The average speed of a Nitrogen molecule in air is about 858 m/s, and its mass is about 4.68 × 10−26 kg.

If it takes 2.6 × 10−13 s for a nitrogen
molecule to hit a wall and rebound with
the same speed but in the opposite direc-
tion, what is the magnitude of the average
acceleration of the molecule during this time interval?

I have a problem with the statement. The molecule only rebounds from the wall if it were moving perpendicular to the wall in the first place. In a gas, this would be most unusual, most are moving in all directions, and "glance" off the wall. Neglecting that reality...

change of velocity=2*original speed

and the direction of the change in vleocity is outward from the wall.

a= change speed/time

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

Average acceleration = (Change in velocity) / (Time interval)

First, let's find the change in velocity:

The molecule hits the wall and rebounds with the same speed but in the opposite direction. Since the magnitude of the velocity remains the same, the change in velocity would be equal to twice the magnitude of the initial velocity.

Change in velocity = 2 × (858 m/s) = 1716 m/s

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

Average acceleration = (1716 m/s) / (2.6 × 10^(-13) s)

To simplify the calculation, it's useful to express the time interval in scientific notation:

Average acceleration = (1716 m/s) / (2.6 × 10^(-13) s)
= 1716 m/s × (1 / 2.6 × 10^(-13) s)
= 1716 m/s × (1 / 2.6) × (10^13)

Now, we can calculate the value:

Average acceleration ≈ 660000000000000 m/s²

Therefore, the magnitude of the average acceleration of the nitrogen molecule during this time interval is approximately 660,000,000,000,000 m/s².