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Paha777 [63]
4 years ago
9

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the par

ticle-in-a-box with mass m and box length L.

Physics
1 answer:
nikdorinn [45]4 years ago
7 0

Answer:

Explanation:

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

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The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w
satela [25.4K]

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

      F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

     U (x) = A x² + B x + C

Let's calculate the derivatives

    dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

      F = - 2 A x - B

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3 years ago
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Lina20 [59]

Answer:

Explanation:

We are not told where A and B are, but I'll assume that they are two points on the orbit of earth about the sun.

As that orbit is an ellipse, the two points likely do not have the same distance between the earth and sun.

As gravity varies with the inverse of the square of the distance (F = GMm/d²), the force at the closer distance will be greater than the force at the longer distance.

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3 years ago
An aurora can be found in the _____.
Tomtit [17]
I believe the correct answer is d. thermosphere. 
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3 years ago
Read 2 more answers
Give Example of mental flexibility
Inessa [10]

Answer:

If you ring the doorbell and no one opens the door, you'll infer that no one is home rather than continuing to ring the doorbell to an empty house. Being able to understand this and look for another solution is another example of mental flexibility.

Explanation:

6 0
3 years ago
An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign
saul85 [17]

Answer:

2.1\times 10^{-9} T

Explanation:

We are given that

Frequency,f=800KHz=800\times 10^{3} Hz

1kHz=10^{3} Hz

Distance,d=4.5 km=4.5\times 10^{3} m

1 km=1000 m

Electric field,E=0.63V/m

We have to find the magnetic field amplitude of the signal at that point.

c=3\times 10^8 m/s

We know that

B=\frac{E}{c}

B=\frac{0.63}{3\times 10^8}=0.21\times 10^{-8} T

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4 years ago
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