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10

It's not likely to effect the tsunami very much. On the open ocean (where you would want to use the bombs), the wave height is not very much, so the air blast from the bombs would blow over the wave. The bombs would also create their own waves which would pass through the tsunami wave. Near to shore, the tsunami wave would increase in height. The bombs ...


7

This is not altogether correct. In particular, it does not matter what the pressure of the room is (unless you reach extreme values), because actually what is holding the water column is not some "high" value of the pressure at the bottom of it, it is the fact that the pressure at the top of the water column (at the water-finger interface, or in the air ...


6

That is right, deeper the pressure is stronger. But the pressure is not just in one direction it is in every direction. So the velocity will decrease in most cases. But also you have to be aware of the density of the object. You could read this classical description of diving objects "Thrust" on wikipedia. This is a classical effect, in real cases the ...


5

This Washington State Department of Transportation page makes it clear that the choppiness is at the very least highly correlated with windstorms and high winds. This page is a good resource as well. The choppiness occurs on the upwind side of the bridge; thus, a south wind (which blows north) will make the south side choppy. The reason for this is that ...


5

Unless I have made a conceptual mistake (which is very possible), surface tension plays essentially no role in the damping of the impact of a fast-moving object with a liquid surface. To see this, a simple way to model it is to pretend that the water isn't there, but only its surface is, and see what happens when an object deforms this surface. Let there ...


5

As best I have been able to tell, vortex air intakes are mostly a scam designed to sell useless car modifications to people, as discussed on this HowStuffWorks article. In case of the inevitable future link rot, I'll paste the article below: The internal combustion engines in cars and trucks are essentially large air pumps: The action of the pistons ...


5

This term looks like Faraday's law that is used in Ideal magnetohydrodynamics (MHD). So yes, it is true. In order to see the mathematical advection, you'll need to apply some vector calculus: $$ \nabla\times\mathbf a\times\mathbf b = \mathbf a\left(\nabla\cdot\mathbf b\right) - \mathbf b\left(\nabla\cdot\mathbf a\right) + \left(\mathbf ...


4

I henceforth assume $c=1$. Consider a 3-volume $\Delta \Sigma_0$ at rest with the dust. For the rest observer (this is its definition) $u^\mu$ has only (unit) temporal component. The energy (i.e the mass) associated with that portion of system is $\Delta \Sigma_0 \rho_0$. The 4-momentum of that portion is therefore $\Delta p^\mu := \Delta \Sigma_0 \rho_0 ...


3

I think the simplest answer to this question would be that the stream of water has a number of forces acting on it (gravity, air drag) from many directions. Some torque is bound to be produced as the stream falls through the air. If you throw a ball or any small object from a height, it rotates, no matter how you drop it. Same logic applies here. As far as ...


3

r is the internal radius. It is clear if you look at the wikipedia page. Think of it like this: if the tube was 5 meters thick, would that affect the drop of water at all? The answer is no: the water droplet is affected only by the diameter of the tube that is in contact with -> i.e the internal radius


3

As far as I remember viscosity of the gases, unlike liquids, increases with increasing temperature. So in order to decrease viscosity, You would have to cool the air (which is already cold at some 10 km). When it comes to it's effect on planes, viscosity is responsible for creating pressure gradient between top and bottom side of an airfoil or wing. That ...


3

If You make a simplified force balance of a sinking box, You can identify two main forces: Force associated with box's weight $F_g$ acting downwards and buoyancy force $F_b$ acting upwards. The formulas are as follows: $F_g=mg$, $F_b=-\rho g V$, where $m$ is the mass of the box, $g$ is the gravitational acceleration, $\rho$ is the density of water, $V$ is ...


3

Let's assume a one litre $1000$ W electric kettle, filled with $0.5$ kilograms of water at $20^o$ C: It takes 4.2 joules to warm one gram of water one degree Celsius. So, to warm the $500$ grams of water $80$ degrees from $20$ to $100$ takes $168,000$ joules. The kettle will supply $1000$ joules per second, so it'll take $168$ seconds for the kettle to ...


2

Helicopter rotor is a rotating wing. It produces lift the same way aircraft wing does, but instead of relying on forward motion of the aircraft it has it's own motion. The lift generated depends on coefficient of lift, air density and forward speed. Formally $L = \frac{1}{2}\rho v^2\alpha C_L$ where $L$ is lift force, $\rho$ is air density, $v$ is forward ...


