A two-stage, solid-propellant sounding rocket has the following properties:
First stage: m0 ¼ 249:5 kg ; mf ¼ 170:1 kg ; _ me ¼ 10:61 kg;sIsp ¼ 235 s
Second stage: m0 ¼ 113:4 kg ; mf ¼ 58:97 kg; _ me ¼ 4:053 kg;sIsp ¼ 235 s

Delay time between burnout of first stage and ignition of second stage: 3 s.
As a preliminary estimate, neglect drag and the variation of earth’s gravity with altitude to calculate the maximum height reached by the second stage after burnout.

Answer:

therefore the exit pressure of air is 331.2 kPa

Explanation:

Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023

Explanation:

Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:

3.3 V ---------> 1023

X  V------------>analogRead(A0)

Then [tex]X=\frac{3.3 \times analogRead(A0)}{1023} Volts[/tex]

Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).

Answer:

The strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex]

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  [tex](K)[/tex] and the [tex](n)[/tex].

[tex]\sigma =K\epsilon^n[/tex]

We will plug the values in the formula.

[tex]250=K\times (0.12)^n\\350=K\times (0.26)^n[/tex]

We will solve these equation.

[tex]K=\frac{250}{(0.12)^n}[/tex] plug this value in [tex]350=K\times (0.26)^n[/tex]

[tex]350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n[/tex]

Taking a natural log both sides we get.

[tex]ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435[/tex]

Now, we will find value of [tex]K[/tex]

[tex]K=\frac{250}{(0.12)^n}[/tex]

[tex]K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625[/tex]

So, the strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex].

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area =  1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft  = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000  × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb  

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

Calculating the maximum external pressure that a cold-drawn AISI 1040 steel tube can withstand involves using the formula for hoop stress and considering 80 percent of the steel's minimum yield strength. The formula relates the difference in pressure across the tube wall to its inner and outer radii.

The question asks for the maximum external pressure a cold-drawn AISI 1040 steel tube can withstand, given that the largest principal normal stress should not exceed 80 percent of the steel's minimum yield strength. The tube's outer diameter (OD) is 50 mm, and it has a wall thickness of 6 mm. To solve this, we must utilize the formula for hoop stress (sigma) in a thin-walled cylinder under external pressure, which is σ = δP(r_o/(r_o - r_i)), where δP is the difference in internal and external pressure, r_o is the outer radius, and r_i is the inner radius. The yield strength of AISI 1040 steel must be considered, and 80 percent of this value is taken as the maximum allowable stress. Without knowing the exact yield strength, it cannot be directly calculated in this response. However, the process would involve deriving δP from the given conditions and substituting into the hoop stress formula to find the maximum external pressure.

Explanation

Question 1:

!StdIn.isEmpty() is false.

So, it does not run the loop body.

So, It does not print anything.

Answer:

none of these

Question 2:

If we print the list starting at x, the result would be: 2 3

Answer:

none of these

To answer the student's question on the exit temperature and rate of entropy generation in an adiabatic nozzle, we need more context-specific information or the assumption that steam behaves as an ideal gas.

The student's question involves determining the exit temperature and the rate of entropy generation in an adiabatic nozzle process, which falls under the subject of thermodynamics, a branch of Physics. Calculating these properties requires knowledge of the first and second laws of thermodynamics, as well as the ideal gas law and flow equations.

For part (a), the exit temperature can be estimated using the ideal gas law and the conservation of energy. For part (b), the Clausius–Clapeyron relation and the concept of entropy would be applied. However, to accurately determine these values, more information would be needed, such as specific heat capacities, or the assumption that the steam behaves as an ideal gas.

Given the provided problem statements refer to various examples, which could lead to confusion, the correct answer would be provided if the precise context and details specific to the student's actual question were known.

Answer:

Depth in the contracted section = 2.896m

Velocity in the contracted section = 2.072m/s

Explanation:

Please see that attachment for the solving.

Assumptions:

1. Negligible head losses

2. Horizontal channel bottom

Answer:

See the attached pictures.

Explanation:

See the attached pictures for explanation.

Answer: =

Explanation:

=    P / (R * T) P- Pressure, R=287.058, T- temperature

From the given that

Sample mean(pressure) = 120300 Pa

Standard deviation (pressure) = 6600 Pa

Sample mean(temperature) = 340K

Standard deviation(temperature) = 8K

To calculate the Density;

Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa

Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa

Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K

Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K

So now to calculate the density:

Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331

Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138

Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345

cheers i hope this helps

Answer:

solution in the picture attached

Explanation:

Foreign keys help define the relationship between tables, which is what puts the "relational" in "relational database." They enable database developers to preserve referential integrity across their system.

