# Tag Info

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I only watched the video to the point (about 15 min after the start) where Susskind he explained that according to Aristotle's "law of motion", the mass in a spring-and-mass system would move towards its equilibrium position with an exponentially decreasing velocity. Therefore, if you observe the mass when it is close to its equilibrium position and not ...

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There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...

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If we are dealing with the gravitational field of the earth, then because of the inverse square law, you should, in theory, need a tiny bit less force to raise an object from 50 m to 100 m, as you do lifting it from ground level to 50 m. But this difference in force is minute, I don't know offhand if we have instruments to measure it. But you can work it ...

0

In a rigid body, according to Goldstein's definition, the distance between any two constituent particles does not change. Work done is force times distance moved in the direction of the force. There is no relative movement in the direction of any force. Therefore, regardless of the form of the internal forces, no work is done by or against them.

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Yes and I think this can be understood easily. When the same object is first lifted to 50m and then to 100 you can conclude that you need the same force but you need to apply it for a longer time(or distance). So, the force you need is same but there will be larger work done(consumption of energy) for larger distances

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You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...

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Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1. ^ F | | 2 x-----------...

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However, Robot A moves at a speed 9 times faster than Robot B. Are you stating this as a given, or do you presume it arises from the consequence of the gearing? If so, it is incorrect because an engine does not deliver a constant speed, but instead a maximum power. The higher geared robot will have a larger load on the engine, reducing the speed it can ...

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For a good reference, try Landau and Lifshitz book on Elasticity. Suppose you have a wheel (approximated by an infinite cylinder) that is coming into contact with the flat ground. By symmetry, the area of contact will be a rectangle that will be very thin in one direction and very long along the length of the cylinder. Call the long length $a$ and the small ...

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I think you are comparing two pretty different cases. Motion of the Earth with respect to the Sun (and vise versa) is different from motion of a block on a table with respect to it. In the former, both of the Earth and Sun experience rotational motion and cannot be assumed as a particle without approximation error. But in the later, the block doesn't ...

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Which way does the y axis point? If the y axis is chosen to point up, (having the positive direction upwards) then you are right, normal force should be positive (it points upwards as well) and weight negative. Is it chosen to point down, then normal force is negative (points in the negative direction along the axis) and weight positive. Remember that ...

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The centripetal acceleration always points toward the center of the circle. In this case, the center of the circle is below the car, so the centripetal acceleration points downward. Now, you'll notice that, in the given solution, the centripetal acceleration term is positive. That means the writer has chosen a coordinate system where positive is downward, ...

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You can't get away from special relativity, which is what "unifies" electric and magnetic phenomena. The electromagnetic field really is a single field (actually a tensor, but don't worry about that), and its components mix together in the Lorentz transform in a similar way to how the x,y,z components of a usual vector field mix together in a rotation. ...

0

Think of an inflated balloon. Pressure inside the balloon is greater than atmospheric pressure. Entire system is in mechanical equilibrium because of tension provided by balloon surface. If you keep inflating the balloon, pressure inside the balloon keeps increasing, and so does tension in balloon surface, until it can take no more and bursts, thus ...

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There is no acceleration in Y direction. If you consider the X and Y axis like this. Maybe this will help. Work done is P*d. Remember the net force will always be zero because there is no acceleration.

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A deodorant can contains a liquid hydrocarbon, typically a propane/butane mixture, and the pressure inside the can is due to the vapour pressure of this hydrocarbon. The pressure can be set to any desired value by varying the composition of the propellant - more propane makes a higher pressure while more butane makes a lower pressure. For a deodorant the ...

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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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The charged accelerometer will register a non vanishing acceleration. The reason why the setup you are proposing (interaction via electric charge) gives a physically distinct result from the setup in the answer you linked (interaction via gravity), even though they can be described by the same mathematical force, is because the Equivalence Principle applies ...

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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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There is indeed a nett force on the body owing to the electrostatic attraction / repulsion. Therefore, there is nonzero four acceleration, and the body will have a different orbit from the ones defined by the spacetime geodesics for the metric describing the massive body's neighborhood. From the standpoint of an observer stationary with respect to the ...