2

Hydrodynamic perturbations = change in pressure due to a flow velocity (particles don't return to equilibrium positions). Acoustic perturbations = change in pressure due to the fact the particles undergo an elastic restoring force (for a compressible fluid) which causes perturbations to travel at the speed of sound. Any change in the pressure/velocity ...


2

The state from $1\text{ atm}$ to $2\text{ atm}$ is normally called decompression or contraction. An equation you can use going from one state to the next is: $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$ Where $P$ is pressure, $V$ is volume and $T$ is temperature. Now if you want to calculate the force you have to know the surface area of what you are ...


2

I depends on the pressure ratio between the chamber and ambient . As long as the ratio is critical, the ambient air will flow through the orifice at the speed of sound (with respect to the state in the chamber). For undercritical pressure ratio the flow can basically be found by the Bernoulli Equation.


2

The motion of the fluid parcel only depends on the gradient of the pressure, not the pressure itself. Think about it this way: we know that pressure increases the deeper you go into water. But if I place a cylinder in the water, the flow will always look like (the colors represent pressure with red being high pressure and blue low pressure; the arrows are ...


2

We are asking how the rod moves through the viscous medium if we apply a force to it. Since the Reynolds number is so low, inertial forces must be small, and the externally applied force must be balanced by a viscous force. Also since we are in the low Reynold's number limit, the viscous force is linear in the velocity of the object. Thus there must be a ...


2

The Torricelli Formula you used is certainly a good way to start if the level of the water source is stationary. In addition there will be losses which depend on geometry of your flow restriction and on Reynolds Number. Loss coefficients $\zeta_{loss}$ for free jet discharge, valves, etc. can be found in literatute, for example here: ...


2

The relevant equation is the kinematics with linear drag. In this case, there is a resistant force that acts opposite gravity (i.e., upwards) and is linear to the velocity at which it travels: $$ \mathbf F_D=-b\mathbf v $$ where $b$ is some fluid- and object-dependent constant. Using Newton's 2nd law, $$ m\ddot{\mathbf x}=m\mathbf g - b\dot{\mathbf x} $$ If ...


2

A gravity-driven free surface flow on an incline is unstable at moderate Reynolds number, as was shown by Yih, 1963, http://deepblue.lib.umich.edu/handle/2027.42/69956 : surface waves form. These waves are, I believe, also unstable in the transverse direction, leading to the criss-cross pattern you mention. See Liu et al 1995, ...


2

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...


1

I recommend Turbulence: The Legacy of A. N. Kolmogorov by Uriel Frisch. It explains how turbulence is a top-down behavior, large scale turbulence in turn causes turbulence at smaller and smaller scales. This continues until a small enough scale is reached where the energy from the turbulence serves to create more heat on the molecular level. Because of this ...


1

Your method seems fine to me. Essentially you're measuring volume by filling it with a fluid, and this is a method commonly used for determining the densities of powders or other materials where it's hard to measure the volume. The method is known as pycnometry if you feel the urge to Google it. To improve the accuracy I would: weigh the tube fill it with ...


1

As long as we are far from compressible effects (low Mach) and the 0 pressure under which cavitation may appear, yes, pressure is defined up to a constant if boundary conditions are given in terms of velocity. If you provide boundary condition in terms of normal stress (e.g. at an inlet), then you provide a pressure value with this boundary condition. But ...


1

If you were to draw out the velocity field, you would see that it is going outward. In fact if you were really astute, you would notice that your velocity has the same form as the electric field from a wire. This should help you visualize the field. Because of the symmetry, it makes sense to work in cylindrical coordinates. The position along the axis will ...


1

Yes, what you have formulated is fine. The pressure acting on the water from the bottom of the straw will be equal to the weight of the water times the cross section area. So $101325\pi r^2$ is the force acting from below also. That is precisely why there is an equilibrium and the water is not falling.


1

When an object moves in a fluid, the fluid molecules just around it get entrained along with it due to friction. In turn, this layer of fluid entrains the next layer at a lower velocity, and so on. This is due to the fact that individual molecules switch layer and thus "diffuse" the momentum (a molecule coming from a fast layer to a slower transfers some ...


1

Not likely. The wind can not transfer enough energy directly into building a wave big enough to cancel a tsunami. The detention would need to be underwater and create a displacement which would build a large wave, however this would be almost impossible to control to the precision required to cancel in incoming wave whose energy is usually only determined ...



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