# Tag Info

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Yes it is possible, though highly unlikely with random position and velocities. The probability goes up with each of the following: They are in a closed space where they bounce off from the edge. Initial direction is towards the general direction of the other particles. If you consider any point on the object instead of only the center of mass. You can ...

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Here's how to intuitively understand that $a=g$. Take a metal ball having mass 1kg and drop it. Its downward acceleration is $9.8m/s^2$, right? Now take a second ball and drop it. Same thing, right? Now drop both at the same time. Same? Now connect them together (with a tiny drop of weld metal) into a single 2kg mass, and drop them. Do they suddenly slow ...

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Acceleration due to gravity remains roughly constant near the surface of the earth. Yes, $a = F/M$, but as mass increases, the force exerted by gravity increases too($F\ \alpha \ m1m2\over r^2$), keeping $F/M$ or $a$ roughly constant around the surface of the earth

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The key point here is to find enough expressions for $\dot m$. Here are some hints: 1: Define some constants first (such as mass density of raindrop, mass density of water droplets in space)! 2: Express $\dot m$ in two different ways, one using attributes of spherical geometry, another using the rate at which the raindrop picks up water droplets as it ...

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Their paper is inconsistent. They filled in $\omega = 264$ with the other quantities in SI units, so ω should be expressed in rad/s (often written $\mathrm{s}^{-1}$). So they assumed ω was already in rad/s. If they say they assumed $\omega = 264~\mathrm{rpm}$, that's not consistent with the values they plugged in. Your value of 69696 is hard to decipher ...

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Since as you're stating humid air ("condensed", even), why not assume net evaporation doesn't occur, so $\dot{m}=0$? Then set up a simple Newtonian equation of motion ($x$-axis is the vertical): $$F_{net}=ma$$ $$mg-\alpha v^2=ma$$ $$mg-\alpha \dot{x}^2=m\frac{d\dot{x}}{dt}$$ Which is separable. And for low velocities you could also use Stokes Law, ...

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As stated above, the mass of the whole system (sugar + water) doesn't change. In addition, with "ideal" mixing, the total volume of the water plus the total volume of the sugar equals the total volume of the mixture. However, this is not a sure bet, and there are many cases of a volume of one material mixed with a different volume of water, and the total ...

