Answer:
Building technology is building technology such as coding an app or a website
Building engineering is making computers or cars or phones
Explanation:
Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.
Answer:
-0.1006Kw/K
Explanation:
The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,
ΔS = Q/T
Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.
S(air) = - Q/T(air) .......1
Where S.air =
Q = 30-kW
T.air = 298k
Substitute the values into equation 1
S(air) = - 30/298
= -0.1006Kw/K
A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.
Answer: attached below is the power triangles
7.13589 MVAR
Explanation:
Power ( P1 ) = 10 MW
power factor ( cos ∅ ) = 0.6 lagging
New power factor = 0.85
Calculate the reactive power of a capacitor to be connected in parallel
Cos ∅ = 0.6
therefore ∅ = 53.13°
S = P1 / cos ∅ = 16.67 MVA
Q1 = S ( sin ∅ ) = 13.33 MVAR ( reactive power before capacitor was connected in parallel )
note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2
cos ∅2 = 0.85 ( new power factor )
hence ∅2 = 31.78°
Qsh ( reactive power when power factor is raised to 0.85 )
= P1 ( tan∅1 - tan∅2 )
= 10 ( 1.333 - 0.6197 )
= 7.13589 MVAR
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
Answer:
0.2
Explanation:
Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.
Let the span of the rectangular wing be 0.225 m
Let the chord of the rectangular wing be 0.045 m.
Then, the area of any rectangular chord is
A = chord * span
A = 0.045 * 0.225
A = 0.010 m²
And using the weight of the glider given to us from the question, we can find the LER for the wing.
LER = Area / weight.
LER = 0.010 / 0.05
LER = 0.2.
Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2
Please mark brainliest...
Answer: 0.2025
Explanation: I got it correct