Madison and Oxford are the same distance from the equator and they are both near the ocean. Use the information on the map to an

Response:

Refer to the explanation

Clarification:

Code:

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void displayNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

displayNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

displayNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

displayNumPattern(num1, num2);

}

}

See attached example output

That's correct -.-.-.-.-.-.-.-.-.-.-.-.-.- Easy.

Answer:

Refer to the attached document

1512 ft

Explanation:

Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:

Acceleration - Time

0 < t < 6: Velocity change = area beneath the a–t graph

V_6 - 0 =  (6 s)(4 ft/s²) = 24 ft/s

6 < t < t2: The velocity rises from 24 to 48 ft/s,

Velocity change = area beneath the a–t graph

48 - 24 = (t2 - 6) * 6

t2 = 10 s

t2 < t < 34: The velocity remains constant, meaning acceleration is zero.

34 < t < 40: Velocity change = area beneath the a–t graph

0 - 42 = 6*a4

a4 = - 8 ft / s²

A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.

Velocity - Time

Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.

Position change = area beneath the v−t graph

0 < t < 6:  x6 - 0 = 0.5*6*24 = 72 ft

6 < t < 10:    x10 - x6 = 0.5*4*(24 + 48) = 144 ft

10 < t < 34: x34 - x10 = 48*24 = 1152 ft

34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft

Summing these position changes yields the distance from A to B:

d = x40 - 0 = 1512 ft

The value obtained is 0.60. From the given conditions, we take k = 1.4 for air and r = 16. We know that for calculations involving diesel engine efficiency, we arrive at 0.60.

Answer:

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in () = 0.12

Product mass flow rate () = 100 kg/min

Product total solids out () = 0.1

Product temperature in () = 50°C

Product temperature out () = 120°C

Steam pressure = 232.1 kPa at () = 125°C

Product specific heat in () = 3.936 kJ/(kg°C)

The mass equation is:

Also

Therefore:

The energy balance equation is:

 kJ/(kg°C)

By substituting values into the energy equation:

From the properties of saturated steam at 232.1 kPa,

= 524.99 kJ/kg

= 2713.5 kJ/kg

% quality =

Any steam quality above 63.5% will result in higher total solids in the heated product.