the yo-yo does not in fact quite get back up to your hand before it stops rotating.

Thereafter, the process repeats, with the yo-yo returning short of its previous height on each cycle. Eventually, the yo-yo comes to rest at the bottom.

Of course, as everyone knows, it is possible to keep the yo-yo going indefinitely by giving it a slight upward pull on each cycle. This pull can be combined with the tug required to initiate the climb back up the string. The pull serves to give the center of mass of the yo-yo a little extra kinetic energy to compensate for frictional losses, so that the yo-yo can be kept going indefinitely.

Yo-yos can also be thrown horizontally, or launched in other directions. The principle of operation is then just the same except that the kinetic energy of the center of mass, which is converted into spin as the string unwinds, results from being thrown, rather than from falling through a gravitational potential."

As an airplane is taking off at an airport its position is closely monitored by radar. The following three positions are measured with their corresponding times:
x₁ = 264.08 m at t₁ = 4.90 s,
x₂ = 312.33 m at t₂ = 5.40 s,
x₃ = 364.72 m at t₃ = 5.90 s.
What is the acceleration of the airplane at t₂ = 5.40 s? (Assume that the acceleration of the airplane is constant.)​

a=_______m/s^2