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Sound leaves the vehicle. When it strikes the wall, it reflects at a 45 degree angle and returns to the car. The triangle you will be using has legs that are length X, with a hypotenuse that is length Xsqrt2. That is for evey 1 foot of length in a leg, I will have 1.414 feet of length in the hypotenuse. What you want to solve for is time based on this ...


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To change the velocity of the ball a force must be applied against it. In your example the force is applied by the wall on the ball and this force does work. And this work which is not same in all frames of reference, according to your example, gets completely converted to heat.To explain it simply, lets say that the ball is deformed and becomes completely ...


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To solve these, it's enough that you know: $$ v = v_0 + at $$ $$ x = x_0 + v_0 t + \frac{1}{2} a t^2 $$ Fill in the values you know, then solve for the value you want.


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Timaeus has given the full technical answer: the kinetic energy of the wall itself changes a tiny bit. Since kinetic energy scales as $v^2$, this is totally negligible in the first case (where the wall starts with $v = 0$) but actually significant in the second case (where the wall starts with $v = 2$), and that's where the missing energy goes. Luckily, in ...


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Imagine a ball ($mass= 1kg$) moving at a velocity $2 m/s$ towards a wall. When it hits the wall, it suddenly stops, thereby liberating all its KE as heat. Here, $Initial K.E. = (1/2)*m*v^2 = 2J$, and final KE is obviously zero. So heat liberated ($Final KE - Initial KE$) equals $2J$. Now, suppose I hop on to a car moving at $2 m/s$ towards the oncoming ball ...


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Initially your ball has some energy ($2J$) and some momentum ($2Ns$). And the wall has some energy ($0J$) and some momentum ($0Ns$). And there is some internal energy, $U=U_0,$ the thing that heat increases. Afterwards the ball has some energy ($0J$) and some momentum ($0Ns$). And the wall has some energy ($0J$) and some momentum ($2Ns$). And there is some ...


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In the car frame the ball initially has 8J of kinetic energy relative to you, however it has only 2J of kinetic energy relative to the wall which is also moving in the cars frame.


1

Engineers usually treat thermal expansion as isotropic, which means the expansion occurs with the same magnitude in every direction. This means that an unconstrained object will have a constant strain and zero stress associated with thermal expansion, it's as if object just scaled up. However, as you suspected, materials with an organized structure can be ...


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On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any ...


4

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...


2

Draw an arrow to represent a vector, with its length representing the vector magnitude. Draw a coordinate system and get the components of the vector. Now draw another coordinate basis, rotated with respect to the first, and get the components with respect to the new basis. The length of the arrow is the same in both systems - i.e length is invariant - ...


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Our equation for work follows from the conservation of energy. If we consider some object then we expect that if we do work $W$ on it then its kinetic energy must increase by $W$. So the requirement for the equation for work is that it must be equal to the change in kinetic energy. Proving this is usually done using integral calculus, but since you give the ...


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The notion of work in physics was first formulated by the French mathematician Gustave Coriolis in Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation published in 1829. Coriolis defined work as "weight lifted through a height". He was concerned with developing a term that could measure the units of work ...


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There are some physical quantities that are usefull (and this is under statement), like energy. It is conserved, it is a function of some other very important quantities that can help you describe the motion of the body etc. If you can justify energy, there should be no problem in justifying work, which is energy transfered to a body by some force. Quantity ...


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They are not related structurally: Configuration space is a manifold which in general has no vector space structure. For example $\mathbb{R}$, the configuration space of a free particle moving on a line can be viewed as a vector space (you can sensibly "add" two configurations to get a new one and so on), but if you constrain it to move on a circle this ...


2

My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space? Think of it as a map. You can read a map and loom around and you can learn how to relate the common points. So of you have 700 particles there are 2100 scalars. So have a giant 2100 dimensional vector ...


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A 60 watt bulb will apply 60 watts of power to the room it is in, minus whatever small amount of visible light gets out the windows. Window glass is largely opaque to infrared and ultraviolet wavelengths (it's coated to be such, these days). Whatever EM energy is emitted by the bulb is absorbed by the surfaces in the room and reradiated or conducted to the ...


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The velocity of the standing wave is the velocity of the incoming and reflecting wave that formed this standing wave. See http://www.physicsclassroom.com/class/waves/Lesson-4/Formation-of-Standing-Waves


3

We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...


