Answer:
1) function of fire doors and making sure theyre properly wired to fire alarms
2) proper water piping and purifaction for water fountains and sinks.
3) falty sprinkler systems/rsuty sprikler systems that wont work
4) weather durable roofing and walls for storms and snow depending on were your located .
5) rsuted pipes in showers or fountains that could give you tetnus or other disaeases
Explanation:
HOPE THIS HELPS good luck!!
Hoiu-10,4000 mm.
<h3>
Is positive pressure good for PC?</h3>
- A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
- Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.
To learn more about it, refer
to https://brainly.in/question/413163
#SPJ4
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
<u>Solution and Explanation:</u>
Volume of gas stream = 1000 cfm (Cubic Feet per Minute)
Particulate loading = 400 gr/ft3 (Grain/cubic feet)
1 gr/ft3 = 0.00220462 lb/ft3
Total weight of particulate matter = 
Cyclone is to 80 % efficient
So particulate remaining = 
emissions from this stack be limited to = 10.0 lb/hr
Particles to be remaining after wet scrubber = 10.0 lb/hr
So particles to be removed = 685.7136- 10 = 675.7136
Efficiency = output multiply with 100/input = 98.542 %
