What is your understanding and opinion on the validity of the stand-by equipment maintenance requirement?
1 answer:
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Answer:
Circular tube
Explanation:
Now for better understanding lets take an example
Lets take
Diameter of solid bar= cm
Outer diameter of tube =6 cm
Inner diameter of tube=2 cm
So from we can say that both tubes have equal cross sectional area.
We know that buckling load is given as
If area moment of inertia(I) is high then buckling load will be high.
We know that area moment of inertia(I)
For circular tube
For circular bar
Now by putting the values
For circular tube
For circular bar
So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.
Answer:
a) 3.607 m
b) 1.5963 m
Explanation:
See that attached pictures for explanation.
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.