Answer:
Explanation:
energy emitted by source per second = .5 J
Eg = 1.43 eV .
Energy converted into radiation = .5 x .12 = .06 J
energy of one photon = 1.43 eV
= 1.43 x 1.6 x 10⁻¹⁹ J
= 2.288 x 10⁻¹⁹ J .
no of photons generated = .06 / 2.288 x 10⁻¹⁹
= 2.6223 x 10¹⁷
wavelength of photon λ = 1275 / 1.43 nm
= 891.6 nm .
momentum of photon = h / λ ; h is plank's constant
= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹
= .0074 x 10⁻²⁵ J.s
Total momentum of all the photons generated
= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷
= .0194 x 10⁻⁸ Js
b ) spectral width in terms of wavelength = 30 nm
frequency width = ?
n = c / λ , n is frequency , c is velocity of light and λ is wavelength
differentiating both sides
dn = c x dλ / λ²
given dλ = 30 nm
λ = 891.6 nm
dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²
= 11.3 x 10¹² Hz .
c )
10 nW = 10 x 10⁻⁹ W
= 10⁻⁸ W .
energy of 50 dB
50 dB = 5 B
I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s
I = I₀ x 10⁵
= 10⁻¹² x 10⁵
= 10⁻⁷ W
= 10 x 10⁻⁸ W
power required
= 10⁻⁸ + 10 x 10⁻⁸ W
= 11 x 10⁻⁸ W.
Answer:
hello your question is incomplete attached below is the missing part
answer : short period oscillations frequency = 0.063 rad / sec
phugoid oscillations natural frequency ( 
) = 4.27 rad/sec
Explanation:
first we have to state the general form of the equation
= 
where :
comparing the general form with the given equation
= 18.2329
hence the short period oscillation frequency ( 
) = 0.063 rad/sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.
A
method of procedure that has characterized natural science since the
17th century, consisting in systematic observation, measurement, and
experiment, and the formulation, testing, and modification of
hypotheses
