An insulated, vertical piston–cylinder device initially contains 10 kg of water, 6 kg of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and 350°C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine (a) the final temperature in the cylinder and (b) the mass of the steam that has entered.

Answer:

Its C. you may proceed, but only if the path is clear

Explanation:

I just gave Quiz and  its correct

Answer:

0.05 mg / gallon

Explanation:

mass of chemecila coming in per minute = 50*10 = 500 mg/min

at a time t min , M = mass of chemical = 500*t mg

conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

d[0] = 11111111

d[1] = 11011101

d[2] = 1111011

Explanation:

Assume that the number of bits is 8. The voltage range input is -8 to 7 volts. The range is thus 15V, and the resolution is 15/2^8 = 0.0586 volts. We will first add +8 to the input to convert it to a 0-15v signal. Then find the equivalent bit representation. For 7.8 volts, the binary signal will be all 1's, since the max input voltage for the ADC is 7 volts. For 4.95, we have 4.95+8 = 12.95 volts. Thus, N = 12.95/0.0586 = 221. The binary representation is 11011101. For -0.8, we have -0.8 + 8 = 7.2. Thus, N = 7.2/0.0586 = 123. The binary representation is 1111011.

Thus,

d[0] = 11111111

d[1] = 11011101

d[2] = 1111011

Answer:

Gc(s) = [tex]\frac{0.1s + 0.28727}{s}[/tex]

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              [tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]

Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]

Then, we should apply the values in the COP.

                           [tex]=\frac{1.15\;T_L}{1.15-1}[/tex]

                           [tex]=7.67[/tex]

The number of heat rejected by the heat pump must then be calculated.

                   [tex]Q_H=COP_{HP}\times W_{nst}[/tex]

                          [tex]=7.67\times5=38.35[/tex]

We must then calculate the refrigerant mass flow rate.

                   [tex]m=0.264\;kg/s[/tex]

                   [tex]q_H=\frac{Q_H}{m}[/tex]

                         [tex]=\frac{38.35}{0.264}=145.27[/tex]

The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   [tex]T_L=\frac{T_H}{1.15}[/tex]

                        [tex]=\frac{64+273}{1.15}=293.04[/tex]

                        [tex]=19.89\°C[/tex]

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

Answer:

  30√2 ≈ 42.426 volts

Explanation:

The RMS value of the signal is the DC equivalent. For a sine wave, the mean of the square is half the square of the peak value. Then the Root-Mean-Square is ...

  RMS = √((Vp)^2/2) = Vp/√2

For Vp = 60 volts, the equivalent DC voltage is ...

  V = (60 volts)/√2 = 30√2 volts ≈ 42.426 volts

Answer:

bye hooking plugs up to it to amp it up

Answer:

Cause and effect

Explanation:

pls mark brainliest

The  structure that makes or turned many smiles to frowns can be regarded as compare/contrast.

The term compare/contrast  is a common terms. The act of comparing is known to be depicting the similarities, and contrasting is said to be showing differences that exist between two things.

Conclusively, from the above, we can see that it is a compare/contrast scenario as it talks about the effects of taking ice cream. It went from  smiles to frowns.

See option below

cause/effect

descriptive

compare/contrast

sequence/process

Learn more about compare/contrast from

https://brainly.com/question/9087023

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

the police officer cruise each streets precisely once and he enters and exit with the same gate.

Explanation:

NB: kindly check below for the attached picture.

The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.

The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.

Here in this question, the door is the vertex and the road is the edge.

The information needed to detemine a Euler circuit and a Hamilton circuit is;

"the police officer cruise each streets precisely once and he enters and exit with the same gate."

Check attachment for each type of circuit and the differences.

the police officer cruise each streets precisely once and he enters and exit with the same gate.

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]

Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

[tex] \sigma max= \frac{pd}{4t} [/tex]

[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]

= 50p

For the minimum normal strain due to pressure, we have:

[tex] E_max= \frac{change in L}{L_g} [/tex]

[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]

The minimum normal stress for a thin pressure vessel is 0.

