An asphalt mixture is placed and compacted using normal rolling procedures. Two tests are taken from the compacted mixture and checked for density. Core 1 had a density of 150 pcf and an air void content of 6.5 %. Core 2 had a density of 151 pcf. What is the air void content of core 2.

The maximum power delivered to the circuit is found to be 3.75W, occurring at the start (t = 0). Calculating the total energy requires integrating the power over time, involving exponentials reducing over time to approach a finite total energy delivered.

The question involves calculating the maximum power and the total energy delivered to a circuit element, where the voltage V and current I are given as time-dependent expressions.

Maximum Power Delivered

Power is calculated using P = IV. Inserting the given expressions for V and I,

P(t) = V(t) × I(t) = (75 - 75e^{-1000t}) × 50e^{-1000t} mA.

To find the maximum power, we would differentiate P(t) with respect to t and set the derivative equal to zero. However, because the setup includes exponential decay functions, their maximum product occurs at t = 0 for these specific functions.

Pmax = 75V × 50mA = 3.75W.

Total Energy Delivered

The total energy delivered can be found by integrating the power over time:

Energy = ∫ P(t) dt.

Considering the specific forms of V and I provided, this becomes an integral of the product of two exponentials, leading to an expression that evaluates the total energy consumed by the circuit over time.

Given the nature of exponential decay in V and I, the energy delivered approaches a finite value as t approaches infinity.

Answer:

Explanation:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

A) Thermal conductivity of wall = 50 W/m.c

B) value of the convection heat transfer coefficient, h = 5 W/m².C

Explanation:

I've attached all explanations

The question involves applying thermodynamics to calculate work and heat transfer for propane in a piston-cylinder assembly undergoing a process with a specific pressure-volume relationship. It requires integrating over the process path for work and applying the first law of thermodynamics for heat transfer, considering the neglect of kinetic and potential energy effects.

A piston–cylinder assembly containing propane undergoes a process where the pressure-volume relationship is given as pV1.1 = constant. To evaluate the work and heat transfer for the propane, we apply the principles of thermodynamics, specifically the first law of thermodynamics, and the properties of processes adhering to specific equations of state. The work done in such processes can be calculated using the integral of p dV, considering the pressure-volume relationship provided. The heat transfer can then be inferred by applying the first law of thermodynamics, which equates the change in internal energy to the net heat added to the system minus the work done by the system.

The initial and final states of the propane provide the necessary boundary conditions to evaluate these quantities. However, without specific values for the molar mass or specific heat capacities of propane at constant pressure and volume, exact numerical answers cannot be provided. Generally, for processes described by a polytropic equation (pVn = constant), the work done is W = (p2V2 - p1V1)/(1-n) for an ideal gas, where p1, V1 are the initial pressure and volume, and p2, V2 are the final conditions. Heat transfer, Q, requires specific thermal properties of propane and can be approached via Q = ΔU + W, with ΔU denoting the change in internal energy of the gas.

For a precise evaluation, one would typically reference thermodynamic tables for propane or apply real gas equations of state considering the polytropic process specifics. It is essential to note that kinetic and potential energy changes are negligible, focusing the analysis solely on the thermodynamic work and heat transfer.

Answer:

807.5N

Explanation:

The combined mass (m) on the back muscle is 55kg + 30kg = 85kg

Acceleration due to gravity (g) = 9.8m/s²

Therefore the force FB = ma = 85*9.8

FB= 807.5N

Answer: 9.9KJ

Explanation: Q = U + W + losses

Q is heat transfered to the water

U is the change in energy of the system

W is work done by the system = 2J

Losses = 80J

Heat into system is 10kJ = 10000KJ

Therefore

U = Q - W - losses

U = 10000 - 2 - 80 = 9990J

= 9.9kJ

Answer:

The net work output from the cycle = 520.67 [tex]\frac{KJ}{kg}[/tex]

The efficiency of Brayton cycle = 0.448

Explanation:

Compressor inlet temperature [tex]T_{1}[/tex] = 300 K

Turbine inlet temperature [tex]T_{3}[/tex] = 1700 K

Pressure ratio [tex]r_{p}[/tex] = 8

For the compressor the temperature - pressure relation is given by the formula,

⇒ [tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒ [tex]\frac{T_{2} }{300} = 8^{\frac{1.4 - 1}{1.4} }[/tex]

⇒ [tex]T_{2}[/tex] = 543.42 K

This is the temperature at compressor outlet.

