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1 year ago

How much does it cost to replace a roof on a 2,200 square foot house.

Engineering

1 answer:

Norma-Jean [14]1 year ago

7 0

Answer:

An estimated amount of $8,200 to $12,000

Explanation:

~~It all really depends on what materials you have purchased for the replacement.

~~If you're to choose lower-quality materials the cost would be less but high-quality ones will be more expensive.

(The answer is if you're most likely to purchase asphalt shingles)

Hope this Helps!

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Why is communication one of the most important aspects of an engineer's job?

xenn [34]

Answer:

Any miscommunication can cause problems with time, cost, and safety

Explanation:

Safety is everyone's number one priority and the only way to stay safe is to communicate to the ones around to ensure everyone knows what is going on. Hope this helped<3 Brainliest please :)

3 0

1 year ago

Solve the inequality. Then graph your solution. -9v – 10 < 7y +6

Cerrena [4.2K]

16

if you add 9+10 you get 18 - 7+6

5 0

2 years ago

At time t the resultant force on a particle, of mass 250kg is (300ti-400tj)N. Initially, the particle is at the origin and is mo

Harlamova29_29 [7]

Answer:

Explanation:

4 0

2 years ago

Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.

Marizza181 [45]

Answer:

55%

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

This is a Carnot system

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - $\frac{Tc}{Th}$

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - $\frac{500}{1100}$

Eff = 1 - 0.45 = 0.55 or 55%

7 0

2 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

2 years ago