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Pachacha [2.7K]
3 years ago
6

What different tests did the team perform to come up with a workable design?

Engineering
1 answer:
I am Lyosha [343]3 years ago
5 0
They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.
You might be interested in
A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and deca
dexar [7]

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

5 0
2 years ago
If OSHA determines that an employer's response to a non-formal complaint is adequate, what options does the employee filing the
Anuta_ua [19.1K]
Fill it out without telling ur employer as that may cause backlash and have an osha certified employee come check out ur work or job site
4 0
3 years ago
Build a 3-input XOR gate using a 3x8 decoder and an Or gate.please Quickly i have final​
lana [24]

Answer:

Take a 3 to 8 decoder with active low outputs

Assuming you are familiar with the functioning of decoders,

The three inputs of decoder of course are the first, second and the carry bit which you feed to the subtractor.

Next we examine the truth table of the full subtractor i formatted in the picture.

Then write the minterms for the difference output and borrow output from the given truth table picture I have mentioned before!!

Explanation:

hopi it to help you!!

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
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
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