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3 years ago

Why is diffraction used in soil profiles?

1 answer:

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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Light with wavelength in air ( lambdaair ) is incident on a oil slick ( noil = 1. 25) floating on the ocean ( nwater = 1. 33). w

crimeas [40]

<u>26mm</u> is the thinnest thickness of oil that will brightly reflect the light.

What is wavelength ?

The distance over which a periodic wave's shape repeats is known as the wavelength in physics. It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

To learn more about wavelength visit:

brainly.com/question/16051869

#SPJ4

4 0

2 years ago

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m

olasank [31]

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

7 0

2 years ago

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560

Alex73 [517]

Answer:

= 5.1 W

Explanation:

time (t) = 30 ms = 0.03 s

mass (m) = 560 g = 0.56 kg

initial velocity (U) = 0 m/s

final velocity (V) = 0.74 m/s

power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v

m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}

m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}

= 5.1 W

7 0

3 years ago

iris [78.8K]

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

$v^{2} = u^{2} +2as$

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

a) Given information from the question,

u = $\frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s$ (Convert km/h to m/s first)

a = $2m/s^{2}$

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

$v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)\\$

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

v = u + at

22.761 = 19.444 + 2t

2t = 22.761 - 19.444

t = $\frac{22.761-19.444}{2}$

t = 1.659s

7 0

2 years ago