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Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

1.

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

2.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

3.

Follow of nullable variable X is Follow (V).

Follow (S) = $

Follow (T) = {z, w}

Follow (V) = ;

Follow (X) = Follow (V) = ;

Follow (C) = , and ;

Explanation:

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

[tex]Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1[/tex]

The Lower Surface Cp is given as

[tex]Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1[/tex]

The difference of the Cp over the airfoil is given as

[tex]\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32[/tex]

Now the Lift Coefficient is given as

[tex]C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192[/tex]

Now the coefficient of moment about the leading edge is given as

[tex]C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306[/tex]

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

The lift coefficient and the pitching moment coefficient about the leading edge due to lift are respectively; 0.3192 and -0.13

We are given;

Distance of upper surface from leading edge to percentage of chord = -0.8

Percentage of chord for both surfaces = 60% = 0.6

Rate of increase at trailing edge for both surfaces = +0.1

Distance of lower surface from leading edge to percentage of chord = -0.6

angle of incidence; α_i = 4° = 4π/180 rad

Let us first calculate the Cp constant for both the upper and lower surface.

Cp for upper surface is;

Cp_u = (-0.8 × 0.6) - 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_u =  (-0.8 × 0.6) - (0.1 × 0.4)

Cp_u =  -0.52

Cp for lower surface is;

Cp_l = (-0.4 × 0.6) + 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_l =  (-0.4 × 0.6) + (0.1 × 0.4)

Cp_l =  -0.2

Change in Cp across the foil is;

ΔCp = Cp_l - Cp_u

ΔCp = -0.2 - (-0.52)

ΔCp = 0.32

Formula for the lift coefficient is;

C_L = ΔCp * cosα_i

C_L = 0.32 * cos (4π/180)

C_L = 0.3192

Formula for the pitching moment coefficient is;

(-0.3 * 0.4 * 0.6) - ((0.6 + (0.4/3)) * 0.2 * 0.4)

C_m,p =  -0.072 - 0.059

C_mp ≈ -0.13

Read more about aerodynamics at; https://brainly.com/question/6672263

Answer:

Explanation:

From the equation:

Power dissipated= square of voltage supplied by battery ÷ Resistance of the load

i.e P= V^2/R

It means that at constant voltage, the the power consumed is inversely related to the resistance. Therefore the 10W bulb which has a higher resistance will consume less power using the sufficiently excess power dissipated to glow brighter than the 250W bulb which has a low resistance. The power dissipated will partly be used to overcome this low resistance making less power available for heating up the 250W bulb .

ANSWER:

A 10W bulb may shine brighter than a 250W bulb, when the two are connected to the same battery. This can happen when the battery is low or not fully charged. Because the 10W bulb has a high resistance more than a 250W bulb, it makes the 10W bulb to be more efficient than the 250W bulb.

When a low current is passed through the fillament of the two bulbs, the 10W bulb which has high resistance, uses almost all the current that enters the filament to emit more light, more than the 250W bulb which has a low resistance and will converts almost all the current that enters the filament into heat energy. In the 250W bulb only about 8% of the current are used to light up the bulb and the rest are converted to heat energy. This explains why a bulb which it's watt is higher will be more hotter when lighted up, than a bulb which watt is lower.

Another reason while the 10W bulb will shine brighter is when the fillament in it are coiled tight round itself ( example is a fluorescent bulb). It will emit more light than a 250W watt bulb that the filament are not coiled (example is an incandescent bulb).

NOTE : On a normal circumstances, a 250W bulb will shine brighter than a 10W bulb, because the higher the electric energy a bulb consumes the brighter light the filament will produce. Watt is the amount of electric energy the bulb can consume, therefore a bulb with 250W is assumed to produce more light than a bulb with 10W.

