Toadfish Monastery

When does one use the -one and when the -body form?
(anyone/anybody, someone/somebody, noone/nobody)

In colloquial usage, I don`t think there`s much of a diference.  The -one form tends to be considered more proper / formal I think, but I can't think of an offhand example of one form being completely inappropriate.

I think I read about it recently but have completely forgotten the actual content.

Googling anybody anyone difference yields 16 Million hits !!!

An interesting proposal is:

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Both are singular pronouns, and use anyone when you are suggesting a person in a known group, as in, "Has anyone in this room been hunting lately?" And anybody when you are unaware of the group, as in "Has anybody ever gone thirty days without food and lived?"

Quote from: Swatopluk on October 22, 2009, 02:00:59 PM
An interesting proposal is:

Quote
Both are singular pronouns, and use anyone when you are suggesting a person in a known group, as in, "Has anyone in this room been hunting lately?" And anybody when you are unaware of the group, as in "Has anybody ever gone thirty days without food and lived?"

I think it works better the other way round !

Agree with Griffin - and if it's in a present group I'd insert 'here' as well for precision, although it's not strictly necessary.

I tend to think -one should be used in the same context / tone of writing that one would use 'one', and -body used when you'd use 'you' instead. 

_{hope that makes sense}


Overall, I think it's a minor point.  -body is less formal, so I'd stick with -one in polite / formal writing, but there's no reason to avoid it in colloquial use.

You mean in the sense of "Could somebody close the door please?" as opposed to "Could anyone defeat that dragon?"

Precicely.  Even if I'm not sure which part of the post (or which post) you are referencing.

When they announce 12 million viewers watched "How to Ruin Your Best Friend's Birthday"* on TV last night, how do they know it was 12 million, switched on, tuned in, and actively watching?

They don't. What they do is to place some boxes in a number of households with certain known statistic references and from their viewing patterns they infer a number.

It is suggested that Tivo and other digital boxes actually do the counting in a more global way, and it is expected that as more digital distribution happens and more TVs are plugged to the internet the statistics will be closer to reality.

Provided that users check one of those opt-in options. 
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As I walked in airports this past two days I noticed that servicemen/women had a mirror image of the flag on their shoulders, does any one know why?

Quote from: Sibling Zono (anon1mat0) on November 14, 2009, 01:04:12 AM
As I walked in airports this past two days I noticed that servicemen/women had a mirror image of the flag on their shoulders, does any one know why?

To confuse any passing aliens?

Thanks for the easy answer on the 12 million.

Usually mirror images are used, so they can be read when seen in the mirror. For example on ambulance cars. If you see the readable word 'ambulance' in your rear mirror, then you know to make room. If it was printed the normal way, you'd be seeing only the less readable mirror image. But why flags on uniforms should be mirrored, I don't know. For most flags it wouldn't actually matter  .

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A mathematical/physics question probably only proving that I am out of school dar to long:

If one would drill a shaft from pole to pole, keep it open (and cool) and remove the air in it, then dropped an object into it.
How could one calculate the time it would need to fall down the core and what the final speed would be?

The problem I have is that the acceleration is not time but space related* going from g=9.81 m/s² at the surface to zero at the core (linearly btw). But if one integrates the acceleration over the way the result would be a squared velocity, i.e. the units be wrong. With a constant accelaration the problem is easy (v=a*t; s=.5*a*t²).
Is it only solvable as coupled differential equations or a single but implicit differential equation? Or am I looking at it from a completely wrong direction?
Or to simplify/generalize the problem: An object is accelerated over a given distance with the acceleration dependent on the position. How fast will it be at the end and how long will it have taken?

*for a rocket the accelaration also changes but over time, so the integration is also done over time and therefore easy.

My physics of gravity never reached that point. My first guess is that the slope of the gravitational curvature would gradually increase to the maximum possible for that mass but I wouldn't be sure of the progression/graph for a given point. If you have the formula to calculate gravity at any given depth an integral should do the trick.

You'll have to ask a professor of physics for a more detailed solution though.

As I said, the gravitational pull at each point is clear (linearly falling from g to zero) but how to I correlate time t and location x?
a=f(x) but to get to v one has to integrate over t not x or the result is not m/s but m²/s².
So I have to somehow replace x with t but for that I'd have to know x(t) which is the thing I want to calculate in the first place.
Interestingly it is easy to calculate the energy needed to move the object in the gravitational field because that is not time-dependent.

For the velocity you could cheat and say that all the gravitational potential energy must have been converted to kinetic energy at the centre and set 0.5mv^2 to it .

Otherwise, if G is Gravitational constant, M mass of earth, R radius of Earth, r distance of object of object mass m  from  centre of Earth, at time t you get

md2r/dt2= -(GMmr)/R^3                  (from Newton)

Which is of form

d2r/dt2+kr=0

Where k is a constant.  Either by recognising it as the equation of a sinusoid from pendula/Hooke etc, or mathematically realising that for a second derivative to be able to cancel itself sin or cos is a good bet we can choose a test solution to be either cos(wt) or sin(wt).

We know at time zero that r is R, so choose the cos form and assume solution is of form
r=Rcos(wt)

For boundary conditions we know at time 0 from Newton above with r at R

d2r/dt2=-GM/R^2

or by differentiation

-Rw^2=-GM/R^2   

So w=1.24e-3
So from wT=2pi we get period as: 5062 seconds.

Now we've got r=Rcos(wt) as the equation of motion we can work out the rest.

At  1/4 the way through the period (90 degrees phase) differentiation gives the velocity will be
-Rwsin(1265.5w)

Where the sin value is at the pi/2 maximum and gives a velocity of 7898m/s.

(apologies for not working in g – Physicists don't like parochial units  ).

If it works, say Hi to DaveL and Bluenose for me.

edit:

Change "speed" to "velocity". Get kicked out the Physicists union for putting a minus sign on a "speed".

I also thought about the kinetic energy solution but was unsure whether that would work.
I think it will take some time to digest your solution.

But what about the more general problem with the acceleration an optional function of x with object accelerated along x?

I hated differential equations in math at school and at the university and have forgotten most about them apart from the most simple problems.