Omplete the following program: [0.5 X 4 = 2]

The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is discovered the lower deck is

leva [86]

Answer:

It will be around 146,27 min since the pump is turned on until the deck is clear of the water.

Explanation:

When the leak is discovered and the pump is turned on, the lower deck is already submerged and the leak is not fixed; then, in order to have the deck clear of water, the bilge pump has to remove the accumulated water ( $V_{0}$ ) and the water that is taking on ( $r_{in}*t$ ) through the leak. We can represent this mathematically as follow:

$V_{0} +r_{in} *t-r_{out}*t=0$ Equation 1

Where:

$V_{0}$ : is the accumulated water when the leak was discovered

$r_{in}$ : is the takes on rate through the leak = 57.5 gal/min

$r_{out}$ : is the removing rate of the bilge pump = 73.8 gal/min

t= is the time since the pump is turned on until the deck is clear of water.

To calculate the accumulated water ( $V_{0}$ ), we will model the lower deck as a flat-bottomed container with a bottom surface area of 510 $ft^{2}$ and straight vertical sides. Knowing that the level submerged is 7.5 inches, and performing the corresponding unit conversions, we obtain:

$V_{0}$ = bottom surface area * lever submerged

$V_{0}= 510ft^{2}*7.5 in*\frac{1ft}{12in}=318.75 ft^{3}*7.48\frac{gal}{1ft^{3}}=2384.25 gal$ Equation 2

Solving equation 1 for time (t), and replacing the value obtained in equation 2, we get:

$t=\frac{V_{0}}{(r_{out}-r_{in})} =\frac{2384.25 gal}{(73.8-57.5)gal/min}=$ 146,27 min

8 0

3 years ago

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro

Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\$

$= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\$

if $P$ flow is chocked

if $P>P^{*} \to$ flow is not chocked

When P= 10 psia < $P^{*}$ $\to$ not chocked

match number:

$\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}$

$= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}$

$M_0=7.625$

$p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}$

$=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}$

$\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\$

$\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\$

$=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\$

R= gas constant=1716

$m=PAV\\\\$

$=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$

5 0

3 years ago

Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?

BigorU [14]

Answer: Maybe A

Explanation:

7 0

1 year ago

Read 2 more answers

Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output tho

Likurg_2 [28]

Answer:

See explanation

Explanation:

//Include the

//required header files.

#include <stdio.h>

//Define the

//main() function.

int main(void) {

//Declare the

//required variables.

char input_char;

int input_int;

double input_double;

char input_string[100];

//Prompt the user

//to enter an integer.

printf("Enter integer: ");

//Read and store

//the integer.

scanf("%d", &input_int);

//Prompt the user

//to enter a double value.

printf("Enter double: ");

//Read and store

//the double value.

scanf("%lf", &input_double);

//Prompt the user

//to enter a character.

printf("Enter character: ");

//Read and store

//the character.

scanf(" %c", &input_char);

//Prompt user to

//enter the string

printf("Enter string: ");

//Read and

//store the string.

scanf("%s", input_string);

//(1)

//Display the values.

printf("%d %lf %c %s\n",

input_int, input_double,

input_char, input_string);

//(2)

//Display the values

//in reverse order.

printf("%s %c %lf %d\n",

input_string, input_char,

input_double, input_int);

//(3)

//Cast the double to

//an integer and display it.

printf("%lf cast to an integer is %d",

input_double, (int)(input_double));

//Return from the

//main() function.

return 0;

}

4 0

3 years ago

Find the Rectangular form of the following phasors?

almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45, 30, 60, 90, -34, -56, 20, -42, -65, -15

P=10, P=5, P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10, θ = 45° rectangular coordinates

x = r cosθ , y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

= 7.07

y =P sinθ = 10 sin 45°

= 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos 30° = 4.33

y = 5 sin 30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos 60° = 12.5

y = 25 sin 60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos 90° = 0

y = 54 sin 90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos (-34°) = 53.88

y = 65 sin (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos (-56)° = 53.12

y = 95 sin (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos 20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos (-42)° = 5.94

y = 8 sin (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos (-65)° = 14.79

y = 35 sin (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos (-15)° = 144.88

y = 150 sin (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0

3 years ago