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stiks02 [169]
4 years ago
11

A liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about

the microscopic cause of this force. Student A says that the force exerted by individual molecules as they bounce off the wall is always perpendicular to the wall. Student B says that the molecules may strike the wall at angles that are not perpendicular. Which student is correct and why?A. Student A . If the molecules hit at nonperpendicular angles, there would be a component of force parallel to the wall exerted on them. The net result would be macroscopic motion of the fluid.B. Student A . The molecules in a liquid are in random motion at the microscopic scale, so they move in every direction. Near the wall the molecules collide with each other in such a way that the components of their motion not parallel to the wall cancel.C. Student B . The molecules in a liquid are in random motion at the microscopic scale. For every atom that hits at an angle to one side of perpendicular, there is likely to be another atom hitting at the same speed at the same angle on the other side of perpendicular. On average the nonperpendicular components cancel.D. Student B . The walls of the container cannot move, so near the walls the parallel component of the motion of the molecules is cancelled by the net force the wall exerts on the liquid.
Physics
1 answer:
Svetllana [295]4 years ago
5 0

Answer:

C

Explanation:

Given that a liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about the microscopic cause of this force. Student A says that the force exerted by individual molecules as they bounce off the wall is always perpendicular to the wall. Student B says that the molecules may strike the wall at angles that are not perpendicular. Which student is correct and why?

The correct answer is option C

Student B . The molecules in a liquid are in random motion at the microscopic scale. For every atom that hits at an angle to one side of perpendicular, there is likely to be another atom hitting at the same speed at the same angle on the other side of perpendicular. On average the nonperpendicular components cancel

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Select the correct answer.
IgorC [24]

Answer:

D. Newton's Third Law of Motion

Explanation:

Newton's law of gravity is definitely not applicable to your hands. So we can cross this bad boy out

Newton's First Law is F=MA (force equals mass times acceleration). This is basically the root of most physics but it isn't the reason for your hand being red after hitting a wall.

Newton's Second law deals with velocities and forces, so even though you are apply a force your are not changing the velocity of the wall much.

Newton's Third Law basically says that for whatever force you apply to an object, that object will apply an equal and opposite force back to you. This is why your hand gets red. When you slap the wall with all your strength, the wall hits your hand back with the same amount of force. The 2nd law can also be seen when you're trying to push a desk and it won't budge. You are pushing on it, but the desk is pushing back. (there are multiple other factors applicable like friction but we physicists like to ignore them :) )

I hope this helps!

7 0
3 years ago
Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.
romanna [79]

Answer:

(a): \rm -5.627\times 10^3\ N.

(b):  \rm 7.626\times 10^2\ N.

Explanation:

<u>Given:</u>

  • Charge on one sphere, \rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.
  • Charge on second sphere, \rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.
  • Separation between the spheres, \rm r=3.59\ cm = 3.59\times 10^{-2}\ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}

where,

k is called the Coulomb's constant, whose value is \rm 9\times 10^9\ Nm^2/C^2.

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, \rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.

The new charges on both the spheres are equal and given by

\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.

The magnitude of the force that each sphere now experiences is given by

\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.

7 0
3 years ago
Parallel conducting tracks, separated by 1.50 cm, run north and south. There is a uniform magnetic field of 1.20 T pointing upwa
Y_Kistochka [10]
I WISH I CAN HELP IM SO
SO
5 0
3 years ago
A wire as a length of 1.50m, diameter 0.60mm and resistance of 2ohms. calculate the resistance R of a wire of the same materials
Otrada [13]
  1. R=ρ×L/A

ρ=R×A/L

A=pie(r²)=pie(0.6²)=pie(0.36)=1.13

ρ=2ohm×1.13mm²/1500mm

ρ=0.0015ohm.mm

2. we were told that the wires were made of the same substance so the resistivity(ρ)is the same for both wires so:

R=ρ×L/A

R=0.0015ohm.mm×500mm/0.09mm²

R=8.3'ohm

so our resistance for the second wire is :

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>

3 0
3 years ago
What sort of element can be found in Group 13 of the periodic table?
frutty [35]

Answer:

Boron group

Explanation:

I googled

Good night

7 0
3 years ago
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