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3 years ago

15

given the classes above, what output is produced by the following code? meg[] elements ={new Lois(), new Stewie(), new Meg(), ne

w Brian()}; for (int i = 0; i

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

<u> Inheritance Mystery </u>

<u />

Lois a Meg a

Lois b

Meg

Lois a Meg a Stewie a

Lois a Meg a Stewie a Brian b

Brian Stewie

Meg a

Meg b

Meg

Lois a Meg a

Lois a Meg a Brian b

Brian

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The popularity of orange juice, especially as a breakfast drink, makes it an important factor in the economy of orange-growing r

tiny-mole [99]

Answer:

A) 22.22 kg

B) 0.1364

Explanation:

When we perform a degree of freedom (DOF) analysis on the bypass subsystem, we will have 2 unknown masses which are; (m1, m2).l with degree of freedom as 1.

Secondly, when we perform a degree of freedom analysis on the overall system, we will have 2 unknown masses namely (m3, m5) with degree of freedom as 0.

Thirdly, when we perform a degree of freedom analysis on the evaporator, we will have 3 unknown masses namely (m1, m3, m4) with Degree of freedom as 1.

Lastky,when we perform a degree of freedom analysis on mixing point, we will have 3 unknown masses namely, (m2, m4, m5) with degree of freedom as 1.

We are given;

Fresh orange = 10.0 wt% solids

Frozen juice concentrate = 45.0wt% solids

Mixture produced = 67.0 wt% solids

Final concentrate = 45.0 wt% solids

Now, making use of the overall solids balance, we have;

0.1 × 100 = 0.45(m5)

m5 = (0.1 × 100)/0.45

m5 = 22.22 kg

Making use of overall mass balance, we have;

100 kg = m3 + m5

m3 = 100 - 22.22

m3 = 77.78 kg

Now, making use of the Mixing point mass balance, we have;

m4 + m2 = m5

So; m4 + m2 = 22.22 - - - (eq 1)

Also, making use of the Mixing point solids balance, we have;

0.67m4 + 0.1m2 = 0.45m5

0.67m4 + 0.1m2 = 9.999 - - - (eq 2)

Solving eq 1 and eq 2 simultaneously, we have;

m2 = 13.64 kg

m4 = 8.58 kg

Final concentrate is m5 = 22.22 kg

Now, fraction = m2/100 = 13.64/100 = 0.1364

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4 years ago

Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2

Iteru [2.4K]

Answer and Explanation:

The answer is attached below

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3 years ago

5. Can you still prepare solutions using available materials at home why

svetoff [14.1K]

Yes, we can by the proper guidance and proper doing. We can create solution by any materials in the house that you need

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3 years ago

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$ , $P_{A1} =y_{A}$ , $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R = $\frac{kJ}{(kmol)(K)} ,$ T = 298 K and $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵ $\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴ $\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

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3 years ago

1. Discuss the benefits of observing good safety measures in relation to an increase in

Aneli [31]

The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.

<h3>What are the benefits of practicing safety in the laboratory?</h3>

A laboratory is known to be one that is known to have a lot of potential risks that is said to often arise due to a person's exposure to chemicals that are corrosive and toxic, flammable solvents, high pressure gases and others.

Therefore, A little care and working in line to all the prescribed safety guidelines will help a person to be able to avoid laboratory mishaps.

Therefore, The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the spills of chemicals and other kinds of accidents, as well as reduce the damage to the environment that is outside of the lab.

Learn more about safety measures from

brainly.com/question/26264740

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2 years ago