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Proper time is the reference time all observers can agree on. Makes for calculating things much much easier. Since tau is attached to the particle the space differentials are zero Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero. Furthermore ...


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4-vectors have invariant length as defined by $$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$ Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.


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Given a path $x(s)$ on a manifold, the velocity with respect to that path is defined taking derivatives with respect to the invariant parameter the path is described with, on the manifold (the arc length). In general relativity such parameter is the proper length $s$ (or proper time $\tau = \gamma s$).


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Hint: By the use of coordinate geometry find the distance between the $2kg$ and other masses. Now first of all use the expression for the gravitation force $F=Gm_1m_2/r^2$ between two particles. And by the use of principle of superposition tackle the particles pair by pair. By this I mean Lets, assume the force between $4kg$ and $2kg$ is $\vec{F_1}$ and ...


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It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...


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If the blade was spinning in a vacuum (and there was no friction and no motor), there would be no torque. The blade could spin by itself forever. Torque comes from spinning in air and pushing air around. Air is viscous. It resists the motion of the blades. It exerts a torque on the blades. The torque on all 4 blades is the torque on the quadcopter. ...


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The torque supplied by $F_m$ results in the torque due to $F_v$, so these torques are equal : $\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either $\vec A \times \vec B = (AB \sin\theta) \hat k$ where $A$, $B$ are magnitudes and $\theta$ is the angle between (b) or $(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ...


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The question asks for the change in the velocity vector, which should be done by vector subtraction, not addition. This would give you pi*sqrt2/30, as you have, but directed at an angle of 225 deg - toward the center of the circle, as it should be. (The first vector is +i, the second is -j, so the subtraction is -j-i, which gives the correct angle.) Your ...


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Your formula is wrong. The correct formula is: $$\bar {\vec v}=\large{\frac{\Delta \vec r}{\Delta t}}$$ $\bar {\vec v}$ is the average velocity.


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A unit vector $v$ is a vector, whose norm is unity: $||v||$. That's all. Any non-zero vector $w$ can define a unit vector $w/||w||$. A basis vector is one vector of a basis, and a basis has a clear definition (it's a linearly independent family of vectors which spans a given vector space). So both have nothing to do. Your confusion may come from the fact ...


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How can I calculate the position of the particle after applying this force? By using Newton's second law. If the mass of the particle is $m$ and it is constant; then we have: $$\vec F=F\cos \theta \;\vec i+F\sin \theta\;\vec j=m\vec a=m(a_x\;\vec i+a_y\;\vec j)$$ $$\Longrightarrow\;a_x=\frac{F\cos\theta}m\quad\textrm{and}\quad a_y=\frac{F\sin\theta}m$$ On ...


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In a complex number $z=a+i b$,the real part is $a$ and abs is $\sqrt {a^2+b^2}$,. I am not an expert of acoustics but in general, when pressure is given in complex quantity and you want to measure its magnitude then take absolute. If you are comparing two complex numbers then equate real with real and imaginary with imaginary.


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Difference between real and absolute value in general: Look at count_to_10 's answer. For acoustics and preasure measurement: Absolute pressure - pressure against perfect vacuum. Real pressure: Usually defined as the pressure against a reference-environment. Also called differential pressure. For example the pressure of the air inside a football against the ...


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Here's an animation which shows how to keep the triangle correct when working incline plane problems. Hope this helps... The red lines are parallel. Inclined Plane Problem


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I'm not sure if it helps you with your students, but maybe gives you some background: I guess the underlying reason for orthogonal basis vectors is that you are implicitly using a euclidean metric that will just have diagonal values. These would e.g. be $$g_\mathrm{\mu\nu, ~euclidean}=I=\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \...


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We don't always use orthogonal coordinate frames. For example working with three phase motors it's sometimes convenient to work with a three axis coordinate system in a plane. Convenience, simplicity set aside, the main reason we most often work with orthogonal reference frames is the concept of dimension. We can express an n-dimensional linear system as a ...


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For an open orientable surface there are two possible, equivalent normals: $\vec n$ and $-\vec n$. The usual convention is that you choose a direction in which the perimeter of the surface is traversed and define the positive direction as the direction given by the right hand rule, as shown in the following picture. This can also be done for non-simply ...


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By convention, for a flat lamina or a plane surface, the area vector is a vector whose magnitude is the area of the surface and whose direction points in a direction perpendicular to the surface. If you have a curved surface, then you have to consider elemental areas, i.e: small patches of area denoted by $dA$ whose direction is perpendicular to the small ...


