# Tag Info

1

This builds partially off of TZDZ's answer. In order to fully understand the moment of inertia, you need to fully understand the idea of torque. The only sensible way to fully understand torque that I've seen is to first understand it as the derivative of rotational work $W_R$ being done with respect to the angle $\theta$ through which the work is done. So: ...

1

For a newtonian fluid, you can write the total stress tensor $\sigma_{ij}$ as $$\sigma_{ij} = -p \delta_{ij} + \tau_{ij}$$ with $$\tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$ $-p \delta_{ij}$ is the classic pressure term and $\tau_{ij}$ is the shear stress tensor with $\mu$ the shear viscosity. When ...

0

As the weight hanging off the table falls, it has some acceleration downwards. The force on the body is entirely due to gravity and is known. The string acts as a linkage between the hanging mass and the mass on the table. Really in this case, we have weights $m_1$ and $m_2$ being pulled by a force $m_1g$, which leads to the acceleration of the system being ...

0

I've seen and studied much on the quantum of electromagnetic energy, which we call photon. There are a lot of contradictory statements concerning interpretation, but there isn't much debate about the theory. My conclusion is that it is not a particle like a small bit of something, and it is not a wave as in the oscillations of a medium, and it is not some ...

0

"The frequency is said to be that of electromagnetic waves. Does it only have a single frequency, or can it have a group?" An individual photon can only have one frequency - but it can be any frequency - like, if you take a test, you can only get one score, but it can be any score from 0 to 100. 2 photons can travel in the same direction close to each ...

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At the present time in physics, the photon is an elementary particle, a member of the table of elementary particles that are the basis for the standard model of physics. Elementary particles included in the Standard Model It has zero mass, zero charge and spin one, and it is the gauge boson of the electromagnetic interactions. In all electromagnetic ...

0

By the one who's exerting the force. It depends. When a system exerts some force on some other system, it does work on it and loses its energy. The lost energy is transferred to the second object, which in turn does work against the frictional force and loses its energy which is transferred into the molecules or atoms of the surface as heat.. There are a lot ...

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Actually, the conditions $$\langle x|A x \rangle \geq a \langle x |x \rangle$$ and $$\langle x|A x \rangle \leq b \langle x |x \rangle$$ with $a,b \in \mathbb R$ fixed and all $x\in D(A)$ (the domain of $A$) refer to boundedness (resp. from below or from above) of the quadratic form associated to the linear operator $A$. This operator can always be supposed ...

0

For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it . Work cannot be done by a body who does not have it's own energy source like an hammer can never do work.

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For your case work is done by the one who applies force and displaces an object in any direction. You take this example "if an object falls from a height then the work is done by gravity (look who is applying force here )" I hope u understood it .

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Strictly speaking, if a body is moved from a point to another by some force, it is said that the force does some work on the body.

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Probably not the correct way to think about it but this is my reasoning if you applied force on the body you had to spend the energy to do that work therefore you did the work the body gained the energy in some form say you lifted it of the ground the body has gained potential energy while you lost some bio-mechanical energy.

0

If you exert a force on a body and it becomes displaced, then it is said that work is done by you or, by the force on that body.

3

The difference is typically the diffusion coefficient: \begin{align} \frac{\partial \psi}{\partial t}&=\nabla\cdot\left(\kappa\nabla\psi\right)\tag{diffusion}\\ \frac{\partial \psi}{\partial t}&=\kappa\nabla^2\psi\tag{heat} \end{align} Under the diffusion equation, we typically take $\kappa$ to be a spatially-dependent variable whereas in the heat ...

1

To understand rolling without slipping, first consider the case of rolling only case about the center of mass.In this case, a point on the top rim will have a speed $v=\omega\cdot R$ and a velocity $v=-\omega\cdot r$, at bottom of the rim, as observed by you. However in case of rolling without slipping, We observe the velocity of center of mass is $v$ (only ...

1

Under common assumptions and ignoring potential energy, static pressure is the expression of the fluid's temperature (internal energy) and dynamic pressure is the expression off the fluid's velocity, so if the fluid is brought to a rest adiabatically, their sum is equal to the stagnation pressure. The stagnation pressure represents the total energy of the ...

0

Each moving part oscillates with the same frequency Every moving part crosses the equilibrium at the same time.

