Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
Answer:
y
Explanation:
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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:
Answer:
The micturition reflex can be voluntarily controlled by the relaxation of the external urethral sphincter.
Answer:
a) 0.018 kg
b) 262 kPa
Explanation:
The volume of the concentric cylinders would be:
V = π/4 * h * (D^2 - d^2)
V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3
The state equation of gases:
p * V = m * R * T
Rearranging:
m = (p * V) / (R * T)
R is 287 J/(kg * K) for air
25 C = 298 K
m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg
After pumping more air the volume remains about the same, but temperature and pressure change.
30 C = 303 K
m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg
The mass that was added is
m1 - m0 = 0.057 - 0.039 = 0.018 kg
If that air is cooled to 0 C
0 C is 273 K
p = m * R * T / V
p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa