The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
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Explanation:
Answer:
radius = 0.045 m
Explanation:
Given data:
density of oil = 780 kg/m^3
velocity = 20 m/s
height = 25 m
Total energy is = 57.5 kW
we have now
E = kinetic energy+ potential energy + flow work
![E = \dot m ( \frac{v^2}{2] + zg + p\nu)](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p%5Cnu%29)
![E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p_%7Batm%7D%20%5Cfrac%7B1%7D%7B%5Crho%7D%29)
solving for flow rate
![\dot m = 99.977we know that [tex]\dot m = \rho AV](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%2099.977%3C%2Fp%3E%3Cp%3Ewe%20know%20that%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdot%20m%20%20%3D%20%5Crho%20AV)
solving for d
d = 0.090 m
so radius = 0.045 m
Answer:
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By being attentive in classes, taking good notes and making sure to put time fourth into your grades.
