Answer 1: minimum required flow rate of fresh air is 0.57 m^3/ses
Explanation: since the minimum requirement per person is 30 L/sec
Converting to m^3 it becomes
30/1000 = 0.03 m^3/sec
For 19 heavy smoker we will require
19 * 0.03 = 0.57m^3/sec
Answer 2: diameter of the duct will be 0.3m
Explanation: since flow rate is
Q =0.57m^3/sec
Also
Q = AV (continuity equation)
Where A is the duct area and V is the velocity of air flow in m/sec
0.57m^3/sec = A * 8m/sec
A = 0.57/8 = 0.071m^2
Area of the duct is that of a circle
A = 3.142 *(d^2 ÷4)
d^2 = (0.017 * 4)/3.142 = 0.09
d is square root of 0.09
d = 0.3m
Answer:
Explanation:
= Gauge pressure = 2.2 atm = 
= Absolute pressure = 
= Local atmospheric pressure
Absolute pressure is given by
The local atmospheric pressure is .
It is when you don’t give up and you stand for your rights
The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays
<h3>How to utilize z-score statistics?</h3>
We are given;
Mean; μ = 15
Standard Deviation; σ = 5
We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.
Formula for z-score is;
z = (x' - μ)/σ
We want to find the value of x such that the probability is 0.95;
P(X > x) = P(z > (x - 15)/5) = 0.95
⇒ 1 - P(z ≤ (x - 15)/5) = 0.95
Thus;
P(z ≤ (x - 15)/5) = 1 - 0.95
P(z ≤ (x - 15)/5) = 0.05
The value of z from the z-table of 0.05 is -1.645
Thus;
(x - 15)/5 = -1.645
x ≈ 7
Complete Question is;
A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.
Read more about Z-score at; brainly.com/question/25638875
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Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em>
</em> therefore 
We have now that
multiplying through by T
multiplying through by 300
-
The temperature T= 648.07k
