008 10.0 points A uniform horizontal rod of mass 1.5 kg and
length 2.1 m is free to pivot about one end
as shown. The moment of inertia of the rod
about an axis perpendicular to the rod and
through the center of mass is given by I
ml2
12
F
570
-pivot
1.5 kg
2.1 m
If a 7.7 N force at an angle of 57° to the hor-
izontal acts on the rod as shown, what is the
magnitude of the resulting angular accelera-
tion about the pivot point? The acceleration
of gravity is 9.8 m/s.
Answer in units of rad/s2.

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

uranium fissions/breaks up giving out energy and neutrons

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.