Answer:
the car to the right
Explanation:
its in the name the RIGHT of way hope it helps good luck
Answer:
I would say do it at an even pace
Explanation:
Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast
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
Answer:
Impossible.
Explanation:
The ideal Coefficient of Performance is:


The real Coefficient of Performance is:


Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.
Answer:
0.008
Explanation:
From the question, the parameters given are:
Velocity V = 5 m/s
Pressure = 10 pa
But pressure = F/A
10 = F/A
F = 10A
Substitute all the parameters into the formula below
Coefficient of viscosity (η) = F × r /[AV]
Where
F = tangential force,
r = distance between layers,
A = Area, and
V = velocity
(η) = 10A × 0.004 /[A × 5]
The A will cancel out
(η) = 10 × 0.004 /[5]
(η) = 0.04 /5
(η) = 0.008
Therefore, the coefficient of viscosity of the fluid is 0.008