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03-02-2003, 03:38 PM
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#31
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wsjb78
should edit this
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Here's a new one:
A man takes a piece of wood and makes it into something... he sells it to a second man... this second man gives it to a third man. The third man uses it everyday but he doesn't know he has it... what is it???
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03-02-2003, 05:06 PM
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#32
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Reverendpoon
should edit this
Citizen X
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A coffin.
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03-02-2003, 05:10 PM
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#33
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StuartD
should edit this
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See... that's just creepy.... clever... but creepy
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"If you are not going to heaven, why miss it by an inch?" - Sam Kinison
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03-02-2003, 05:31 PM
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#34
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wsjb78
should edit this
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You're too clever, Reverendpoon!
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03-02-2003, 06:20 PM
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#35
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barryf
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Edit
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Hi all,
Ok I got one too:
A man leaves his cabin with a gun to go hunting. He walks exactly one mile due south from his cabin and spots a bear. He tracks the bear exactly one mile due west and then shoots it. Finally he drags the bear one mile due north back to his cabin.
What colour was the bear?
This is not a joke! It really is possible to determine the bear's colour from my story. If you ALREADY KNOW the answer don't spoil it for others! But if you want to GUESS go for it!
B
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03-02-2003, 06:36 PM
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#36
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StuartD
should edit this
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white... right?
It has got to be a polar bear.... cause the only way you could walk north to your cabin was if it was at the north poll... in which case, you could walk south, west, east... any direction you want and still have to go north to your cabin
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"If you are not going to heaven, why miss it by an inch?" - Sam Kinison
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03-04-2003, 03:44 AM
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#37
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wsjb78
should edit this
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Add the necessary math symbols to the left side of the equation to make the following correct:
3 4 5 = 90
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03-04-2003, 10:48 PM
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#38
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Feynman
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Edit
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Quote:
Originally posted by XXXPhoto
wsjb78:
LOL, that is original... however, no; not answer I was looking for...
The correct answer was: He juggles the bundles.
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Please let me cross the bridge before you cross it.
Juggling them means that he has to throw them upwards.
The bigger the bundles, larger the movements he has to make and the longer the bundles have to remain into the air.
The longer the bundle have to remain into the air, the faster he has to throw them upwards.
the faster he has to throw them upward, the more reaction will be exerted by his throwing them of into the air.
I.e. the very act of throwing them into the air increases the force he exerts on the bridge (his weight plus the downward reaction from the accelerating of the bundle upward).
I won't get into calculations, but my intuition is that the average weight increase caused by the accelerating the three bundles to launch them into the air is exactly the weight of the three bundles. Assuming you would have constant acceleration and zero delay between the load switch. Accounting for friction and other losses and inefficiencies, it is actually more. And if you have the slightest delay between switching of the loads, the peak reaction will be more than three times the weight.
Here's a physics teaser that I never tried to solve, (nor anybody I know).
You have a fly resting inside a sealed large jar, the jar is on a sensitive balance.
If the fly starts to fly inside the jar (assume constant-altitude flight), what will the balance register ?
Will the reading on the scale vary according to the fly's altitude (methink yes, because of a ground effect; and it will probably take place if the fly is at an altitude lower than her wingspan).
Otherwise, is all the momentum imparted to the air lost into vortices that have no major vertical component ? (i.e. the weight registered will be that of the empty jar minus and all air momentum will be converted to heat).
I entirely forgot my fluid mechanics...
Anybody who ever tried the experience, come back to me.
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03-04-2003, 11:00 PM
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#39
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Feynman
should edit this
Edit
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Quote:
Originally posted by Reverendpoon
A coffin.
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AFAIK, dead have no capacity to use, nor to "receive".
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03-04-2003, 11:14 PM
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#40
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Feynman
should edit this
Edit
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Quote:
Originally posted by wsjb78
Add the necessary math symbols to the left side of the equation to make the following correct:
3 4 5 = 90
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You have to use the factorial symbol ( ! )
3! = 1 x 2 x 3 = 6
4! = 1 x 2 x 3 x 4 = 3! x4 = 24
5! = 4! x 5 = 120
-3!-4!+5! = -6-24+120 = 90
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03-05-2003, 12:19 AM
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#41
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XXXPhoto
should edit this
The Bad Man Touched Me
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Quote:
Originally posted by Feynman
Please let me cross the bridge before you cross it.
