11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75
m away from the pivot point on the opposite side. The second child's mass is _____ kg.
1 answer:
Answer:
<h2>40 kg</h2>Explanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
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Answer:
b. varies inversely with the square of the distance from the center of Earth.
Explanation:
Comparing the Newton's law of universal gravitation and second law of motion;
from Newton's second law of motion,
F = ma ............. 1
from New ton's law of universal gravitation,
F = ........... 2
Equating 1 and 2, we have;
mg =
g =
Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.