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Surface tension is plausible : it implies that, at the level of the hole, the pressure in water will be less than 1 atm. Thus the pressure isolines will look like this: Quantitatively, curvature $c$ will be of the order $1/(1$mm$)$ because the hole looks about 1 mm size in your video, which with pure water would lead to a pressure difference drop of the ...

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I believe that the answer you gave ("Now I found this out...") is incomplete. It seems probable that you need to find the individual forces from two trestles, not just the sum, which you have correctly calculated. You can find the size of one of the trestle forces by taking the other trestle as the center of rotation and finding setting the sum of all the ...

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For part d: distance between CG of plank and point C = 1.5 -1 = .5 m To net momentum = 700*1-1000*.5 = 200 Nm clockwise So he will topple as you said.

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The total force at the centre of the see-saw must equal the total downward force (due to the masses) of the people on the see-saw. i.e. 500N + 400N + 50N = 950N I can see why this is slightly confusing, it is nothing to do with moments, just equal and opposite forces. Since the forces are balanced, you can now ignore the fact that it is a see-saw, and ...

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There's no such thing as minimum force for equilibrium. There is no range. Just a single value. For equilibrium, there should be no net external forces on the particle, so $\vec{F}_{net} = \vec{P} + \vec{T} + m\vec{g} = 0$. This would mean, in the horizontal direction: $P \cos \alpha = T \sin \theta$, or $P = T \sin \theta / \cos \alpha$ In the vertical ...

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In this problem you get two equations. One for forces in equilibrium along the horizontal direction and the other for the vertical direction. With the help of horizontal equation $T \sin \theta = P \cos \alpha$ You can insert $T$ from above equation in the vertical force equation thus eliminating $T$. Then your answer gives $P$ as a function of $\alpha$ ...

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What you have written down is just the force balance. For any body to be in equilibrium, the net moment acting on it must also be zero. So use the moment balance relation also: Moment about $A$ should be zero for equilibrium, $$|\vec{P}|\times \cos (\theta-\alpha)=mg\times a$$ Now solve the above equation along with the force balance equation you have to ...

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A charged particle will always travel along a field line - negative particles in one direction and positive particles in the other. So to have a point of stable equilibrium you would need converging field lines a bit like this: So the charged particle, like the red dot, will be attracted to the point where the lines meet then held there. The trouble is ...

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Charged particles can be trapped for long periods of time in pureley electrostatic ion traps like the one described in this publication. The principle of operation of these electrostatic traps is similar to optical 'traps' in cavity ring down spectroscopy. Two electrostatic lenses are placed opposite each other and if I remember correctly it is organized so ...

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It depends. The time evolution stays the same: $V(t)=V_0 e^{-5}e^{-t/\tau}$, the "non stabilized" part of the total voltage is just smaller. Often this part is considered constant for practical purposes.

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After $5 \tau$, the voltage across the capacitor is about $.7$% of what it was originally. There are definitely situations where this $.7$% could be significant, so when the problem says you can consider this voltage to be zero, it probably means that the accuracy of any tool you would use to measure the system is low enough that it wouldn't be able to ...

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The shape is caused by normal-field instability. This is the condition where these small 10nm droplets of ferrofluids are described by Maxwells equations where the divergence of the B field is zero and the curl of H is zero. The imposed magnetic field leads to a stress condition mismatch at the interface between the internal of the droplet and the outside. ...

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Yes, and I think it is even better. This might not be a full proof, but I think it gives a good idea. I will consider $n$ point sources. We know that the potential is harmonic, so there is no maximum and no minimum in the space we are interested in. All critical points, if they exist are saddle points. We are looking for those saddle points, or for points ...

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