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joja [24]
3 years ago
13

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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Answer:

5E22 atoms/cm³

Explanation:

We need to find the number of moles of silicon per cm³

number of moles per cm³ = density/atomic weight = 2.33 g/cm ÷ 28.09 g/mol = 0.083 mol/cm³.

Since there are 6.022 × 10²³ atoms/mol, then the number of atoms of silicon per cm³ = number of atoms per mol × number of moles per cm³

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= 0.4995 × 10²³ atoms/cm³

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≅ 5 × 10²² atoms/cm³

= 5E22 atoms/cm³

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2 years ago
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A controlled process is described by the closed-loop transfer function G(s).
MissTica

Answer:

The answer is "Option B".

Explanation:

Given equation:

G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\

if

\to 2s^2 + (K-1)s + (K-1)=0

Calculating by the Routh's Hurwitz table:

\to s^2  \ \ \ \ \    2  \ \ \ \ \ \  K-1 \\\\\to s^2  \ \ \ \ \    K-1  \ \ \ \ \ \   \\\\\to s^0 \ \  ( \frac{(K-1)(K-1)(-2) (0)}{K-1}  \\\\    \ \ \ \  = (K-1) )

Form the above table:

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jek_recluse [69]

Answer:

1170 km/hr

Explanation:

Let's first state the formula to be used

c = √(KRT)

The temperature at an altitude of 15km is -56.5° C

Let's not convert this to °K, we have

-56.5° + 273.15 = 216.65° K

Also, the temperature at 8km is -36.9° C, on converting to °K we have

-36.9° + 273.15 = 236.25° K

Then again, we look for the speed at both 15 km and 8 km both of which are 295 m/s and 308 km

Finally, we use the mach similarity formula

(V/c) of 15km = (V/c) of 8km

V of 8km = c of 8km * (V/c) of 15km

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