They actually DO have velocity limits. There are legal restrictions on both speed and direction.
-- Speeds are limited according to the black numbers on white signs that you see on sign-posts everywhere.
-- Directions are limited by the layout of the pavement and curbs on all the highways, avenues, roads, boulevards and streets, as well as the countless signs that say "One Way", "No Left Turn", "Keep Right", "Keep Left", etc. Violate one of these, and you get nailed as sure as if you had exceeded a posted speed limit.
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
Answer:

Acceleration, in m/s, of such a rock fragment = 
Explanation:
According to Newton's Third Equation of motion

Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:

So Equation will become:

Acceleration, in m/s, of such a rock fragment = 
B. Interdependence among specie