Same relation for turbine we can write this as,

⇒ [tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒[tex]\frac{1700 }{T_{4} }[/tex] = [tex]8^{0.2857}[/tex]

⇒ [tex]T_{4}[/tex] = 938.5 K

This is the temperature at turbine outlet.

Now the work output from the turbine [tex]W_{T}[/tex] = [tex]m C_{p} (T_{3} - T_{4} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{T}[/tex] = 1 × 1.005 × ( 1700 - 938.5 )

    [tex]W_{T} = 765.3 \frac{KJ}{kg}[/tex]

This is the work output from the turbine.

Now the work input to the compressor is [tex]W_{C}[/tex] = [tex]m C_{p} (T_{2} - T_{1} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{C}[/tex] = 1 × 1.005 × ( 543.42 - 300 )

⇒ [tex]W_{C}[/tex] = 244.63 [tex]\frac{KJ}{kg}[/tex]

This is the work input to the compressor.

Net work output from the cycle [tex]W_{net} = W_{T} - W_{C}[/tex]

⇒ [tex]W_{net}[/tex] = 765.3 - 244.63

   [tex]W_{net} = 520.67\frac{KJ}{kg}[/tex]

This is the net work output from the cycle.

The thermal efficiency is given by

[tex]E_{cycle} =1 - \frac{1}{r_{p}^{\frac{\gamma - 1}{\gamma} } }[/tex]

[tex]E_{cycle} =1 - \frac{1}{8^{\frac{1.4 - 1}{1.4} } }[/tex]

[tex]E_{cycle} = 0.448[/tex]

This is the efficiency of Brayton cycle.

(b). the graph between plot the net work developed per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50 is shown in the image below.

Answer:

The code is as attached here.

Explanation:

The code is as given below

theta = input(' Enter the value of theta?');

y = sin(theta*pi()/180);

z = cos(theta*pi()/180);

if z < 0.01

fprintf('The value of theta is very low')

else

t=round(y/z,2);

disp(['The value of tangent theta is ',num2str(t)]);

end

In MATLAB, evaluate tangent of theta if the magnitude of cosine is

[tex]> = 10[/tex]⁻²; otherwise, display an error message.

Here are the MATLAB statements to evaluate [tex]\( \tan(\theta) \)[/tex] as long as the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], and display an error message if the magnitude of [tex]\( \cos(\theta) \)[/tex] is less than [tex]\( 10^{-2} \)[/tex]:

```matlab

% Define theta in degrees

theta_deg = input('Enter the value of theta in degrees: ');

% Convert theta to radians

theta_rad = deg2rad(theta_deg);

% Calculate cosine of theta

cos_theta = cos(theta_rad);

% Check if the magnitude of cos(theta) is greater than or equal to 10^-2

if abs(cos_theta) >= 1e-2

   % Evaluate tangent of theta

   tan_theta = sin(theta_rad) / cos_theta;

   disp(['tan(theta) = ', num2str(tan_theta)]);

else

   % Display error message

   disp('Error: The magnitude of cos(theta) is too small. Cannot evaluate tan(theta).');

end

```

This script prompts the user to enter the value of [tex]\( \theta \)[/tex] in degrees. It then converts [tex]\( \theta \)[/tex] to radians and calculates [tex]\( \cos(\theta) \)[/tex]. If the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], it evaluates [tex]\( \tan(\theta) \)[/tex] using the given formula and displays the result. Otherwise, it displays an error message indicating that the magnitude of [tex]\( \cos(\theta) \)[/tex] is too small to evaluate [tex]\( \tan(\theta) \).[/tex]

Complete Question

Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.

