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1

The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...


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The most straightforward answer is no, you cannot consider a component to be a vector. A vector is something which has an associated direction, but a component has no direction. It's just a number. For example, if $\vec{F}$ is a vector, the component of $\vec{F}$ in the $\hat{x}$ direction is $\vec{F}\cdot\hat{x}$, and hopefully you know that the value of a ...


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If you only take the component along 45 degrees, you are neglecting the component perpendicular to it. This force counteracts the first component in the x direction. This then results in no net force in the x direction, as expected.


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In general the vector equation $$\mathbf{A}+\mathbf{B}+\mathbf{C}=0 $$ means $$A_x+B_x+C_x=0,\quad A_y+B_y+C_y=0 $$ or in terms of your angles and magnitudes of the vectors $$A \cos\alpha+B \cos\beta=0,\quad A\sin\alpha+B\sin\beta-C=0 $$ since $C$ is along a negative $y$. You know $\alpha,\beta, C$ so you can solve for $A$ and $B$.


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Draw figures of the vector in a paper and then use resolution of vectors using the angles you have got. And you will get the solution, by making sum of components of vectors in x and y- direction zero. Or better: use Lami's theorem.


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What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+ $$ The force on the negatively charged particle is opposite the direction of the field from the positively ...


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The electric field lines show the direction of the electric force acting on a unit positive charge at a particular point in space. So, therefore, the force acting on a negative charge, as is in your question, will act in the opposite direction shown by the electric field lines. I believe that the fact that field lines are defined in terms of a unit positive ...


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It doesn't imply that directly, you still have gauge freedom. I think what they mean is that if you set V1=V2 at one point on the boundary, it will hold true for all the boundary because the tangential component is identical.


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Short answer, you can get the result in electrostatics, for simple (path) connected surfaces, if the electric field isn't singular on the surface. How? For simple surfaces were can adjust the whole potential in one surface by a uniform amount and it doesn't change the electric field in that region. So pick a point A on the interface surface and why not ...


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Square any quaternion, not necessarily one involving space and time: $$(a, b, c, d)^2 = (a^2 - b^2 - c^2 -d^2, 2ab, 2ac, 2ad)$$ The first term in invariant under a Lorentz transformation. The next three terms were omitted by the OP. Squaring a quaternion generates another quaternion. Under a Lorentz transformation, the "space-times-time" terms will change. ...


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I don't know the context, but I assume there is some sort of typo in the "correct" equation. I think it should read $$(\rho\nabla_\mathbf{v}\mathbf{v})_\phi=\rho\frac{u}{r}\frac{\mathrm{d}}{\mathrm{d}{r}}(rv_\phi)$$ To prove this, we use the formula found here for spherical coordinates. It gives ...


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To solve for T at at an angle x, you would need to resolve for the collinear mg component (because they're equal but opposite). So, T = mgcosx. Tcosx isn't equal to mg.


4

The equation you quote is an approximation that is only valid if the horizontal force on the string remains the same. In practice that is not the case - and your concern is valid. The increase in length $\Delta \ell =\ell(\frac{1}{\cos\theta}-1)$; how much additional force that generates depends on the unstretched length of the string (or equivalently on ...


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Firstly, tension force itself depends on the elastic/rigid property of the string. So, the properties are implicit in the mathematical representation of the tension. Let a small part of the string be acted by tension $T_0$ from both sides of the string & hence it remains in static equilibrium as in the first picture. Now, when that part is stretched, ...


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I think in this example, you have to find the tension "in terms of". To do so, you have to do it using the static equilibrium equation which gives you the solution that you wrote. So, here it works the other way around. You first find the tension in terms of $T_0$ and THEN you find the new length of the wire. Why do you have to do it this way? Well, you ...


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Speed has no direction while velocity does. For example, if I say that I'm running at 10 mph, I have given you my speed. If I say that I'm running 10 mph north, then I have given you my velocity. Acceleration is the rate of change in velocity. Imagine this: I am in my car and you look at me before I even press the gas pedal. You close your eyes then open ...


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Vectors do have dimensions. Specifically, the dimension of a vector is (and always must be) the same as the dimension of its components. This also means that al the components of a vector must have the same dimension. In your example, the position vector $\vec r$ does indeed have units of length. The vector $\hat r$ is defined as $\hat r = \vec r / |\vec ...


2

Yes it does have the dimension $[L]$. You can, however, be more specific and assign a different length dimension to each component. So that the $x$ component has length dimension of $[L_x]$ and analogous for $y$ and $z$. These are called direction dimensions, and can be helpful in dimensional analysis.


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You are missing the bit under-braced here: $$ \dot{m}=\rho\underbrace{\mathbf A\cdot\mathbf v} $$ This is a dot product, which takes two vectors and returns a scalar: $$ c=\mathbf a\cdot\mathbf b=a_xb_y+a_yb_y+a_zb_z $$ So the mass flow rate is indeed a scalar value.


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You are right, the vector decomposition doesn't care about the medium. You decompose your vector into unit length basis vectors in a particular coordinate system, e.g in the Cartesian system, $$\mathbf{v} = c_1\mathbf{i} + c_2\mathbf{j} + c_3\mathbf{k}$$ This decomposition was done with no reference to any medium or its transformations. In this way you ...



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