The number 100 and that’s the only one I can think of
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.
Hey there! The sum of (3x - 6) and (5x + 8) is 8x + 2.
just a quick clarification, tis usually -4.9 and that's a rounded number to reflect earth's gravity on an object in motion, but -5 is close enough :)
![\bf ~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20meters%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-4.9t%5E2%2Bv_ot%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf h(x)=-5(\stackrel{\mathbb{F~O~I~L}}{x^2-8x+16})+180\implies h(x)=-5x^2+40x-80+180 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x)=-5x^2+\stackrel{\stackrel{v_o}{\downarrow }}{40} x+\stackrel{\stackrel{h_o}{\downarrow }}{\boxed{100}}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20h%28x%29%3D-5%28%5Cstackrel%7B%5Cmathbb%7BF~O~I~L%7D%7D%7Bx%5E2-8x%2B16%7D%29%2B180%5Cimplies%20h%28x%29%3D-5x%5E2%2B40x-80%2B180%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%3D-5x%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B40%7D%20x%2B%5Cstackrel%7B%5Cstackrel%7Bh_o%7D%7B%5Cdownarrow%20%7D%7D%7B%5Cboxed%7B100%7D%7D~%5Chfill)
Answer:
a redlecrion across the x axis results nvm in a negative x value: -x
