Answer:FALSE
Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.
Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.
Answer:
The minimum diameter for each cable should be 0.65 inches.
Explanation:
Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,
(1/2)(Weight/Cross Sectional Area) = Allowable Stress
Here,
Weight = 1000 lb
Cross-sectional area = πr²
where, r = minimum radius for each cable
(1/2)(1000 lb/πr²) = 1500 psi
500 lb/1500π psi = r²
r = √1.061 in²
r = 0.325 in
Now, for diameter:
Diameter = 2(radius) = 2r
Diameter = 2(0.325 in)
<u>Diameter = 0.65 in</u>
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = 
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = 
· 
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × 
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
The mass flow rate of steam m=5.4 Kg/s
Explanation:
Given:
At the inlet of turbine P=10 MPa ,T=500 C
AT the exit of turbine P=10 KPa ,x=0.9
Required power=5 MW
From steam table
<u> At 10 MPa and 500 C:</u>
h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)
<u>At 10 KPa:</u>
=2675.1 KJ/Kg, 
=417.51 KJ/Kg
= 7.3 KJ/Kg-K ,
=1.3 KJ/Kg-K
So enthalpy of steam at the exit of turbine
h= 
Now by putting the values
h= 417.51+0.9(2675.1- 417.51) KJ/Kg
h=2449.34 KJ/Kg
Lets take m is the mass flow rate of steam
So 
m=5.4 Kg/s
So the mass flow rate of steam m=5.4 Kg/s
Answer:
Helps to accurately calculate job costs
Explanation:
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