I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
Answer:
a. metallic bond
b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
c. due to the presence of free electrons in its outer energy levels
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

Rearranging the equation, we can write

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:
