# Tag Info

## New answers tagged estimation

1

Welcome to StackExchange! The area that is being referred to here is the area over which the power is "passing through". Using the Germany example, the simplest way to explain what I mean is that if the power is emitted uniformly in all directions, which is at least somewhat close to how speakers usually operate due to diffraction and their small size. The "...

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For safety it is advisable not to use capacitors that store more than 1 milli-joule. So no large Leyden jars in schools. A suitable size were those film canisters in the era of analog photography. The energy stored in a capacitor is $E = CV^2/2.$ So at $10$ kV the capacitance should not be larger than about $2.10^{-11}$ F = $20$ pF.

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I would rather compare it with the speed of the neural activity. A signal in your body would move by around 10 m/s if not some tricky feature allowing it to jump, increasing its speed tenfold. Typically a signal travels on a 100 m/s speed, except a few of them like pain, which is slower. Since those are also electric signals (just fueled by Kalium/Natrium ...

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It is typically difficult to determine the duration of the discharge in this kind of situation. It would be easier to work out how much energy is initially stored on the generator. This sets an upper limit on how much energy could be delivered to your body when you contact it.

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First, the other answers are correct that your understanding of the processes inside the sun is flawed. Secondly, I believe that the other answers are fundamentally wrong regarding shooting iron into the sun to cause it to explode. I am confident that shooting a large enough amount of iron into the sun would cause it to explode. Basically, an object with ...

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Yes, indeed it does. The astronaut, in this scenario, is above a fixed (I'm going to ignore the Sun's rotation for simplicity) point on the Sun's surface, even if the "height" is profound (150 gigameters) and so begins to fall. For the same reason that if you are placed at a fixed point of height about the Earth's surface, and are released, you would begin ...

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Without a perpendicular velocity, an object should fall directly towards the sun, because the only force pulls an object towards the center of the sun. This means that at the distance Earth is from the Sun, an astronaut would also experience acceleration towards the Sun.

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You should check your math. Convert all units to kg,m,and sec. Your basic understanding is correct. Unless there is some velocity perpendicular to the line from the object to the sun, the object will fall straight toward the sun. Although the acceleration is relatively small, the speed will build up pretty quickly.

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The thick hollow moon is an interesting case. On the outside it will have mass $M=(4\pi/3)(7/8)R^3\rho$ and corresponding surface gravity, but inside the shell there is nothing below, so the gravitational force (by Newton's shell theorem) is zero. That may look like the shell should be stable. However, every part of the inner surface will be subject to the ...

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but is it possible to ionize an insulator, say wood, with an extremely strong electric field? Yes it is possible. It's called dielectric breakdown. If it is theoretically possible, how strong would the electric field have to be in order for this to happen (orders of magnitude would be fine)? I don't know about wood, but most pure plastics (which ...

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As you can see from the comments, the net change is probably close to zero. Though, a photon does have stress-energy, and as it gets absorbed by the atmosphere or the land or waters on the surface, it transfers its energy to the energy of the parts of Earth. All internal energy such as thermal, rotational, and internal potential energy contributes to the ...

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The relevant equations are (Heat equation)$$\frac{\partial T}{\partial t} = \alpha\left(\frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2}+\frac{\partial^2T}{\partial z^2}\right)$$ and (fouriers law)$$\mathbf{q} = - k {\nabla} T$$ Thus you can see the thermal conductivity(k) tells us how much heat is transferred for a given temperature ...

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Yes, but it’s very small rate of change, and it is almost perfectly balanced with an equal loss. The total power of all the sunlight striking the earth is $\pi r_{earth}^2 {solarirradiance} = 173.9 PW$, divide that by $c^2$ and we get $1.935 kg/s$. However, a lot of this is simply reflected, and even that which is absorbed is almost perfectly balanced by ...

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While composing the question and much reflection afterwards, I couldn't help asking myself if there are objects in our Milky Way galaxy that are 1 kg or greater in mass that are traveling at speeds of 0.1c or greater. The typical energy released in a supernova is ~1-100 foe (1 foe = $10^{44}$ J = $10^{51}$ ergs). Roughly 1% of this energy goes into ...

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That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths. ...

