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When we decompose vector into it's rectangular components, we get two vectors along x-axis and y-axis. So the vector in the second diagram may have its x-component as a vector also whose magnitude is $F\cos\theta$ and direction is same as $\vec{S}$ i.e. along positive x-axis. And hence $\vec{F}$ is the resultant vector of $\vec{Fx}$ and $\vec{Fy}$. When ...


1

If $\theta$ is obtuse then $\cos\theta$ is negative and thus, $W=\vec{F} \cdot \vec{s}=Fs \cos\theta$ is negative.


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Yes. Work $W$ is found as: $$W=\vec{F} \cdot \vec{s}=Fs \cos(\theta)$$ The angle will from the dot product take care of the sign for you. A perpendicular force e.g. is not doing any work, since $\cos(90^\mathrm{o})=0$. When $\theta < 90^\mathrm{o}$, $\cos(\theta)>0$, and when $\theta > 90^\mathrm{o}$ you change sign, $\cos(\theta)<0$. ...


2

The answer is no, we cannot rule out forces not directed along the line joining particles with current theory. As you point out, magnetic monopoles are a counterexample, and, as CuriousOne points out, magnetic monopoles both fit into a classical framework, are consistent with the Standard Model and are actively, experimentally sought. Some historical ...


1

$F1 - F2$ is the same as turning $F2$ around 'head to tail' and you get $-\Delta F$ or a vector, which is the same magnitude, but the opposite direction to $\Delta F$. To get $\Delta F$ you need to reverse $F1$ as in the diagram I have modified from your diagram below


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That's because you want $F_1 + \Delta F = F_2$ by additivity of vectors (for a more rigorous approach, see the formalization of affine spaces). Hence, $\Delta F = F_2 - F_1$ PS: I couldn't comment because of my low reputation, so I made an answer for so little


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I'm going to assume you have some familiarity with linear algebra, as the math becomes much less tedious than trying to do three-dimensional trigonometry with the $x$, $y$, $z$-coordinates directly. You are looking for a function describing the line of the horizon. Since it is a circle and thus a one-dimensional object, I'm going to call it $\vec{h}(t)$, ...


1

I have helped some school students through a course which was taught with Matter and Interactions which focussed on using physics principles to program computer simulations. Because of the programming, my students were comfortable with vectors from the first week. The computer made the vectors visual and as simple to manipulate as variables containing ...


3

It seems to me that it's not 'vectors' or 'vector algebra' which these students aren't grasping. Its the connection between a given 'physical phenomenon' and a corresponding 'mathematical representation'. I suspect this has something to do with conceptualising the physics rather than the mathematics. To put is simply: Physics $\neq$ Mathematics When ...


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I can't comment on the up-till-now, but here's something to try for the "now-and-henceforth". I suspect that many student difficulties have in the past been left unhelped by the fact that lecture courses were "linear" (no pun meant here): there was a set coursework and a set way of thinking about concepts that the lecturer or teacher chose that students ...


0

When vectors are introduced, it should be empathized that a vector has 2 components, a magnitude and a direction. You can make a list of non-vectors and vectors (mass, speed velocity, acceleration, etc) which you can put on a quiz. Also, any equation that utilizes vectors is incorrect if the vector symbol is not included.


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Momentum is a vector quantity. For a particle with mass, the momentum equals mass times velocity, and velocity is a vector quantity while mass is a scalar quantity. A scalar multiplied by a vector is a vector. A moving body would be a particle with a mass. If the body moves through space, relative to an observer, it will have a velocity, momentum, and ...


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The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...


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Put the arrows after one another. Then draw a new arrow to the point they reach. The net force vector is the new arrow. The direction its' angle. The maginute is its' lenght (just as the magnetude of the two original forces were the lenghts of each).


0

I'll leave the problem for you to solve, but here's a hint. Remember that forces are vectors, and "net" means "sum of components". You can certainly use angles when summing vectors - just be careful with the signs so that you cancel what chould get canceled and add what should get added.


3

A general vector is written as $v = v^\mu \partial_\mu$. Its norm is defined as $$ g(v,v) = g_{\mu\nu}(\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(v^\mu\partial_\mu,v^\nu\partial_\nu) = g_{\mu\nu}v^\mu v^\nu (\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(\partial_\mu,\partial_\nu) = g_{\mu\nu}v^\mu v^\nu$$ where we have used linearity of the duals and ...


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The answer to the above question is the definition of work i.e. how we define it. When we apply force on a body, it is displaced with respect to its position. Work is said to be on the body. The definition of work says, "Work is the product of the component of force in the direction of displacement or vice versa". So the case I stated above has it's force ...


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You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...


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The vertical component of the force doesnt do work, so a force with an obtuse angle can be considered to be oppossed to the diplacement. In any case, whenever work is negative, the system is losing energy.


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Restricting ourselves to just vector spaces without any extra structure, the theorem is true. One way to see this is to note that any member $f$ of the dual space is uniquely defined by the value it returns acting on the basis $\{\psi_n\}$, say $f(\psi_n) = z_n$ for complex numbers $z_n$. Then $V^*$ is isomorphic to $\mathbb{C}^\mathbb{N}$, the set of ...


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There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...


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Acceleration is simply a rate of change of velocity. So the magnitude tells you, how quickly velocity changes.


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In physics, magnitude is the size of a phusical object, a property by which the object can be compared as larger or smaller than other objects of the same kind. More formally, an object's magnitude is an ordering (or ranking) of the class of objects to which it belongs.


3

When one says that "kinetic energy is conserved in an elastic collision" that means that the total kinetic energy of the system of particles involved in the collision doesn't change. It does not mean that the kinetic energy of each particle is unchanged. For a two particle system, the kinetic energy of each will change, but the sum won't. Also, your ...