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Your logic is actually correct. The discordance between the conservation of phase-space volume according to the Liouville theorem and the Second Law is known as the Ergodic Problem. Heuristic explanations as the one provided by Ross Millikan, or course graining the dynamics for another example, do not hold under closer formal examination, since the math ...


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Why is your answer incorrect? Because at first you're calculating velocity of $C$ assuming point $A$ is at rest. Then the velocity you find for point $O$ is its velocity relative to the point $A$. Similarly, when you're calculating the velocity of point $A$ what you find is actually its velocity relative to $O$. They are the same thing, only with a sign ...


1

Liouville's theorem says the accessible volume in phase space does not increase, but it tends to become narrow filaments that "fill up" a much larger volume. If you think of a particle in a reflecting box, you might start it with a known position $\pm 1$ mm in all three axes and a known velocity $\pm 1$ mm/sec in all three axes. This is a phase space ...


0

This sounds like ;) it's because the object that the TV is resting on is acting as a sound conductor. Solids conduct sound better then gasses. On the 1st floor, sound goes from the TV, through the platform, through to the floor and to the ceiling of the floor below without losing much energy before being transmitted through the air. On the ground floor, ...


3

The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ...


1

First you need to define the orientation of the cube relative to the axis you want to measure. Typically a 3×3 rotation matrix $E$ does the job transforming local coordinates along the principal axes to the world coordinates. The use the transformation $E I_{body} E^\intercal$ Example: A single rotation $\theta$ about the world $z$ axis is $$E = ...


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Two theorems worth knowing about when it comes to calculations like this: 1) Parallel axis theorem. For any object of mass $m$, the moment of inertia $I_A$ about an axis A that is parallel to but displaced by a distance $x$ from an axis $C$ through the center of mass, is given by $$I_A = mx^2 + I_C$$ From this, it follows immediately that the moment of ...


0

Air resistance or air drag is basically proportional to square of the velocity of the moving object ,that is, F=bv^2 since air has turbulent motion where b is the air drag constant.Drag constant depends on the dimensions of the object. So air drag is independent of length or volume of object, rather it depends on the surface area of the object. The ...


3

The fastest point of sail depends on the boat (both its hull shape and its sail plan), the wind strength, and the sea state. In general, a beam reach is not the fastest point of sail. For instance, in very light wind some boats will go fastest on a close reach due to the increased apparent wind from going toward the wind. For boats that sail faster than ...


0

The thrust a rocket can generate is proportional to the mass flow (thus $\frac{dm}{dt}$) and the velocity at which this mass leaves the rocket, often called the effective exhaust velocity, $v_e$. So the sum of all forces acting on the rocket, when also neglecting the fictitious Coriolis and centrifugal forces and assuming a pure vertical ascent, will be ...


3

The Lagrangian formalism treats $x$ and $\dot{x}$ as independent variables. In particular, you cannot write $\frac{\mathrm{d}}{\mathrm{d}t}x$ because $x$ is not dependent on time. What is dependent on time is a particular trajectory $x(t)$ that is the solution to the equations of motion (the Euler-Lagrange equations). Prior to solving the equations of ...


1

The thing is that when you write the Lagrangian, you don't know the particle's trajectory yet. If you had a specified function $x(t)$, then of course $\dot{x}$ is not independent. But if you only know the particle's position at a given time, its velocity can be anything, because you are free to set position and velocity as initial conditions how you ...


2

With regards to intuition, it might help to think about situations of mechanical advantage. For example, consider a simple pulley system. modified from "Pulley1a". Licensed under Public Domain via Commons - https://commons.wikimedia.org/wiki/File:Pulley1a.svg#/media/File:Pulley1a.svg You can work out using force that the weight $W/2$ balances the weight ...


1

Work does depend on frame of reference, but so does change in kinetic energy. Work done and changes in kinetic energy should either both bother you or neither bother you. To know how much the kinetic energy changes from one location to another you need to know the force (if constant) and how far apart the locations are: ...


1

As you point out, work done is a function of the frame of reference. More specifically, if you apply a force on an object, that force typically connects two different objects, and it's the relative velocity of these two objects that really concerns us. Example: you are walking in a train, and pull a suitcase behind you. The friction between the suitcase and ...


