Newton's third law states "for every action, there is an equal and opposite reaction."
What this is pretty much saying is that for every action, there is a consequence. One force connects and triggers another.
The energy is 3.06 electronvolts, E = 3.06eV
1eV = 1.6 * 10^-19 J
3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J
To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency


At the same time we have that



Therefore there is not have same units and both are not consistent and the correct answer is B.
If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas
4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:
