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2 years ago

Choose the three questions that an engineer should ask himself or herself when identifying the need of a problem.

Engineering

2 answers:

boyakko [2]2 years ago

5 0

Answer:

What is the goal? What will a successful solution look like? What is it that the client expects me to accomplish?

Explanation:

alexira [117]2 years ago

3 0

Answer:

What is the goal? What will a successful solution look like? What is it that the client expects me to accomplish?

Explanation:

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Which permission do you need to shoot on the owner’s property?

Elena L [17]

Answer:

filming permit,

( MARK ME BRAINLIEST!!)

4 0

2 years ago

7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute

Crank

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

V = 28 KN

From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

#SPJ1

4 0

1 year ago

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

yanalaym [24]

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

$Re=\dfrac{F_i}{F_v}$

For plate can be given as

$Re=\dfrac{\rho\ v\ l}{\mu }$

Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

$Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.$

$Reynolds\ number\is \$

7 0

2 years ago

Programming Assignment 2 Decimal and IEEE-754 ConversionsObjective: To write a C program (not C++) that converts numbers between

kondaur [170]

Answer:

// Program is written in C Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include<stdio.h>

#include<math.h>

//Function to Convert to float

void To float(int num, int I)

{

//Create a kount variable

int kount;

// Start an iteration

for(kount=i-1; kount>0; kount--)

{

if((num>>kount) && 1) {printf("1");}

else { printf("0"); }

}

// Create a user defined variable

typedef union {// Definition

float Number;

struct

{

// Mantissa

unsigned int mant : 23;

// Exponent

unsigned int exp : 8;

// Sign

unsigned int sign: 1;

} raw;

} myfloat; // Variable name

// Create print segment

void printsegment(myfloat var)

{

printf("%d |", var.raw.sign);// Sign

To float(var.raw.exp,8); // Exponent

printf("|");

To float(var.raw.mant,8); // Mantissa

printf("\n");

}

// Function to Convert to Real

unsigned int ToReal(int* dig[], int l, int h)

{

unsigned int f = 0, I;

Start an iteration

for(I = h; I>=l;I--)

{

// Calculate individual value

f = f + dig[I] * pow(2,h-1);

}

return f;

}

// Main method start here

int main()

{

printf("Floating Point Conversion\n");

printf("Select any of the following options\n");

printf("1. Decimal to IEEE754 Conversion\n");

printf("2. IEEE754 to Decimal Conversion\n");

printf("3. Quit");

// Declare integer variable for option

int opt;

// Prompt to select option

printf("Select an option; Option 1 to 3: ");

scanf("%d", $opt);

if(opt == 1)

{

printf("You have selected option 1");

// Declare a user defined variable and a system defined variable

myfloat var; float number;

// Accept input

scanf("%d", number);

// Check for special cases

if(isnan(number/0.0))// Not a number

{

printf("Not a Number");

}

else

{

var.f = number;

// Print Sign

printf("%d | ", var.raw.sign);

// Print Exponent

ToFloat(var.raw.exp,8);

printf(" | ");

// Print Mantissa

ToFloat(var.raw.mant,23);

}

}// End of option 1;

// Beginning of option 2

else if(opt == 2)

{

printf("You have selected option 2");

// Declare an array and two integer variables

unsigned int number[32];

int ctrlno, I = 0;

// Accept input by through an iteration

for(int k = 0; k < 32; k++)

{

// Create a label

label: scanf("%d", ctrlno);

// Check for special cases

if(isnan(ctrlno/0.0))// Not a number

{

printf("Not a Number"); I++;

break;

}

else if(ctrlno>1 || ctrlno < 0)

{

printf("Invalid Number\n Please enter a valid digit");

goto label;

}

else {

// Assign number to array

number[k] = ctrlno;

}

// Check validity of number

if(I != 0)

{

printf(" Invalid Number Representation");

}

else

{

// Declare user defined variable

myfloat var;

// Get sign

var.raw.sign = number[0];

// Get mantissa; From to 31

unsigned f = ToReal(number,9,31);

var.raw.mant = f;

// Get exponent; 1 to 8

f = ToReal(number,1,8);

var.raw.exp = f;

// Print Output

printf("The converted digit is ");

printf("%f", var.f);

}

else

{

// Quit Application

break;

}

return 0;

}

3 0

3 years ago

I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th

SIZIF [17.4K]

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

4 0

2 years ago