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larisa [96]
2 years ago
12

Explain under what circumstances earthworm channels might increase downward saturated water flow, but not have much effect on th

e leaching of soluble chemicals applied to the soil.
Engineering
1 answer:
mezya [45]2 years ago
6 0

Answer: soul formation is a critical on soil fertility,Earth worm improves soul,through humus formation, earthworm contribute to soil structure. Porosity and structural stability and reduces run-off. Earthworm modifies nutrient cycling and soil organic matter.it induce substance that improves plant health and growth.

Earthwormth changes the soil structure,nutrient and water.cultural practices like tillage and organic matter regulate the size and activities of the earthworm community.

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Q4. What happen when a steady potential is connected across the end points of the [3] conductor? Illustrate with an example.
den301095 [7]

Answer: Flow of current and the resistance to the flow of the current.

Explanation:

When a steady potential is connected across the end points of the conductor, there will be a of current through the conductor and the internal resistance in the conductor will tend to resist the flow of the current

A good example is a simple circuit comprises of a battery, wires and a touch bulb.

When the battery is connected to the two ends of the torch bulb, current flow through and the bulb is lit up.

4 0
2 years ago
Which statement concerning symbols used on plans is true?
Zielflug [23.3K]

Answer:

The symbols are to be noted on the title sheet or other introductory sheet of the plans.

Explanation:

in able to be understood for example a map key is always on a map so the reader can use it efficently without confusion

8 0
2 years ago
4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and
hodyreva [135]
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
6 0
2 years ago
At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut
professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

thus,

on substituting the respective values, we get

Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0
2 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
2 years ago
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