Use Coral Programming Language

Answer:

// Program is written in Coral Programming Language

// Comments are used for explanatory purpose

// Program starts here

// Variable declaration

int currentprice

int prevprice

int change

float mortgage

Put "Enter current price to output" to output

currentprice = Get next input

Put "Enter last month price to output" to output

prevprice = Get next input

// Calculate Change since last month

change = currentprice - prevprice

// Calculate Monthly Mortgage

mortgage = currentprice * 0.045 / 12

// Print Results

Put "This house is $" to output

Put currentprice to output

Put "\n" to output

Put "This change is $" to output

Put change to output

Put "\n" to output

Put "This house is $" to output

Put currentprice to output

Put "since last month\n" to output

Put "This estimated monthly mortgage is $" to output

Put mortgage to output

// End of Program

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

LI = Lifting Index = 0.71

No lifting risk is involved

Explanation:

NIOSH Lifting Index

LI = Load Weight / Recommended Weight Limit

LI = L / RWL                                                     ............. Eq (A)    

NIOSH Recommended Weight Limit equation is following,

RWL = LC * HM * VM * DM * AM * FM * CM   ........... Eq (B)

Where,

 LC = Load constant

 HM = Horizontal multiplier

 VM = Vertical multiplier

 DM = Distance multiplier

 AM = Asymmetric multiplier

 FM = Frequency multiplier

 CM = Coupling multiplier

Given data

V = 30 + (8/2) = 34 in

H = 7 in

D = 55 - 30 = 25 in

A = 0

F = 10 boxes/min

C = 1 = Good coupling

According to NIOSH lifting guide

 LC = 51 lb

 HM = 10/H

 VM = 1 - {0.0075*(v-30)}

 DM = 0.82 + (1.8/D)

 AM = 1 - (0.0032*A)

 FM = 0.45         (Table 5 from NIOSH lifting guide)

 CM = 1               (Table 7 from NIOSH lifting guide)

Solution:

RWL = 51 * (10/7) * [1-{0.0075(34-30)}] * (0.82+ 1.8/25) * (1-0.0032*0) * 0.45 * 1

RWL = 51 * (10/7) * 0.97 * 0.892 * 1 * 0.45 * 1

RWL = 28.37 lb

Using equation A

LI = L / RWL

LI = 20 / 28.37

LI = 0.71

According to NIOSH lifting guide LI <= 1

So No lifting risk is involved

Answer:

Explanation:

AVERAGE: the average value is given as

[tex]\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}[/tex]

=[tex]\frac{1}{0.1} ([2.5-10]-10)=-175[/tex]

RMS= [tex]\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }[/tex]

[tex]y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t[/tex]

[tex]\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }[/tex]

=[tex]3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92\\[/tex]

therefore, RMS = [tex]\sqrt{2845.92} = 53.3[/tex]

To achieve an effluent concentration of 20 mg/L in a sewage lagoon receiving sewage with 100 mg/L of a contaminant, a precise biodegradation reaction rate coefficient, determined by the mass balance equation under steady-state conditions, must be achieved.

Calculating the Biodegradation Reaction Rate Coefficient

The question involves determining the biodegradation reaction rate coefficient necessary to reduce the concentration of a contaminant in a sewage lagoon, illustrating principles of environmental engineering. Given a lagoon with a surface area of 100,000 m2 and a depth of 1 m, receiving 8,640 m3/d of sewage that contains 100 mg/L of biodegradable contaminant, the goal is to lower the effluent concentration to no more than 20 mg/L.

To find the required biodegradation reaction rate coefficient (d-1), we must apply the mass balance concept in a steady-state condition, assuming the lagoon is well mixed. The mass balance equation for a contaminant undergoing a first-order degradation reaction can be expressed as: Input = Output + Decay. By substituting the given values and solving for the decay rate, we can find the coefficient that ensures the specified effluent concentration.

The calculation involves deriving relationships between the influent and effluent concentrations, the volume of the lagoon, and the decay process characterized by the reaction rate coefficient. For the lagoon described, achieving an effluent concentration of 20 mg/L from an influent concentration of 100 mg/L through biodegradation requires precise control of the treatment process and understanding of the kinetics of contaminant degradation.

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

COP = 13.31

Explanation:

We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.

Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg

While the enthropy(s1) = 0.92234 Kj/KgK

Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.

Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of  273.18 Kj/Kg.

Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.

Thus, h3=h4 = 107.34 Kj/kg

Formula for COP = QL/w = (h1- h4) / (h2 - h1)

COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

Answer:

δ_AB = 0.0333 in

Explanation:

Given:

- The complete question is as follows:

" The rigid beam is supported by a pin at C and an A−36

steel guy wire AB. If the wire has a diameter of 0.5 in.

determine how much it stretches when a distributed load of

w=140 lb / ft acts on the beam. The material remains elastic."

