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July 4

Trivalve moluscs

Do these actually exist? It's just that I seem to remember reading something about them years and years ago somewhere...^{[dubious – discuss]} --Kurt Shaped Box (talk) 00:30, 4 July 2008 (UTC)[reply]

I see trivalves mentioned fairly often in fantasy fiction and science fiction. There's also this story about Constantine Samuel Rafinesque concerning a discovery of a trivalve mollusk, which was a joke played on him by John James Audubon. There's also a page from Google books about the incident. ~Amatulić (talk) 00:52, 4 July 2008 (UTC)[reply]

Is tungsten ferromagnetic?

I checked the article, and all it said was no data... ScienceApe (talk) 06:12, 4 July 2008 (UTC)[reply]

I am pretty sure the answer is no. Graeme Bartlett (talk) 06:48, 4 July 2008 (UTC)[reply]

No, it isn't. Though some tungsten-containing compounds are. 131.111.228.15 (talk) 07:44, 4 July 2008 (UTC)[reply]

Well I know they use tungsten projectiles in railguns, so if it isn't ferrous, then how do they propel it? ScienceApe (talk) 16:32, 4 July 2008 (UTC)[reply]

The projectile (and rails) of a railgun only need to be good conductors; no ferromagnetism required. Algebraist 16:40, 4 July 2008 (UTC)[reply]

Seems like it. Thanks. ScienceApe (talk) 17:52, 5 July 2008 (UTC)[reply]

where do herring gulls and black back gulls sleep at night?

Does anyone know what sort of places they usually sleep in? They seem to all leave the town and the local landfill area at night and all go off in the same direction. I'd like to be able to drive out and see them sleeping one night. —Preceding unsigned comment added by 84.67.233.220 (talk) 07:29, 4 July 2008 (UTC)[reply]

I believe Gulls are naturally cliff dwellers, I imagine they would roost on cliff faces if they're near the sea. For urban gulls, buildings would present a similiar landscape. —Preceding unsigned comment added by 62.25.96.244 (talk) 09:42, 4 July 2008 (UTC)[reply]

Just to note that I am not a Gull and am not called Cliff. --Dweller (talk) 15:05, 4 July 2008 (UTC)[reply]

Battery-powered microphone amplifier

Can someone suggest a battery-powered circuit for amplifying the output of an electret mike to drive a headphone (the type commonly used for portable media players)? The circuit should draw very little power, require no more than 2 AA batteries, and be buildable using easy-to-find parts. —Preceding unsigned comment added by 71.175.20.73 (talk) 11:45, 4 July 2008 (UTC)[reply]

You might look at [1], they are a huge producer of ICs and sell all kinds of audio amps. You would probably be best to purchase from [2]. Depending on your requirements, this could be a good starting point for selecting them [3]. You should know the impedance of your headphones. The data sheets of all these audio amplifiers should have sample schematics of simple circuits that you can start from. Depending on your input, you might need a preamp. It is also important to note that if it is a condenser mic, you will need to bias the microphone as well. Gjmulhol (talk) 12:13, 4 July 2008 (UTC)[reply]

On a side note, be careful about grounding. Audio signals are very sensitive to bouncing ground planes. Gjmulhol (talk) 12:17, 4 July 2008 (UTC)[reply]

"Grounding" is important, as is shielding of microphone leads, to prevent hum from influence of the powerline frequency electricity, but certainly no earth-ground connection would be needed. Edison (talk) 14:04, 4 July 2008 (UTC)[reply]

Termite control in Scarborough, Ontario, Canada

Dear Wikipedians:

Does anyone know some termite control shops that are located in Scarborough, Ontario, Canada?

Thanks.

74.12.39.232 (talk) 15:31, 4 July 2008 (UTC)[reply]

PCO Pest Control used to sponser my T-Ball team, but a phonebook or a google are better bets. Wily D 15:34, 4 July 2008 (UTC)[reply]

Hi. Have you tried yellowpages.ca? Thanks. ~A H 1^(TCU) 15:41, 4 July 2008 (UTC)[reply]

Overlapping tidal radii

Hi. On an issue of SkyNews magazine, I read that stars can have large zones where an object will orbit it, 1.5 - 15 light-years in radius. It said that the Sun's zone, called the tidal radius, is 3.5 light-years. It also said that they can overlap. This made me wonder.