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

Answer:

time = 4.89 min

Explanation:

given data

volume = 15,000 ft³

ventilation rate of oven = 2,100 cubic feet per minute

safety factor (K) = 3

evaporates at a rate = 0.6 cubic feet per minute

solution

we get here first solvent additional rate in oven that is

solvent additional rate = [tex]\frac{0.6}{1500}[/tex]

solvent additional rate = 4 × [tex]10^{-5}[/tex] min

solvent additional rate == 4 × [tex]10^{-5}[/tex] × [tex]10^{6}[/tex]

solvent additional rate =  40 ppm/min

and

solvent removal rate due to ventilation will be

removal rate = [tex]\frac{2100}{1500}[/tex] × concentration in ppm

removal rate = 0.14 C  ppm/min

and

net additional rate is

net additional rate (c) = m - r

so net additional rate (c) is = 40 - 0.14C

so here

[tex]\frac{dc}{dt}[/tex] = 40 - 0.14C

so take integrate from 0 to t

[tex]\int\limits^t_o {dt}[/tex] = [tex]\int\limits^{425/3}_0 \frac{dc}{40-0.14C}[/tex]    ....................1

here factor of safety is 3 so time taken is [tex]\frac{425}{3}[/tex]  

solve it we get

time = [tex][\frac{-50}{7} \ log(40-\frac{7c}{50} ]^{425/3} _0[/tex]

time = 4.89 min

Answer and Explanation:

The answer is attached below

Answer:

Explanation:

The answer to the above question is given in attached files.

Answer:

Q_total = 1431 W

Explanation:

Given:-

- The dimension of the engine = ( 0.5 x 0.4 x 0.8 ) m

- The engine surface temperature T_s = 80°C  

- The road surface temperature T_r = 25°C = 298 K

- The ambient air temperature T∞ = 20°C

- The emissivity of block has emissivity ε = 0.95

- The free stream velocity of air V∞ = 80 km/h

- The stefan boltzmann constant σ = 5.67*10^-8 W/ m^2 K^4

Find:-

Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation

Solution:-

- We will extract air properties at 1 atm from Table 15, assuming air to be an ideal gas. We have:

                        T_film = ( T_s +  T∞ ) / 2 = ( 80 +  20 ) / 2

                                   = 50°C = 323 K

                        k = 0.02808 W / m^2

                        v = 1.953*10^-5 m^2 /s

                        Pr = 0.705

- The air flows parallel to length of the block. The Reynold's number can be calculated as:

                       Re = V∞*L / v

                            = [ (80/3.6)*0.8 ] / [1.953*10^-5]

                            = 9.1028 * 10^5

- Even though the flow conditions are ( Laminar + Turbulent ). We are to assume Turbulent flow due to engine's agitation. For Turbulent conditions we will calculate Nusselt's number and convection coefficient with appropriate relation.

                       Nu = 0.037*Re^0.8 * Pr^(1/3)

                             = 0.037*(9.1028 * 10^5)^0.8 * 0.705^(1/3)

                             = 1927.3

                       h = k*Nu / L

                          = (0.02808*1927.3) / 0.8

                          = 67.65 W/m^2 °C

- The heat transfer by convection is given by:

                     Q_convec = A_s*h*( T_s -  T∞ )

                                        = 0.8*0.4*67.65*(80-20)

                                        = 1299 W

- The heat transfer by radiation we have:

                      Q_rad = A_s*ε*σ*( T_s -  T∞ )

                                        = 0.8*0.4*0.95*(5.67*10^-8)* (353^4 - 298^4)

                                        = 131.711 W

- The total heat transfer from the engine block bottom surface is given by:

                      Q_total = Q_convec + Q_rad

                      Q_total = 1299 + 131.711

                      Q_total = 1431 W

Following are the calculation to the given question:

Calculating the average film temperature:

[tex]\to T_f = \frac{T_s + T_{\infty}}{2} =\frac{80+20}{2}= 50^{\circ}\ C\\\\[/tex]  

At  [tex]T_f = 50^{\circ}\ C[/tex], obtain the properties of air:

[tex]\to k=0.02735\ \frac{W}{m \cdot K} \\\\\to v=1.798 \times 10^{-5} \ \frac{m^2}{s}\\\\ \to Pr =0.7228 \\\\\to Re_{L} =\frac{VL}{v} =\frac{80 (\frac{5}{18}) \times 0.8}{1.798 \times 10^{-5}} = 988752.935 \\\\[/tex]

Assume the engine block's bottom surface is a flat plate.

