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3 years ago

What is a Flame Front Generator?

Engineering

1 answer:

Inessa [10]3 years ago

7 0

Answer and Explanation:

Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.

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What is the best countermeasure against social engineering?

Mkey [24]

Answer:

Hello Monk7294!

Answer:

Employee education

Explanation:

The most important countermeasure for social engineering is employee education. All the employees should be trained to keep confidential data safe. As a part of security education, organizations have to provide timely orientation about their security policy to new employees. The security policy should address the consequences of the breaches.

- I Hope this helps Have an awesome day!

~ Chloe marcus <3

3 0

3 years ago

If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b

zalisa [80]

Answer:

h=32.1 km

Explanation:

solution:

using newton law of gravitational attraction and newton second law:

$W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\$

$m_{E}= mass of earth$

r= distance between two masses

at sea level

a=g

$r=R_{E}$

$a=\frac{Gm_{E}}{r^{2}}$ .............................(1)

$Gm_{E} =gR_{E}^2$ .........................(2)

by substituting (2) and (1) $a=g\frac{R_{E}^2 }{r^{2} }$ acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0

3 years ago

You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the

siniylev [52]

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

5 0

3 years ago

Given a reservoir watershed of 22 square miles, and assuming a

saveliy_v [14]

Answer:

308 acre-ft of water

Explanation:

Given:

Area of the watershed = 22 square miles

Depth of Rainfall = 0.75 in = $\frac{\textup{0.75}}{\textup{12}}$ =0.0625 ft

Percentage rainfall falling in reservoir as runoff = 35%

Now,

1 square mile = 640 acre

Thus,

22 square miles = 22 × 640 = 14,080 acres

Thus,

The total volume of rainfall = Area of watershed × Depth of the rainfall

The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft

also,

only 35% of the total rainfall is contributing as runoff

thus,

Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water

8 0

3 years ago

2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to

Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

$P_1=150\ KPa$

$P_2=600\ KPa$

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

So now putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=2.5\times 0.287\times 285\ln \dfrac{150}{600}$

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0

3 years ago