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3 years ago

1. se reception. It's In remote areas, your GPS device a good idea to have a _

Engineering

1 answer:

marysya [2.9K]3 years ago

5 0

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.

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A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a

Xelga [282]

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) $in^u$

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum

Ts is torque transmitted by steel

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅ $_{alu$ = TaLa/GaJa

for steel rod ∅ $_{steel$ = TsLs/GsJs

but we know that, ∅a = ∅s = ∅ $_B$

[TaLa/GaJa] = [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa] = [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

∅ $_{steel$ = TsLs / GsJs

we substitute

∅ $_{steel$ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅ $_{steel$ = 83.26 / 276460.1535

∅ $_{steel$ = 0.000301

∅ $_{steel$ = 3.01 × 10⁻⁴ rad

∅ $_{steel$ = ∅ $_B$ = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad

5 0

2 years ago

The difference in quantity between the add and full marks on an engine oil dipstick is typically

svetoff [14.1K]

Answer:

1 quart (0.9 liters).

Explanation:

A proper inspection of various systems and components in a vehicle at regular intervals is very important and necessary because it helps to ensure that the vehicle is in a safe and reliable condition.

Generally, these inspection includes tyres, lighting systems, fan belts, shock absorbers, fluid (oil and water) level, etc. If any fault or concern is detected in the course of an inspection, it should be noted for quick repair or servicing by an expert technician.

All automobile engine requires an adequate amount of engine oil as a lubricant so as to mitigate friction and enhance proper functionality of the vehicle. Thus, the proper functionality of an engine is largely dependent on the level of the engine oil; it shouldn't be too low or high.

Basically, the engine oil should be checked at regular intervals (periodically) and should be on the level indicated or chosen by the manufacturer of the vehicle.

A dipstick is designed to be used for checking the engine oil level in a vehicle and it is marked with lines indicating minimum and maximum, low and high or add and full.

The difference in quantity between the add and full marks on an engine oil dipstick is typically 1 quart (0.9 liters).

5 0

2 years ago

A. 50

Licemer1 [7]

Answer:

love

Explanation: you

5 0

2 years ago

the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when

pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

$v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2$

acceleration is a derivative of velocity with respect to time.

$a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5$

evaluate acceleration at 2 seconds

$a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}$

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

$=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2$

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

$x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\$

3 0

2 years ago

This problem demonstrates aliasing. Generate a 512-point waveform consisting of 2 sinusoids at 200 and 400-Hz. Assume a sampling

aalyn [17]

Answer and Explanation:

clear all; close all;

N=512;

t=(1:N)/N;

fs=1000;

f=(1:N)*fs/N;

x= sin(2*pi*200*t) + sin(2*pi*400*t);

y= sin(2*pi*200*t) + sin(2*pi*900*t);

for n = 1:20

a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))

b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))

c(n) = sqrt(a(n).^2+b(n).^2)

theta(n) =-(360/(2*pi))*atan(b(n)./a(n));

end

plot(f(1:20),c(1:20),'rd');

disp([a(1:4),b(1:4),c(1:4),theta(1:4)])

8 0

3 years ago