# Tag Info

1

No it is still zero. Take $\theta$ to be the angle between the point at the bottom of the cylinder, C and P. Take the radius of the cylinder to be $R$ and max length of rope $L$. The vector from C to P is $$r_{CP} = R(\sin \theta,-\cos\theta).$$ The vector from P to the particle is $$r_{T}=(L-R\theta)(\cos\theta,\sin\theta).$$ So the position vector of ...

0

My suspicion is that the output of your first capacitor still has a 2.8 V peak-to-peak 60 Hz ripple on it. There is always some level of ripple out of the usual rectifier + capacitor DC power supply. Putting a diode between the first and second capacitors creates a "peak detector" that samples the highest points of this ripple at +1.4 V above your 160 V ...

1

I am puzzled at your leap: quantum bound state of electron to hydrogen, to the earth potential classical bound state. Bound classically and bound quantum mechanically are two different frameworks. The electron is bound quantum mechanically to the hydrogen atom and does not "see" the gravitational coupling quantum mechanically due to the very small value of ...

0

One usually goes to the continuum because of its nice mathematical properties; lattice QFT is a hint of how hard a quantum theory becomes if we break the symmetries. For scattering theory, you usually want to be able to apply Lorentz invariance, or its classical counterpart, Galilean, and this already implies you are working with the continuum spectra of ...

0

I think the real reason is because if you change your velocity then everything that used to be at rest should now continue just as they did before you changed your velocity. You can call it indifference, nothing cares how you move unless you interact with it. Or you can call it relativity. What this does is reduce the case 1) of things at constant motion ...

3

For example, the escape velocity of a particle from the galaxy is about 400 km/s and in most conceivable circumstances (unless you are basically on top of the event horizon of a black hole or on the surface of a neutron star), escape velocities will be far, far below relativistic speeds (here defined as $3\times10^4$ km/s). So basically, if a particle has a ...

0

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 ...

2

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 ...

-1

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 ...

0

It's a bit unfortunate that you called it $J$ because energy is stored in electromagnetic fields and so it moves around but sometimes the net flow of electromagnetic field energy into a region is not completely balanced by an increase in the field energy at that point and one time that happens is when the fields exchange energy with electric charges. And ...

1

The expression for the total potential energy stored in the fields is given by $$\frac{\epsilon_0}{2} \int \left| \mathbf{E}_1 + \mathbf{E}_2 \right|^2 d\tau = \frac{\epsilon_0}{2}\left( \int \left| \mathbf{E}_1 \right|^2 d\tau + \int \left| \mathbf{E}_2 \right|^2 d\tau + 2 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau \right)$$ Notice that the first and ...

1

To start with the law of increasing entropy applies to isolated systems. The system you describe is isolated if one considers the total entropy of both the paramagnetic material and the permanent magnet, including any radiation. The order introduced in the paramagnetic material is balanced by a disorder in the permanent magnet plus any radiation from ...

3

There's two main things to consider - energy and absorption charasteristics of different photon wavelengths. The Sun emits a lot of energy, obviously. Even at Earth's distance from the Sun, the energy concentration is still far from negligible - when this energy impacts your body and is absorbed, it mostly causes heating (a bit complicated by wavelength, ...

1

There are multiple "kinds" of photons - different wavelengths have different effects on you. X-ray works somewhere around the 1nm range of the spectrum. It is ionizing radiation which can mostly go through soft materials but can harm cells when passing. So you usually get only the minimal needed amount of photons to create the image and not much more. The ...

20

In addition to the answer from @MichaelS, you need to consider where the energy from each source is deposited: Sunlight energy is deposited on/in the skin where there are numerous nerve endings. An increase in skin temperature is "measured" and your brain is aware of it. X-ray energy which is absorbed by the body is mainly absorbed by bones and some ...

28

X-rays do warm you up. It's just that the X-rays are more dangerous per photon (they can do major damage to cells and DNA, and are known to cause tumors and cancer), so they limit the amount of time you're exposed to the bare minimum needed for a clear image. The total energy from standing in the sunlight for several seconds is much higher than the energy ...