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Depends on how much ya weigh there Feynman... )
Quote:
Juggling them means that he has to throw them
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Actually it doesn't mean that at all; one could do a drop juggle (lifting and releasing bundles) and never have to throw them, merely lift and release...
Quote:
The bigger the bundles, larger the movements he has to make and the longer the bundles have to remain into the air.
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Bigger bundles do not matter, I think you might be referring to mass. As gravity is fairly constant across the length of said bridge (and rest of the Earth for that matter), bundles of similar shape would fall (accelerate downward) at pretty much the same change in velocity (delta v)... ;o)
Quote:
The longer the bundle have to remain into the air, the faster he has to throw them upwards.
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Not sure what you are referring to here, when juggling 3 objects, you are never in contact with more than one object at any one time. Yes, if you want a 5 stone bundle to reach the same position in the air as a 1 stone bundle, it will require more delta v exerted on it.
Quote:
the faster he has to throw them upward, the more reaction will be exerted by his throwing them of into the air.
I.e. the very act of throwing them into the air increases the force he exerts on the bridge (his weight plus the downward reaction from the accelerating of the bundle upward).
I won't get into calculations, but my intuition is that the average weight increase caused by the accelerating the three bundles to launch them into the air is exactly the weight of the three bundles. Assuming you would have constant acceleration and zero delay between the load switch. Accounting for friction and other losses and inefficiencies, it is actually more. And if you have the slightest delay between switching of the loads, the peak reaction will be more than three times the weight.
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Yes, to throw them upwards would increase the force (f=m*delta v) he acts on the bridge, but also the force the bridge acts on him; however, there are many ways that this force could be handled... he could get on his knees or lay on his back and skootch across the bridge... He could time his throws when both of his feet are off the bridge... any number of ways this could be addressed... The MASS of him plus one (or even two) bundles though would never be as great as that of him plus all three bundles... Reading the original question you will see it was geared toward weight (m*g) solution rather than a force equation.
As for the increase of force being too great, perhaps this quick demonstration will paint a clearer picture for you... Step on a bathroom scale holding an object (or two or three if you can juggle), watch the reading, toss an object into the air or begin juggling while looking at the needle... I don't believe it should ever go over your original reading... ;o)
Oh... from information given (at constant temp), the scale would decrease the further the fly moved away from the base of the jar always approaching the true weight (mass * gravity) of the sealed jar.
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03-05-2003, 11:00 AM
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#42
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Evil Chris
is drinking Heineken
Clone of myself
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Quote:
Originally posted by barryf
Hi all,
Ok I got one too:
A man leaves his cabin with a gun to go hunting. He walks exactly one mile due south from his cabin and spots a bear. He tracks the bear exactly one mile due west and then shoots it. Finally he drags the bear one mile due north back to his cabin.
What colour was the bear?
This is not a joke! It really is possible to determine the bear's colour from my story. If you ALREADY KNOW the answer don't spoil it for others! But if you want to GUESS go for it!
B
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Was Stuart's answer correct?
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03-05-2003, 11:32 AM
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#43
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Feynman
should edit this
Edit
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Quote:
Originally posted by Evil Chris
Was Stuart's answer correct?
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Yes.
He makes three 90 degrees turn and three equal lenghts legs. That isn't 360, so on a flat surface, he could not possibly be back at his point.
It works on a sphere.
Note that when he walks "due east", this leg is not a straight line but a circle of 1 mile of radius.
So the answer is "white".
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03-05-2003, 12:04 PM
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#44
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Feynman
should edit this
Edit
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For your suggested experiment to yield decent results, you should be able to take readings at least ten times faster than a typical "event" since you want to integrate, i.e sum up the real-time multiply of force for each time slice, by the duration value of each time slice.
If you touch a bundle for, say, 1/10th of a second, I suggest you take at least 100 force measurement per second.
Alternatively, try to perform your experiment with a medicine ball or some heavy object (20 lbs) and tell me how your scale behave.
Do [INTEGRAL f . dt ] / t over a complete cycle with whatever number of bundles you wish.