2

Normal modes are the separable solutions to the string's (linear) partial differential equation $$y(t) = X(x)T(t)$$ that arise from applying the solution method of separation of variables. These solutions form an orthogonal (normal) basis for any solution. Due to the form, a function of space only multiplied by a function of time only, the shape of the ...

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A normal mode of a system is a pattern of motion (to borrow Wikipedia's term) where every point of the system oscillates with the same frequency and are in phase with each other (with the caveat that some points of the system may have a negative amplitude, which is equivalent to having a positive amplitude but being 180 degrees out of phase). Your second ...

1

To fluid dynamicists, Bernoulli's equation is better known as the 'Energy Equation' since it does indeed account for the energy changes that occur along a fluid path. The energy equation says that the energy is constant along any given streamline. Static or stagnation pressure can exist in the absence of fluid velocity creating a potential energy component. ...

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The quantity $\frac{1}{2}\rho v^2$ is called dynamic pressure for two reasons: because it arises from the motion of the fluid, and because it has the dimensions of a pressure. It is not really a pressure at all: it is simply a convenient name for the quantity (half the density times the velocity squared), which represents the decrease in the pressure due ...

1

Technically, a system is in thermodynamic equilibrium if: Every part of the system has a well-defined temperature. For example: If there's an ideal gas, the velocities satisfy the Maxwell-Boltzmann distribution (for some temperature); If there are photons, their intensity and spectrum satisfies the blackbody radiation formula (for some temperature) If ...

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Yes, any molar quantity it is considered intensive. You can find a list at this wikipedia page: http://en.wikipedia.org/wiki/Intensive_and_extensive_properties

2

You can express energy in Joules. Then it is extensive (it will scale with the system). You can also call it specific energy and express it in Joules per kg. Then it is intensive. Or you could consider "molar" energy and have Joules per mole and it would also be intensive. Same goes for entropy, which is Joules per Kelvin (extensive), or enthalpy (Joules), ...

2

TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia. In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$. It follows ...

1

When one says that an operator is bounded, you can think of it as being bounded from above. This is different from being bounded from below. An operator can be bounded (from above) and bounded from below, or perhaps just bounded, or just bounded from below. Observe that $(Af,Af)$ and $(\psi,B\psi)$ are slightly different: the former is always positive for ...

1

This is explained in the Wikipedia article on Hubble's Law: http://en.wikipedia.org/wiki/Hubble%27s_law In particular, "Dimensionless Hubble parameter Instead of working with Hubble's constant, a common practice is to introduce the dimensionless Hubble parameter, usually denoted by h, and to write the Hubble's parameter $H_0$ as 100 h km s −1 Mpc−1, all ...

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This goes back to the history of the Hubble constant. It's easy enough to write most cosmological formulas in terms of this value, but measuring it was something of a challenge for a while. For several decades, we were confident it was between $50$ and $100\ \mathrm{km/s/Mpc}$. Many results scale with this value (to some power), so what people did was write ...

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From the wikipedia article: The total enthalpy, H, of a system cannot be measured directly. The same situation exists in classical mechanics: only a change or difference in energy carries physical meaning. Enthalpy itself is a thermodynamic potential, so in order to measure the enthalpy of a system, we must refer to a defined reference point; therefore ...

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I have contributed this issue with this arXiv:1411.2425 and previous works. I stress that Gibbs entropy is not "defined" but rather "constructed". It is constructed on the expression of thermodynamic forces in the microcanonical ensemble, and is constructed in such a way that it reproduces them always and exactly. The construction is actually unique. ...

2

Heat is a type of energy transfer from one system to another, rather than an internal energy of any given system ('thermal energy' might be a good term for what you're thinking of). In An Introduction to Thermal Physics by Daniel Schroeder, p. 18 says: Heat is defined as any spontaneous flow of energy from one object to another caused by a difference in ...

0

As the other answer states, objects don't posses heat. Heat is the actual amount of energy transferred from an object to another. The typical case is when you have two objects $A,B$ at different temperature in contact, $T_A>T_B$. Then, it means as you say that the average energy of $A$ is bigger than the average energy of $B$. This induces some of the ...

3

Object don't posses heat. They posses internal energy. Heat, like work, is a transfer of energy and is a property of a process or interaction not of an object.

3

The energy stored in a field is the energy required to create it. In your case of the inductor there is no field when no EMF is applied. When we apply an EMF a current flows and does work, and the work goes into creating the field. When we talk about the energy of e.g. a charge in an electrostatic field, we normally assume the charge is small enough that ...

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