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The mass doesn't change at all, it will be just the sum of the water mass and the mass Added, what happens is the change of density because the mixture, in general the molecules get closer to each other ( through the intermolecular forces) and, this way, the volume become lower to the same mass quantity, what increase the density by the equation $$\rho = ... 2 I am not sure if Thompson ever determined the charge-to-mass ratio of a proton, but currently, the most precise measurements of the charge-to-mass ratio of a proton still use a magnetic field like Thompson, but rely on measuring (cyclotron) frequencies rather than deflection. As frequencies are the quantities that can be determined most accurate (see the ... -1 like the quantum mechanics if you could not measure something so that not exists/ how could you detect a mass if there is no force even if you get close to that to touch it you have a mass yourself so the force would appear and if not you are contradicting newtons law of gravity so the argument is not that simple mass and force are defined by each ... 0 In simplest terms and using Newton's mathematics: F = m * a. or Force F = mass m * acceleration a. Example #1 - A body on Earth. Now on the planet Earth, the gravitational acceleration "g" is about equal to 9.8 meters/second^2. So let's substitute a=g in the above equation. Then the force required to keep an object of mass m AT REST near the surface of ... 0 It is possible in theory, but there are challenges. You need the mass of the empty balloon, which presumably you can measure before filling it with helium. Then you need the volume of the inflated balloon, which you can measure by a 3D laser scanner. Then if you allow for the volume occupied by the material of the balloon (which will be close to 1 in ... 1 People have certainly measured the electron's charge and mass more than once in the last 100 years. See for example this table from the Particle Data Group, where you can find the constants you want to around 8 significant digits, much more than what was possible for Millikan. For comparison, Wikipedia claims that Millikan and Fletcher measured e to be ... 6 There are various reasons why water isn't a good basis for length measurements. In the long term, the most important is that water evaporates. To appreciate this, keep in mind that you can't just weigh out a kilogram of water and then somehow make a perfect cube container, then measure the sides. At the very least, dealing with the meniscus at the top would ... 9 When we pick standards for units it's important to pick standards that are as percisely defined as possible and measureable to very high precision in the lab. So for example the second is defined as 9192631770 times the period of the radiation emitted from a specific transition of the caesium atom. This is easily measured and indeed thousands of atomic ... 2 Well that depends. As always, you should be very careful with such reasoning as Due to the fact that gravity is related to the square of the distance should not the gravitational sum of every particle exceed the force when calculated by the center of mass. because this is a problematic statement. In general your (Newtonian) gravitational potential is ... 0 We know that is not regular energy nor is regular matter. Dark energy must clump as per space is distorted from its flat symetry (space is dragged like a blanked would be so more cubic space dragged by a massive object than by one less massive) in places where you have concentrated mass like neutron stars and blackholes. There seems to be the posibility ... 2 Set up a device to measure the upward pull of a helium balloon. The upward pull in Newtons is the net buoyant force. The net buoyant force is the weight of the displaced air, with the weight of the helium and the weight of the balloon subtracted from it. Using the ideal gas law, calculate the weight of the displaced air and the weight of the helium. This ... -1 According to the special relativity, there is a formula describing the relation of particle mass with its speed:$$m = {m_0 \over \sqrt {1- ({v \over {c}})^2}}$$where m_0 is the mass of the particle when it is still----rest mass. So as long as a particle moves, no matter how large or small the velocity is, the mass of the particle m will vary and be ... 1 To give a qualitative answer, the equivalence of gravitational and inertial mass is not a coincidence. Mass has inertia and this resists its motion through space-time (we're moving into the future at the speed of light!). The resistance leads to a bending of space-time and it is this that we interpret as gravity. Sort of... 0 Thank you for your help. Here is a new model built by my friends. Even though this model is not perfect (don't consider the table and hand will also vibrate with phone), I think it is good enough for me to explain it. In this mode, phone equilibrium changes according to applied force. However, it is impossible to go below the table since the phone is ... 2 Why m^2 in front of \phi^2 and why is m the mass? Fist of all, from dimensional analysis the prefactor to the \phi^2 term in the Lagrangian must have mass-dimension^1 2 in 3+1 dimensions since the Lagrangian has mass-dimension 4 and \phi has mass-dimension 1. This just tells us that we can write the term as m^2\phi^2 where m is ... -2 Wow 'll try and sort this out. Light travels at a velocity that is dictated by a relationship of properties of 'freespace' (empty) called permittivity and permeability. it is the ability to reach and pull for lack of a simplier explaination ... Lasers 'cut' because they heat up and either melt or disintegrate complex molecules. Once melted the excess can be ... 0 thanks for the reply. Here are two candidate answers I got from my friends and the other forum: Vibration amplitude decreases because the system damping factor is increased when the force of hand is applied Effective mass is increased when force is applied (because the table is vibrating also with the phone) Both of those factors are not considered on ... 0 I think the reason is that you are just adding these two forces i.e. Force by hand and Force by the vibrating mechanism. But you have to understand that the hand comes in contact to the mobile surface only when the surface is going up(towards the hand) i.e. when the positive cycle of the vibration is taking place. When the vibration is moving on the other ... 0 When the neutron binds gravitationally to the neutron star, it loses a significant fraction of its mass. Not immediately. When things falls they don't gain or lose energy. What really happens is more space is created above them and this makes them look (to people far away) a bit similar to falling. But the people far away don't think they are more ... -1 Checkout John Williamson's work, for example being discussed here: https://www.physicsforums.com/threads/is-the-electron-a-photon-with-toroidal-topology-what-is-that.614799/ He derives Maxwell's equations for the electron with the Lorenz transformation taken account for relativity. It's quite good IMHO; carrying on with the work that Dirac was doing with ... -1 Mass is energy, not being able to go at the speed of light in a direction. If the photons creating the electrons and positrons are going at the speed of light in a tight loop as John Williamson describes in a half dozen papers, then it can't also go at the speed of light in the perpendicular direction. 1 The textbook writer is referring to the concept of relativistic mass, which is the idea that accelerating a body tends to become harder and harder as its speed approaches the speed of light. This is sometimes thought of in terms of an increase in the object's mass as the speed increases. However, you should think of this as a deprecated concept that most ... 3 The cause for neutrino oscillations is that the flavour eigenstates are not the same as the mass eigenstates. Therefore, once you know the flavour of a neutrino, i.e. whether it is a electron, muon, or tau neutrino, the mass is not well defined. And the other way around: Once you know the mass, the outcome of a flavour measuring experiment is uncertain. The ... 2 Mass is the scientific term. It's a measure of how much inertia a body has; that is, how hard it is to push around if it was sitting floating in space. It is a fundamental quantity and has units of kilogram. Weight is not really a scientific term. It's a common-speech term that means Force due to gravity. So strictly speaking, a weight should be in units of ... 2 Short answer: no. Longer answer: In Newtonian gravity: definitely no. The gravitational attraction between two bodies depends only on their masses and separation. There is nothing else for the "amplifier" to couple to. In general relativity: practically speaking, no. For one thing in all "mundane" situations the predictions will be identical to those of ... 1 Keep in mind that the equation$$ E^2 = p^2c^2 + m^2c^4 $$is derived from the relations$$ \begin{align} E = \gamma mc^2,\qquad p = \gamma m v. \tag{1} \end{align} $$Therefore$$ p = E\frac{v}{c^2}.\tag{2} $$Although (1) is only defined for massive particles, it turns out that (2) remains valid when v=c, i.e. for massless particles. Indeed, we get$$ E= ...