  For

[tex]\to 5\times 10^{5} \leq Re_{L} \leq {10^7}, 0.6 \leq Pr \leq 60 \\\\[/tex]

[tex]\to Nu=0.0296 \ \ Re_{L}_{0.8} Pr^{\frac{1}{3}}\\\\[/tex]

          [tex]= 0.0296(988752.935)^{0.8} (0.7228)^{\frac{1}{3}}\\\\ = 1660.99[/tex]

But,  

[tex]\to Nu=\frac{hL}{k} \\\\ \to h= Nu \frac{k}{L}\\\\[/tex]

      [tex]=\frac{1660.99 \times 0.02735}{0.8}\\\\ = 56.7850 \frac{W}{m^2 . K}\\\\[/tex]

[tex]\to A_s= L \times w\\\\[/tex]  

         [tex]= 0.8 \times 0.4\\\\ =0.32\ m^2\\\\[/tex]

[tex]\to Q_{com} = h A_s (T_s -T_{\infty})\\\\[/tex]

              [tex]= 56.7850 \times 0.32 \times (80-20)\\\\ = 1090.272\ W\\\\[/tex]

[tex]\to Q_{rad} = \varepsilon A_s \sigma (T_{s}^{4} -T_{surr}^4) \\\\[/tex]

            [tex]= (0.95)(0.32 )(5.67 \times 10^{-8}) [(80+273)^4-(25+273)^4]\\\\ = 131.710\ W\\\\[/tex]

Then the total rate of heat transfer is  

[tex]\to Q_{total} = Q_{corw} + Q_{rad}\\\\[/tex]

              [tex]=1090.272 +131.710 \\\\= 1221.982\ W\\\\[/tex]

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Answer / Explanation:

To answer this question, we first define the parameters which are,

Volume: This can be defined or refereed to the quantity of three-dimensional space enclosed by a closed surface. For the purpose of illustration, we can say that the space that a substance or shape occupies or contains. The SI used in measuring volume is mostly in cubic metre.

Therefore,

Volume, where Volume = base area x height x 1/3

where,

Base area = base length x base width

However, we also watch out for the division integer.

So moving forward to write the code solving the question, we have:

double Pyramid Volume (double baseLength, double baseWidth, double pyramid Height)

{

double baseArea = baseLength * baseWidth;

double vol = ((baseArea * pyramidHeight) * 1/3);

return vol;

}

int main() {

  cout << "Volume for 1.0, 1.0, 1.0 is: " << PyramidVolume(1.0, 1.0, 1.0) <<

endl;

  return 0;

}

Answer:

The problem demands a function PyramidVolume() in C language since printf() command exists in C language. Complete code, output and explanation is provided below.

Code in C Language:

#include <stdio.h>

double PyramidVolume(double baseLength, double baseWidth, double pyramidHeight)

{

return baseLength*baseWidth*pyramidHeight/3;

}

int main()

{

  printf("Volume for 1.0, 1.0, 1.0 is: %.2f\n",PyramidVolume(1, 1, 1));

  printf("Volume for 2.5, 5, 2.0 is: %.2f\n",PyramidVolume(2.5, 5, 2.0));

return 0;

}

Output:

Please also refer to the attached output results

Volume for 1.0, 1.0, 1.0 is: 0.33

Volume for 2.5, 5, 2.0 is: 8.33

Explanation:

A function PyramidVolume of type double is created which takes three inputs arguments of type double length, width and height and returns the volume of the pyramid as per the formula given in the question.

Then in the main function we called the PyramidVolume() function with inputs 1.0, 1.0, 1.0 and it returned correct output.

We again tested it with different inputs and again it returned correct output.

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times.  When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

Answer:

True

Explanation:

Because in quasi static process, process occurs very slowly so it is treated as reversible process.

Complete question

The complete question is shown on the first uploaded image

Answer:

a The table is shown on the second uploaded image

b The simplest possible sum product is shown on the fifth uploaded image

Explanation:

The explanation is shown on the third fourth and fifth uploaded image

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with  water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol  = 0.8 g/cm³

solution

we get here water content that is express as

water content = [tex]S_s \times \gamma _w[/tex]   ....................1

here soil is full saturated so [tex]S_s[/tex] is 100% in both case

so put here value for water

water content = 100 % ×  1

water content = 1 g

and

now we get for alcohol that is

water content  = 100 % × 0.8

water content  = 0.8 g

so here water sample have more water content

The soil sample saturated with water has a larger water content than the one with alcohol because water's unit weight is greater than that of alcohol and both soils have the same void ratio.

The soil sample saturated with water will have a larger water content compared to the soil saturated with alcohol. This is because the unit weight of water is greater than that of alcohol. Since both soils are fully saturated and have the same void ratio, the volume of liquid in both cases is identical. Therefore, the soil with the denser liquid (water) will contain more mass of liquid per unit volume, leading to a higher water content.

