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Think of it in the context of classical mechanics, or more intuitively your everyday environment. For example, take a cup of coffee on a table. It has a certain mass. This is just a pure number, we don't think of it as having 'direction' - what would it mean for mass to have direction? This is just a physically intuitive idea of a scalar quantity. Now push ...


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Space is 3D quantity. So anything that is coupled with space lineraly is 3D quantity. I am intentionally using a beginner information without touching any relativistic phenomena. Acceleration is 2 times coordinate derivative, hence coupled linearly. Force is acceleration times mass, hence coupled linearly. Electromagnetic field is also coupled linearly, ...


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When one drives a car, one generally wants to be aware of the direction in which they're moving. This information, along with the speed of the car, can be expressed as a vector. When one wants to know how far they've gone, or how long it took to get there, direction doesn't matter. All they need is a number. This is a scalar. This distinction is ...


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In short, you are to think of the direction of the torque as pointing along the axis of the rotation it would induce in a rigid body initially at rest. But if the conception of torque as a vector out of the page seems artificial, that's because it is. Torque is not fundamentally a quantity that is a vector but a directed plane or directed area. Such an ...


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There are probably lots of duplicates, so my apologies, but for clarity I will try a short answer, as the graphic from Wikipedia is particularly illustrative. The torque is perpendicular, ( orthogonal) to the other two vectors, so it could be the line where the hinges are located, depending on the direction of the other two forces. From Wikipedia ...


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Gradient is covariant. Let's consider gradient of a scalar function. The reason is that such a gradient is the difference of the function per unit distance in the direction of the basis vector. We often treat gradient as usual vector because we often transform from one orthonormal basis into another orthonormal basis. And in this case matrix transpose and ...


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Your question is as follows: Why centripetal force does not increase the value of tangential velocity? Answer: Assume you have circular motion as in the case where a person, with her hand, twirls a ball on a string. The string connects her fingers to the ball as the ball travels in a circle around her hand. In this case, the force in the string is ...


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I think the OP has made a mistake in applying second law of Newton. The law (about a particle) says: $$\Sigma \vec F=m\vec a$$ As it is seen, this is a vector equation. This means that corresponding components of both side of the equation must be equal. Although it is not said in the law's body, but it is obvious that we must write the equation above with ...


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Centripetal force is the name of the force that points towards the centre. This is in the radial direction. Tangential velocity is, as the name suggests, a velocity direction tangent to the circle. The radial and tangential directions are by definition always perpendicular - in the same way that the x and y axes are. You are of course right that if any ...


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This used to be covered in textbooks. A fairly recent article about it is "Why do forces add vectorially? A forgotten controversy in the foundations of classical mechanics" by Marc Lange in the American Journal of Physics 79(4) 380-388 (2011); http://dx.doi.org/10.1119/1.3534836 And there are two common answers. In dynamics you can used Newton's Second ...


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I don't think that "satisfying the laws of vector addition" is necessary for something to be a vector, depending on what you mean by that. Take velocities in special relativity. They are vectors; the vector sum of velocities is well defined. But it's rarely useful. More commonly, when you have two velocities and need to combine them somehow, the combination ...


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1) Usually special relativity is taught at the end of the semester, after the class got through rotations, which they, on average, don't understand; torques, which are pseudo vectors and for this reason blow their minds up... By the time they get to relativity they are done! If you teach the same kind of course, it is unavoidable, that they will get confused....


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Draw a diagram. As you understand Trigonometry, you know that $$Cos(\theta)=(\frac{Adj}{Hyp})$$ and $$Sin(\theta)=(\frac{Opp}{Hyp})$$ for any right angled triangle. In this case, your "Hypotenuse" is the $6.0N$, and your $\theta$ is what you have above. This should help you confirm whether you are right or wrong in your attempt.


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I want a more mathematical way to see it. Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes ...


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I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions: Velocity is a vector. Speed is it's magnitude. Acceleration is a vector. It's magnitude has no new name. We agree that acceleration is present if there is a change in velocity: $$\vec a= d \vec v /dt$$ That is, any change. So if magnitude (speed) and/...


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Your substitution is almost correct. You can have multiple forces each supplying their own acceleration vector, but it doesn't make a lot of sense to include the mass in the sum because it would mean you have to distribute the mass among all the constituent forces just to add them again. $$\sum _{i}{\vec F_i}=\vec F_{total}$$ $$\vec F_{total}=m\vec a_{total}=...


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Say $\Sigma F=F_1+F_2+...=0$. Then your substitution, which is mathematically correct, would physically imply the following: if $F_1$ alone had acted on the body then it would accelerate to $a_1$, if $F_2$ alone had acted on the body then it would accelerate to $a_2$, and so on. So $F_1+F_2+...=0$ implies $a_1+a_2+...=0$, while individual components ...


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You cannot substitute the force $\vec F$ with $m\vec a$. This is second law of Newton. The law talks about equivalency not sameness. Indeed we have $\Sigma \vec F\equiv m\vec a$. For substitution of the force, you should consider to the first law of Newton that defines the force.


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You can't meaningfully make that substitution. Newton's second law (for constant mass) is actually: $$ \sum F = ma $$ In other words, the acceleration is proportional to the total force.


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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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