To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.
<h3>What is a Class D fire?</h3>
A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.
Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.
By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.
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Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
= 
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:
We can now replace 
in the above equation.
Therefore, = × CρAv²
The A which stands as the area of the jet is given by the formula:
We can now have a new equation after substituting our A into the previous equation as:
= × Cρ 
Substituting our data from above; we have:
= × 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:
a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.
b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods
c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.
d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.
Explanation:
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Answer:
Maximum Normal Stress σ = 8.16 Ksi
Maximum Shearing Stress τ = 4.08 Ksi
Explanation:
Outer diameter of spherical container D = 17 ft
Convert feet to inches D = 17 x 12 in = 204 inches
Wall thickness t = 0.375 in
Internal Pressure P = 60 Psi
Maximum Normal Stress σ = PD / 4t
σ = PD / 4t
σ = (60 psi x 204 in) / (4 x 0.375 in)
σ = 12,240 / 1.5
σ = 8,160 P/in
σ = 8.16 Ksi
Maximum Shearing Stress τ = PD / 8t
τ = PD / 8t
τ = (60 psi x 204 in) / (8 x 0.375 in)
τ = 12,240 / 3
τ = 4,080 P/in
τ = 4.08 Ksi
Answer:
Explanation:
B. you would grab the plug closest to the outlet