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The definition of momentum isn't $\gamma m \dot x$. The proper definition of momentum is that it is the generator of translations. Then you find that for massive representations of the Lorentz group (~timelike curves), $p = m \gamma \dot x$, while for massless representations (~lightlike curves), $p$ is arbitrary, as long as $E = pc$. Another way of ...

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If one considers that the deBroglie relationship holds for photons we have $$p=\frac{h}{\lambda} = \frac{hf}{c} = \frac{E}{c}$$ which immediately gives us $$E=pc.$$ This is consistent with the Lorentz invariant energy four-vector magnitude which yields the mass of a particle: $$mc^2=\sqrt{E^2-(pc)^2}=0.$$

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I found the answer to my question in this publication: http://arxiv.org/pdf/hep-ph/0309075v2.pdf The mass formula is given by $$m = \sqrt{\frac{2m_1m_2}{\alpha}J+(m_1^2 + m_2^2)},$$ where $m_1$ and $m_2$ are the masses of the partons, $\alpha$ is the fine structure constant and $J$ is the angular momentum of the hadron.

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I found a 1988 paper by Mitrofanov et al which describes a Cavendish style experiment where the "big" mass was 706 mg - where Cavendish used balls of over 150 kg. The "small" mass ( the one on the torsion pendulum) was only 59 mg. This experiment was done to examine possible deviations of Newton's law at extremely short distances, and established a lower ...

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You don't need to worry about any spherically symmetric distribution of matter, dark or otherwise, that lies outside the radii you have measurements for. It has no effect (see the shell theorem). Your velocities tell you something about the mass interior to your velocity measurements. Unless you have lots of measurements as a function of radius, there's ...

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Sometimes people talk about "relativistic mass" which depends on your reference frame... this is perhaps what you're thinking of when you talk of your mass increasing as you approach the speed of light. However, more typically if a physicist says "mass" they mean the "invariant mass" or "rest mass" which is just your energy content (divided by the speed of ...

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The mass of your spaceship and your mass is not affected in your frame of reference. However from an outter stationnary observer measuring your mass in some kind of way would see it increase, they would see your spaceship contract along its direction of motion and they would even see, if they could look at a clock in your spaceship, that your clock tics ...

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Gravitation effect on neutrons have been demonstrated. Bouncing neutrons To obtain neutrons with quantized gravitational energy states, the team used a technique first described in 2011, in which a nuclear reactor produces neutrons travelling at 2,200 metres per second. These are then slowed to less than 7 metres per second and cooled to just a ...

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The classic gravitational measurement is the Cavendish Experiment, and the masses involved were a pair of 0.73 kg lead weights. So that forms an accessible reference. Other versions of the experiment may have used smaller weights, though.

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