What is the next measurement after 2' -6" on the architect's scale?

Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500

Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

return 0;

}

Explanation:

7 0

2 years ago

Air in a piston–cylinder assembly, initially at 3 bar, 142 K, and a volume of 2 m3. The air undergoes a process to a state where

Archy [21]

Answer:

mass of the air = 14.62kg

Workdone = 415.88 kJ

Heat transfer = 415.88 kJ

Explanation:

Please find the attached files for the solution

6 0

2 years ago

Read 2 more answers

**Please Help ASAP**

RoseWind [281]

Answer:

look up the assignment number. its in the left side of the screen. its what i did when i had problems.

Explanation:

6 0

2 years ago

7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21

Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

$g = (\frac{v}{s}) (\frac{C}{X})$

Let's first find the fourth stage critical lane group ratio $\frac{v}{s}$ , using the formula:

$C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}$

Solving for $(\frac{v}{s})$ , we have:

$(\frac{v}{s}) = 0.185$

Let's also calculate the volume capacity ratio X,

$X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}$

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

$g_4 = (\frac{v}{s}) (\frac{C}{X})$

Substituting figures in the equation, we now have:

$g_4 = (0.185) (\frac{60}{0.704})$

$g_4 = 15.78 seconds$

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0

2 years ago

A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhea

AURORKA [14]

Answer:

A)

D = 158.42 kmol/h

B = 191.578 kmol/h

B) Rmin = 1.3095

Explanation:

a) Determine the distillate and bottoms flow rates ( D and B )

F = D + B ----- ( 1 )

Given data :

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

B) Determine the minimum reflux ratio Rmin

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )

q - line ; y = ( 9 / 9-1 ) x - xf/9-1

therefore ; x = 0.45

Finally Rmin

= (( 0.97 / (Rm + 1 )) = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

8 0

2 years ago