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They are linked by the "law of the right hand": The preferred direction of dℓ⃗ dℓ→ along the loop is that from the palm to fingertips of your right hand when it surrounds the loop. Then, the associated preferred direction of dA⃗ dA→ is indicated by the thumb.


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All vectors, except $\:\mathbf{r}\:$, are infinitesimals. I wonder if the author (Irodov) makes use of this result anywhere in his textbook. EDIT The infinitesimal rotation of a vector $\:\mathbf{r}\:$ around the direction of a unit vector $\:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right)\:$ by an infinitesimal angle $\:\mathrm{d}\theta\:$ may be represented ...


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The converse is not a "theorem" in that you can't prove that it is true, even assuming Maxwell's equations as axioms. It is simply a "hunch" that $|\vec{S}|$ represents the power intensity and $U=\frac{1}{2}\,\epsilon\,|\vec{E}|^2+\frac{1}{2}\,\mu\,|\vec{H}|^2$ the energy density. What you can prove (and what you already understand) from Maxwell's equations ...


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Taking $\phi = \tan^{-1}\left( R_y / R_x \right)$ gives the angle between the vector and the $x$-axis, on the interval $[-90^\circ,90^\circ]$, with positive angles in the upper-right quadrant. Your favorite implementation of $\tan^{-1}$ might return angles on the interval $[0^\circ, 180^\circ)$ instead. In the problem, however, your $\theta$ is drawn to ...


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A vector is a quantity described by a magnitude and a direction. The magnitude is always +ve or zero. A -ve sign in front of a vector indicates the same magnitude but in the opposite direction. The - sign is part of the direction rather than the magnitude. Like all vectors, a "resultant vector" is neither +ve or -ve. It has a magnitude (which is >= 0) ...


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Here are a few examples of simple notation you could adopt. Rotations in $d$ dimensions can be represented using $d\times d$ matrices (call them $R$) such that $R^TR$ is the identity matrix. If your initial vector that describes a 'positive' direction is $v$, then you can express directions relative to $v$ as $R v$, where $R$ is one of these rotation ...


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I'm not aware of any such convention. You can always show the net force vector acting on a free body as long as it is clearly labeled as such, to avoid confusion with any other applied forces. I personally wouldn't include a net force vector unless there was a good reason to, like to illustrate some accompanying discussion.


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Actually, all of your derivation is correct. But, potential energy is the energy due to gravitational field with respect to point at infinity. By your derivation, u at infinity is zero. u at (r,0) is GMm/r . Thus, potential energy at (r,0) is U(r,0) = u(0,inf) - u(0, r) = 0 - GMm/r = -GMm/r


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Generally, a magnetic field is produced either by a moving charge or a magnet. This magnetic field's strength and direction at a point are described by magnetic field vector. The closer we move the point to magnet, the denser is the field at the body. Thus, the length of magnetic field vector increases. magnetic field vector is also used in determining the ...


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See this and links therein. In general, $A_\mu B^\mu$ and $A^\mu B_\mu$ are the same thing, so you can use either $\partial_\mu$ or $\partial^\mu$, as long as the thing you're contracting it with has the index on the opposite position. That is, $\partial_\mu j^\mu$ is fine, $\partial^\mu j_\mu$ is fine though weird, and $\partial^\mu j^\mu$ is nonsense. By ...


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Your explanation is almost exactly the same as the one which you say that you do not understand. So what exactly is confusing you? If $v_p-v_q$ is due south it does not necessarily follow that $r_p-r_q$ is also due south. For example, Q could be stationary while P moves due south, but the initial position of P could still be east or west or even south of Q ...


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The head to tail does work here and the answer is the difference in magnitude. Try imagining a triangle and then collapsing it so that the vertex lies on the opposite side. You'll understand why it is difference in magnitude of the vectors.


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Relative to Q, Q is not moving, and P is north of Q. $\vec v_p - \vec v_q$ is the velocity of P relative to Q. So basing on that, and knowing that P is exactly north of Q initially, we only need to know if P is going south relative to Q. Also, this means that $r_p-r_q$, the position of P relative to Q, is due north. Then, we should expect the velocity of P ...


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Yes : see Parallelogram of Forces. https://www.youtube.com/watch?v=LYXGlXMi6_c



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