- Properties for A-36 steel guy wire:

       Young's Modulus E = 29,000 ksi

       Yield strength σ_y = 250 MPa

- The diameter of the wire d = 0.5 in

- The distributed load w = 140 lb/ft

Find:

Determine how much it stretches under distributed load

Solution:

- Compute the surface cross section area A of wire:

                            A = π*d^2 / 4

                            A = π*0.5^2 / 4

                            A = π / 16 in^2

- Apply equilibrium conditions on the rigid beam ( See Attachment ). Calculate the axial force in the steel guy wire F_AB

                            Sum of moments about point C = 0

                            -w*L*L/2 + F_AB*10*sin ( 30 ) = 0

                             F_AB = w*L*L/10*2*sin(30)

                             F_AB = 140*10*10/10*2*sin(30)

                             F_AB = 1400 lb

- The normal stress in wire σ_AB is given by:

                            σ_AB = F_AB / A

                            σ_AB = 1400*16 / 1000*π

                            σ_AB = 7.13014 ksi

- Assuming only elastic deformations the strain in wire ε_AB would be:

                            ε_AB = σ_AB / E

                            ε_AB = 7.13014 / (29*10^3)

                            ε_AB = 0.00024

- The change in length of the wire δ_AB can be determined from extension formula:

                            δ_AB = ε_AB*L_AB

                            δ_AB = 0.00024*120 / cos(30)

                            δ_AB = 0.0333 in

Answer:

Conventional single-level page table  [tex]2^{20}[/tex]  pages

Inverted page table are [tex]2^{17}[/tex] frame

Explanation:

given data

logical address = 32-bit = [tex]2^{32}[/tex]  Bytes

page size = 4-KB =   [tex]2^{12}[/tex] Bytes

physical memory = 512 MB  = [tex]2^{29}[/tex]  bytes

solution

we get here number of pages that will be

number of pages = [tex]\frac{logical\ address}{page\ size}[/tex]   ..............1

put here value

number of pages =   [tex]\frac{2^{32}}{2^{12}}[/tex]

number of pages = [tex]2^{20}[/tex]  pages

and

now we get number of frames  that is

number of frames  = [tex]\frac{physical\ memory}{page\ size}}[/tex]   ............2

number of frames  = [tex]\frac{2^{29}}{2^{12}}[/tex]

number of frames  = [tex]2^{17}[/tex] frame

so

Conventional single-level page table  [tex]2^{20}[/tex]  pages

and

Inverted page table are [tex]2^{17}[/tex] frame

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

Answer:

Explanation:

1) true ( reason is due to polarizability given in statement)

2)false ( nucleophile does not involve in rate determining step)

3)false ( it is an inverse attack , attacks from backside)

4) true

5) false ( it inverts stereo centres which are being attacked)

6) false (not H-)

7) true ( strong bases are good nucleophiles)

8) true ( because of steric hindrance)

9)false ( ther are stabilized by hyper conjugation)

10) false

Answer:

1) True

2) False

3) False

4)

5)

6) True

7)True

8) True

9) False

10) True

Explanation:

1) Generally, polarization increases down the column of the periodic table.

Answer:

a = 30.1 kj

b = 115 kj

Explanation:

To determine the mass we use the formula m = V/v1

v1 =1.08m3/kg, and V = 20L

m = 20/1000 × 1.08 = 0.0185kg

Next we determine the initial specific internal energy, u1.

Using softwares and appropriate values of T1 and p1, we get

u1 = 2650kj/kg.

After this we determine the final specific internal energy, u2 using the formula u2 = uf + x2 × ufg

Therefore we need to find x2 first.

x2 = u2 - uf/ug - uf

x2 = 1.08 - 0.001029/3.4053 -0.001029

x2 = 0.3180

But u2 = uf + x2× uf=334.97 + 0.3180×2146.6 = 1017.59 kj/kg

Now heat transfer Q= DU

Q = m x (u1 - u2)

Q = 0.0185(2650-1017.59

Q = 30.1 kj

Calculating the b part of the question we use the formula

W = m( u1-u2) - m. To. (s1 - s2)

Where s1 = 7.510kj/kgk

And s2 = 3.150 kj/kgk

We need to convert To and Ta to k values by adding 273 to 0 and 21 respectively.