Answer:

a) A = 1667 and B = 2353

b) Oven A

c) Oven A

d) Below 13,333 pizza: Oven A

Above 13,334 pizza: Oven B

Explanation:

We have the following data:

                             Oven A:                             Oven B:

Capacity                 20 p/hr                              40p/hr

Fixed Cost            $20,000                            $30,000

Variable Cost        $2.00/p                             $1.25/p

Selling Price: $14

a) Break-even point  →   Cost = Revenue

([tex]x[/tex] refers to the number of pizza sold)

Oven A:

20000 + 2[tex]x[/tex] = 14[tex]x[/tex]

20000 = 14[tex]x[/tex] - 2[tex]x[/tex]

[tex]x[/tex] = 20000/ 12

[tex]x[/tex] = 1666.67 ≈ 1667 pizza

Oven B:

30000 + 1.25[tex]x[/tex] = 14[tex]x[/tex]

30000 = 14[tex]x[/tex] - 1.25[tex]x[/tex]

[tex]x[/tex] = 30000/ 12.75

[tex]x[/tex] = 2352.9 ≈ 2353 pizza

b) Comparing both oven for 9,000 pizza

Profit = Selling Price - Cost Price

Oven A:

Profit = (9000 x 14) - (20,000 +  2 x 9000)

Profit = 126000 - 38000

Profit = 88000

Oven B:

Profit = (9000 x 14) - (30,000 +  1.25 x 9000)

Profit = 126000 - 41250

Profit = 84750

Oven A is more profitable.

c)

Oven A:

Profit = (12000 x 14) - (20,000 +  2 x 12000)

Profit = 168000 - 44000

Profit = 124000

Oven B:

Profit = (12000 x 14) - (30,000 +  1.25 x 12000)

Profit = 168000 - 45000

Profit = 123000

Oven A is more profitable.

d) Using the equation formed in a):

20,000 - 12[tex]x[/tex] < 30,000 - 12.75[tex]x[/tex]

12.75[tex]x[/tex] - 12[tex]x[/tex] < 30000 - 20000

0.75[tex]x[/tex] < 10000

[tex]x[/tex] < 10000/0.75

[tex]x[/tex] < 13333.3

Hence, if the production is below 13,333 Oven A is beneficial.

For production of 13,334 and above, Oven B is beneficial.

Answer:

a.       Find the break even points in units for each oven.

Breakeven for type A pizza x =  = 1,666.6 units of pizza need to be sold in order to obtain breakeven for Type A

Breakeven for type B pizza x =  = 2,352.9 units of pizza need to be sold in order to obtain breakeven for Type B

b.      If the owner expects to sell 9000 pizzas, which oven should she purchase?

Type B: because the profit will be twice what will be obtainable from type A considering the fact that it produces pizza at the ration of TypeB:TypeA, 40:20 or 2:1

Profit for type a = 9000/20 x 14 = 6,300 – 1,666,6units ($23, 3332) = 4366.4 units

Profit for type B = 10,247.1 units of pizza - which makes it justifiable

Explanation:

The question involves calculating free moisture content, plotting it against time, determining the drying rate, and predicting the total drying time using numerical integration during the falling rate period. The constant and falling rate periods of drying are crucial in understanding the sample's drying behavior.

The student is tasked with calculating the free moisture content ( extit{X}) for various data points during the drying of a foodstuff in a tray dryer, plotting  extit{X} versus time (t), determining the drying rate ( extit{R}), and predicting the total drying time using numerical integration for the falling rate period of drying.

The falling rate period is characterized by a decreasing drying rate which requires numerical integration to find the total drying time. The constant rate period is defined by a uniform drying rate and typically occurs before the falling rate period.

The detailed answer provides calculations for free moisture content, drying rate, and prediction of total drying time. It also includes instructions for plotting the moisture ratio vs. time.

Calculating the Free Moisture Content:

To calculate the free moisture content (X) for each data point, subtract the bone dry weight from the wet weight. Then, divide by the bone dry weight.

Calculating Drying Rate and Predicting Total Drying Time:

Determine the drying rate using the slope of the moisture ratio vs. time plot. Then, use numerical integration to predict the time to dry the sample from X = 0.20 to X = 0.04. Identify the drying rate in the constant rate period and X in that period.

Moisture Ratio vs. Time Plot:

Plot the moisture ratio (MR) versus time to observe the constant rate and falling rate periods during the drying.

