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brilliants [131]
2 years ago
7

Why is diffraction used in soil profiles?

Physics
1 answer:
nekit [7.7K]2 years ago
7 0

Answer:X-ray diffraction (XRD) is the technique most heavily relied on in soil mineralogical analysis. X-ray diffraction is a technique that provides detailed information about the atomic structure of crystalline substances. It is a powerful tool in the identification of minerals in rocks and soils.

Explanation:

XRD is used to identify the minerals composing clay-rich, hydrothermally altered rocks that occur on several Cascade volcanoes. Such rocks are believed to play an important role in the generation of large landslides and mudflows. XRD is used to analyze saline minerals, including borates.

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A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars?
zhannawk [14.2K]
Field strength =force/unit mass
= 14.8N/ 4kg = 3.7N/kg
8 0
3 years ago
Read 2 more answers
HELP PLEASE!!!
Leya [2.2K]

For the first one 320

second

1200W

Data

R = 12 Ω ∆V = 120V I =? P =?

Solution:

According to Ohm’s law,

∆V = I R

I = ∆V / R  

= 120 / 12  

= 10 A

Power P = I ∆V  

= 10 x 120  

= 1200 W

Third

∆V = 120 V P = 60 W I =? R =?

Use the formula, P = I ∆V

I = P / ∆V = 60 / 120 = 0.5 A

∆V = I R

R = ∆V / I = 120 / 0.5 = 240 Ω

3 0
3 years ago
You are studying a population of flowering plants for several years. When you present your research findings you make the statem
Anit [1.1K]

Answer:

Option A

Explanation:

The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."

Hence, the survivor ship must be on the Y axis and the resources on the X axis.

Here the resources include the number of seeds produced.

hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)

Hence, option A is correct

6 0
3 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

3 0
3 years ago
How can an object overcome static friction?
larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0
3 years ago
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