Putting the values into the formula, we get W = 30.1 - 0.0185 × 273 (7.510-3.150)

W = 8.179kj

Finally maximum heat transfer

Qm = W/1 - to/ta

Qm = 8.179/1 - 273/294

Qm = 115kj

To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.

To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:

ρQVj = kV^2

Solving for V will give us the steady speed expression we are looking for:

V = √(ρQVj / k)

To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:

Q = kV^2 / (ρVj)

With given values, we would have:

Q = k * 10^2 / (ρ * 15)

The constant k can be determined using the known conditions and solving for k.

k = ρQVj / V^2

For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.

Answer:

The code is given which can be pasted in the Javascript file

Explanation:

The code is given as below

package Sorters;

public class javasort

{

private int[] arr=new int[13];

public void bubbleSort(int[] a){

int c,d,temp;

for (c = 0; c < ( 13 - 1 ); c++) {

for (d = 0; d < 13- c - 1; d++) {

if (a[d] > a[d+1]) /* For descending order use < */

{

temp = a[d];

a[d] = a[d+1];

a[d+1] = temp;

}

}

System.out.print("\n[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void insertionSort(int[] a){

int temp;

for (int i = 1; i < 13; i++) {

for(int j = i ; j > 0 ; j--){

if(a[j] < a[j-1]){

temp = a[j];

a[j] = a[j-1];

a[j-1] = temp;

}

}

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void shellSort(int[] a){

int increment = a.length / 2;

while (increment > 0)

{

for (int i = increment; i < a.length; i++)

{

int j = i;

int temp = a[i];

while (j >= increment && a[j - increment] > temp)

{

a[j] = a[j - increment];

j = j - increment;

}

a[j] = temp;

}

if (increment == 2)

increment = 1;

else

increment *= (5.0 / 11);

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void MergeSort(int[] a, int low, int high){

int N = high - low;  

if (N <= 1)

return;

int mid = low + N/2;

// recursively sort

MergeSort(a, low, mid);

MergeSort(a, mid, high);

// merge two sorted subarrays

int[] temp = new int[N];

int i = low, j = mid;

for (int k = 0; k < N; k++)

{

if (i == mid)

temp[k] = a[j++];

else if (j == high)

temp[k] = a[i++];

else if (a[j]<a[i])

temp[k] = a[j++];

else

temp[k] = a[i++];

}

for (int k = 0; k < N; k++)

a[low + k] = temp[k];

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

printM(a);

}

public void quickSort(int[] a,int low,int high){

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(a[c]+" ");

  }

  System.out.print("]");

int i =low, j = high;

int temp;

int pivot = a[(low + high) / 2];

/** partition **/

while (i <= j)

{

while (a[i] < pivot)

i++;

while (a[j] > pivot)

j--;

if (i <= j)

{

/** swap **/

temp = a[i];

a[i] = a[j];

a[j] = temp;

i++;

j--;

}

}

/** recursively sort lower half **/

if (low < j)

quickSort(a, low, j);

/** recursively sort upper half **/

if (i < high)

quickSort(a, i, high);

printM(a);

}

public void printM(int[] a){

arr=a;

}

public void fPrint(){

  System.out.println("\nSorted list:");

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(arr[c]+" ");

  }

  System.out.print("]");

}

}

package mani;

import java.util.Random;

import java.util.Scanner;

public class javasorttest

{

public static void main(String[] args){

int[] a=new int[13];

Random r=new Random();

for(int i=0;i<13;i++){

a[i]=r.nextInt(99)+1;

}

System.out.print("[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

javasort j=new javasort();

System.out.println("\nSelect the sorting algo.\n1.bubbleSort\n2.insertionSort\n3.shellSort\n4.MergeSort\n5.QuickSort.");

Scanner s=new Scanner(System.in);

int opt=s.nextInt();

switch(opt){

case 1:

j.bubbleSort(a);

break;

case 2:

j.insertionSort(a);

break;

case 3:

j.shellSort(a);

break;

case 4:

j.MergeSort(a, 0, 13);

j.fPrint();

break;

case 5:

j.quickSort(a ,0, 12);

j.fPrint();

break;

}

}

}

Answer:

Explanation:

The bolt is under double shear because it connects 3 plates.

