Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find the Fourier series coefficient of x(t). Then find its Fourier transform and sketch the magnitude and phase spectra.b)Sketch the magnitude and phase spectra of the output (i.e., |Y(????)|and∡Y(????)) if the system is a low-pass filter with H(????)={1|????|<3????020other????ise, where ????0=2πT0.c)Sketch the magnitude and phase spectra of the output(|Y(????)|and∡Y(????)) if the system is a high-pass filter with H(????)={1|????|>5????020other????ise, where ????0=2πT0.d)Sketch the magnitude and phase spectra of the outputif the system is a filter with H(????)=11+j????.

Answer:

# list of 10 zip codes assigned to zip

zips = ["12789", "54012", "54481", "54982", "60007", "60103", "60187", "60188", "71244", "90210"]

# user is prompt to enter zip code and assigned to user_zip

user_zip = input("Enter your zip code: ")

# if else statement to check if user input is available for delivery

# if statement check if user zip is in zip, if it is, it display

# delivery is okay to specified zip

if user_zip in zips:

   print("Delivery to {} ok.".format(user_zip))

# else it display no delivery to such zip code

else:

   print("Sorry - no delivery to {}.".format(user_zip))

Explanation:

The question doesn't specify programming language to use. Since no programming language was stated, the problem was solved using Python3. List structure is the equivalent of array in Python.

Assumption was also made on the array holding 10 zip codes of areas to which the company make deliveries.

The program first initialized a list of 10 zip codes and assigned it to zips. Then it prompt the user to enter a zip code which is assigned to user_zip.

Then if-else statement is used to check if user inputted zip is available with the zips variable.

If it is available, "Delivery ok" is displayed to the user else "no delivery" is displayed to the user.

Answer:

Check the explanation

Explanation:

defs.h

#ifndef DEFS_H

#define DEFS_H

#include <stdio.h>

#include <stdlib.h>

typedef struct

{

   char item[20];

   char quantity[10];

   int calories;

   float protein;

   float carbs;

   float fats;

} food;

int size;

void printArray(int size, food arr1[]);

#endif

arrayProcessing.c

#include "defs.h"

void printArray(int size, food arr1[])

{

   int i = 0;

   printf("\nFOOD ITEM\t\tQUANTITY\tCALS\tPRO\tCARBS\tFAT");

   for (i = 0; i < size; i++)

   {

       printf("\n%i.%s", i + 1, arr1[i].item);

       printf("\t\t%s", arr1[i].quantity);

       printf("\t\t%i", arr1[i].calories);

       printf("\t%.2f", arr1[i].protein);

       printf("\t%.2f", arr1[i].carbs);

       printf("\t%.2f\n", arr1[i].fats);

   }

}

lab12.c

#include "defs.h"

int main(int argc, char *argv[])

{

   int i = 0;

   FILE *inFile;

   inFile = fopen(argv[1], "r");

   if(inFile==NULL){

       fprintf(stderr, "File open error. Exiting program\n");

       exit(1);

   }

   fscanf(inFile, "%i", &size);

   food *arr = (food *)malloc(sizeof(food)*size);

   for (i = 0; i < size; i++)

   {

       fscanf(inFile, "\n%[^\n]s", arr[i].item);

       fscanf(inFile, "%s", arr[i].quantity);

       fscanf(inFile, "%i", &arr[i].calories);

       fscanf(inFile, "%f", &arr[i].protein);

       fscanf(inFile, "%f", &arr[i].carbs);

       fscanf(inFile, "%f", &arr[i].fats);

   }

   printArray(size, arr);

   return 0;

}

Kindly check the Output below,

Answer:

1.5258m

Explanation:

Please see attachment

Answer:

diameter of the pipe = 0.4932ft.

Explanation:

assuming d = 0.4932

Re = 3.16 x 10∧5/0.4932

= 6.4 x 10 ∧5

E/d = 0.0005/0.4932

= 0.0010 from moody chat t = 0.02

if 0.02 is beign substituted in equation 2 we will get the same required diameter of the pipe which is 0.4932ft.

check the attachment  for better explanation.thanks

Answer:

Average power = 108.25W

Complex power = 125W

PF = 0.866

Explanation:

Check attachment for step by step instructions.

Answer:

Average power: 108.25 W

Complex power: 125 VA

Power factor: 0.866

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

Answer:

theoretical fracture strength  = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength  =   [tex]s_{0} \times \sqrt{\frac{L}{r} }[/tex]   .............................1

put here value and we get

theoretical fracture strength  =    [tex]1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }[/tex]  

theoretical fracture strength  =  [tex]16919.98 \times 10^6[/tex]  

theoretical fracture strength  = 16919.98 MPa

A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. Technician A is correct while Technician B is wrong.

