# 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 ...

31

Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ...

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 ...

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 ...

14

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). ...

9

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.

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 ...

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

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 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

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

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

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

Let's turn this around. In an inelastic collision, some of the energy, instead of remaining with the center of mass of the objects colliding, is dissipated as heat - an increase in the random motion of the atoms and molecules of the objects colliding. At the (sub) atomic level, the two particles involved in a simple collision are the same two objects that ...

5

Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ... 5 No mater the inertial referential: The pain is linked to the energy dissipated by the change of speed of the 2 objects in any inertial referential. Remarque 1: a part of the energy can be absorbed by the ball by deformation or heating. So a soft ball would make less pain. Remarque 2: The pain depends on what is static behind you and what is static behind ... 5 I would reason along these lines. Put the system in the centre of mass. Then, at collision, consider the plane through the centre of mass and perpendicular to the line through the centres of the balls. Since this plane is fixed in the frame of the centre of mass, this behaves like a wall the balls smash against, and therefore the velocity vectors after ... 5 In answer to the main question, matter does, in fact, "pass" through other matter. Starting from the macro scale (stars , galaxies), down to the micro scale (atoms), it happens all the time. The "free" movement of matter starts to get impeded, as the atoms start making latices (solids, crystals). But even at this scale, as Rutherford demonstrated, matter ... 4 I'll set up the problem in the following way. Let's assume the line of centers is along the x-axis. Initial speed of$m_1$is$u_1$at an angle$\theta$to the x-axis. Final speed of$m_{1}$is$v_1$at an angle$\alpha$to the x-axis. The final speed of$m_2$is$v_2$and is along the line of centers (the x-axis), such that$\alpha$is the angle between the ... 4 The reference frame doesn't matter. You cannot define a reference frame the way you did. Their deformation should be the same. The reason for this is that you can always transform to a different reference frame, i.e the centre of mass frame and physics should be the same as in any other inertial frame. Having established this, in the centre of mass frame ... 4 I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ... 4 In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$\sum F = \frac{\Delta mv}{\Delta t}$$ Think of$\int F \mathrm{d}t$being equal to the change in momentum, that is:$$\Delta mv = \int ... 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 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 ...

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