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3 years ago

A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.

Engineering

1 answer:

Monica [59]3 years ago

6 0

Answer: (a). E = 3.1656×10³⁴ √k/m

(b). f = 9.246 × 10¹² Hz

(c). Infrared region.

Explanation:

From Quantum Theory,

The energy of a proton is proportional to the frequency, from the equation;

E = hf

where E = energy in joules

h = planck's constant i.e. 6.626*10³⁴ Js

f = frequency

(a). from E = hf = 1 quanta

f = ω/2π

where ω = √k/m

consider 3 quanta of energy is lost;

E = 3hf = 3h/2π × √k/m

E = (3×6.626×10³⁴ / 2π) × √k/m

E = 3.1656×10³⁴ √k/m

(b). given from the question that K = 15 N/m

and mass M = 4 × 10⁻²⁶ kg

To get the frequency of the emitted photon,

Ephoton =hf = 3h/2π × √k/m (h cancels out)

f = 3h/2π × √k/m

f = 3h/2π × (√15 / 4 × 10⁻²⁶ )

f = 9.246 × 10¹² Hz

(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to 400 THz in the electromagnetic spectrum.

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3 years ago

In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh

astraxan [27]

( 12 17 18 19 25 )

Explanation:

First Pass:

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

Second Pass:

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

Third Pass:

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

Fourth Pass:

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

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( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

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3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit

8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, $A_1 = 0.0044 m^2$

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

$KE = \frac{V_2^2 - V_1^2}{2} \\$ .........(1)

Where Inlet velocity, V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

$KE = \frac{50^2 - 80^2}{2} \\$

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a steady state flow.

$q - w = h_2 - h_1 + KE + g(z_2 - z_1)$

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, $W = \dot{m}w$ ..........(3)

Mass flow rate, $\dot{m} = 12 kg/s$

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, $h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg$ , x₂ = 0.92

specific enthalpy at the outlet, h₂ = $h_1 + x_2 h_{fg}$

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

$A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2$

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3 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

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3 years ago