Answer:
B. G = 333 mS, B = j250 mS
Explanation:
impedance of a circuit element is Z = (3 + j4) Ω
The general equation for impedance
Z = (R + jX) Ω
where
R = resistance in ohm
X = reactance
R = 3Ω X = 4Ω
Conductance = 1/R while Susceptance = 1/X
Conductance = 1/3 = 0.333S
= 333 mS
Susceptance = 1/4 = 0.25S
= 250mS
The right option is B. G = 333 mS, B = j250 mS
Answer:
... spilling water or getting anything cascading onto the floor
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:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω
Answer:
I think true
Explanation:
Well I mean...we cant see the future. Certain things will be achieveable in different ganerations like going on mars