Answer: P = 0.416 kW
Explanation:
taken a step by step process to solving this problem.
we have that from the question;
the amount of heat rejected Qn = 4800 kJ/h
the cooling effect is Ql = 3300 kJ/h
Applying the first law of thermodynamics for this system gives us
Шnet = Qn -Ql
Шnet = 4800 - 3300 = 1500 kJ/h
Next we would calculate the coefficient of performance of the refrigerator;
COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2
COPr = 2.2
The Power as required gives;
P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416
P = 0.416 kW
cheers i hope this helps!!!!1
Answer:
R min = 28.173 ohm
R max = 1.55 × 
ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = 
...........1
so R will be
R = 
....................2
put here value
so for t min 11.5 μs
R = 
R min = 28.173 ohm
and
for t max 6.33 ms
R max = 
R max = 1.55 × ohm
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.
dude thx for the points lol
An o ring intended for use in a hydraulic system using MIL-H-5606 (mineral base) fluid will be marked with a blue stripe or dot.
