This would be an illustration of Newton's first law of motion. Law of inertia, an object moving at constant velocity will keep moving at constant velocity until a force is acted upon it. Inertia is the tendency of an object to keep moving in a particular direction resisting to change, unless a force acts upon the object.
The _quamtum mechanical_ model of the atom states that an electron's exact location within an atom can not be determined, but its probable location can be estimated within a three-dimensional region called an atomic orbital and that an electron's properties within an orbital can only be described by a set of mathematical values called a quantum number.
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
brainly.com/question/14669575
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Answer:
a.) 1567.2 m/s
b.) 149.4 m/s
Explanation:
Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.
The x-component of the third part can be calculated by assuming that it moves in a positive x axis.
The third mass = 26 - ( 7.8 + 8.8)
The third mass = 26 - 16.6
The third mass = 9.4kg
since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion
26 x 350 = -8.8 x 640 + 9.4V
9100 = -5632 + 9.4V
9.4V = 9100 + 5632
9.4V = 14732
V = 14732/9.4
V = 1567.2 m/s
(b) y-component of the velocity of the third part will be
7.8 x 180 = 9.4 V
1404 = 9.4V
V = 1404/9.4
V = 149.4 m/s