1

There's a small error here: you say, "...these two vectors $v$ and $\mathbf{v}$" (emphasis mine). The problem there is that $v$ is not a vector. Rather, it's the magnitude of a vector; specifically, it is the magnitude of the velocity vector $\mathbf{v}$. This is actually implicit in your derivation: you created a unit vector in the direction of the motion ...


2

It's a general consequence of Newton's Laws of Motion that an object is stationary when no net forces or momenta act on it. Avoiding net forces or momenta is precisely what makes perfectly balancing a rectangular block on a triangular one so difficult. Above is a rectangular block with mass $m$, perfectly balanced on a triangular block. I've added the ...


2

Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...


1

If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.


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As @rmhleo pointed out in his answer, the frictional force doesn't depend on the surface area, because no matter which part of the object is in contact with the other surface, the total normal force (and thus the total frictional force) is unchanged. However, that assumes a couple of simplifying conditions: namely, that the two surfaces are consistent in ...


1

Yes, frictional force does not depend on the area. This is clear from a simple mental experiment, think of a block resting on a surface. Assume it is a prism whose faces have different areas. Whichever face it is resting on, the friction force will be the same: when on the smaller surface, the contact is reduced compare with the case where the resting face ...


0

A conventional car work with high energy density fuel. But the drive works efficient in some revolutions only. That's way there is a gear box. Working with electric drives is not new. It was invented at the same time (to be precise a little before) the Benz developed his car. But there are weak points too. Accumulators are heavy and the energy density is ...


1

You say that you "can actually feel a substantial difference when I lift her, until her feet are not in contact with the ground anymore". That makes perfect sense i.e. she is also participating in the lifting process by pushing off the ground except when she's being difficult. Once she's off the ground, as long as you're holding her in the same way and she ...


0

From Making yourself heavier Yes, it is possible to make yourself harder to lift. By shifting your center of gravity , which is usually referred to as your hips, down you can make it much harder to be lifted. It depends on which kind of lift, but this usually does the trick. Shifting your center of gravity does not make you heavier. It shifts the point ...


1

The key point here is that, any dynamical system that is not completely integrable will exhibit chaotic regimes1. In other words not all orbits will lie on an invariant torus (Liouville's torus is the topological structure of a fully integrable system), in principle a chaotic system can even have closed stable periodic orbits (typical for regular/integrable ...


3

First some terminology: A non-degenerate 2-form $\omega$ is called an almost symplectic structure. A closed 2-form $\omega$ is often called a pre-symplectic structure. If the 2-form $\omega$ is both non-degenerate and closed, it becomes a symplectic structure. In the non-degenerate case, the closedness condition $$\mathrm{d}\omega~=~0\tag{C}$$ is ...


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Did you see section 4 of the calculator "warnings"? "Your index is too large" (index = ratio of diameter of spring to outer diameter. You have a ratio of 40 - a very thin wire wound on a very large diameter. And since length inside hook needs to be decreased by the diameter in order to get the length of the "actual spring", you basically have "almost no ...


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Here is a solution for a spring force that varies directly with displacement. It thus varies with time implicitly, but has no explicit dependence on time or any other variable. Givens and Assumptions oscillator with mass $m$ amplitude of oscillation $A$ oscillator displacement, $x$, varies with time, but $x(t)$ is unknown spring applies force varying with ...


0

To obtain some kind of practical answer, you have to determine how k varies. For example, if k varies with temperature, I would determine its value at -50, 0, and 50 degrees, then use those values and calculate T (which varies inversely as the square root of k). I would Use more points if a higher accuracy is required. If a formula is required, I would ...


1

From Newton's second law we have (whether $k$ is constant or not) that: \begin{equation} m\ddot{x}+kx=0 \end{equation} The only difference is whether or not $k$ is a function of $t$ or not. If it is a function of $t$, the only general way to solve this differential equation is by using Taylor expansions. Let us take: \begin{equation} ...


2

Because atoms do not show the linear Stark effect. As atoms have no permament electric dipole moment the leading order effect is the quadratic Stark effect, which is supressed by another factor of the field strength (which is already small compared to the atomic field strengths). Early precision spectrometry was based entirely on atoms in electric discharge ...



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