1

Careful between $\hat{r}$ and $\vec{r}$. The trajectory $\vec{r}(t)$ is a function that maps $t$ to a position vector $\vec{r}$. But the tangent to the trajectory is not the same as $\vec{r}$. Thus, $\vec{v}\cdot \vec{r}$ tells you nothing about the tangent vector. To be tangent to something means to be going in the same direction at exactly one point. The ...


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This is no different than the linear motion case. The generator moves because torque is applied to it. In response it applies torque to the shaft turning it. The net torque on the system is zero but individual components feel (opposite) net torques. This is absolutely fundamental to understanding any Newtonian mechanics.


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the velocity is tangential to the curve and given by $$\vec{v} =\frac{d\vec{r}}{dt}$$ and the position vector $\vec{r}$ is given as $$\vec{r}=x~\hat{i}+y~\hat{j}$$ Given $y=f(x)$, $$\vec{r}=x~\hat{i}+f(x)~\hat{j}$$ So, $$\frac{d\vec{r}}{dt}=\frac{dx}{dt}~\hat{i}+\frac{d}{dt}f(x)~\hat{j}$$ Simplifying, ...


0

According to your example s = d|r|/dt = 1, and r(t = 0) = 0, s.t. (1) |r| = t Now, since x^2 = y, r^2 = y + y^2 , s.t. you get a simple equation of 2nd degree in y, y^2 + y - r^2 = 0 . Solve this equation, (2) y = [-1 +- sqrt(1 + 4r^2)/2. Substituting here (1), (3) y = [-1 +- sqrt(1 + 4t^2)]/2, and therefore, (4) x = sqrt(y) = sqrt{[-1 +- sqrt(1 ...


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You can derivate each component separately: $r'(t)=<x'(t),y'(t)>$ However your example is different, you do not give x(t) and y(t), but only the shape of curve and the constraint of constant speed. This contraint is $r(t)^2=x(t)^2+y(t)^2=s^2$ using that you should be able to get the solution after some algebraic work


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Two issues: 1) the equation was derived assuming that the initial speeds were zero, so both masses started at rest. To get a more general expresion yuo need to integrate again (I'll check later if this is easily doable) 2) you can apply this equation in 3d, the equatiosn assumes that the masses start at rest and follow a staight path until they collide. ...


0

Because: $$\vec p = \vec r_2 - \vec r_1 -r_3,$$ where $\vec r_3$ is the unlabelled side of the triangle that is parallel to $\vec H$. $$ \vec r_2 - \vec r_1 = \vec p + \vec r_3$$ $$ (\vec r_2 - \vec r_1) \times \vec H = \vec p \times \vec H + \vec r_3 \times \vec H$$ but $\vec H$ and $\vec r_3$ are parallel so $\vec r_3 \times \vec H = 0$, and $$ (\vec r_2 ...


2

The notation $\nabla_1$ refers to the gradient with respect to the first coordinate $\mathbf{r}_1$. I think the most transparent way to do the derivation is to switch to the notation $\partial/\partial\mathbf{r}_1$, then expand the derivative using the multivariable chain rule, and then switch back to the nabla notation: $$\begin{align}\nabla_1 &\equiv ...


-1

There are two types of vector.One is polar,the other is axial.Angular velocity is an axial vector.So,no displacement is needed along its direction.


2

The planet will follow an elliptical path and as an ellipse is a 2D figure it can only be traced to a plane and from two points infinite planes can pass, so you need to decide first which plane you want which is the missing information, because of which you are not able to do so. If finally you form a plane $\vec r.\vec N=p$ then the direction of the ...


0

In the context of e.g. a pseudo-orthogonal Lie group $$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$ of pseudo-orthogonal matrices $\Lambda$ for the metric $$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{1,\ldots,1}_{p~\text{times}},\underbrace{-1,\ldots -1}_{q~\text{times}}), \qquad n~=~p+q,$$ ...


3

The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more. The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical ...


0

Yes, he defined the vector as behaving that way (a vector rotation is equivalent to a change of basis), otherwise it would not be a vector. A tensor is a different kind of object, it has at least two indexes and behaves different that a vector under transformation of coordinates (as defined in your book). You might probably read more about linear algebra up ...


0

A function of $p_{\mu}$, $T^{\mu\nu...}p_{\mu}p_{\nu}...$is invariant under Lorentz transformation if and only if $T^{\mu\nu...}$ is a invariant tensor. However, there is only one invariant tensor for Lorentz group. That is $\delta^{\mu}_{\nu}$. Thus, each saclar function of $p_{\mu}$ must be a function of $p^2=\delta^{\mu}_{\nu}p_{\mu}p^{\nu}$. ...


2

What you want is that your function does not transform under Lorentz transformations that take $$ p^\mu \to {\Lambda^\mu}_\nu p^\nu.$$ To build invariants from one vector there is only the possibility to construct the invariant product with itself $$ p^2 \equiv p^\mu p_\mu \to p^\mu p_\mu.$$ There is one more thing though. The Lorentz group has two branches: ...


2

ACuriousMind's Answer pretty much summed up the reasons, which are essentially mathematical. If you want to grasp the "physical significance", then I suggest you should work through an example: think of two quantum systems, each with three base states: $\left.\left|1\right.\right>$, $\left.\left|2\right.\right>$ and $\left.\left|3\right.\right>$. ...


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The "Kronecker product", better known as the tensor product, is the natural notion of a product for spaces of states, when these are considered properly: A space of states is not a Hilbert space $\mathcal{H}$, but the projective Hilbert space $\mathbb{P}\mathcal{H}$ associated to it. This is the statement that quantum states are rays in a Hilbert space. ...



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