Answer:
Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation
, Cooling fans
, Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat
, Insulation of hot surfaces Elimination of steam leaks
, Cooled seats or benches for rest breaks
, Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.
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Explanation:
I think it’s rationalization.
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Answer:
a)6.8 KPa
b)0.264 gallon
c)47.84 Pa.s
Explanation:
We know that
1 lbf= 4.48 N
1 ft =0.30 m
a)
Given that
P= 1 psi
psi is called pound force per square inch.
We know that 1 psi = 6.8 KPa.
b)
Given that
Volume = 1 liter
We know that 1000 liter = 1 cubic meter.
1 liter =0.264 gallon.
c)
Answer: Design a solution or problem
Explanation:
When engineers develop a model, the step that is taking place in the engineering process is referred to as the design a solution or problem step.
This is when engineers tackles and solved a problem by finding solutions and this is done by incorporating their analytical, and synthetic thinking and using their skills an experience.
Answer: a) -5 ft/s²
B) 4.5 ft
Explanation: Radius= 250ft
Velocity V = 3(t-t²)ft/s
A). When t= 3 s, the acceleration is
dv/dt = 1-2t
dv/dt = 1- 2(3)
= 1-6
= -5 ft/s²
B. How far it traveled in 3 sec
Distance= 3t²/2 -t³/3 ft
Substituting 3
Distance = 27/2 - 9
= 13.5 - 9
= 4.5 ft