A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. It is linked in parallel. It generally has a high resistance and draws very little current from the circuit.

Given the technician connects the voltmeter in parallel across two points in a circuit. Therefore, the technician A is correct the meter will provide a reading of the potential difference in volts.

Hence, Technician A is correct while Technician B is wrong.

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Answer:

60 rpm

Explanation:

At t = 0,

Angular speed = 0

At t = 10 sec

Angular speed = 20/10 = 2 rev/s

Average speed = (2 - 0)/2 = 1 rev/s

= 1 x 60 = 60 rpm

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

Explanation:

Please kindly go through the attached file for a step by step approach to the solution of this problem

Answer:

Inlet and exit velocities are 143.71 m/sec and 465.697 m/sec

Explanation:

At inlet of the pipe

[tex]P_1=150kPa[/tex] and [tex]T_1=150^{\circ}C[/tex]

At this pressure and temperature from steam table.

[tex]h_1=2870kj/kg[/tex] and [tex]s_1=7.508kj/kgK[/tex]

At pressure [tex]P_2=200kPa[/tex] and [tex]T_2=200^{\circ}C[/tex]

By steam table from interpolation method.

[tex]h_2=2776.38+(2768.80-2776.38)(\frac{150-100}{200-100})[/tex]

[tex]h_2=2772.59kj/kg[/tex]

[tex]Q=A_1V_1=A_2V_2[/tex]

[tex]D_1^2V_1=D_2^2V_2[/tex]

We have given [tex]\frac{D_1}{D_2}=1.80[/tex]

[tex]\frac{V_1}{V_2}=\frac{D_2^2}{D_1^2}[/tex]

[tex]\frac{V_1}{V_2}=(\frac{1}{1.80})^2[/tex]

[tex]V_1=0.3086V_2[/tex]

Now energy equation in the pipe

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}[/tex]

[tex]2870.7\times 10^3+\frac{0.3086V_2^2}{2}=2772.59\times 10^3+\frac{V^2}{2}[/tex]

[tex]V_2=465.697m/sec[/tex]

[tex]V_1=0.3086\times 465.697=143.714m/sec[/tex]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks =  512 blocks

(b) Tag is 18

(c)  Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

                                  = 223

                                  =8388608 bytes

Answer:

Explanation:

first convert difference equation to transfer function form,

apply laplase transform to difference equation

s2Y(s) + 6 * s * y(s) + 2 * y(s) = s * X(s) + 4 * X(s)

(s2 + 6s+2) * y(s) = (s+4)*X(s)Y s)

Lets write below code in matlab command prompt

lets use lodspace to create values from 10^-1 to 10^5 and use freqs to plot frequency response of above system with frequency w

>> n=[1 4];

>> d=[1 6 2];

>> w = logspace(-1,5);

>> freqs(n,d,w)

Attached is the written solution and the MATLAB diagram

Answer:

Weed Mapping  Application is the most reasonable approach for surveying.

Answer:

The most reasonable approach to survey the property would be a GPS survey using a dual-frequency GPS system.

Explanation:

Surveying makes use of coordinates to determine positions of objects, it is the first step towards developing a property. There are different types of survey which are specific to various needs and sizes of a property.

GPS SURVEY

GPS survey makes use of the global positioning system to determine positions of a property or object. It is quite accurate and also an efficient type of survey based on its flexibility.

In the illustration, the property size is 180 acres and 20 volunteers are expected to work on it. To carry out this survey, a station whose coordinate is known normally referred to the master station is used to get the coordinates of the remote stations. The volunteers are stationed at 20 remote stations and the surveyor at the master station makes use of dual-frequency GPS system to pick the coordinates of the remote stations. The dual-frequency GPS system would be used because it covers more distance and for a survey as massive as 180 acres this is reasonable.

For the survey to be carried out seamlessly by the 20 volunteers, a dual-frequency GPS system would be used for the GPS survey.

Answer:

R = 7.854 x 10⁵ anpert turns / Wb

Relative permeability = 405.3

Explanation:

Detailed explanation is given in the attached document.

Answer: 85398.16, 405.28473.

Explanation:

We are given that the number of turns on the core is 500 2 , the cross sectional area is :

A=5cm^{2}({1meter}/{100cm})^{2} =0.0005meters^{2}

and the length of the core is l=20cm ({1meter}/{100cm})= 0.2meters .

