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timurjin [86]
3 years ago
10

Explain what happens to the particles in a substance during a physical change.

Physics
2 answers:
Bingel [31]3 years ago
5 0
During Physical Change there would be a re-arrangements of atoms or molecules, changes of the arrangement may be change in the distance between atoms or molecules, change in the crystal form, .....etc 

for example: water when heated it undergoes a Physical Change and turn into vapor, this means the heat cause the distance between water molecules to increase, so it transferred from the liquid form to the gas form. 

NOTE that in Physical Change there is no change in the chemical structure and the material retains all its chemical properties, and no new compounds are produced. 

again, A physical change is any change not involving a change in the substance's chemical identity. Matter undergoes chemical change when the composition of the substances changes: one or more substances combine or break up (as in a relationship) to form new substances.Physical changes occur when objects undergo a change that does not change their chemical nature. A physical change involves a change in physical properties. Physical properties can be observed without changing the type of matter. Examples of physical properties include: texture, shape, size, color, odor, volume, mass, weight, and density. 

BUT in Chemical Change ( or Chemical Reaction ) there would be change in the chemical nature of the material undergoing a Chemical Change with the production of new compounds.
Simora [160]3 years ago
5 0

Explanation:

Physical change is a change in which there is no change in chemical composition of the substance.

For example, when we freeze liquid water then it changes into solid, that is, only physical state of water has changed and no change in chemical composition.

Therefore, in a physical change there is change only in the arrangement, energy and movement of particles.

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

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