The average speed of a nitrogen molecule in air is about 686 m/s, and its mass is 4.85×10-26 kg. If it takes 3.78×10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in the opposite direction, what is the magnitude of the average acceleration of the molecule during this time interval?

Acceleration is the rate of change of velocity.

Initial velocity, v0 = -686 m/s
Final velocity, v1 = 686 m/s
Time, Δt = 3.78*10-13 s
average acceleration = (v1-v0)/Δt

To find the magnitude of the average acceleration of the nitrogen molecule, we can use the formula for average acceleration:

acceleration = (change in velocity) / (time interval)

First, let's calculate the change in velocity. When the nitrogen molecule hits the wall and rebounds, its velocity changes from positive 686 m/s to negative 686 m/s. So the change in velocity is:

change in velocity = (-686 m/s) - (686 m/s) = -1372 m/s

Next, we can substitute the values into the formula for average acceleration:

acceleration = (-1372 m/s) / (3.78×10^-13 s)

Now, let's calculate the magnitude of the average acceleration:

magnitude of acceleration = |acceleration| = |(-1372 m/s) / (3.78×10^-13 s)|

To calculate the magnitude, we take the absolute value of the acceleration. So:

magnitude of acceleration = 1372 m/s / (3.78×10^-13 s)

Simplifying this further, we have:

magnitude of acceleration ≈ 3.625 × 10^15 m/s²

Therefore, the magnitude of the average acceleration of the nitrogen molecule is approximately 3.625 × 10^15 m/s².