Which of the following placards allows a driver to park in a handicapped parking space.

A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo

Natalija [7]

Answer:

For $20^{\circ}$ - 5.556 lb/s

For $15^{\circ}$ - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = $20^{\circ}$

Temperature rise, T' = $15^{\circ}$

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, $C_{p} = 0.24 btu/lb ^{\circ}F$

Now, for a rise of $20^{\circ}$ in temeprature:

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s$

Now, for $15^{\circ}$ :

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s$

4 0

2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

Driving Distraction Brainstorming Session

Leto [7]

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give <u>BRAINLIEST ANSWER └[T‸T]┘</u>

5 0

2 years ago

In this exercise, you will write a Point structure that represents a space in two-dimensional space. This Point should have both

Afina-wow [57]

Answer:

Check the explanation

Explanation:

Points to consider:

We need to take the input from the user

We need to find the manhatan distance and euclidian using the formula

(x1, y1) and (x2, y2) are the two points

Manhattan:

$|x_1 - x_2| + |y_1 - y_2|$

Euclidian Distance:

$\sqrt{(x1 - yl)^2 + (x2 - y2)^2)}$

Code

#include<stdio.h>

#include<math.h>

struct Point{

int x, y;

};

int manhattan(Point A, Point B){

return abs(A.x - B.x) + abs(A.y- B.y);

}

float euclidean(Point A, Point B){

return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));

}

int main(){

struct Point A, B;

printf("Enter x and Y for first point: ");

int x, y;

scanf("%d%d", &x, &y);

A.x = x;

A.y = y;

printf("Enter x and Y for second point: ");

scanf("%d%d", &x, &y);

B.x = x;

B.y = y;

printf("Manhattan Distance: %d\n", manhattan(A, B));

printf("Euclidian Distance: %f\n", euclidean(A, B));

}

Sample output

8 0

3 years ago

What happens to the duty cycle for a GMAW Gun when 75Ar/25COzgas

skad [1K]

So what happens is the host will not kill the y no se que hacer para no one can see it in

6 0

2 years ago