<span>(20 cm)/(5 sec) = (0.20 meters)/(5 seconds)
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I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part AFor point A we have:

In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
Part BAt the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
Part CThe child will stay in place at point A when centrifugal force and force of gravity are in balance:
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)
Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ
Hope I helped!! xx
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As the distance from a charged particle, "q", increases, the electric potential decreases.
<h3>
Electric potential between particles</h3>
The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.
V = Kq/r
where;
- K is Coulomb's constant
- q is the magnitude of the charge
- r is the distance between the charges
Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.
Learn more about electric potential here: brainly.com/question/14306881
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Fortunately, 'force' is a vector. So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.
When we talk in vectors, one newton forward is the negative of
one newton backward. Hold that thought, while I slog through
the complete solution of the problem.
(100 N forward) plus (50 N backward)
= (100 N forward) minus (50 N forward)
= 50 N forward .
That's it.
Is there any part of the solution that's not clear ?