Answer:
For the two you haven't answered: (Drag greater than thrust, lift greater than weight) It will accelerate backwards (decelerate) and upwards
(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.
Answer:
The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.
Explanation:
The CO2 requirement for the plant is:
Amount of CO2 per year = (2.4 lb / KWh)(300,000 KWh)
Amount of CO2 per year = (720000 lb)(1 ton/ 2204.62 lb)
Amount of CO2 per year = 326.59 ton
The perceived economic impact of CO2 generated per year will then be:
Economic Impact = ($25 / ton)(326.59 ton)
<u>Economic Impact = $8164.67</u>
I think the coach wants Brittany and her team to use more shooting *when the time is right*.
Answer:
T1 = 625.54 K
Explanation:
We are given;
T_α = Tsur = 25°C = 298K
h = 20 W/m².K,
L = 0.15 m
K = 1.2 W/m.K
ε = 0.8
Ts = T2 = 100°C = 373K
T1 = ?
Assumption:
-Steady- state condition
-One- dimensional conduction
-No uniform heat generation
-Constant properties
From Energy balance equation;
E°in - E°out = 0
Thus,
q"cond – q"conv – q"rad = 0
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)
Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)
Thus;
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0
This gives;
(8T1 - 2984) - (1500) - 520.31 = 0
8T1 = 2984 + 1500 + 520.31
8T1 = 5004.31
T1 = 5004.31/8
T1 = 625.54 K
Answer:
[a]. 0.49.
[2]. 0.536
[c]. 4.15 kmol; 5.84 kmol.
Explanation:
Without mincing words let's dive straight into the solution to the question above.
[a].
The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].
The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.
[b].
The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].
Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.
[c].
The amount of distillate product that is withdrawn:
4 = 0.85 H(t) + 0.8 - 0.08.
H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.