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Yes, there was a quite significant solar activity event in 1859, when the electrification of our cities was still in its infancy. The event heavily ionized the upper atmosphere, and induced strong currents in the ground, to the degree that telegraph operators could communicate over landline with their batteries disconnected (the power being supplied by the ...

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It’s more of an engineering question than a pure science one, so a place to look for research is the IEEE. There’s a lot there. For example, the first hit on my search was “Unaddressed threat to power grid infrastructure: Electromagnetic pulse” (IEEE 2016) with abstract: Detonation of a nuclear weapon at high altitude, produces high magnitude ...

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Phonons are lattice vibrations. The distance between two consecutive phonons is of the order of 1/N where N is the number of atoms in the lattice. At room temperature a phonon travels approx 10 to 100 lattice constants before scattering. In this article https://www.nature.com/articles/srep17131 they say that a phonon travels $< 1 \mu m$ before ...

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The dimensionless amplitude of a gravitational wave (known as "the strain") is given approximately by $$h \sim \frac{2G}{rc^4}\ddot{Q},$$ where $\ddot{Q}$ is the second derivative of the mass quadrupole moment and $r$ is the distance to the potential source of gravitational waves. It is this strain amplitude that is detected by a gravitational wave ...

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Studying General Relativity (GR) one learns that what we call gravity is space time distortions due to the energy and momentum vector carried by massive objects. It is describe by specific tensors connecting energy momentum with space time curvature.. This is a classical physics system. It can be mathematically shown within GR that time varying ...

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To get the same fusion rate as in the Sun, you need not only the high temperature but also the high density, which requires high pressure. According to the cited wiki article the pressure at the core is 265 billion bar. The deepest oceans are 1,000 bar and Diamond anvil cells Schematics of the core of a diamond anvil cell. The culets (tip) of the two ...

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Think about how you got the formula. We had, $g$=$GM\over R^2$ and $g\prime$=$GM\over (R+h)^2$ . Now from them we get, $g\prime$=$R^2\over (R+h)^2$$\times g$ $\implies$ $g\prime$=$g\over(1+h/R)^2$ From here, by an approximation, $h<<R$ we get your formula. But if you want to get $g\prime$=0, the denominator must go to infinity, which implies $h$ ...

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Range of the gravitational field of any body, not just the earth, extends upto infinity. The formula you are using is an approximation that only applies when $h<<R$ which is definitely not the case when $h=R/2$. The more general formula for $g$ is $g = GM_e/r^2$ where $G$ is the universal gravitational constant, $M_e$ is the mass of the earth and $r$ ...

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The "storage time" of a photon in the cavity is given by the cavity round trip time, $\tau_{RT} = \frac{2L}{c}$, multiplied by the cavity Finesse $\mathcal{F}$. $c$ is the speed of light and $L$ is the length of the cavity. The cavity Finesse is essentially the inverse of the photon survival probability on a single round traip. That is, what is the ...

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Considering a radioactive nucleus, let the initial number of nuclei be $N_o$ with its disintegration constant $\lambda$. From it's first order decay/radioactive decay law, $N(t)=N_o e^{-\lambda t}$( let this be equation $\boxed{1}$) is the number of nuclei left at a general time t. The lifetime of each $dN$ nuclei which decays at a given instant is NOT the ...

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It was many times discussed here that the interaction between a photon and a mirror always is accompanied by the pressure of the photons momentum to the mirror. Or the mirror moves back, or some amount of the energy gets dissipated into inner vibrations (this is a local movement of a part of the mirror with fooling vibrations). The re-emission (as Bill ...

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Each unstable nucleus, in a given sample of radioactive matter consisting of nuclei of the same type, decays after a certain period of time. If you add up all those times and then divide the sum by the number of unstable nuclei in the beginning you obtain the average time for an unstable nucleus to decay. This average time is often called the (average) ...

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The answer for light, by Lubos Motl here gives an estimate of the time it takes for electromagnetic waves to be absorbed, $1/1,000$ of a second When we are talking of photons, we are in the quantum regime, and in the quantum regime there may be surprises not existing in the classical solutions. A photon hitting a perfect mirror will undergo elastic ...

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