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

 int data;

 Node next;

 public Node()

 {data = 0; next = null;}

 public Node(int x, Node n)

 {data = x; next =n;}

}

public void Displaylist(Node q)

 {if (q != null)

       {  

        System.out.println(q.data);

         Displaylist(q.next);

       }

 }

public void Buildlist()

  {Node q = new Node(0,null);

       head = q;

       String oneLine;

       try{BufferedReader indata = new

                 BufferedReader(new InputStreamReader(System.in)); // read data from terminals

                       System.out.println("Please enter a command or a line of text: ");  

          oneLine = indata.readLine();   // always need the following two lines to read data

         head.data = Integer.parseInt(oneLine);

         for (int i=1; i<=head.data; i++)

         {System.out.println("Please enter another command or a new line of text:");

               oneLine = indata.readLine();

               int num = Integer.parseInt(oneLine);

               Node p = new Node(num,null);

               q.next = p;

               q = p;}

       }catch(Exception e)

       { System.out.println("Error --" + e.toString());}

 }

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

 mylist.Buildlist();

 mylist.Displaylist(head);

}

}

Answer:

critical stress required is  18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × [tex]10^{-3}[/tex] m

so we get here critical stress that is

[tex]\sigma _c = \sqrt{\frac{2E \gamma }{\pi a}}[/tex]     ...............1

put here value we get

[tex]\sigma _c[/tex] =   [tex]\sqrt{\frac{2\times 225\times 10^9 \times 1 }{\pi \times 0.4\times 10^{-3}}}[/tex]

[tex]\sigma _c[/tex] =  18923493.9151 N/m²

[tex]\sigma _c[/tex] =   18.92 MPa

Answer:

(C) ln [Bi]

Explanation:

Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.

Answer:

Option C. ln[Bi]

Explanation:

The nuclear equation of first-order radioactive decay of Bi to Po is:

²¹⁴Bi₈₃  →  ²¹⁴Po₈₄ + e⁻

The radioactive decay is expressed by the following equation:

[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]      (1)  

where Nt: is the number of particles at time t, No: is the initial number of particles, and λ: is the decay constant.                

To plot the variation of the quantities in function of time, we need to solve equation (1) for t:

[tex] Ln(\frac{N_{t}}{N_{0}}) = -\lambda t [/tex]      (2)

Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:

[tex] [Bi] = \frac {N particles}{N_{A} * V} [/tex]                            (3)  

[tex] [Bi]_{0} = \frac {N_{0} particles}{N_{A} * V} [/tex]               (4)  

where [tex]N_{A}[/tex]: si the Avogadro constant and V is the volume.

Now, introducing equations (3) and (4) into (2), we have:

[tex] Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t [/tex]  

[tex] Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t [/tex]      (5)

Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:

[tex] Ln ([Bi]) = -\lambda t + Ln([Bi]_{0}) [/tex]      

Therefore, the correct answer is option C. ln[Bi].

I hope it helps you!          

Answer:

Speed with which the box hits the truck at B relative to the truck = 5.29 ft/s

Explanation:

First of, we calculate the deceleration of the truck+box setup using the equations of motion.

x = distance tavelled during the deceleration = 350 ft

u = initial velocity of the truck = 60 mph = 88 ft/s

v = final velocity of the truck = 0 m/s

a = ?

v² = u² + 2ax

0² = 88² + 2(a)(350)

700 a = - 88²

a = - 11.04 ft/s²

But this deceleration acts on the crate as a force trying to put the box in motion.

The motion of the box will be due to the net force on the box

Net force on the box = (Force from deceleration of the truck) - (Frictional force)

Net force = ma

Frictional Force = μmg = 0.3 × m × 32.2 = 9.66 m

Force from the deceleration of the truck = m × 11.04 = 11.04 m

ma = 11.04m - 9.66m

a = 11.04 - 9.66 = 1.4 ft/s²

This net acceleration is now responsible for its motion from rest, through the 10 ft that the box moved, to point B to hit the truck.

x = 10 ft

a = 1.4 ft/s²

u = 0 m/s (box starts from rest, relative to the truck)

v = final velocity of the box relative to the truck, before hitting the truck's wall = ?

v² = u² + 2ax

v² = 0² + 2(1.4)(10)

v² = 28

v = 5.29 ft/s

Answer:

184.077N

Explanation:

Please see attachment for step by step guide

Answer:

Answer is 184.077

Refer below for the explanation.

Explanation:

As per the question,

V 5 15 ft/s,

D 5 4 in,

d 5 1 in

Refer to the picture for detailed explanation.