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7

He is going normal (perpendicular) to the river current so the time it takes him to cross it doesn't depend on the river current, he will just be further down the river.

6

If you're seeing web sites disagreeing about something very basic like this, why not just look it up in a reliable source like a textbook? The relation is $v=\omega\times r$. You can verify this using the right-hand rule.

6

It would be misleading to write $$\vec P=(r, \theta, \phi)$$ or $$\vec P=\begin{pmatrix}r \\ \theta \\ \phi \end{pmatrix}$$ because the curvilinear coordinates $r,\theta,\phi$ don't behave like vector components with the usual simple rules for vector addition and multiplication. And it would easily lead to nonsensical conclusions (as you already noticed). ...

5

The Equation, $$\nabla^2 \mathbf{A}- \frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} = - \frac{4 \pi \mathbf{J}}{c}$$ , is actually three seperate equations in three dimensions. In Cartesian coordinates this Equation expands to, $$\nabla^2 A_x- \frac{1}{c^2} \frac{\partial^2 A_x}{\partial t^2} = - \frac{4 \pi J_x}{c}, \\ \nabla^2 A_y- \frac{... 4 For me this is easiest to see by introducing some explicit coordinates. Work on a cartesian reference frame whose origin is at the center of the loop, with the z axis along the loop axis and the x-z plane going through the point A. Thus, you have four points of interest: \vec A = r(\sin(\theta),0,\cos(\theta)) the origin, \vec 0 \vec P = a (\... 3 There are two triangles in this picture. The one starting at the origin of the circular plane connecting to point A and consisting of the two dashed lines. The second triangle starts once again at the origin and consists of line r to point A, then from point A along line R to the edge of the disk and inward along line a to the center. It's just ... 2 to solve this problem, first choose coordinate system and draw a "good" picture from hear you can calculate what ever you need.$$\vec{a}=a \begin{bmatrix} \cos(\phi') \\ \sin(\phi') \\ 0 \\ \end{bmatrix}\vec{r}=r \begin{bmatrix} \sin(\theta) \\ 0 \\ \cos(\theta) \\ \end{bmatrix}$$and \vec{R}=\vec{r}-\vec{a} to calculate the angle \... 2 Try to prove the more general relation in Figure-02. It will be very useful, as for the composition of two rotations around not collinear axes in Figure-04. 2 1) Why do we use Taylor Expansion, what does it do? That is not a Taylor's series expansion. That is just simple multiplication into a quadratic, i.e. (x+a)(x+b)=x^2 + ax + bx + ab, but with the "ab" part subtracted out. Try applying that to your \left[ r(t) + \Delta t \frac{dr}{dt}\right]\left[ \hat {\vec r} (t) + \Delta t \frac{d\hat {\vec r}}{dt}\... 2 You want to resolve the force of gravity F=mg into two components, F_x perpendicular to the string and F_y parallel to the string. \theta is the angle that the string makes with the vertical. So when \theta=0 you want to have F_x=0 and F_y=mg. And when \theta=90^o you want to have F_x=mg and F_y=0. And at all values of \theta you want ... 2 Why are the components the way they are instead of being the other way around? Because if the components (sin and cos) were reversed it would not make physical sense. Think about the tension in the pendulum string and the vertical acceleration of the bob. Based on the first diagram, at any given instant the tension in the string is mg cos θ and the ... 2 What you've found is the difference between affine spaces and vector spaces. Affine spaces are just a bunch of points with no origin (though your coordinates may have a bunch of zeros, it doesn't mean anything physically, unlike the null vector in a vector space, with is the addition identity). In affine spaces, vectors come in as the difference between ... 1 It means that your field exists in 3D space (ie it has an x, a y, and a z component). To get the force, you need to know the derivative of a field that points along x, the derivative that points along y, and the derivative that points along z. The generalization of that concept is that it is the 3D vector force is conservativ if it can be written as a ... 1 I'm getting the same result, which is not correct since this is a left-handed system. What am I missing here? You are not missing anything. It is correct, you should be getting the same result. You have changed both the handed ness of the coordinate system and the roles of the vectors. Those two changes cancel out so that in the end you get the same result. ... 1 In a left-handed system (which is the one on the right), the relation that connects your basis vectors \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 (that signify \textbf{i}, \textbf{j} and \textbf{k} respectively) is:$$\textbf{e}_i \times \textbf{e}_j = \sum_{k=1}^3\epsilon_{ijk} \textbf{e}_k where $\epsilon_{ijk}$ is the \$\textbf{Levi - Civita Symbol}...

1

I'm not sure what you mean by one dimensional, because one dimensional is a literal point, but the closest we can get to one dimensional to explain this is a line. Let's say there is a two dimensional space, the xy plane, but you can only see along the y axis, and your position is (0, 0) facing the North. The light beam is travelling parallel to the x axis ...

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