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nikitadnepr [17]
3 years ago
11

Mr. Blue lives in a blue house, Mrs. Pink lives in a pink house and Mr. Red lives in a red house. Who lives in the White House?

Engineering
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

the president and mr.white my history teacher lol

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A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature T_\infity = 15 degree celcius

T_S =  220 + 273 = 473 K

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

Q = hA(T_S - T_{\infty}) \eta

= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8

Q = 125.538 W

5 0
3 years ago
If superheated water vapor at 30 MPa iscooled at ​constant pressure​, it will eventually become saturated vapor, and with suffic
nirvana33 [79]

Answer:

False.

Explanation:

False. The pressure is above pressure at critical point (22.064 MPa.), the limit where pressure can prevent boiling.

3 0
3 years ago
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0
3 years ago
Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat
Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0
3 years ago
If they opened up the International Space Station to tourism, would you go? Why? answer in 2 sentences
Arlecino [84]
Personally, I would go to the space station. The space station has extreme different levels of technology and abilities, plus who doesn’t want to go to space.
6 0
3 years ago
Read 2 more answers
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