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Gravity is acting upon every object on the earth, and it is pulling everything towards its center. But everything does not collapse to its center because the surface on which the object stands provides the reaction force upward (away from the center) and balances the weight of the object. If it can't balance the weight of the object, then certainly the ...

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I think the electromagnetic force plays an important rule,which prevent you falling into the centre of earth..

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Let's try to make the answer as simple as possible. Static weight is written as $w=mg$.(Notice Newtons 2nd law looks similar $F=ma$.) This is $$\mathrm{mass} \times \mathrm{one\ unit\ of\ gravity}$$. On earth, $g=9.8\frac{m}{s^2}$. On a smaller planet, $g$ is less. On a larger planet, $g$ is more. It is important to understand that despite acceleration in ...

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Somewhere in between $T_1$ and $T_2$ -- but exactly what depends on details about how the friction between the string and the pulley varies, how the string stretches under tension and the pulley deforms while being accelerated by the string ... All of these are things that cannot be deduced from an idealized picture such as this. In one extreme, if the ...

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power = force × velocity so at 10 km/h , power is 4000 N × 10 km/h while at 100 km/h, power is 4000 N × 100 km/h this way fuel will b consumed approx 10 times faster at 100 km/h than at 10 km/h

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At the point P, resolve mg into mgcosx in the radially outward direction and mgsinx in the negative angular direction(if anticlockwise sense is positive). Now, from the reference point of moving P, radially outward forces on it is mgcosx + mv^2/r (second term is the pseudoforce). Radially inward force from the diagram is T. Since there is no radial motion in ...

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The weight of the mass $m$ is resolved into two components. The angle between the direction of the weight vector and the radial vector is $\theta$. It might help if you draw a vector diagram with the weight as the hypotenuse of a right angled triangle as shown below? A tangential component $mg \sin \theta$ and a radial component $mg \cos \theta$. The ...

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Tension $T$ is the reactive force in the string when you pull at both ends with force $F$. If the string has stopped stretching or accelerating then $T=F$. The action-reaction pair of forces in Newton's 3rd Law are two equal but opposite forces with the same cause which act mutually on two different bodies : object A exerts force $F$ on object B, and B ...

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To simplify this answer assume that the string is made up of a line of molecules so each molecule bar the end ones have only two nearest neighbours. When there is no tension force in a string then on average the molecules which make up the string are at their equilibrium spacing and have no net force acting on them. Imaging that you apply a force $F$ on ...

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If the rope is in static equilibrium then $T=F$. If $T\neq F$ then that section of the rope (where tension is $T$) must be accelerating, which may happen if the rope is slack or if it is extensible.

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I think the difficulty is that you are not stating the whole of the problem as given by Morin. You are only giving us part of it, out of context. That is why it is confusing. The puzzling force $Mg$ acting horizontally to the right (which I shall call $P$) is probably the applied force mentioned in the title. Giving $P$ the value $Mg$ seems designed to ...

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As a general rule, whenever you are setting sudden discontinuities (as in your example with the Heaviside function) it is not a surprise that these may reflect in discontinuous distributions, or derivatives, or infinities here or there. Do keep in mind that it is just a result of the mathematical simplifications that we are introducing (although ...

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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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Setting up Newton's 1st law along the slope: $$\sum F=0\\ F_f-(Mg)_{x, left} +(Mg)_{x, right}=0 \\ F_f-Mg\sin(\theta) +Mg\cos(\theta) =0 \\ F_f=Mg\sin(\theta) - Mg\cos(\theta)$$ I don't see the confusion. Maybe you tried to solve this in your head, which usually makes one stuck in believing the signs are wrong along the way because some final rearranging ...

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The equation is built up from considering considering the mass $M$ moving with velocity $v$ splitting into $M+\delta M$ moving at ${\bf v}~+~\delta\bf v$ and a smaller piece with mass $\delta M$ moving with velocity ${\bf v}-\bf V$, for $\bf V$ the velocity of hot gases or plume from the rocket. The diagram below illustrates this Conservation of momentum ...