In this solution, we are meant to neglect the resistance of the coil , and the current through the coil is I=1amPrms when the voltage applied across it is:

V=120voltsrms at f=60Hz. From this, We can calculate the inductance(L) whiof the coil (a coil have an inductance value of one Henry when an electromotive force of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second).

The natural frequency of the applied voltage is:

ω =2π f=2π(60)=120π{radians}/{second} .

The inductive reactance of the coil is equal to X=ω L=120π L . We then know that current is :

I=V/X

=I =20/120πL

L=120/120π

=1/π henries .

For reluctance (R) (which is a unit measuring the opposition to the flow of magnetic flux within magnetic materials and is analogous to resistance in electrical circuits). Looking at the relationship between inductance and reluctance . You will note that it is :

L=n^2/R .

We can use this relationship to find reluctance for our closed iron core coil :

L=1/π = 500^2/R

R=500^2π =785398.16{amps}/{volt-seconds}}

We can therefore use the other equation for reluctance.

R=1/μ A= 1/μA or μRA

To calculate the relative permeability of the core :

R=500^2π

=0.2/(4π ×10^-7)/ μR(0.0005)

μR=0.2/(4π ×10^-7)/ 500^2π(0.0005)

= 405.28473

The IRS and the U.S Department of the treasury declared that social security recipients are not required to file a simple tax return to receive stimulus payments under the CARES Act.

Explanation:

Due to the impact of corona virus problem, the CARES Act calls for stimulus payment to be sent to Americans based on their gross income.

The Social security recipients are not required to file a tax return and do not take action and they will receive the payments directly to their bank accounts.

The Automatic payments will begin by next week. The eligible taxpayers who filed tax returns for 2019  or 2018 and chose direct deposit for their refund will automatically receive a stimulus payment of  $1,200 for individuals or $2,400 for married couples and $500 for each qualifying child.

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

Since the area is  

Time (min)           Runoff (cm)       Volume of runoff m³

0                   0                                     0

10                  4                                     1000000 m³

20                 2.5                                  625000 m³

30                 2                                      500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

For the 1st  10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

Answer:

The resulting strain is [tex]1.39\times 10^{-3}[/tex].

Explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

[tex]\epsilon=\dfrac{F}{AE}[/tex]

E is elastic modulus of Cu is are of cross section

[tex]\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}[/tex]

So, the resulting strain is [tex]1.39\times 10^{-3}[/tex].

Answer:

The mass of the hydrogen for one hour would be 0.0039168 kg/hr

Explanation:

The concentration and the distance at concentration point is calculated before calculating the mass of hydrogen. The attached images show a clear explanation;

The mass of the hydrogen for one hour would be 0.0039168 kg/hr. The derivation of the diffusion coefficient has been attached in the image below:

According to Fick's law, the mass of a substance dM is proportional to the concentration gradient grad c of this substance as it diffuses in time dt over a surface dF normal to the diffusion direction: dM = D grad c dF dt.

Physically, the diffusion coefficient, therefore, suggests that for a concentration gradient of unity, the mass of the substance diffuses through a unit surface in a unit of time. A square meter per second corresponds to D in the SI system. Physical constants such as temperature, pressure, and the size of the dispersing substance's molecules all affect the diffusion coefficient.

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Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

[tex]tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43[/tex]

The torque is:

[tex]T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm[/tex]

ii) The torque to lowering the load is:

[tex]T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm[/tex]

iii)

[tex]T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}[/tex]

Clearing u:

u = -0.016

Answer:

1)

Static scoping

A=0

B=5

C=3

D=4

2) For dynamic scoping

A=2

B=5

C=1

D=8

Answer:

length of cylinder can not calculated

Explanation:

given data

tensile stress = 10,000 psi

Original length = 1

Original cross-sectional area = 0.1 in²

Yield strength, σy = 9 ksi

Young’s modulus, E = 1000 ksi

solution

we can see that here that applied stress is greater than yield stress of material  that is express

1000 ksi  >  9 ksi

so here hooks law and strain relation is not working

so length of cylinder can not calculated

as stress applied 10000 psi

Under 10,000 psi tensile stress, the metal cylinder elongates by 0.01 inches, resulting in a final length of 1.01 inches.

