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3 years ago

10

Wells drilled by a nonprofit called Water for South Sudan use a pump that can provide up to 5,500 gallons of water per day. Use

this total amount to calculate the following:
How many gallons per day per person could the well produce for a village with a population of 850 if every gallon went to domestic use?

1 answer:

Soloha48 [4]3 years ago

7 0

Answer:

6.47 gallon per person per day

Explanation:

Given

The total amount of water available for domestic use is 5500 gallon

Let us consider that the entire available water goes to domestic use.

The total population of the village is 850.

The amount of water available to each villager is

$\frac{5500}{850}\\6.47$ gallon per person per day

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Ideias de nomes para um gato (macho), foto no anexo.

Rzqust [24]

So cute! xxxxxxxxxxxxxxxxx

4 0

3 years ago

A square steel bar has a length of 8.4 ft and a 2.1 in by 2.1 in cross section and is subjected to axial tension. The final leng

nikitadnepr [17]

Answer:

Poissons ratio = -0.3367

Explanation:

Poissons ratio = Lateral Strain / Longitudinal Strain

In this case, the longitudinal strain will be:

Strain (longitudinal) = Change in length / total length

Strain (longitudinal) = (8.40392 - 8.4) / 8.4

Strain (longitudinal) = 4.666 * 10^(-4)

While the lateral strain will be:

Strain (Lateral) = Change in length / total length

Strain (Lateral) = (2.09967 - 2.1) / 2.1

Strain (Lateral) = -1.571 * 10^(-4)

Solving the poisson equation at the top we get:

Poissons ratio = -1.571 / 4.666 <u>( 10^(-4) cancels out )</u>

Poissons ratio = -0.3367

6 0

3 years ago

Extra Credit: The Linc (parking lot and stadium)In celebration of the upcoming Super Bowl, for a maximum 10 points of extra cred

Sidana [21]

Answer:

Explanation:

// Below is the code to draw parking lot only , i have also put the o/p of code.

import java.util.*;

import java.lang.*;

import java.io.*;

public class Linc

{

static int size=4;

static int numBoxes = 1; // one row.

static int height =(int) Math.pow(size, 2); //Just how many | will it have.

static int width = 2; // two boxes in a row

public static void main (String[] args) throws java.lang.Exception

{

top();

printHeight();

}

public static void top(){

for(int i = 0; i <= numBoxes*width;i++)

{

if(i%width == 0) {//When this happens we're in a new box.

System.out.print(" ");

System.out.print("____________");

}

else

System.out.print(" ");

}

System.out.println(); //Move to next line.

}

public static void printHeight(){

for(int j = 0; j < height;j++){

for(int i = 0; i <= numBoxes*width;i++)

{

if(i%width == 0) {

System.out.print("|"); //Whenever this happens we're in a new box.

System.out.print("____________");

}

else

System.out.print(" ");

}

System.out.print("|");

System.out.println(); //Move to next line.

}

}

}

8 0

3 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago