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30

The explosion certainly is hemispherical, see, for instance, this explosion caused by the Trinity bomb: The gas cloud that you posted, and what many would consider is synonymous to the nuclear weapons, comes after the explosion. Nuclear bombs are actually usually ignited above ground for "maximum destruction." Since the nuclear reaction is immensely hot ...


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

Formally, the incompressibility of a fluid is defined by the compressibility, $$ \beta=\frac1\rho\,\frac{\partial\rho}{\partial p} $$ where $\rho$ is the mass density and $p$ the gas pressure. This means that, the compressibility is the measure of how much the density (volume) changes when a pressure is applied. For water at STP, this works out to be on ...


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

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 ...


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

First of all, anything which raises water is by definition a machine, and further unless you wish to defy the laws of physics, will require external energy input. Now, if you want a simple machine, I'd recommend the Archimedes Screw, http://en.wikipedia.org/wiki/Archimedes'_screw . It's a simple thing to build and operate.


5

Matter is made up from point like fundamental particles, like electrons and quarks, that have zero volume. This puts us in the interesting position where the true volume of all matter is zero, and the only reason that everything doesn't instantly collapse into a point of zero volume is that the pointlike fundamental particles maintain a finite distance from ...


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 ...


4

You are mistaken. Actually, you can melt ice by applying pressure. This is why ice is so slippery, when you step on a frozen lake, you are melting the very first layer of water, and thus creating a very good instant lubricant for you to slide on. It is a common knowledge false fact, see comments. Ok, granted, at very high pressures water does become solid. ...


4

Exotic designs of atomic devices have directionality because of interaction with the material surrounding the core. This was documented by George Dyson in his book PROJECT ORION, when he revealed the 'pusher capsule' bombs would have had a filler which turned into plasma and 'pushed' on the plate on the bottom of the spacecraft. However, as seen in the ...


4

This is a very good question! Drag due to viscous effects can be broken down into 2 components: $$D = D_f + D_p$$ where $D$ is the total drag due to viscous effects, $D_f$ is the drag due to skin friction, and $D_p$ is the drag due to separation (pressure drag). The equation above demonstrates one of the classic compromises of aerodynamics. As you ...


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

The truck will have in its wake some unknown mass of air almost moving with a speed $v$ comparable to the truck's speed $\bf V$. The pressure behind the truck will be lower than the pressure at the sidewalk because air pressure follows the Bernoulli equation, $$ P_\mathbf{P} = P_\text{road} + \frac{1}{2}\rho v^2, $$ where $\rho \approx 1~$kg/m$^3$ is the ...


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 ...


3

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 ...


3

I found the original text at http://babel.hathitrust.org/cgi/pt?id=uc1.b2619178 (page 5 onwards). The relevant pages and key translation (each time I place the page before the translation - and I will repeat snapshots of equations in the translation where appropriate): 1) Contrary to the work by H Blasius, in this work we treat the problem of a body of ...


3

The power of the water stream can be used to pump it up continuously, if it is reliably flowing as on the picture. Have a look at results in Google from "watermill irrigation" keywords, you get e.g. http://www.panoramio.com/photo/59986317 Maybe then you have to worry about how much work it is compared to your water needs, and would it be worth doing if ...


3

What is meant is that $$\frac{d\rho(q(t), p(t), t)}{dt} = 0$$ when $q,p$ are solutions to Hamilton's equations. While it is notationally convenient and space-saving to not write everything out in this detail, it is as you noted confusing. This particular confusion actually has a name -- it's the first fundamental confusion of calculus. (There's a second ...


3

The Basic Idea Physically the total derivative tells you how a quantity changes when it is subjected to a space and time dependent velocity field. In physics we usually call it the material derivative. An Intuitive Example Suppose $\rho(\mathbb{x},t)$ measures the temperature of a fluid, according to a thermometer immersed at the point $\mathbb{x}$ and ...


3

It seems to me that you are making confusion between a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb ...


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

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

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 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 ...



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