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fredd [130]
2 years ago
8

1. Which type of fit implies that a piece will never fit? a. interference fit b. construction fit c. transition fit d. impeding

fit
Engineering
1 answer:
VladimirAG [237]2 years ago
6 0

Answer:

d

Explanation:

interference fit: the dimensions are larger and high force like heating is required to fit in the object

transition fit: the dimensions are fractionally larger and slight force like hammering is required to fit in the object

impeding fit: the dimensions are larger and the object will not fit at all

construction fit: activities making a commercial space sutiable for tenant occupation.

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Assume the impedance of a circuit element is Z = (3 + j4) Ω. Determine the circuit element’s conductance and susceptance.
djyliett [7]

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

8 0
3 years ago
Sinks must be used for the correct intended purpose to prevent
irakobra [83]

Answer:

... spilling water or getting anything cascading onto the floor

8 0
2 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 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:

F_{drag} = \frac{1}{2} × 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, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 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.

5 0
3 years ago
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

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

5 0
3 years ago
Timescale limits knowledge for scientists because it is difficult for them to see much beyond their lifetimes. Question 1 option
vaieri [72.5K]

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

8 0
2 years ago
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