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Phantasy [73]
3 years ago
8

What would be the most likely scale factor to use for an n-gauge model train setup? (An n-gauge layout uses locomotives that are

typically about five centimeters long, on a train platform that may be one meter by one meter.)
A. 1:100
B. 1:500
C. 100:1
D. 500:1
Engineering
1 answer:
slega [8]3 years ago
8 0
A or b for sure brother
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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
For a turbulent flow of a fluid in 0.6 m diameter pipe, the velocity 0.15 m from the wall is 2.7 m/s. Estimate the wall shear st
MAVERICK [17]

Answer:

If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.

Explanation:

hope it helps

3 0
2 years ago
1.<br> se reception. It's<br> In remote areas, your GPS device<br> a good idea to have a _
marysya [2.9K]

GPS device  details are given below.

Explanation:

Even a simple GPS unit has a wide range of settings and features. Because every unit’s operation varies, this article won’t provide step-by-step details. Read the owner's manual to familiarize yourself with it.. If you’d like additional help, you can also sign up for a GPS navigation class at an REI store.

Though steps vary, all GPS receivers do the following basic functions:

Display position: A GPS tells you where you are by displaying your coordinates; it also shows your position on its base map or topo map.

Record tracks: When tracking is turned on, a GPS automatically lays down digital bread crumbs, called “track points,” at regular intervals. You use those later to retrace your steps or to evaluate the path you traveled.

Navigate point-to-point: A GPS directs you by giving you the direction and distance to a location, or “waypoint.” You can pre-mark waypoints by entering their coordinates at home. In the field you can have the unit mark a waypoint at a place you'd like to return to, such as the trailhead or your campsite. A GPS unit provides the bearing and distance “as the crow flies” to a waypoint. Because trails don’t follow a straight line, the bearing changes as you hike. The distance to travel also changes (decreasing, unless you’re heading the wrong direction) as you approach your goal.

Display trip data: This odometer-like function tells you cumulative stats like how far you’ve come and how high you’ve climbed.

GPS and your computer: GPS units come with a powerful software program that lets you manage maps, plan routes, analyze trips and more. Invest the time to learn it and to practice using all of its capabilities.

5 0
3 years ago
We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V
Angelina_Jolie [31]

Answer:

51.4 Ohms

Explanation:

By applying voltage division rule

V_f=v_i\times \frac {R_l}{R_l+R_m} where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then

7=10*\frac {120}{120+R_m}\\7R_m+840=1200\\R_m={1200-840}{7}=51.428571\approx 51.4 Ohms

3 0
3 years ago
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge
tangare [24]

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

E=\dfrac{1}{2}mv^2\ J

That is why mechanical energy per unit mass will be

e=\dfrac{1}{2}v^2\ J/kg

Now by putting the values in the above equation we get

e=\dfrac{1}{2}\times 10^2\ J/kg

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

5 0
3 years ago
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