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Work must also be done on the system (at the input end) to push the
entering fluid into the tube and by the system (at the output end) to
push forward the fluid that is located ahead of the emerging fluid.
This pressure can be estimated, or at least understood, by means of the Bernoulli Principle (so for an ideal, inviscid fluid).
$$p_1+\frac12 \rho v_1^2+\...
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I suspect it has to do with the edge of the cup you were using to transfer the liquid and the adhesive forces between the liquid and the cup. When the cup is tilted at an angle lesser than, say, $\theta_\text{max}$, the relevant component of the weight of the liquid moving out from the cup is not sufficient to outweigh the adhesive forces between the cup and ...
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There are long papers published just on this topic! The rate, and height, of a wave's transition from smooth sinusoidal form to breaker depends on its speed, wavelength, and crucially on the rate of change of water depth as the wave approaches land (or reef, or whatever). Waves only break because the deeper portions of the wave are slowed down due to ...
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In ground effect, the moving vehicle is going so fast and is so close to the ground that it is riding on a bubble of air that is trapped between the vehicle and the ground. The ground is still supporting its weight, through the action of the compressed air in between the vehicle and the ground.
It takes less work to move a vehicle through the air when it is ...
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The answer is ultimately due to viscosity. Viscosity serves as an energy "sink" – it constantly dissipates energy out of the system. Viscosity is due to the friction between two layers of fluids moving at different speeds and hence "dragging" each other. With ideal fluids (ones with no viscosity), kinetic energy is conserved, assuming no ...
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The question is a bit unclear, but here is my response. I will delete if it's not on-topic.
Assume no losses due to friction. Then note that every small volume of fluid that we wish to push into the pipe must be given a certain flow velocity, after which that fluid element will coast its way down the length of the pipe with that velocity, and leave the end ...
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You will find tables of this sort of data within the materials science literature under key words wettability, spreading, capillary rise, contact angle, and surface energy.
You will also find surface energy test fluid sets consisting of ranked series of specially-prepared chemical solutions with distinct and well-controlled surface energies. placing droplets ...
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When you write down $Q = vA$, it's implicit that the velocity profile is uniform over the cross-section A (and purely perpendicular to it).
In general,
$$Q = \int \mathbf{v} \cdot \mathrm{d}\mathbf{A}$$
This no longer implies that $v \propto \frac{1}{R^2}$.
If we assume $\mathbf{v}$ has purely radial dependence and is aligned with $\mathrm{d}\mathbf{A}$ as ...
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With the Reynolds Number:
$$\text{Re}=\frac{\rho v L}{\mu}$$
the characteristic length $L$ that can be used here is (e.g.):
$$L=\frac{l+h}{2}$$
assuming that $L \gg w$
Why is the formula for the characteristic length $L$ what you have
written it to be and why does $L$ need to be much bigger than $w$?
In many cases, the choice of a 'characteristic length' ...
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The short answer is that the flow coming into a stagnation point gets pushed to the side, rather than piling up against the fluid ahead of it.
The simplest way to see this mathematically is via the Cauchy-Riemann equations,which were derived from complex-analysis and apply for 2-D functions satisfying Laplace's equation (including inviscid flows in fluid ...
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Ok let me explain... First of all I regret for not being able to upload an image of siphon... So please follow my words. I hope I can clear your doubts...
So consider a working siphon. We know that the flow is very much steady in the siphon tube. Ok. Now I am asking you to consider the longer right portion of this u-tube. As we can see the diameter of the ...
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