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melisa1 [442]
4 years ago
13

Why does it matter if a material is amorphous or crystalline? How does the arrangement of atoms affect the properties of materia

ls?
Physics
1 answer:
Rashid [163]4 years ago
8 0

Answer:

<em>It matters because crystalline and amorphous materials have different properties. The arrange affects the melting point (defined in crystals and a larger range in amorphous) and shape (geometrical in crystals, no geometrical in amorphous). </em>

Explanation:

The particles that compose a solid material are held in place by strong tractive forces between them when we analyze solids we consider the position of the atoms (molecules or ions) rather than their motion (which is important in liquids and gases). This positioning can be arranged in two general ways:

  • Crystalline solids have internal structures that in turn lead to distinctive flat surfaces or face, these faces intersect at angles that are characteristic of the substance, crystals tend to have sharp, well defined and high melting points because of the same distance from the same number and type of neighbors. They generally have geometric shapes, some examples are diamonds, metals, salts.
  • Amorphous solids produce irregular or curved surfaces when broken and they have poorly defined patterns when exposed to x rays because of their irregular array. In contrast with crystal solids, amorphous solids soften over a wide temperature range due to the different amounts of thermal energy needed to overcome different interactions. Some examples of these solids are gels, plastics, and some polymers.

I hope you find this information useful and interesting! Good luck!

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
Explain why the energy produced by burning wood in a campfire is energy from the sun
Makovka662 [10]
Because trees get thier nutrients and energy from the sun. They absorb it through the leaves and it makes them grow strong. then someone cuts down the tree and makes the wood for your fire. So, the fire burns the wood that was made from the tree that got nutrients and grew from the sun.
3 0
3 years ago
When you accelerate in a car and are pushed back into your seat, you are experiencing _______, which is registered by your _____
il63 [147K]

Answer:

E) linear motion; otolith organs 100%

Explanation:

7 0
3 years ago
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.
Aleks [24]

By definition, centripetal acceleration is given by:

a = \frac{v ^ 2}{r}

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

a =\frac{3.3^2}{0.13}\\a = 83.76923077

Rounding the result we have:

a = 83.8 \frac{m}{s^2}

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

a = 83.8 \frac{m}{s^2}

3 0
4 years ago
Read 2 more answers
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
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