A2.9 gram bullet is shot into a tree stump. it enters at a velocity of 304 m/sec and comes to rest after having penetrated the stump in a straight line. it takes 0.00043 sec to come to a stop after striking the stump. what was the force on the bullet during impact?

First convert grams to kilograms (so that the answer will be N) 1000g=1kg so 2.9g=.0029kg.

To solve for force we use the equation F=ma, so we need to find acceleration (a) .Next use the equation a=Δv/Δt. Δt is given as being .00043 seconds. Δv is the difference in velocities - final minus initial (this is important to discern the direction of the force vector); 0-304=-304. Now plugin and solve for a. The result is -706976.7442 m/s², or -710000 m/s² if you are using significant figures. 

Now use the values for mass and acceleration to solve for force. F=ma=-2050.232558N or F=- 2100N (if you are using significant figures).

Question was not very clear if there are specific answer choices tell me and i'll try to you more. hope this though2 laps is 1 min. we know that for each min she does two times the amount of laps. for 12 laps we can do 12/2 which does give us 12 laps in 6 mins not 16 mins.