You'll need to estimate how long bundles remain in the air to evaluate the exact instantaneous force. (hint: can be computed from the height you throw them up)
Methink the average force has to even out to the total weight of all the bundles.
Otherwise, you could levitate an arbitrary weight with a limited force. A Chevette could carry 5000 tons of gravel if you'd rig it with a juggling mechanism (just don't drive it under a bridge ). Don't tell me that it does not exist all over the streets only because the DOT did not approve of the vehicle...
Physical laws applies at every end of the spectrum. Take a case, and within conceptual similitude limits, push it to the extreme and see if the hypothesis still holds.
But you were right about letting me cross the bridge: it also depends on how much *I* weight.
You're just a too clever boy ! Dayum!
Forgetting what we've learned is not a sin, it's the way of Nature.
Knowledge: use it or loose it.
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03-05-2003, 05:03 PM
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#45
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XXXPhoto
should edit this
The Bad Man Touched Me
Join Date: Feb 2003
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Quote:
Originally posted by Feynman
You're just a too clever boy ! Dayum!
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No, just a simple inbred country boy actually... I can't even afford decent artwork for my avatar like other boys...
Yes, had one access to more sensitive equiptment and the ability to plot numerous split second readings over duration of the event then one could figure the area under the curve and know the force exerted. As long as this force was always less than the force needed to cause the bridge to collapse (as well as less than the force of the fellow carrying all 3 across the bridge) then all is right with the world...
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An astronaut is working on a satellite and looses his safety line to the shuttle. He grabs his tool bag and readies himself to return to the shuttle only to discover that his jet pack has also malfunctioned... With no line to pull him back to the shuttle and no jet pack; how can he return?
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03-05-2003, 06:08 PM
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#46
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wsjb78
should edit this
Local Shadow Agent #1
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He just needs to throw his toolbag away from the shuttle... Newton's second law I think!
Each force applied to an object results in an equal strong force in the opposite direction!
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03-05-2003, 07:03 PM
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#47
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XXXPhoto
should edit this
The Bad Man Touched Me
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Quote:
Originally posted by wsjb78
He just needs to throw his toolbag away from the shuttle... Newton's second law I think!
Each force applied to an object results in an equal strong force in the opposite direction!
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Yes; would be better to throw individual tools one at a time, lest he have no easy way to correct an error... but giving the whole shebang a big shove would work... ;o)
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03-05-2003, 07:47 PM
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#48
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Feynman
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Edit
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Quote:
Originally posted by XXXPhoto
Yes; would be better to throw individual tools one at a time, lest he have no easy way to correct an error... but giving the whole shebang a big shove would work... ;o)
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And with nothing to throw away, I always thought I would attempt to disconnect a hose or puncture a small hole in my suit and jet myself back home. All tricks are good to prevent being "Major Tom-ed"
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03-05-2003, 07:58 PM
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#49
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XXXPhoto
should edit this
The Bad Man Touched Me
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Quote:
Originally posted by Feynman
And with nothing to throw away, I always thought I would attempt to disconnect a hose or puncture a small hole in my suit and jet myself back home. All tricks are good to prevent being "Major Tom-ed"
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Actually, even if you had 'nothing' to throw away, you could still make a rapid throwing motion... would not work nearly as well though...
Peeing down the leg of your suit would work as long as you could maintain the output/stream... But it would collect in bottom of your boot and cause you to both stop and have stinky/wet piggies... ;^)
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03-05-2003, 08:51 PM
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#50
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Ounique
should edit this
Content, Fetish and Fun
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Okay, here's a brain teaser! FIND ME A DATE!!!!
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03-05-2003, 09:16 PM
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#51
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Feynman
should edit this
Edit
Guest
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Quote:
Originally posted by XXXPhoto
Actually, even if you had 'nothing' to throw away, you could still make a rapid throwing motion... would not work nearly as well though...
Peeing down the leg of your suit would work as long as you could maintain the output/stream... But it would collect in bottom of your boot and cause you to both stop and have stinky/wet piggies... ;^)
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It would move your center of gravity by zero.
