Answer:
A) Cancer of the Lungs
B)Larynx and Urinary Tract, as well as nervous system and kidney damage
Explanation:
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Answer:
- hoop stress
- longitudinal stress
- material used
all this could led to the failure of the garden hose and the tear along the length
Explanation:
For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :
hoop stress ( which acts along the circumference of the pipe):
αh = 
EQUATION 1
and Longitudinal stress ( acting along the length of the pipe )
αl = 
EQUATION 2
where p = water pressure inside the hose
d = diameter of hose, T = thickness of hose
we can as well attribute the failure of the hose to the material used in making the hose .
assume for a thin cylindrical pipe material used to be
≥ 20
insert this value into equation 1
αh = 
= 60/2 = 30 psi
the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose
The examples of engineering controls is Biohazard waste containers and Spill clean up kits.
What is engineering controls?
An engineering controls is a workplace process that protect workers by removing hazardous conditions or by placing a barrier between the worker and the hazard.
An example of engineering controls is installation of exhaust ventilation to remove airborne emissions to shield the worker.
Hence, the examples of engineering controls is Biohazard waste containers and Spill clean up kits.
Therefore, the Option C and D is correct.
Given Information:
Static Pressure = P = 1 atm
Temperature = T = 70°F
Velocity = v = 150 mi/hr
Required formation:
Pressure of Pilot Tube = P₀ = ?
Answer:
Pressure of Pilot Tube = P₀ = 2172.36 lb/ft²
Solution:
Units Conversion:
(1 atm = 2116.22 lb/
ft²)
Static Pressure = P = 1 atm = 2116.22 lb/ft²
(1° F = 459.67° R)
Temperature = T = 70°F = 70 + 459.67 = 529.67° R
( 1mi/hr = 1.467 ft/sec)
Velocity = v = 150 mi/hr = 220 ft/sec
Density of Test Section
Density = ρ = P/RT
Where R is the idea gas constant 1717 (ft-lbf/slug-°R)
ρ = 2116.22/1717*529.67 = 0.00232 slug/ft³
Pressure of Pilot Tube
P₀ = P + (1/2)*ρ*v²
P₀ = 2116.22 + (1/2)*0.00232*220²
P₀ = 2172.36 lb/ft²
