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3 years ago

What is the locating position of the land field?

Engineering

2 answers:

BaLLatris [955]3 years ago

5 0

Answer:

Look at a map or globe. Where are places located? Every place has a "global address" that tells exactly where in the world it's located, just as your home has a street address. There are two numbers in a global address--a number for latitude and one for longitude. If you know these numbers and how to use them, you can find any place in the world and give its absolute location. (For definitions, see the glossary at the end of this booklet.)

Why are things located in particular places and how do these places influence our lives? Location can describe how one place relates to another. For example, the Panama Canal was cut across an extremely narrow strip of land in Central America. It provides a shipping lane between the Atlantic and Pacific Oceans, eliminating the need for long, dangerous journeys around South America.

Explanation:

<h2><u>hope</u><u> it</u><u> helps</u></h2>

inessss [21]3 years ago

3 0

Above one answered by my frnd is right .....

I won't like to answer it.....

sorry me

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Unwanted resistance is being discussed.

krok68 [10]

I don’t know if I’m right but I’m guessing B

6 0

4 years ago

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

$W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ$

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0

3 years ago

(25 points).The heat is turned on in a building in a cold morning, but the heater does not burnthe natural gas properly and emit

nydimaria [60]

Answer: Yes, it will be around 40ppm(at steady state)

Explanation:

The above question can be solved by performing material balance of CO in the building. The material balance equation is given by, Rate of change of CO in the building = Rate of inflow of CO into the building - Rate of outflow of CO from the blood

5 0

3 years ago

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correct

Zanzabum

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).

<h3>What is the distribution of pressure at some stage in bending?</h3>

Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.

The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.

What is the locating position of the land field?​

What is the locating position of the land field?