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____ [38]
3 years ago
15

The diesel exhaust fluid is injected​ ________.

Engineering
2 answers:
Maurinko [17]3 years ago
6 0

Explanation:

The diesel exhaust fluid is injected c just in front of the muffler.

tiny-mole [99]3 years ago
5 0

Answer:

C

Explanation:

The diesel exhaust fluid is injected​ just in front of the muffler.

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You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
3 years ago
how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng
Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

  • <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
  • <u><em>Reduce the surface roughness of the pipes</em></u>:  By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
  • <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

brainly.com/question/13348561

#SPJ4

3 0
1 year ago
Given a force of 72 lbs at a distance of 15 ft, calculate the moment produced.​
Elis [28]

Answer:

1425.78 N.m

Explanation:

Moments of force is calculated as ;

Moments= Force * distance

M= F*d

The S.I unit for moment of force is Newton-meter (N.m)

Given in the question;

Force = 72 lbs

1 pound = 4.45 N

72 lbs = 4.45 * 72=320.4 N

Distance= 15 ft

1ft= 0.3048 m

15 ft = 15*0.3048 = 4.57 m

d= 4.57 m

M= F*d

M=320.4*4.57 =1425.78 N.m

5 0
3 years ago
An airliner is flying at 34,000 ft cruise altitude on a standard day. Calculate the pressure difference between the cabin and th
nadya68 [22]

Answer:

\Delta P=61,952.8\ lb/ft^2

Explanation:

Given

Airline flying at 34,000 ft.

Cabin pressurized to an altitude 8,000 ft.

We know that at standard condition ,density of air

\rho =0.074\ lb/ft^3

We know that pressure difference    

ΔP=ρ g ΔZ

Here ΔZ=34,000-8,000  ft

        ΔZ=26,000 ft

g= 32.2\ ft/s^2

ΔP=0.074 x 32.2 x 26,000

\Delta P=61,952.8\ lb/ft^2

So pressure difference will be \Delta P=61,952.8\ lb/ft^2.

7 0
3 years ago
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