The nearest star system, Alpha Centauri/Rigil Kentarus/Toliman, is 4.3 light-years away from us. If the sun's tidal radius is 3.5 light-years, I'd expect Alpha Centauri's radius to be a bit larger as it has higher total mass, say 4.0 light-years. This leaves 3.2 light-years of overlapping tidal space at the horizontal plane diametre. Well, is it plausable that AlphaCen has an oort cloud of comets, just like the sun?

I would like to know (and this is not homework, and I don't know if I'm able to calculate this):

The volume of overlapping space, in cubic light-years;
The number of Oort cloud comets estimated to be orbiting the Sun in that space;
The approximate percent of the Sun's tidal sphere surface that is within the overlapping zone;

Or, is the tidal zone not a sphere, but an elliptoid, just like the Heliophere? Oh no!!! Is the spherical model a good approximation? Or is it possible to calculate the same parameters for an elliptoidical model? Would we know in which direction the tidal elliptoid extends farthest and its shape, if it is indeed an elliptoid? Why is the Heliosphere elliptoidical, does its direction change over time, what direction does it point in, does it have any effects on comets just entering the Heliosphere, and could its shape be caused by Nemisis?

Consider the known non-returning comets, with eccentricities of ≥1. Do we know any which have the open end of the parabola(/hyperbola?) pointing towards the AlphaCen direction? Might they be from/going towards the AlphaCen system? Might some be coming back? Are there any other star systems which have tidal radii overlapping ours? Three or more overlapping together? Could this explain some Oort cloud collisions that send the comets plunging towards the sun?

How will AlphaCen's negative radial velocity change this zone over time? What about Gliese...(something, forgot its number, in Ophiuchus) and Barnard's Star as they approach our Sun? Is there an article concerning the subject of orbital tidal radii (and I'm not referring to the Roche limit, which "orbital radius" redirects to)? Might this explain some extinctions?

Using current extimates, about how many comets are sent towards our sun due to the overlapping tidal zone from AlphaCen per each specified specific period of time (eg. per year, per century, per millenium, and how many actually arrive)? Or is it not possible to answer some of these questions due to limited knowledge data? Thanks. ~A H 1^(TCU) 16:17, 4 July 2008 (UTC)[reply]

I've never heard of "tidal radius" before, and our article on it redirects to Roche limit, which has nothing to do with what you're talking about. The closest term to what you describe that I've heard is Hill sphere, but that's for one object orbiting another, rather than two separate stars. So, I'm not really sure what a tidal radius is meant to be (I can't see what your description has to do with tidal forces, for a start!). The maximum distance at which you can have a stable orbit around a star is going to depend on the other stars around it, and there won't be any overlap (they might go right up to eachother, though, I'm not sure if there would be a region of instability inbetween or not). It's certainly not going to be a sphere. An orbit around the sun in a plane perpendicular to the direction of Alpha Centauri could be larger than the distance to Alpha Centauri, I suppose, but one in the same plane as Alpha Centauri would be unstable if it got too close to the star. --Tango (talk) 16:47, 4 July 2008 (UTC)[reply]

A (crude) diagram used to vaguely illustrate the concept of stellar "tidal radii". ~A H 1^(TCU) 01:54, 5 July 2008 (UTC)[reply]

Hi. What I (and the magazine) mean by tidal radius might have nothing to do with tidal forces. It states that close orbits around a star can be stable, but the tidal radius, according to the magazine, is, beyond which, an object will no longer be within the gravitational reaches of the star.

Within the "tidal radius", objects like planets, comets, and other bodies are close enough to the star to be able to orbit it. Beyond that distance, the gravitaional forces of the rest of the galaxy cause any object to be lost to the star's grip. It also says that wandering comets might enter the Sun's tidal radius, perhaps from another star's orbital field, and be captured by the sun's gravity and form an orbit.

I see how a star's gravitational field can be affected by other stars, but perhaps it is referring to the zone in which a minor body (and perhaps another star) can be affected by/orbit the star. It also refers it to as the "tidal limit". It also says that some stars' tidal limits can overlap. The tidal limits may, indeed, go right up to another star, but I don't think it does in the case of Sol/AlphaCen.

The magazine goes on to approximate the tidal zone as a sphere, saying Altair's stretches 30 deg across the sky. It says that some comets may have entered the tidal limits of other stars or were flung out to empty space to become "vagabonds", and that those might enter our Sun's tidal radius and possibly may have been seen from Earth, but that is speculation.