To calculate the elongation (change in length) of the metal cylinder under tensile stress, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] is directly proportional to the applied tensile stress [tex](\( \sigma \))[/tex] and the original length [tex](\( L_0 \))[/tex], and inversely proportional to the Young's modulus [tex](\( E \))[/tex]:

[tex]\[ \Delta L = \frac{\sigma \cdot L_0}{E} \][/tex]

Given:

- Original length [tex](\( L_0 \))[/tex] = 1 in

- Applied tensile stress [tex](\( \sigma \))[/tex] = 10,000 psi

- Young's modulus [tex](\( E \))[/tex] = 1000 ksi = 1,000,000 psi

Substitute the values into the formula:

[tex]\[ \Delta L = \frac{10,000 \times 1}{1,000,000} \][/tex]

[tex]\[ \Delta L = \frac{10,000}{1,000,000} \][/tex]

[tex]\[ \Delta L = 0.01 \, \text{in} \][/tex]

So, the elongation of the metal cylinder under the given tensile stress is 0.01 inches.

To find the final length, we add the elongation to the original length:

[tex]\[ \text{Final length} = \text{Original length} + \Delta L \][/tex]

[tex]\[ \text{Final length} = 1 + 0.01 \][/tex]

[tex]\[ \text{Final length} = 1.01 \, \text{in} \][/tex]

Therefore, the final length of the metal cylinder under the given tensile stress is 1.01 inches.

Answer:

The process is possible:

Explanation:

We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.

[tex]P_{1} = 3 MPa[/tex]

[tex]x_{1} = 50 % = 0.5[/tex]

[tex]V_{1} = 20 L = 0.02 m^{3}[/tex]

[tex]P_{2} = 1.2 MPa[/tex]

[tex]T_{H} = 300^{0} C = 573 K[/tex]

Received heat energy, [tex]Q_{12} = 600 kJ[/tex]

Work done, [tex]W_{12} = 124 kJ[/tex]

At state 1, using the steam table:

[tex]T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K[/tex]

[tex]v_{1} = v_{f1} + x_{1} * v_{fg1}[/tex]

[tex]v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg[/tex]

[tex]M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg[/tex]

[tex]u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg[/tex]

[tex]s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg[/tex]

[tex]Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\[/tex]

Solving for u₂

[tex]u_{2} = 2612.3 kJ/kg[/tex]

Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,

then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K

The entropy generated will be:

[tex]\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K[/tex]

Since ΔS > 0, this process is possible

Answer:

Yes it is possible

Explanation:

Attached is the solution

Answer:

See explaination

Explanation:

Rolling resistance which in some occassions can be called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.

In order to calculate our rolling resistance, there should be a force.

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

Answer:

Diodes consists of two-terminal electronic parts which conducts current mainly in one direction. It has low resistance in one direction and high resistance in the other direction. These are correct statements about a Diode.

A. Diodes are used in voltage regulators and limiters.

C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD

Explanation:

Diodes

Answer:

435.032 kj

Explanation:

We can describe Heat transfer as a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.

Please refer to the attached file for the detailed step by step solution of the given problem

Answer:

See explaination

Explanation:

We can say that that the The Darcy Friction factor or Equation is a theoretical equation that predicts the frictional energy loss in a pipe based on the velocity of the fluid and the resistance due to friction. It is used almost exclusively to calculate head loss due to friction in turbulent flow.

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

Answer:

(a) BOD removal efficiency = 89.51%

(b) Aeration tank volume = 4732m³

(c) MLSS = 1706.669 mg/L

(d) Secondary clarifier surface area 6250ft²

(e) Air requirement =  12930.284 lb

Explanation:

See the attached file for explanation.  This is continuation from page 3 of the attached file.

Air required = 172403.792 ft3

since, density of air = 0.075lb/ft3

air required = 0.075*172403.792 lb

                    = 12930.284 lb

Answer:

a. 0.95 in

b. 1.19 in

c. 1.137 in

Explanation:

Express the factor of safety equation for maximum-normal-stress theory as:  

S SF = Eau  

Here, the factor of safety is SF, the yield strength is S„ and the maximum stress :I  

Modify the above equation for shear stress acting on the solid rod as:  

S. SF = To  

Here, the combined shear stress on solid rod s  

Calculate the combined shear stress for solid rod.  

16T r2:1 =  trd3  

(1)  

Here, the torque is T, and the diameter of the solid rod is d.  

Substitute 5,000 'bin. for T.  

— 16(5,000 lb • in ) v  ird 80 000lb - in. _  

rd3  

60ksi 2 —  80,000lb -in. trd  

Solve the above equation for d.  

60 x 1031bAn?  80,00016 -in. ird3 —  2(80, 000 lb in.) d3  rrt60x103Ibfln?)  

v3 d —[  2(80,000lb-in.) rz-(60 x103112,An?) = 0.8488 in.3r3 =0.95 in.  

check the attached files for clear cut details