[tex] \sigma _min = 0 [/tex]

i) Let's use Hookes law to calculate the pressure causing this deformation.

[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]

Substituting figures, we have:

[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]

[tex] 120 * 10^6 = 35p [/tex]

[tex] p = \frac{120*10^6}{35}[/tex]

[tex] p = 3.429 * 10^6 [/tex]

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]

[tex] = \frac{50p - 50p}{2} = 0 [/tex]

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]

[tex] = \frac{50p - 0}{2} = 25p [/tex]

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

[tex]m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\[/tex]

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

Answer:

<E1, E2>.

Explanation:

So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;

E = |E| cos θi + |E| sin θj. -------------------(1).

Therefore, if we make use of the components division rule we will have something like what we have below;

x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ

​

Therefore, we will now have;

E = x |E2| i + y |E1| j.

The base vectors is then Given as <E1, E2>.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

Answer:

The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day

Explanation:

Check the file attached for a complete solution.

The volume of the aeration tank was first calculated, V = 5000 m^3 / day.

The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.

Answer:

  see attachments

Explanation:

A Karnaugh map for the output is shown in the first attachment. The labeled and shaded squares represent the cases where Y = HIGH. The associated logic can be simplified to

  Y = A3 xor A0

when the don't care at 1110 gives an output of HIGH.

__

The second attachment shows a logic diagram using a 4:1 multiplexer to do the decoding. A simple XOR gate would serve as well. If AND-OR-INV logic is required, that would be ...

  Y = Or(And(A3, Inv(A0)), And(A0, Inv(A3)))

Answer:

mass of chemical coming in per minute = 60 × 20 = 1200 mg/min

at a time t(min) , M = mass of chemical = 1200 × t mg

concentration of chemical = 1200t / 600 = 2t mg / gallon

Explanation:

Since it is only fluid that is leaving and not chemical.

Answer:

Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount

Explanation:

Solution

Code:

Main.java

public class Main {

public static void main(String[] args) {

  BulkDiscount bd=new BulkDiscount(10,5);

BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);

CombinedDiscount cd=new CombinedDiscount(bd,bnd);

System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));

  System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));

 System.out.println("Combined discount :"+cd.computeDiscount(20, 20));    

  }

}

discountPolicy.java

public abstract class DiscountPolicy

{    

public abstract double computeDiscount(int count, double itemCost);

}    

BulkDiscount.java  

public class BulkDiscount extends DiscountPolicy

{    

private double percent;

private double minimum;

public BulkDiscount(int minimum, double percent)

{

this.minimum = minimum;

this.percent = percent;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if (count >= minimum)

{

return (percent/100)*(count*itemCost); //discount is total price * percentage discount

}

return 0;

}

}

BuyNItemsGetOneFree.java

public class BuyNItemsGetOneFree extends DiscountPolicy

{

private int itemNumberForFree;

public BuyNItemsGetOneFree(int n)

{

  itemNumberForFree = n;

}

at Override

public double computeDiscount(int count, double itemCost)

{

if(count > itemNumberForFree)

return (count/itemNumberForFree)*itemCost;

else

  return 0;

}

}

CombinedDiscount.java

public class CombinedDiscount extends DiscountPolicy

{

private DiscountPolicy first, second;

public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)

{

first = firstDiscount;

second = secondDiscount;

}

at Override

public double computeDiscount(int count, double itemCost)

{

double firstDiscount=first.computeDiscount(count, itemCost);

double secondDiscount=second.computeDiscount(count, itemCost);

if(firstDiscount>secondDiscount){

  return firstDiscount;

}else{

  return secondDiscount;

}

}  

}

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Answer:

The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K

Explanation:

The detailed solution is attached as files below.

However, the steps followed are highlighted:

1) The average temperature was calculated as 380.5 K

2) The properties of air at 380.5 K was highlighted

3) The Prandti number was calculated. Pr = 0.693

4) The Reynold number was calculated, Re = 28716.77

5) The Nusselt umber was calculated, Nu = 75.94

6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.