Also, some orbits of comets are not a complete ellipse, but have two parallel or angled ends, forming a parabola. My question is, might some of those parabolic comets' orbits have come from or are heading towards the orbital boundaries of other stars? What about AlphaCen in particular?

The estimate of 3.5 light-years for our Sun's tidal boundaries would suggest that an orbit leaving this zone, probably already parabolic as opposed to elliptical, would be lost from the Sun's gravitational force and become an interstellar "vagabond". If it enters AlphaCen's zone of gravitational attraction but have already exited ours, might it form a new orbit, perhaps still parabolic due to its great distance of entry, around AlphaCen?

The distances in space are great, but if say AlphaCen had an Oort cloud similar to ours, and our Oort clouds overlapped, could a select few comets be influenced by other comets' orbits and gravitational attraction, so that their own orbits end up destabilised and plunge towards one of the stars?

If a comet was heading towards or around the AlphaCen system, could Proxima, itself more massive than Jupiter and (currently) closer to us than AlphaCen itself, act like Jupiter does in our solar system, flinging comets' orbits towards AlphaCen or our own Sun? If a comet approaches AlphaCen, could interactions between stars A and B cause a star-grazing comet to act differently than one near the Sun, and have its orbit drasticly changed (if it survives)?

Also, and this is speculation: if comets were theorised to have been one of the harbingers of life on Earth, might some of those same comets, containing similar organic compounds, have survived an interstellar journey and struck a planet in another star system, producing life there? Could the same have happened to Earth, from other star systems? Thanks. ~A H 1^(TCU) 01:16, 5 July 2008 (UTC)[reply]

I can't see anything to stop comets from our Oort cloud being perturbed and ending up orbiting another star, but I expect it would be very unlikely due to the distances involves (the perturbation would have to be very precise, or it would miss the other star). If this kind of thing did happen, it could possibly transfer life between stars, this is called exogenesis. However, I really don't see how the regions in which you can have stable (closed) orbits around two stars could possibly overlap - how would an object in that overlap "know" which star to orbit? If you had an object orbiting one star and an object orbiting the other such that their orbits intersect at one point, their apoapsis (further distance from the star they are orbiting), directly between the two stars. Then, at that intersection point, both objects would be travelling in the same direction (perpendicular to both stars), although possibly at different speeds. One object would then need to fall towards one star and the other towards the other star, but why would they go in different directions if they started out going in the same direction? The objects could "choose" to fall towards either star, so the orbits would be unstable. --Tango (talk) 01:32, 5 July 2008 (UTC)[reply]

convert -

A hole guage of 10-32 -> diameter(inches) —Preceding unsigned comment added by 75.60.90.55 (talk) 19:50, 4 July 2008 (UTC)[reply]

I'm assuming that 10-32 is refering to a screw size. Take a look at [4] for dimensional information. -- Tcncv (talk) 20:02, 4 July 2008 (UTC)[reply]

Or here: Unified Thread Standard. --Heron (talk) 21:02, 4 July 2008 (UTC)[reply]

Simple mechanics.

Let's say there's a stick lying on the ground and I kick it. Well depending on where I hit it and at what angle, the stick will gain some linear momentum and some angular momentum. If I always kick it staight on (i.e. perpendicular to the stick), then only the position matters. So my question is, how can I determine how much angular and linear momentum the stick will possess as a function of the kick's position? Now I gave this question a shot, and my guess is that because the stick has a bit of thickness, the angle between the force and the radius of rotation changes depending on rotation, and that product of the sin of the angle and the force would give the angular acceleration and so on. Is this right? A second and somewhat related question, why does the stick always spin around the center? —Preceding unsigned comment added by 65.92.4.238 (talk) 22:43, 4 July 2008

This is actually a bit difficult to work out (and easily confuses me still)

Basically it depends on amongst other things, the way the weight in distributed in the stick, I'll assume that the stick is straight and has equal thickness - so that the weight is evenly distributed along the length of the stick.

You can work this out using newtonian mechanics, you need to know about conservation of momentum, conservation of angular momentum (see also torque)

Let's say the stick is L long and you kick it at distance d from the middle (ie d<L/2)

(You also need to be able to calculate the Moment of inertia of the stick by integration, this is in fact mL²/12 (from List_of_moments_of_inertia) IF you want to calculate the exact numbers)

Suppose your kick at distance d gives an impulse 'S' to the stick, this impulse can be split into rotational and translational (straight line) momentums..

