4. In the common collector amplifier circuit, which of the following options is the relationship between the input voltage and the output voltage? (10points) A. The output voltage > The input voltage

The normalization constant A can be determined by integrating the absolute value squared of the wave function over the entire domain and setting it equal to 1, which represents the normalization condition. In this case, the wave function is given by:

ψ(x) = A cos (2πx/L) for -L/4 ≤ x ≤ L/4, and ψ(x) = 0 everywhere else.

To find A, we integrate the absolute value squared of the wave function:

∫ |ψ(x)|^2 dx = ∫ |A cos (2πx/L)|^2 dx

Since the wave function is zero outside the range -L/4 ≤ x ≤ L/4, the integral can be written as:

∫ |ψ(x)|^2 dx = ∫ A^2 cos^2 (2πx/L) dx

The integral of cos^2 (2πx/L) over the range -L/4 ≤ x ≤ L/4 is L/8.

Thus, we have:

∫ |ψ(x)|^2 dx = A^2 * L/8 = 1

Solving for A, we find:

A = √(8/L)

The probability of finding the particle in a specific region can be calculated by integrating the absolute value squared of the wave function over that region. In this case, if we want to find the probability of finding the particle in the region -L/4 ≤ x ≤ L/4, we integrate |ψ(x)|^2 over that range:

P = ∫ |ψ(x)|^2 dx from -L/4 to L/4

Substituting the wave function ψ(x) = A cos (2πx/L), we have:

P = ∫ A^2 cos^2 (2πx/L) dx from -L/4 to L/4

Since cos^2 (2πx/L) has an average value of 1/2 over a full period, the integral simplifies to:

P = ∫ A^2/2 dx from -L/4 to L/4

= (A^2/2) * (L/2)

Substituting the value of A = √(8/L) obtained in part (a), we have:

P = (√(8/L)^2/2) * (L/2)

= 8/4

= 2

Therefore, the probability of finding the particle in the region -L/4 ≤ x ≤ L/4 is 2.

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To solve this problem using energy considerations, we can equate the potential energy of the ball at its maximum height (touching the roof) with the initial kinetic energy of the ball when it is released.

The potential energy of the ball at its maximum height is given by:

PE = mgh

Where m is the mass of the ball (190 g = 0.19 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height (11 m).

The initial kinetic energy of the ball when it is released is given by:

KE = (1/2)mv^2

Where v is the initial velocity we need to find.

Since energy is conserved, we can equate the potential energy and initial kinetic energy:

PE = KE

mgh = (1/2)mv^2

Canceling out the mass m, we can solve for v:

gh = (1/2)v^2

v^2 = 2gh

v = sqrt(2gh)

Plugging in the values:

v = sqrt(2 * 9.8 m/s^2 * 11 m)

v ≈ 14.1 m/s

Therefore, the minimum speed at which the ball must be tossed straight up to just touch the 11 m-high roof of the gymnasium is approximately 14.1 m/s.

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The Galilean transformations are a set of equations that describe the relationship between the space-time coordinates of two reference systems that move uniformly relative to one another with a constant velocity. The aim of this question is to demonstrate that the free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is invariant under Galilean transformations.

The free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is represented as:$$\frac{\partial \psi}{\partial t} = \frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$Galilean transformation can be represented as:$$x' = x-vt$$where x is the position, t is the time, x' is the new position after the transformation, and v is the velocity of the reference system.

Applying the Galilean transformation in the Schrodinger equation we have:

[tex]$$\frac{\partial \psi}{\partial t}[/tex]

=[tex]\frac{\partial x}{\partial t} \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial t}$$$$[/tex]

=[tex]\frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$[/tex]

Substituting $x'

= [tex]x-vt$ in the equation we get:$$\frac{\partial \psi}{\partial t}[/tex]

= [tex]\frac{\partial}{\partial t} \psi(x-vt, t)$$$$\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x} \psi(x-vt, t)$$$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]

Substituting the above equations in the Schrodinger equation, we have:

[tex]$$\frac{\partial}{\partial t} \psi(x-vt, t) = \frac{-\hbar}{2m} \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]

This shows that the free-particle one-dimensional Schrodinger equation is invariant under Galilean transformations. Therefore, we can conclude that the Schrodinger equation obeys the laws of Galilean invariance.

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