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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
= 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
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
