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Sergio039 [100]

3 years ago

11

Suppose that the wave function for a system can be written as and that are normalized eigenfunctions of the operator with eigenv

alues E1, 2E1, and 4E1, respectively. a. Verify that is normalized. b. What are the possible values that you could obtain in measuring the kinetic energy on identically prepared systems

1 answer:

7nadin3 [17]3 years ago

6 0

Answer:

(a) = 1. Thus the wave function is normalized.

(b) E, 2E₁, 4E₁

Explanation:

The detailed solving of the question is in the attach document.

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The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an

slava [35]

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by $\frac{\pi}{2}$ . Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by $\frac{\pi}{2}$ .

6 0

3 years ago

A container has a 5m^3 volume capacity and weights 1500 N when empty and 47,000 N when filled with a liquid. What is the mass de

Helen [10]

Answer:

927.62 kg/m³ and 0.9275.

Explanation:

Density: This can be defined as the ratio of the mass of a body and its volume.

The S.I unit of density is kg/m³.

Mathematically, Density can be expressed as

D = m/v......................... Equation 1

Where D = density of the body, m = mass of the body,v = volume of the body

Also,

m = W/g.................... Equation 2

Where W = weight of the body, g = acceleration due to gravity.

Given: W = 47000 - 1500 = 45500 N, g = 9.81 m/s²

Substitute into equation 2,

m = 45500/9.81

m = 4638.12 kg.

Also given: v = 5 m³

Substitute into equation 1,

D = 4638.12/5

D = 927.62 kg/m³

Hence the mass density of the liquid = 927.62 kg/m³

Specific gravity: This is the ratio of the density of a body to the density pf water.

R.d = D/D'................................ Equation 3

Where R.d = specific gravity, D = density of the liquid, D' = density of water.

Given: D = 927.62 kg/m³, D' = 1000 kg/m³

Substitute into equation 3

R.d = 927.52/1000

R.d = 0.9275.

Hence the specific gravity of the liquid = 0.9275.

6 0

3 years ago

Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)

rusak2 [61]

Answer:

$F_2=2.43\times10^{-19}\ N$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

$m_A=200\ kg$

$m_B=250\ kg$

$m_C=1700\ kg$

$m_D=100\ kg$

Now their respective coordinate positions (in cm) are as given below:

$P_A=(0,50)$

$P_B=(0,0)$

$P_C=(-80,0)$

$P_D=(40,0)$

<u>Now force on B due to A:</u>

$F_{BA}=G\times \frac{200\times 250}{50^2}$

$F_{BA}=20G\ N$

<u>Now force on B due to C:</u>

$F_{BC}=G\times \frac{1700\times 250}{80^2}$

$F_{BC}=66.406G\ N$

<u>Now force on B due to D:</u>

$F_{BD}=G\times \frac{100\times 250}{40^2}$

$F_{BD}=15.625G\ N$

<em>We observe that the forces due to masses C&D act opposite in direction.</em>

<u>So, the net force in the x-direction:</u>

$F_x=F_{BC}-F_{BD}$

$F_x=66.406G-15.625G$

$F_x=50.781G\ N$ in the positive x-direction

<em>We have only one force in y-direction due to mass A.</em>

So,

$F_y=20G\ N$ in the positive y-direction.

<u>Now the net force:</u>

$F_2=\sqrt{F_y^2+F_x^2}$

$F_2=\sqrt{(20G)^2+(50.781G)^2}$

$F_2=54.5776G^2\ N$

$F_2=2.43\times10^{-19}\ N$

<u>Now the direction of this force with respect to x-axis:</u>

$tan\ \theta=\frac{F_y}{F_x}$

$tan\ \theta=\frac{20G}{50.781G}$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

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Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

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3 years ago