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Why Wasn't (0,-3) Included in the Initial Graph of a Traveling Sinusoidal Wave?
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The discussion centers on the absence of the point (0, -3) in the initial graph of a traveling sinusoidal wave. Participants question why this point was not included and why a y versus time graph was not utilized. Clarifications are sought regarding the significance of units in the context of the problem. The conversation highlights the importance of understanding the wave equation and its parameters. Ultimately, the initial omission of (0, -3) raises questions about the completeness of the graph representation.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
Here we know that if block B is going to move up or just be at the verge of moving up
##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ##
Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
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