Answer:
The required mechanical work is required to reduce each day by 1.05×10^8 Joules.
Explanation:
Coefficient of Performance (COP) = Q/W
Q is thermal energy absorbed by the air conditioner
W is mechanical work done
Q = 3.9×10^8 J
COP of old air conditioner = 2.3
W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J
COP of new air conditioner = 6
W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J
Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J
Answer:
Explanation:
Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz
Zero at -200MHz, zero at infinity.
A) A(s) = 100FH(s)
B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz
(2): 2pi × 20 × 10^6= 4pi × 10^6 MHz
Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.
T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)
Answer:
With increased technological knowledge and consequent decreased factors of ignorance, the structures have less inert masses and therefore less need for such decoration. This is the reason why the modern buildings are plainer and depend upon precision of outline and perfection of finish for their architectural effect.