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You're measuring the wrong thing. Measure the momentum of a punch, not the "force". The force will be different if you hit something soft or hard, so you can't depend on that. The momentum is just the mass of hand+glove+forearm times velocity of punch. How to measure that? Hit the punching bag and see how far it swings.

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If you think of a magnet as being made up of individual dipoles all aligned with one another (this is what happens in Iron or Neodymium bar magnets), then you can ask "how does each dipole transform under Lorentz transformation?" Under Lorentz transformation magnetic dipoles become electric dipoles. The electric field from the electric dipoles causes the ...

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To measure anything, first you need a reference. For example, suppose you want to measure length in terms of meter. There is a particular length which is accepted by everyone as a meter. You say something is so many meters long by calculating how many meter-lengths fit into it, i.e. you take a ratio of length to be measured to standard-meter and express the ...

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Yes it does. If you fix a (non inertial) frame of reference whose origin is on Earth's surface, at latitude $\lambda$, then a freely falling particle of mass $m$ has the equation of motion, $$m\frac{d^2\vec r}{dt^2}=m\vec g_{ef}-2m\vec\omega\times\vec r,$$ where $$\vec g_{ef}=\vec g-\vec\omega\times\left[\omega\times(\vec R+\vec r)\right]\approx \vec g-\vec\... 1 If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors. In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch ... 3 If you mean the mass of the hadrons, it is because of the energy/mass contributed by the gluons carrying the SNF and virtual quarks and antiquarks, which is high compared the three valance quarks. The valance quarks are the three "permanent" quarks that we think make up the proton and neutron. The Higgs field has a much lower effect on the mass of these ... 3 It is not incorrect, it is incomplete. A physical value consists of: Magnitude - this is "the number" Unit - "type" of the value Physical values are not only magnitudes, they also have a unit. It is inseparable. Unit defines a physical meaning to a value. You could formally treat it as a pair (magnitude, unit). All calculations in physics are done this ... 3 The basic fact is units are a demarcation of the quantity of something, so they can never be used as a fundamental equation. As you have mentioned F=ma can also be written as N=kg ms^-2 , then how can you say that the equation is of Newton's Law only? The equation can be used to define force dimensionally which is [M L T^-2]. 4 If a physics equation is to be valid, it is necessary for the units to work out, but it is not sufficient. The handling of dimensionality only provides part of the information. Units and dimensionality are good checks to make sure you did the equation right, but the mere fact that the units were right does not automatically mean the equation was right. ... 0 A "naked" number such as 3 does not denote anything in the real world. In order to represent on paper a real world collection of items, we need a numerator and a denominator, e.g. 3 Apples. But to represent on paper a real world amounts of substance we need in addition a unit to, so to speak, "discretize" the amount into a collection of units, e.g. 3 ... 5 Disclaimer: I have no idea what question everyone else is answering; none of the other answers seem to address the question as I understand it. We don't use units in formula because not everything we want a variable to represent (like a for acceleration) has its own nice unit. Acceleration is a perfect example: it's just too long to say "meters per ... 7 For one, the laws of physics are the same whether you're working in SI or Imperial. F = ma, regardless if your m is in kg, pounds mass, or solar masses. Actually, in college we had a bonus challenge to give the answer to a problem in craziest units of energy that we could come up with that actually worked. "Slug lightyears" was a pretty good one, but the ... 4 The purpose of units is to assign numbers to measurements. They are necessary but of secondary importance to the thing being measured. Scientists want to describe the real world with their equations, not just their measurement tools. 46 A specific parameter might correspond to a specific (SI) unit, but not all units correspond to a specific parameter! Kinetic energy is$$\begin{align} K&=\frac{1}{2} mv^2 \\ \text{Joules}&=\frac{1}{2} \text{kilograms}\times\text{meters}^2/\text{seconds}^2 \end{align}$$We also have gravitational potential energy:$$\begin{align}U&=mgh \\ \...

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Writing equations using only units would not work at all for dimensionless equations. For example the Snell's law $$n_1 \sin\theta_1=n_2 \sin\theta_2.$$ You would also lose many of the dimensionless (but usefull) parameters in physics such as the Lorentz factor $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}.$$ Also consider equations whose all variables have the same ...

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