Now the shear force resisted by each shear joint(V):

V=P/(No. of shear area)=185/2=92.5N.

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

[tex]y = ax^{2} +bx+c\\dy/dx=2ax+b\\[/tex]

At highest point of the curve [tex]dy/dx=o[/tex]

[tex]2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275[/tex]

[tex]-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\[/tex]

Station PVT

[tex]Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)[/tex]

There is a part of the question missing and it says;

The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.

Answer:

A) At (1,5),angle is -11.31°

B) At (5,2),angle is -68.2°

C) At (1,0), angle is -90°

Explanation:

From the question ;

V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s

Let us assume that u and v are the flow velocities in x and y directions respectively.

Thus we have the expression;

u = [20y/(x² + y²)^(1/2)]

and v = - [20x/(x² + y²)^(1/2)]

Thus, V = √(u² + v²)

V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²

V = √[400y²/(x²+y²)] +[400x²/(x²+y²)

V = √(400x² + 400y²)/(x²+y²)

Now for the angle;

tan θ = opposite/adjacent

And thus, in this question ;

tan θ = v/u

tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]

Simplifying this, we have;

tan θ = - 20x/20y = - x/y

so the angle is ;

θ = tan^(-1)(-x/y)

So let's now find the angle at the various coordinates.

At, 1,5

θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)

θ = -11.31°

At, 5,2;

θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)

θ = -68.2°

At, 1,0;

θ = tan^(-1)(-1/0) = tan^(-1)(-∞)

θ = -90°

At,

Answer:

Technician B says that a ratchet is used to loosen fasteners that are very tight.

Explanation:

A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.

Answer:

A ratchet wrench is usually used to loosen and tighten parts like steering linkages, tie rod end clamps and muffler clamps. Basically when nut is used on long thread a ratchet wrench is being used.

hence the technician B (option b) is correct.

Explanation:

Answer:

#include<stdio.h>

void main()

{

// using file pointer to print output to txt file

FILE *fptr;

float assessedValue, taxableAmount, taxRate = 1.05, propertyTax;

/* open for writing */

fptr = fopen("output.txt", "w");

if (fptr == NULL)

{

printf("File does not exists \n");

return;

}

// prompting user to enter assessed value and storing it in assessedValue variable

printf("Enter the Assessed Value of property : ");

scanf("%f", &assessedValue);

//writing assessed value to output.txt file using fprintf file i/o function

fprintf(fptr, "AssessedValue : $ %.2f\n", assessedValue);

//calculating taxableAmount based on given condition in the question

taxableAmount = (assessedValue * 0.92);

//writing taxable Amount to output.txt file using fprintf file i/o function

fprintf(fptr, "TaxableAmount: $ %.2f\n", taxableAmount);

//writing tax Rate to output.txt file using fprintf file i/o function

fprintf(fptr, "Tax Rate for each $100.00: $ %.2f\n", taxRate);

//calculating propertyTax based on given condition in the question

propertyTax = ((taxableAmount/100)*taxRate);

//writing property Tax Amount to output.txt file using fprintf file i/o function

fprintf(fptr, "propertyTax: $ %.2f\n", propertyTax);

//closing file using fclose function

fclose(fptr);

}

Explanation :

I used Turbo C compiler to compile and run the C program. The below program compiles and at the run time, automatically, prints output to a file called output.txt.

When you compile the program, remember to check the BIN folder in Turbo c folder of C drive where your turbo c has been installed.

Output:

Assessed value: $200000

Taxable amount: $184000

Tax Rate for each $100.00: $1.05

Answer:

639.4psi

Explanation:

Pressure at the closed valve = Air pressure+ (density*gravity*height)

=23psi+(49.27lb/ft^3*32.17ft/s^2*56ft^3)

=23psi+88760.89psft

=23psi+(88760.89/144)psi

=23+616.4

=639.4psi

Answer:

Ai=2300 in² , Ao=Ai*1.44=3312 in²

m=10 lbm/s

Pi=5 psia , Po=5.4 psia , Pa=5.5 psia

Vi=120 m/s , Vo=78 m/s

a) Force =m(Vo-Vi) = -190.5 N = -42.82 lbf (towards the inlet)

b) since force is negative it will slow down the system.

Explanation: