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zubka84 [21]
3 years ago
7

What i the relationship of space between particles in soil and the flow of water through soil ?

Physics
1 answer:
Luden [163]3 years ago
7 0
Space between particles in the soil is "porosity". When these spaces are interconnected (which they are not always, or not fully) then there is "permeability".

Darcy's Law describes the relation between flow, and porosity and permeability. However viscosity of the fluid plays a part as well. It is a complicated subject.
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mass and weight are similar, but not the same thing. In which of the following examples would the objects weight change, but mas
OverLord2011 [107]

When an astronaut travels from the earth to the moon, her weight changes, but her mass remains constant.  <em>(C ).</em>

7 0
3 years ago
calculate the period of a wave whose frequency is 5 Hertz and whose wavelength is one centimeter give your answer in a decimal f
olga2289 [7]
The period of the wave is the reciprocal of its frequency.

       1 / (5 per second)  =  0.2 second .

The wavelength is irrelevant to the period.  But since you
gave it to us, we can also calculate the speed of the wave.

Wave speed = (frequency) x (wavelength)

                   = (5 per second) x (1cm)  =  5 cm per second
4 0
3 years ago
17-<br> Find the magnitude of vector product \BxĀ| for A=– 23 +3Â and B = 2î – 3+ Å<br> vectors.
aksik [14]

Answer:

alam ko sagot pero mataas

8 0
3 years ago
A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity
nevsk [136]
We can use the kinematic equation
(v_f)^2 = (v_i)^2 + 2*a*d
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
8 0
2 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
2 years ago
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