Assuming that the stick rotates about the centre for simplicity.

The rotational momentum can be viewed as one end of the stick moving forwards at speed V2 and one end moving backwards at speed V2, the translational motion can be viewed as the stick moving at speed V1 from the middle (combining all these gives the overall motion)

The linear motion is therefor mV1 (m is the mass of the stick) (ie the momentum)

The speed of rotation about the centre is given by V2=L/2 x A (A is the angular speed of rotation in radians per second.

The rotational torque is given by speed of angular rotation x angular moment of inertia = A x m x L² /12

since A = 2 x V2/L the torque = V2 x m x L /6

If you supplied an impulse of 'S' to the stick at distance d the torque about the centre was S x d

So S x d = V2 x m x L /6

and S = mV1

So m x V1 = V2 x m x L /6d

ie rearanging gives V1/V2 = L/6d

This means the ratio of the angular speed and linear speed can be calculated and it depends on how far from the centre it was kicked.

Additional work gives a general formula for collisions of two objects using similar methods.

(Apologies for any mistakes I've made whihc I'm sure other will point out..) The method is similar to what I've tried to describe above.87.102.86.73 (talk) 01:46, 5 July 2008 (UTC)[reply]

Your second and related question is one I find difficult to explain.. BUT if you consider the stick mentioned above, and consider it rotating about somewhere other than the centre of mass - you'll find that the stick would be violating the 'laws of conservation of momentum' since it's centre of mass would be moving about.. there for its momentum would be changing over time without an external force (ie newtons law that a body moves in a straight line unless acted upon by another force - in this case the body would be effectively 'spiraling about a point') - in general if you try to make a stick do this it will tend to rotate about it's centre and any extra speed will simply turn into linear momentum. Sorry if I haven't explained that very well.87.102.86.73 (talk) 01:52, 5 July 2008 (UTC)[reply]

Bounce

Why does a ball bounce when it hits the floor? —Preceding unsigned comment added by 65.92.4.238 (talk) 22:44, 4 July 2008 (UTC)[reply]

When it impacts the floor, it deforms by flattening at the bottom, this absorbs the kinetic energy of its motion and converts it into elastic potential energy. That elasticity then causes the deformation to undo, the ball returns the being spherical and is propelled upwards. Does that help? --Tango (talk) 23:27, 4 July 2008 (UTC)[reply]

Why does a steel ball bounce higher than a tennis ball when the tennis ball deforms more then? --antilived^{T | C | G} 01:15, 5 July 2008 (UTC)[reply]

Because more energy is lost to heat and sound when the tennis ball deforms and reforms than when the steel ball does. I'm not really sure why... --Tango (talk) 01:34, 5 July 2008 (UTC)[reply]

I have never seen a steel ball bounce higher than a tennis ball... Plasticup ^T/C 03:47, 5 July 2008 (UTC)[reply]

You haven't bounced it on the right surface. Try bouncing it off of another piece of steel. Someguy1221 (talk) 09:38, 5 July 2008 (UTC)[reply]

There's an excellent demonstration of this at the Exploratorium. There, the steel ball bounces off a very large piece of steel and each bounce of the ball is nearly as high as the previous bounce (so very little energy is being lost on each bounce).

Atlant (talk) 20:55, 5 July 2008 (UTC)[reply]

Or off another steel ball, as in the Newton's cradle. There's a good reason why they make those things out of steel balls and not tennis balls - steel is more elastic than a rubber bladder. --Heron (talk) 13:41, 5 July 2008 (UTC)[reply]

Okay cool. So theoretically, an object which is perfectly rigid would not bounce, right? —Preceding unsigned comment added by 65.92.4.238 (talk) 04:09, 5 July 2008 (UTC)[reply]

That's an excellent question. I think it depends on how you take limits. A "perfectly rigid" body should have an infinite elastic modulus, because it never has any strain, regardless of its stress. So one logical way to end up with a "perfectly rigid" body is to take an elastic body and let its elastic modulus go to infinity. When this body is at rest on the ground, the gravitational potential energy it started out with is completely in the form of elastic energy, which I believe is

