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aleksklad [387]
3 years ago
11

The term variation describes the degree to which an object or idea differs from others of the same type or from a standard.

Engineering
2 answers:
DedPeter [7]3 years ago
8 0
The answer is true mainly because they function
AfilCa [17]3 years ago
5 0
The answer is true. Thank me later<3
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All air-conditioning units must be grounded electrically to
Tpy6a [65]

Answer:

Prevent electrical shock

Grounding is a non current carrying conductor mainly used to guard against hazards due to leakage in electric circuits

Explanation:

The grounding refers to the connection of an electrical equipment exposed metallic parts to the ground to serve as a source of current flow in the event of an insulation failure will cause the fuses to trip thereby isolating or removing electric power from the device

Grounding also prevents the accumulation of static electricity which can be a source of fire in inflammable areas.

8 0
3 years ago
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

7 0
3 years ago
three string are attached to a small metal ring, two of the strings make and angle of 35° with the vertical and each is pulled w
Julli [10]
No clue sorry man I would help but I need help too
6 0
2 years ago
Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–
Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

C_p = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]

Substitute all parameters in the equation

Δs = 5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]

Δs = 5 kg × 0.14629 kJ/kg.K

    = 0.73145 kJ/K

b)

Δs = m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

           T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = 5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]

    = 5 kg (0.13795 kJ/kg.K)

    = 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0
3 years ago
Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels o
aalyn [17]

Answer:

See attached images

8 0
3 years ago
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