# Tag Info

44

Is it even possible to hit 350Gs of force to a hard drive? Sure is. Drop it on the floor. You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level. Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in ...

42

The assumption that you are making here is that with the same motion of a punch, that you are applying $\text{100 N}$ of force to both a wall and to the air. However, you should think about the most fundamental equation of Newton's laws, namely, $F=ma$ The most important part of this in relation to what you are talking about is that the force applied, ...

20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...

17

Here's an application where an ability to withstand high shock is important. Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research). We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a ...

16

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV). ...

14

Elastic collisions do happen at the LHC. The TOTEM experiment measures the differential cross section (rate as a function of angle) for proton-proton elastic scattering at the LHC. Here is their latest result. They don't publish an estimate of the elastic cross section, but according to their data it must be at least 25 mb (millibarns) (my first version of ...

11

The frequency of a sound wave cannot change as it crosses the water-air boundary. The wavelength can, and does, change but the frequency cannot because if it did there would be no way to match the two waves at the interface. This means that the higher frequency is not some quirk of the sound propagation, but that the colliding stones emit a higher frequency ...

9

Photons don't directly interact with each other, but if one photon pair produced an e+/e- then the second photon could interact with that pair. The interaction has to conserve the energy of the two photons and conserve their momentum as well of course. But yes they could (and most probably depending on their energy) just pass right "through" each other.

8

Anything that is not forbidden must happen. That's an important statement to keep in mind when approaching quantum physics. It doesn't mean that anything that can happen always happens, but it must happen at some time or another just like someone eventually has to win the lottery. That said, some protons do go through the LHC, ram into each other and ...

8

Your question highlights a common misconception. A satellite in orbit around the Earth is accelerating towards the Earth right now. Any object moving in a circular path has an acceleration towards the center of the circle because the direction, and therefore the velocity, of the object is constantly changing. This acceleration, called centripetal ...

7

TL;DR: If you have to choose either "near the handle" or "near the tip", the tip will work better. But there's a point in between these two that works even better; exactly where that point is depends on how you swing the sword, and how its weight is distributed. UPDATED now I am near a computer and can draw diagrams etc. If cutting off the zombie's head ...

6

The problem is equivalent to 4 spheres colliding simultaneously, where top sphere center is at $60^o$ relative to the $x'x$ axes (same goes for bottom sphere): We'll name them: sphere A (dark blue), and spheres 1, 2, and 3. During the collision the spheres will behave like springs with an infinite hook constant. The forces on the spheres will be ...

6

elementary particles (e.g. protons) Protons aren't elementary particles, they're made of partons (quarks and gluons) in "soup". Below, $\lambda$ is the wavelength corresponding to the energy of the interaction via the usual de Broglie relation and $r_p$ is the radius of the proton. At low energy with $\lambda >> r_p$ the interactions are just ...

6

The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses. If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed. You ...

5

There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...

5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...

5

Yes , the normal to the surface is the direction of reaction force. And the direction doesnt depend on the material of the object . But note that if friction is considered , direction of net reaction force changes

4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...

4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...

4

You are not the first person to ask this question. http://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't ...

4

No, Newton's third law is not violated. According to Newton's Second Law, we have that force is the rate of change of momentum with time, i.e. $$F=\frac{\Delta p}{\Delta t}$$ where $p$ is momentum and $\Delta t$ is time elapsed. When Punch Strikes Wall Initial momentum of fist = $mv$ Final momentum of fist = $0$ Force applied = $\frac{mv-o}{t} = ... 4 This is the same problem as the famous question states: which is heavier, 1k of feathers or 1kg of iron? It requires many more feathers to get 1kg of them, so it confuses most people. In this case, it requires you to have a big wing to be able to apply a force of 100N on air. 4 When two objects collide, they transfer momentum because they exert an equal and opposit force on each other (Newton's third law), and$\Delta p = \int F dt$. In order to know how fast an object moves after a collision, we need to know the velocities and mass of the objects before the collision and how elastic the collision is (conservation of kinetic ... 3 The assumptions that are being made here are perfectly elastic collisions with the walls where the atom will rebound with the same speed as it came it with. Its trying to calculate the average force based on the number of interactions per unit time with the wall, and is therefore fundamentally related to how long it takes for successive collisions between ... 3 I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the$dt$or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ... 3 Let's revisit "force on impact" for a moment. I will consider a "sticky" ball of mass$m$traveling at velocity$v$at a stationary wall. When it hits the wall (and sticks), what is the maximum force felt by the ball / wall? That is actually not a trivial question to answer. The ball has momentum$p=mv$, and if the impact time (time for the ball to come to ... 3 Usually absolutely nothing. Electromagnetism is linear, which means that the result of doing something with two photons is the superposition of the results of doing something with each one individually. By that reasoning, since one photon by itself just goes on its own merry way, then two photons, even if they go near each other, just go along on their ... 3 In a collision it's often the case that it's hard to measure exactly how long the collision lasts and exactly how the force between the objects changes during the collision. Squishy objects like nerf balls will collide relatively slowly while hard objects like billard balls will have a short collision time. However there is a well defined quantity called ... 3 In principle, you don't need to tune to 'exactly' this energy but having less than$m_h + m_Z$suppresses this diagram. Having more should typically give you a higher cross section because there is 'more phase space' the final state particles can be in, i.e. the 'excess' energy will just be used as kinetic energy of the final state particles ($Z$and$h\$ ...

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