{\frac {1}{2}}\,{\text{stress}}\times {\text{strain}}

. If the elastic modulus continues to increase, the stress must also increase to keep the elastic energy the same, so our "perfectly rigid" body would end up having infinite stress (in order to have a finite elastic energy with zero strain). If you accept this, then the body should bounce perfectly elastically (i.e. it should return to the same height it was dropped from and never lose any energy), because inelasticities are caused by strain. —Keenan Pepper 05:23, 5 July 2008 (UTC)[reply]

An alternative way of looking at this would be to say that a perfectly rigid body has no way of absorbing/losing the energy on impact (since it cannot deform)- therefor it would bounce perfectly.87.102.86.73 (talk) 12:07, 5 July 2008 (UTC)[reply]

For illustration, see the ball bouncer demonstration in the references of the Liquidmetal article, showing how steel balls bounce against three different materials: steel, titanium, and a metallic-glass compound called LiquidMetal. =Axlq 18:11, 5 July 2008 (UTC)[reply]

From Collision: "Collisions can be elastic, meaning they conserve energy and momentum, inelastic, meaning they conserve momentum but not energy, or totally inelastic (or plastic), meaning they conserve momentum and the two objects stick together." I'm not sure what perfectly rigid would mean, but the type of ball that doesn't bounce at all is called totally inelastic (the energy is absorbed by making the floor move down). --Tango (talk) 12:27, 5 July 2008 (UTC)[reply]

Well not really. The ball could have collided with the earth in a perfectly elastic manner, but the mass of the earth made its movement imperceptible. —Preceding unsigned comment added by 65.92.4.238 (talk) 17:02, 5 July 2008 (UTC)[reply]

It's not actually possible for something perfectly rigid to exist. Because of that, it's impossible to say what would happen if it did. If you were to magically fix the atoms in position with relation to each other, the electron orbitals in the atoms would distort as they get closer, so the atoms themselves wouldn't be perfectly rigid. If you were to throw two such bodies against each other, my best guess is that the enormous pressure would cause the two bodies to pass through each other somewhat, and the parts inside each other would be forced in every direction, canceling it out. The circle where the two surfaces overlap would still exert a huge pressure and force the two apart. More simply, they'd bounce off of each other perfectly elastically and extremely quickly, but still slowly enough that for a short time, they'd be partially inside each other. If only one body was "perfectly rigid" as I defined it, I have no idea what would happen. If you want to make the object more rigid by not having the electron orbits distort, their waveforms would take up the whole universe, and the Pauli exclusion principle would mean that no electron could have the same state as any in the rigid body, and they couldn't be the same as each other. Besides the ridiculously high energy states of the electrons this would require, it would also mean that said principle wouldn't be keeping the atoms from occupying the same space, so the rigid body probably wouldn't have any resistance from falling through the ground. Disclaimer: I'm not only not a physicist, but I only just finished high school, and the physics I learned there really doesn't apply here, so don't be surprised if I'm wrong. — DanielLC 16:29, 5 July 2008 (UTC)[reply]

Just because it can't exist, that doesn't mean we can't address the question. Usually when we deal with things like objects bouncing, we focus on the arangement of atoms indirectly by assigning certain constants, which were determined experentally, and don't take quantum mechanichs into account, just like we may, in classical mechanics, we cometimes treat an object as a single point, even though that would violate Heisenberg's uncertainty principle. So what my question is really asking is, using the laws of physics that we normally use to adress such phenomena as bouncing, how would a perfectly rigid body act? —Preceding unsigned comment added by 65.92.4.238 (talk) 17:13, 5 July 2008 (UTC)[reply]

A perfectly rigid body also cannot exist from the standpoint of Special Relativity. When one part of the ball is accelerated/stopped by the collision, the information of the collision cannot travel faster than the speed of light to other parts of the ball, so they cannot stop immediately. Icek (talk) 16:12, 7 July 2008 (UTC)[reply]

Relativity and communication speed

If a spaceship traveling at near light speed were equipped with a radio-wave Internet connection, how would special relativity affect measurements of its bandwidth and latency? Neon Merlin 23:05, 4 July 2008 (UTC)[reply]

The latency depends on the distance. As the latency changes quickly, the time-out would need to be rather large or change with the predicted distance of the spaceship.

If the spaceship travels towards Earth, then the bandwidth is larger, if it travels away from Earth, the bandwidth is smaller (see Relativistic Doppler effect). This just means that the spaceship traveling towards Earth will receive the bits at a higher rate, and the one traveling away from Earth will receive the bits at a lower rate. Icek (talk) 07:48, 5 July 2008 (UTC)[reply]

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