# Tag Info

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The conventional method of nuclear waste disposal is just fine. Just bury it deep into some geologically inactive rock. Consider this: If you dig a deep hole into ground in some geologically inactive area, you will find rocks and minerals that have been there for millions of years. If you put your nuclear waste there, there is no reason why it wouldn't ...

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In the conical container, the downforce is certainly the same. It can be found via calculus, taking into account the following: Since the inside surface of the container is touching the fluid, the calculation is a surface integral. The force on a surface due to a pressure is exerted perpendicular to the surface. The downward component will have to ...

2

As stated above, the mass of the whole system (sugar + water) doesn't change. In addition, with "ideal" mixing, the total volume of the water plus the total volume of the sugar equals the total volume of the mixture. However, this is not a sure bet, and there are many cases of a volume of one material mixed with a different volume of water, and the total ...

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The mass doesn't change at all, it will be just the sum of the water mass and the mass Added, what happens is the change of density because the mixture, in general the molecules get closer to each other ( through the intermolecular forces) and, this way, the volume become lower to the same mass quantity, what increase the density by the equation $$\rho = ... 3 To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here$$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$Recall g = - \vec{\nabla} \Psi, and \vec{u} \cdot ... 0 Air pressure is equal to the weight of the air column per unit area above you. However, air pressure is not like a stack of blocks above you. Like water, air is a fluid, and any local pressure differences will cause rapid movement of the air to equalize the pressure. That means that air pressure isn't directional; the air will be pushed into any cavities ... 0 The obvious interpretation of black hole density is the mass of the black hole divided by the volume inside the event horizon. We need to be a bit cautious about taking this too literally because the volume inside the horizon is not coordinate independant so different observers will measure different densities. However we can easily calculate the density ... 1 Suppose you have two masses M1(=M) and M2(=M) with volumes V1 and V2, respectively. Then the total density is the total mass divided by the total volume. So \rho_{mix}=2M/(V1+V2). V1=M/\rho_1 and V2=M/\rho_2 so \rho_{mix}=2M/(M/\rho_1+M/\rho_2), which after canceling the M's and simplifying the expression is equal to what you wrote above. ... 2 1) The relation \frac{dF}{dS}=d\left(\frac{F}{S}\right) is certainly incorrect as @Floris has mentioned in the comment. As the simplest counter-example, consider a linear function, F(S) = \alpha \, S, with \alpha \neq 0 as a proportionality constant. Then, one could easily see that$$ \frac{dF}{dS}= \alpha \neq 0 = d\left(\frac{F}{S}\right) = ...

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Mine is broken too and water is not heavy/dense enough. The bulbs sinks to the bottom. So Ethanol will be too light too. The original liquid smells "oily". I created my own liquid with enough and high(!) density by solving more and more sugar in water. And you need a lot of sugar before it starts working more or less. Problem now is that the fluid is quit ...

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The reference values come from the physical system you are modeling, i.e. oil in a pipeline will have the density, viscosity of oil, diameter of the pipe and the cross-sectional velocity as reference values. Often the exact physical situation doesn't matter as long as you keep your dimensionless numbers constant (geometric and dynamic similarity) See also ...

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Because conventionally we assume constant temperature, and length and density are also assumed to be constants for a given resistor. Of course, this is not true. In some circuit designs I have to pay very careful attention to resistance changes with temperature, and indeed this is sometimes used to provide temperature measurements in the form of RTD ...

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The continuity equation is $\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{\rho \mathbf{v}} = 0$, Now you can substitute directly for $\nabla \cdot \mathbf{\rho \mathbf{v}}$ with the expression for divergence in spherical co-ordinates \$ \nabla \cdot \mathbf{A} = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over ...

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