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Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.17 m on a side. What is the average force the molecule exerts on one of the walls of the container?

Respuesta :

Answer: The last part of the question has some details missing which is ; (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) molecule v=482 m/s molecule momentum=2.56 x 10^(-23)

Explanation:

  • The momentum of the molecule is 2.56 x 10^(-23) .
  • Particle hits the wall and bounces.
  • Momentum is reversed. Change in momentum = impulse
  • This is Force x time.
  • Momentum change happens at a wall after each trip.

  • time required = distance /speed

  • = 0.17 X 2/(482 m/s)

  • Average force = impulse / time

  • = 2 x 482 x 2.56 x 10^(-23) / (0.17 x 2)

  • = 7.76 x 10^20N, is the average force the molecule exerts on one of the walls of the container.
adefioyelaoye

The average force the molecule exerts on one of the walls of the container is 7.76 x 10⁻²⁰N .

The additional information is that the molecule's velocity is perpendicular

to the two sides that it strikes with the velocity given as 482 m/s and the

momentum given as 2.56 x 10⁻²³kgm/s

The momentum of the molecule is 2.56 x 10⁻²³kgm/s.

Momentum  = impulse as a result of the momentum being reversed due to the particle hitting the wall and bouncing back.

Impulse = Force x time

Time = distance /speed

= 0.17 X 2/(482 m/s)

Average force = Impulse / time

                        = 2 x 482 x 2.56 x 10⁻²³ / (0.17 x 2)

                        = 7.76 x 10⁻²⁰N

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