For man only.

.... and a riddle (maybe some of you have known this)

This useful tool, is commonly found in the range of 7 inches long.
The functioning of which is enjoyed by memebers of both males and females.

It is usually found hung, dangling loosely, ready for instant action.
It boats of a clump of little hairy things at one end and a small hole at the other.

In use, it is inserted, almost always willingly, sometimes slowly, sometimes quickly,
into a warm, fleshy.moist opening where it is thrust in and drawn out again and again
many times in succession, often quickly and accompanied by squirming bodily
movements. Anyone found listening in will most surely recognize the rhythmic,
pushing sound, resulting from the well lubricated movements.

When finally withdrawn, it leaves behind a juicy, frothy, sticky white substance,
some of which will need cleaning from the outer surfaces of the opening and
some from its long glistening shaft. After everything is done and the flowing and
cleansing liquids have ceased emanating,
it is returned to its freely hanging state of rest,
ready for yet another bit of action, hopefully reaching its bristling climax twice or
three times a day, but often much less.

So, WHAT IS THAT TOOL ???

Don't think too far, it's a very familiar tool.



Tao_lao

A tooth-brush.

Well done.
Anybody know a more suitable answer(until now i dun know a ánswer like that).
Here is a new question(i took from the net also).

Rest Frame Description.
There is a large array of lights spaced at one meter intervals along a segment of the x-axis. The lights are labeled from 0 to N in the positive x direction, and are labeled from -1 to -N in the negative x-direction. There are two targets T1 and T2. When at rest relative to the array of lights, T1 and T2 each have a length equal to 1000 meters. Now let T1 and T2 move in opposite directions along the x-axis. Let T1 move in the positive x-direction with velocity V relative to the array of lights and let T2 move in the negative x-direction with velocity -V relative to the array of lights. Let the array of lights be one light-hour in length.
T11111111111111 ->
<- T222222222222222
... -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 ...
When the end points of T1 and T2 meet at the light labeled 0, the experiment begins which I'll call t0. At some random time between t0 and t0+100 seconds as measured in the rest frame of the light array, one of the lights on the light array turns on and off (we can make the on and off time interval arbitrarily small).
Modification to problem: In response to some replies, instead of making the random time interval between t0 and t0+100 as originally stated, let the time interval be between t0-100 and t0+100.
As viewed in the rest frame of the lights, the probability of target T1 being at the light that is randomly turned on and off is equal to the probability of target T2 being at the light that is randomly turned on and off (its a symmetrical problem and T1 and T2 have equal lengths). When this experiment is repeated an arbitrarily large number of times, we find that this is indeed true.
The Problem
A consequence of Special Relativity is that length contraction and time dilation result when measurements are made in reference frames that have different velocities. This problem removes the time dilation aspect from the experiment, so that we can examine the length contraction consequence by itself. However, when this is done, when the probabilities are computed from the viewpoint of either the T1 frame or the T2 frame, SR does not predict that T1 and T2 have an equal probability of being at the light that is randomly turned on and off. As viewed in the T1 frame, at any point in time, T1 spans more of the lights than T2 does (SR length contraction consequence). If T1 spans more of the lights than T2 does, how can the probability of T1 being at the light that is randomly turned on be the same as the probability of T2 being at the light that is randomly turned on?

Just for fun.Enjoy yourself.
Tao_lao