0

Yes it is possible to have water coming out be colder than the water going in. Imagine a big pipe connecting a big container of ice water to a big container of boiling water. Then one end starts out colder than the other. Now raise the container of boiling water higher up so that water starts to flow from the boiling water to the freezing water. Heat ...

0

Write down a few dimensionally correct equations. E.g.: $$E = m c^2$$ $$E = m^2 G/r$$ $$E = h f = h c/r$$ Divide the last two: $$1 = m^2 G/(h c)$$ So, we have that: $$m = \sqrt{\frac{h c}{G}}$$ Therefore: $$E = m c^2 = c^{\frac{5}{2}}\sqrt{\frac{h}{G}}$$

0

Say, This implies m=-1/2 n=1/2 p=5/2 So this way you can formulate equations by dimensional analysis.

1

"I don't know any equations..." is the point of dimensional analysis! Let's make a table of the quantities you listed, and their dimensions: M L T G -1 3 -2 \ c 1 -1 +- given these inputs... h 1 2 -1 / ------------- E 1 2 -2 - I need to get this output If we assume there is an expression $$E \propto G^A c^B h^C$$ then ...

9

Before everyone freaks out, no, you don't use petroleum oil. You use vegetable, fish or animal oil. In earlier times, whale oil would be used. The OP's picture looks like a fuel oil leak, not an attempt at wave calming. I have seen references of this technique being used since at least the early 1800s, probably much earlier. Ernest Shackleton made use of ...

25

Yes it works. But let's not use it on a massive scale, lest we damage the ecosystem (tip of the hat to @phi1123). A hint to the mechanism can be found in Behroozi et al (Am J Phys, 2007) They state in the abstract: From the attenuation data at frequencies between 251 and 551Hz, we conclude that the calming effect of oil on surface waves is principally ...

-1

May be when number of proton and number of neutron exceeds the magic number like 2,8,20,50,82,126 etc by one ,the binding energy of extra nucleon is less than the average value. Which is known as first excitation energy.

1

When you connect a battery to a capacitor, a "real" circuit has at least four components in series: the voltage source (battery) the capacitor series resistance series inductance Any wire has inductance, since current flowing through it will induce a magnetic field. It is possible to have wires without resistance - we call them superconductors. So let's ...

3

I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about. On the contrary, it's crucial. Assuming an ideal voltage source (can supply unlimited current) of voltage $V_S$, an ideal resistor of resistance $R$, and an ideal uncharged capacitor of capacitance $C$, are ...

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 ...

0

Lubos's answer covers very well and in a general way the different possibilities for reactants and products in the reactions of subatomic particles and nucleons. The basic equation is Energy of photons = energy of reactants - energy of products. Due to the enormous energies involved it is actually unlikely that visible light will be released, most ...

6

This is a special example of "what will happen" under given circumstances. Almost all of physics – and natural science – is about answering such questions. But they're really very many very different questions and one must be a little bit more specific about what the question is. Your general question "what forms of energy will result" is so general that it ...

1

Just to clarify the comment above a little. You can treat the velocity of your plane, neglecting air resistance, as completely independent of it's height, if you have studied vectors you may know how this can be drawn out on a graph. The plane has a potential energy, mgh (mass by g by height) and this potential energy does not depend in any way on how fast ...

4

When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for $J=\frac{1}{2}$ with $L=0$, and those are $J_z=\pm\frac{1}{2}$ (Or some linear combination of them) As to why this is, ...

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

First, when you rewrote the first formula using the "explicit notation", it is very hard to see what you gained. On one hand, you indicated that $y$ is a function of $x,t$ which would be fine (although it's clearly redundant to write this long expression instead of $y$ all the time) but I can't understand why you wrote $x_0,t_0$ instead of $x,t$. Even more ...

0

1. Prior to entering the water: Assume that just prior to entering the water the object has a Kinetic Energy KE of: $KE = \frac{1}{2}mv^2$ 2. During impact: During impact and entering the water the object abruptly loses some of its KE. We know this because: The impact is accompanied by sound (which contains energy). Water is being 'kicked up', which ...

1

In physics "principles" are statements that can be derived from the basic laws . Basic_laws/postulates in physics are like axioms in mathematics, they tie up observations to the mathematical model, and one cannot ask "why" since the only answer is "because this is/fits what we observe". Your statement is partially reflected in this principle: The ...

