Answer:
a) q = 7671 W
T0 = 47.6°C
b) ΔP = 202.3 N/m²
P = 58.2 W
c) hDarray = 2 times hD of an isolated element.
Explanation:
see the image for the solution.
estimate the area for a landfill for 12000 p producing waste for 10 y. assume that the national average is
1 answer:
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Answer:
Here is the JAVA program:
import java.util.Scanner; // to take input from user
public class VectorElementOperations {
public static void main(String[] args) {
final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i;
//two arrays origList[] and offsetAmount[] get assigned their values
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
String product=""; // variable for storing the product results
for(i = 0; i <= origList.length - 1; i++){
/* iterates from 0 to the end of origList */
/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */
product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }
System.out.println(product); }}
Output:
80 180 80 400
Explanation:
If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:
import java.util.Scanner;
public class VectorElementOperations {
public static void main(String[] args) {
final int NUM_VALS = 4;
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i;
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
for(i = 0; i <= origList.length - 1; i++){
origList[i] *= offsetAmount[i];
System.out.println(origList[i]);}
}}
Output:
80
180
80
400
The program is shown with the output as a screenshot along with the example's input.
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
Answer:
Total bandwidth: 8 kHz
Explanation:
Data provided:
Transmitter frequency: 3.9 MHz
Modulation up to: 4 kHz
Solution:
For the upper side frequencies:
Upper side frequencies = 3.9 × + 4 × 10³
Upper side frequencies = 3.904 MHz
For the lower side frequencies:
Lower side frequencies = 3.9 × - 4 × 10³
Lower side frequencies = 3.896 MHz
Consequently, the total bandwidth is computed as:
Total bandwidth = upper side frequencies - lower side frequencies
Total bandwidth = 8 kHz
Answer:
Volume change percentage is 2.60%
Water level increase is 4.138 mm
Explanation:
Provided data
Water volume V = 500 L
Initial temperature T1 = 20°C
Final temperature T2 = 80°C
Diameter of the vat = 2 m
Objective
We aim to determine percentage change in volume and the rise in water level.
Solution
We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.
It can similarly relate to density changes.
Thus,
E = ................1
And ............2
Here, ρ denotes density. The density at 20°C = 998 kg/m³.
The density at 80°C = 972 kg/m³.
Plugging in these values into equation 2 gives
dV = 0.0130 m³
Therefore, the percentage change in volume will be
dV % = × 100
dV % = × 100
dV % = 2.60 %
Hence, the percentage change in volume is 2.60%
Initial volume v1 = ................3
Final volume v2 = ................4
From equations 3 and 4, subtract v1 from v2.
v2 - v1 =
dV =
Substituting all values yields
0.0130 =
Thus, dl = 0.004138 m.
Consequently, the water level rises by 4.138 mm.