Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:

T₁ = 300 K


= 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ +
(T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;

= 1400×
= 1007.6 K
T₅ = T₄ + (
- T₄)/
= 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ +
(T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K

= 1400×
= 1007.6 K
T₇ = T₆ + (
- T₆)/
= 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a.
= cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. 
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Answer: Application.
Explanation:
The question on wether to contine the use of cadavers in the lab for test is being centered around its application. Cadaver which is same as a corpse or dead body is used in crash site during automobil test in lab, some of this cadavers are been disrespected with their applications in the automobile industries because many didn’t consent to be used in those experiments or test.
In the United States, fire codes are developed primarily by two model code organizations, the International Code Council (ICC) and the National Fire Protection Association (NFPA).
Answer:
how are supposed to help when you can't do anything?
Explanation:
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