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2 years ago

Which of the following is NOT associated with Urban Sprawl?

Engineering

1 answer:

pochemuha2 years ago

8 0

The option that is not associated with the given term called urban sprawl is; Option A: Blocking high views

What is Urban Sprawl?

Urban sprawl is defined as the rapid expansion of the geographic boundaries of towns and cities which is often accompanied by low-density residential housing and increased reliance on the private automobilefor movement.

Looking at the given options, "blocking high views" is the option that is not typically a problem associated with urban sprawl because urbanization usually takes place on relatively flat levels.

The missing options are;

a. blocking high views

b. destroying animal habitats

c. overrunning farmland

d. reducing green space

Read more about urban sprawl at; brainly.com/question/504389

You might be interested in

The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards

lesya [120]

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

$a_r = \frac{V^2}{R}$

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

7 0

2 years ago

A steam pipe passes through a chemical plant, where wind passes in cross-flow over the outside of the pipe. The steam is saturat

valina [46]

Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

steam is saturated at 17.90 bar.

the pipe is stainless steel and has an outside diameter of 6.75 cm

length = 34.7 m

Air flows over the pipe at 7.6 m/s

Bulk fluid temperature of 27°C

we know that

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

Outside diameter of pipe = 6.75 cm

length of the pipe = 34.7 m

velocity of air = 7.6 m/s

Cp of air = 1.005 kJ/Kgk

viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)

thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k

so as

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 0.0414 w/m².k

Now to find the rate of heat transfer Q

Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 6.1036 w/m².k

so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C

so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

Therefore the rate of heat transfer changes to 3518.61 watt

4 0

3 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

1 year ago

Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of

rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

if ( values[month] > largest ){

largest = values[month];

return largest;

}

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

if ( values[month] < smallest ){

smallest = values[month];

return smallest;

}

int main()

{

double rainFall[NUM_MONTHS];

for (int month = 0; month < NUM_MONTHS; month++)

{

cout << "Enter the rainfall (in inches) for month #";

cout << (month + 1) << ": ";

cin >> rainFall[month];

while (rainFall[month] < 0)

{

cout << "Rainfall must be 0 or more.\n"

<< "Please re-enter: ";

cin >> rainFall[month];

}

cout << fixed << showpoint << setprecision(2) << endl;

cout << "The total rainfall for the year is ";

cout << getTotal(rainFall, NUM_MONTHS)

<< " inches." << endl;

cout << "The average rainfall for the year is ";

cout << getAverage(rainFall, NUM_MONTHS)

<< " inches." << endl;

int subScript;

cout << "The largest amount of rainfall was ";

cout << getLargest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl;

cout << "The smallest amount of rainfall was ";

cout << getSmallest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl << endl;

return 0;

}

8 0

3 years ago

You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen

VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

6 0

3 years ago