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.
Answer:
Space velocity = 30 hr⁻¹
Explanation:
Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.
It is given mathematically as
Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)
Volumetric flowrate of the reeactants
= (molar flow rate)/(concentration)
Molar flowrate of the reactants = 300 millimol/hr
Concentration of the reactants = 100 millimol/liter
Volumetric flowrate of the reactants = (300/100) = 3 liters/hr
Reactor volume = 0.1 liter
Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹
Hope this Helps!!!
Answer:
Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal 
= 0.9
Radial load 
= 2600 lbf
To find C10 value by using the formula:
where;
The Weibull parameters include:
∴
Using the above formula:
![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)
Recall that:
1 kN = 225 lbf
∴
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