2

No. The question is based on a false assumption. Nothing "seeks ground" in the sense the expression implies. Current flows in an electrical circuit. If part of the circuit is connected to "ground" then current will flow into or out of ground as part of flowing in the circuit as a whole. If you look at the "circuit diagrams" of electric circuits, in many ...

0

I can't answer your question specifically. But I suspect between the continuous combustion of 4 to 8 cylinders and all the moving parts of the typical internal combustion engine, if one could expediently redirect that vibrating energy away from the motor, one could greatly improve the engine's trueness of operation. Which should provide far greater ...

1

In terms of radiography / radiology, kVp is the tube voltage / tube potential between the cathode and the anode, set by the operator. The unit eV (or keV as used in the range of general radiography, MeV as used by Radiation Therapy) describes the energy of the particles - in this case, the electrons in the x-ray tube, and the x-ray photons coming off the ...

1

$$m \ddot{\vec{r}} = \vec{F}$$ multiply by $\dot{\vec{r}}$ and integrat over t: $$m \int_{t_0}^t \ddot{\vec{r}} \cdot \dot{\vec{r}}~ dt = \int_{t_0}^t \vec{F} \cdot \dot{\vec{r}}~ dt$$ With $\frac{1}{2} \frac{d}{dt} (\dot{\vec{r}}^2) = \ddot{\vec{r}} \cdot \dot{\vec{r}}$ it follows: $$\frac{1}{2} m v^2 + \left( - \int_{t_0}^t \vec{F} \cdot d\vec{r} ... 0 What about using Galilean free-fall? From S= \frac 12 g t^2 and v = g t you get that velocity after a fall h follows$$h= \frac 1{2g} v^2$$We conclude that if the ball is consuming some essence to get velocity from the line of fall, this essence must be "stored" in space as the square of the velocity. The idea works because if we have already a body ... 1 When you write: Derivations (or at least, convincing arguments) of the kinetic energy formula that didn't require the work formula required relativity to make sense, which is unbelievable considering that Newtonian mechanics were established well before relativity. I assume you are referring to arguments like Ron's argument. Although such ... 3 To get an understanding on quantum field theory issues, you have to understand the difference between virtual particles and real particles. Virtual particles, in contrast to real particles, are a mathematical construct inspired by the Feynman diagrams used to describe interactions. These diagrams start with real particles, i.e. particles that have the mass ... 0 I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, V, is constant: What remains constant is the charge on each plate. So the equation becomes:$$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}} Since $C=\epsilon_0A/d$ we obtain ...

0

From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d. That's not the ordinary understanding and, from the electrical circuit perspective, the energy stored equals the work done by the external circuit separating electric charge in the ...

0

...in what instances is the kinetic energy of a free particle... Remark: The temperature dependent expression for the kinetic energy is not a property of a single particle, but an ensemble. The $\frac{3}{2}$ comes from a statistical consideration of degrees of freedom of a mechanical particle scheme. The $\pi$ comes form the wave picture. Using a ...

0

The calculation is not right, but the result is true. You need infinite work to pull apart two oppositely charged infinite plates, they have each infinite charge! The electric field due to one infinite plate is $\frac{\sigma}{2 \epsilon_0}$. The force on an area $A$ of the second plate will thus be $\frac{- \sigma^2 A}{2 \epsilon_0}$, which is infinite for ...

0

As John Rennie marked, the result should be $\tfrac{1}{2}CV^2$. Let me deduce this for you; Let's start with an uncharged condenser & by some means you remove one an electron from one plate & transfer it to the other plate. You have to do hardly any work to transfer the first electron but as you gradually continue the process, the field that is ...

3

Is it possible to measure the temperature of something using sound...? Yes, it is not only possible, it is available commercially. It is especially useful in harsh environments where conventional temperature probes might not survive. For example, TMT makes an acoustic system for measuring 2-D temperature distributions in blast furnaces: The ...

0

In principle, you could detect changes on the speed of sound when passing between different zones of a fluid which have different thermodynamic conditions (i.e. different temperature and pressure as well, or density). Indeed sound propagates differently in the same fluid when different temperature, pressure or density, or any combination of them is present. ...

Top 50 recent answers are included