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2 years ago

What refers to a soils ability to transmit water

Physics

1 answer:

Keith_Richards [23]2 years ago

6 0

Answer:

Permeability is also called as hydraulic conductivity, which refers to the ability of a soil to conduct water. ... Composition is of four basic categories: water, air, organic matter and minerals.

Explanation:

idk, i knew a bit and looked it up to make sure i was correct.

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A company has been in business for 40 years. It has tens of thousands of customer addresses in three different computer systems

alexira [117]

The person in charge of information management is the database administrator.

<h3>What is a database?</h3>

The term database refers to the arranging of the data that concerns clients in relevant entry point such as computers and file cabinets. This is the way by which companies are bale to keep track of their customers.

The person that is most helpful in the management of the tens of thousands of customer addresses in three different computer systems and written on paper in filing cabinets is the database administrator.

Learn more about database:brainly.com/question/6447559

#SPJ1

7 0

1 year ago

A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri

zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

Here

n = Number of node

T = Tension

$\mu$ = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 281.2Hz}$

Similarly plug in 2 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 562.4Hz}$

Similarly plug in 3 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)$

$\mathbf{f= 843.7Hz}$

4 0

2 years ago

Differences Between light year and astronomical unit in two points .

Marina CMI [18]

Answer:

A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc.

7 0

3 years ago

Read 2 more answers

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

3 years ago

If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o

tangare [24]

The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
</span>
Solution:
18 / x = 12 / 4
12x = 72
x = 6mm

7 0

2 years ago