If you piss 0.5 kg (that's a two cup piss man) and you weigh 100 KG (with suit an all), and you shift the piss weitght by 1 meter, your body would move 1/200th of a meter, or 0,5 cm. Your center of gravity remains unaffected. Then end velocity after the piss lands into the boot would be ZERO. Action-reaction would propell you while the piss flies, but the piss hitting your boot would bring your CG back to point zero.
I'd rather die not in my own piss. Personnal choice, only personnal choice.
:-)
Rapid throwing motion does not work for after the throwing, a force is exerted upon your body to STOP your arm.
There are violation of action-reaction law, but they are not recognized by mainstream physics. And they don't happen with linear motion. They require transfer between variable speed rotational movement and linear movement. It is alledgedly produced by compounded multirotational-linear movement.
There are patents and claims of inertial thrusters that eject no mass.
Nasa is apparently working on one.
A fantastic source of info for that is Infinite Energy Magazine.
Google it and you'll hit right on it.
Have fun !
IMO, the best way to get out of it would be to puncture a cooling fluid line (providing you have your Space Swiss Penknife handy) . It would get ejected at the suit pressure, and have more weight than the gas. Gas has higher theoretical yield, but in that case, since you probably don,t carry a large mass of gas, I think it better to eject coolant (but not all of it unless you want to turn into a half-frozen half sizzling barbecue.
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03-05-2003, 10:07 PM
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#52
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XXXPhoto
should edit this
The Bad Man Touched Me
Join Date: Feb 2003
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Quote:
Originally posted by Feynman
Rapid throwing motion does not work for after the throwing, a force is exerted upon your body to STOP your arm.
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Yes, you would only gain benefit for the brief period your arm is in motion. A 'punching' motion would be more helpful that a throwing motion. And both would cease to help once your arm stoped moving.
Better to drown in your own piss than someone else's amber stream...
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03-05-2003, 11:01 PM
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#53
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Feynman
should edit this
Edit
Guest
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Quote:
Originally posted by XXXPhoto
Yes, you would only gain benefit for the brief period your arm is in motion. A 'punching' motion would be more helpful that a throwing motion. And both would cease to help once your arm stoped moving.
Better to drown in your own piss than someone else's amber stream...
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LOL ! You have a point !
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03-05-2003, 11:05 PM
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#54
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Feynman
should edit this
Edit
Guest
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Another kewl one... The Liar's Paradox
The paradox goes like this:
1. Epimenides is a Cretan.
2. Epimenides states, "All Cretans are liars."
Solution at http://david.tribble.com/text/liar.htm
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03-05-2003, 11:27 PM
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#55
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XXXPhoto
should edit this
The Bad Man Touched Me
Join Date: Feb 2003
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Quote:
Originally posted by Ounique
Okay, here's a brain teaser! FIND ME A DATE!!!!
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Um... ok... will take a stab at this... July 25, 2003... That sounds like a HOT DATE to me...
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03-05-2003, 11:34 PM
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#56
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Phoenix
should edit this
Senior Member
Join Date: Oct 2002
Location: Toronto
Posts: 236
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Quote:
Originally posted by wsjb78
Add the necessary math symbols to the left side of the equation to make the following correct:
3 4 5 = 90
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how about (3^2)(sqr(4))(5)=90
sqr means square root of
QED
Last edited by Phoenix; 03-05-2003 at 11:40 PM.
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03-06-2003, 01:29 AM
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#57
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AdultInsiderJen
should edit this
Member
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Location: Canada
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Wow we've really got some smart cookies here!
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03-06-2003, 05:40 AM
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#58
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wsjb78
should edit this
Local Shadow Agent #1
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Quote:
Originally posted by Phoenix
how about (3^2)(sqr(4))(5)=90
sqr means square root of
QED
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Well, you did add a "2" and therefore not only adding symbols!
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03-06-2003, 09:31 AM
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#59
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Phoenix
should edit this
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Quote:
Originally posted by wsjb78
Well, you did add a "2" and therefore not only adding symbols!
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damn rules
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03-06-2003, 01:03 PM
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#60
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Phoenix
should edit this
Senior Member
Join Date: Oct 2002
Location: Toronto
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Brian dies in Texas, while Cathy dies at sea. Cathy's death was a cause for much celebration, after Brian.
Why?
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