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5

It's easiest to see if you start with the definition of the inverse fourier transform $$f(\mathbf x) = \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x}$$ and take a laplacian of both sides $$\nabla^2 f(\mathbf x) = \nabla^2 \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x} = \int d\mathbf k\, \hat f(\mathbf k) ... 3 Hint: Recall that$$\tag{A} q\cdot (q+2p) ~=~(p^{\prime}-p)\cdot (p^{\prime}+p) ~=~p^{\prime 2}-p^2~=~m^2-m^2~=~0 .$$So$$\tag{B} \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$2 i think \nabla^{2}\phi(r)=\rho(r) is the same as$$ \frac{d^2}{dr^2} \phi(r) = \rho(r) $$. the convention i am used to in electrical engineering is to use large case letters for the Fourier Transform and use \omega for angular frequency instead of "k". in that notation, Fourier transforming both sides is$$ (i \omega)^2 \Phi(\omega) = ...

1

The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ...

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I have used the Darcy Formula together with the following formulas for a quick numeric solution (only a few iterations needed) $$h_f = \frac{\Delta P}{\rho g}$$ $$f = {\rm Darcy}(Re)$$ $$h_f = f\,\frac{L}{D}\,\left( \frac{v^2}{2 g} \right)$$ Solve above for $v$ $$Re = \frac{\rho D\,v}{\mu}$$ Go to step 2 until $f$ converges to a value.

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First assume that $h$ doesn't change very much because you have a large body of water (we can relax this condition later). Let's also assume that the hole is small compared to the depth ($d \ll h$) - we'll relax this too. For this case, the answer is straightforward, you'd use Bernoulli's equations and simply set the static pressure ($\rho g h$) equal to ...

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When I did the derivation I got this for delta: $\Delta = -y^2q^2 +yq^2 -2zpyq +(1-z)^2m^2$ In the text it says this: $\Delta = -xyq^2+(1-z)^2m^2$ Is there some information missing? I recommend looking up "how to use Feynman Parameters" on google to get more detail, but basically it looks like the derivation is given on this webpage, in example 2. It ...

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Hint in a parallel circuit, the voltage across each resistor is the same. What is the voltage across the $4 \Omega$ resistor?

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