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1 year ago

I need help with this question

Engineering

1 answer:

Roman55 [17]1 year ago

6 0

Answer:

last one 8;2per leg

Explanation:

sorry if I got it wrong

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Describe with an example how corroded structures can lead to environment pollution?

raketka [301]

According to EonCoat, corrosion is the process of decay on a material caused by a chemical reaction with its environment. Corrosion of metal occurs when an exposed surface comes in contact with a gas or liquid, and the process is accelerated by exposure to warm temperature, acids, and salts.” (1)
Although the word ‘corrosion’ is used to describe the decay of metals, all natural and man-made materials are subject to decay, and the level of pollutants in the air can speed up this process.

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3 years ago

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor

tensa zangetsu [6.8K]

Answer:

HUH?

Explanation:

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2 years ago

A home electrical system is joined to the electric company's system at the junction of the

aleksandrvk [35]

That would be B, I hope this helps!

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2 years ago

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

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3 years ago

What is the governing ratio for thin walled cylinders?

Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

if $\frac{radius}{thickness} >10$ then you have a thin walled cylinder

or using the diameter:

if $\frac{diameter}{thickness} >20$ then you have a thin walled cylinder

3 0

3 years ago