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2 years ago

9

What do u call a bad bird

1 answer:

statuscvo [17]2 years ago

4 0

Answer:

A buzzerd

Explanation:

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Pouring molten aluminum into a mold and allowing it to cool forms?

neonofarm [45]

Answer:it forms a molten mold that makes it hard to be able to smash something into it then make something like a key

Explanation:

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3 years ago

26 V is measured across a 220 Ω resistance. This means that resistor current equals ________. Group of answer choices 1.2 A 57 A

zysi [14]

You can use ohm’s law
I=V/R

3 0

3 years ago

Which one of the following questions about population growth is the only TRUE statement?A) The size of a population can never ex

tatyana61 [14]

Answer:

Explanation:

5

6 0

3 years ago

A steady flow adiabatic turbine accepts gas at conditions T1, P1 and discharges at conditions T2 and P2. Assuming ideal gas, det

Nimfa-mama [501]

Answer:

W = -3753.8 J

$W_{ideal}$ = - 5163.14 J

$W_{lost} = - 1409.34 \ J$

$S_G = 4.70 \ \ J/K$

Explanation:

Given that:

$T_1 = 500 \ K \\T_2 = 371 \ K \\T_{\sigma} = 300 \ K\\P_1 = 6 \ bar\\P_2 = 1.2 \ bar\\$

$\frac{Cp}{R} = \frac{7}{2} \\n = 1$

Consider the relation

$\frac{Cp}{R} = \frac{7}{2} \\$

$Cp = \frac{7*R}{2}$

$Cp = \frac{7*8.314 \ kJ/kmol.K}{2}$

$Cp = 29.099 \ kJ/kmol.K$

Actual work W = ΔH = n Cp (T₂ - T₁)

= 1 × 29.099 (371 - 500)

= -3753.8 J

Change in entropy

ΔS = $n[Cp In \frac{T_2}{T_1} - R In \frac{P_2}{P_1}]$

ΔS = $1*[29.099 In(\frac{371}{500}) - 8.314 In (\frac{1.2}{6})]$

ΔS = 4.6978 J/K

Ideal Work

$W_{ideal} = \delta H - T_{\sigma} \delta S$

= -3753.8 - 300 ( 4.6978)

= - 5163.14 J

Work lost

$W_{lost} = |W_{ideal}-W|$

$W_{lost} = |-5163.14 -(-3753.8)|$

$W_{lost} = - 1409.34 \ J$

Entropy Generation rate:

$S_G = \frac{W_{lost}}{T_{\sigma}}$

$S_G = \frac{1409.34}{300}$

$S_G = 4.70 \ \ J/K$

5 0

3 years ago

The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar rad

Korolek [52]

Answer:

$T_{surf} =$ 3.9°C

Explanation:

solution:

we express the surface temperature from the quality of the emitted and absorbed heat rates:

$Q_{abs} =Q_{em} \\\$

$\alpha Q_{inc}=$ εσA $T^{4} _{surf}$

$T_{surf} =$ $\sqrt[4]{}$ $\alpha$ /εσ* $\frac{Q_{inc} }{A}$ ........................(1)

where

$\alpha$ =0.2

ε=0.6

A=5.67*10^-8 W/m^2K^4

$Q_{inc}$ = 1000 W/m^2

now putting these value in eq 1

$T_{surf} =$ 3.9°C

6 0

3 years ago