Given :
Capacitor , C = 55 μF .
Energy is given by :
.
To Find :
The current through the capacitor.
Solution :
Energy in capacitor is given by :
Now , current i is given by :
( differentiation of cos x is - sin x )
Therefore , the current through the capacitor is -11.18 sin ( 377t).
Hence , this is the required solution .
Answer:
461.65 KJ/Kg
Explanation:
In this question, we are asked to calculate the values of heat transferred in the process.
Please check attachment for complete solution and step by step explanation
Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj -
a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N