When the diameter is 10 m the volume is <span>523.6 m^3</span>
8x-2= -5+7x
2x= 2x
So you start with one side of the problem and solve it by writing it below. Then do the other side. In the end, they should equal each other
Answer:
13.695 m
Step-by-step explanation:
The assumption made here is that the boat/water interface is essentially frictionless, so that the center of mass of the system remains in the same place as the occupant of the boat moves around.
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We can find the sum of the moments of boat and child about the pier end:
(46 kg)(7.6 m) + (80 kg)((7.6 +9.6/2) m) = 1341.6 kg·m
After the child moves, the center of mass of boat and child is presumed to remain in the same place. If x is the new distance from the pier to the child, the sum of moments is now ...
46x +80(x-4.8)* = 1341.6
126x -384 = 1341.6
x = (1341.6 +384)/126 = 13 73/105 ≈ 13.695 . . . meters
The child is about 13.695 meters from the pier when she reaches the far end of the boat.
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* The center of mass of the boat alone is half its length closer to the pier than is the child, so is located at x-4.8 meters.
