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OlgaM077 [116]
4 years ago
9

Do physical changes rearrange the atoms?

Physics
2 answers:
WARRIOR [948]4 years ago
4 0

Answer:

Explanation:

A physical change, such as a state change or dissolving, does not create a new substance, but a chemical change does. ... In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

CAN I HAVE BRAINIEST

Pani-rosa [81]4 years ago
4 0

Answer:

No, physical changes don't rearrange the atoms

Explanation:

Only chemical changes rearrange the atoms, that is, in chemical reactions, atoms of the molecules of the reactants are rearranged to produce the molecules of the products.

On the other hand, in physical changes atoms remains the same. For example, during boiling of water, the liquid is transformed into steam, but water molecules (H2O) are the same for both liquid water and steam.

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by creating body heat so she can stay warm

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3 years ago
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Energy slowly leaks outward through the diffusion of photons that repeatedly bounce off ions and electrons
stepan [7]

Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.

<h3>What is radiative diffusion?</h3>

A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.

As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

Hence,radiative diffusion is correct answer.

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3 years ago
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

3 0
4 years ago
Consider a point charge qqq in three-dimensional space. Symmetry requires the electric field to point directly away from the cha
bonufazy [111]

Answer:

 E = \frac{1}{4\pi  \epsilon_o } \  r^2

Explanation:

For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere

           Ф = ∫ E. dA = \frac{x_{int} }{\epsilon_o}

the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product

           Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}

           E ∫ dA = \frac{x_{int} }{\epsilon_o}

           E A = \frac{x_{int} }{\epsilon_o}

the area of ​​a sphere is

            A = 4π r²

the charge inside the sphere is q = + q

           

we substitute

           E 4π r² = \frac{x }{\epsilon_o}

           E = \frac{1}{4\pi  \epsilon_o } \  r^2

4 0
3 years ago
You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using e
Mashutka [201]

You are working for a manufacturing company, which is mathematically given as

  • m=3\sqrt{2}
  • m=\frac{15\sqrt{5}}{16}
  • x=0.747a
  • m/n=\frac{(x^2+a^2)3/2}{x^3}

<h3>What is the value of m that will place the movable bead in equilibrium at x-a a ....?</h3>

a)

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3

m=3\sqrt{2}

b)

By force equilibrium

K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}

Therefore

n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\

m=\frac{15\sqrt{5}}{16}

c)

K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

d)

By force equilibrium

K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}

m/n=\frac{(x^2+a^2)3/2}{x^3}

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3 0
2 years ago
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