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eduard
2 years ago
12

The building areas where loose fill insulation is most commonly used are

Engineering
1 answer:
Effectus [21]2 years ago
5 0

Answer:

Loose-fill insulation can be installed in either enclosed cavities such as walls, or unenclosed spaces such as attics

Explanation:

Answer in one of my classes

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Which professional draws maps and plans for projects involving structures other than buildings, such as bridges?
kolbaska11 [484]

Answer:

I believe it is a civil drafter

Explanation:

O*NET site

3 0
4 years ago
Read 2 more answers
An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output \dot W = \dot m×c_p×(T₂ - T₁)

Where;

c_p = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

\dot m =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

\dot W = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

\dot W = -396.17 kJ/s

7 0
4 years ago
What are the relative volume requirements for a CM reactor compared to a PF reactor if first-order kinetics apply and treatment
Tanzania [10]

Answer:

look it up

Explanation:

6 0
3 years ago
If aligned and continuous carbon fibers with a diameter of 9.90 micron are embedded within an epoxy, such that the bond strength
Zielflug [23.3K]
I have no idea what it is
8 0
3 years ago
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor
Naya [18.7K]

Answer:

4.536hp

Explanation:

The decrease in the heat gain of the room is determined from difference in electrical inputs:

Q = W_{shaft} (\frac{1}{n_{1} } - \frac{1}{n_{2} })\\Q = (75hp)*(\frac{1}{0.91 } - \frac{1}{0.963 })\\\\Q = 4.536 hp

8 0
3 years ago
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