A 0.02kg bullet is traveling with a speed of 300m/s when it hits a wall. Assuming it comes to rest after traveling 0.1m with constant deceleration into the wall. 1. Calculate the work done by the wall on the bullet as the bullet comes to a stop.
2. Calculate the deceleration of the bullet as it comes to a stop.

1. W = 900 J

2. a = -450000 m/s²

Explanation:

2.

First, we calculate the deceleration of the bullet by using 3rd equation of motion:

where,

a = deceleration = ?

s = distance traveled = 0.1 m

Vf = Final Speed = 0 m/s (it eventually stops)

Vi = Initial Speed = 300 m/s

Therefore,

a = -450000 m/s² (negative sign shows deceleration)

1.

Now, we calculate the force using Newton's Second Law:

Negative sign shows decelerating force.

For work done:

W = 900 J

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

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