Kedas

Found a link for you: http://www.scitoys.com/scitoys/scitoys/electro/electro4.html

http://www.directindustry.com/prod/relays_58481.html They aren't very expensive you have lot's of diff. ones you need a simple old one with coil (that blue part in picture) not the advanced electronic versions.

To get a changing current you can just feed the relay with the battery and make sure the circuit runs through the closed contact of the relay when it has enough current the relay will break the current this will on its turn close the relay again and so on. (its not very healthy for you relay but its the easied way to get an changing current) you will probably already have sparks in your relay due to the EMF

The amount of energy can be compared with the energy stored in a capacitor. clouds and earth are the two parts with air in between. it builds up until the air between can't hold the two potential differences apart anymore. The energy in a capacitor is: C: capacitor value (Farad) V: The voltage (Volt) Q: charge (represents difference in amount of electrons) (Q=C.V) E: Energy stored in capacitor E = Q.V / 2 or C.V² / 2

uhhmm you do know that the gasses will be water again the moment that you let it spark/burn? To get the spark with a relay and a transfo (or two) this should work maybe you should ask YT2095 I'm pretty sure he has experiance in making sparks.

just a thought: Life begins when it can generate information (for self use) that is based on stored data AND new data. Then we should make different levels of life based on how 'good' they can do that. (this can be a variable during your life) I shouldn't include 'self-preservation' because someone that is trying to kill himself is still a lifeform, or not?

I wasn't asking what could be needed I wanted to know what he/she had to work with

If you want star maps and much more for any location and time you can try the following program: http://www.starrynight.com/support/softwareupdates/update.php?product=Digital+Download&Submit=Search http://www.starrynight.com/download/DigitalDownload-Win.zip (56MB) a good graphics card with OpenGL driver is a must. (Geforce 4 or later) to add stars to mag14 you can find them here. (just copy them in the right directory) http://www.starrynight.com/en/backyardfull.shtml You do need a serial number to make it work You can get a 15 day trial key here: http://www.starrynight.com/digitaldownload/trial_download.php (if 15day's isn't long enough to test it there are places to get a less limited key) BTW the program isn't really cheap but I like it.

Also the frequency of the voltage/current through it will change it but I guess it will only be DC or low frequencies. How much wire do you have and how can you measure resistance?

The direction will depend on the directions of the accelerations of the container and if the balloon meet the condition that I defined a previous post. You can include gravity effect that is also expressed with an acceleration value. if you want to know the direction of the balloon then you have to calculate if the condition is met to go in the opside direction or same direction of the acceleration. so the shape of container or balloon have notting to do with it it's only that condition and the direction of the acceleration that will tell where the balloon is going.

To answer the question: It's an easy question "but this year all answers are different."

saying how you can add two vectors doesn't tell anything about those vectors. (we only needed the horizontal components) did I miss other math? whatever, you can turn it like you want maybe your thoughts are good but you'r sure bad in explaining them.

if you are the physicist then how come that I have to come up with the little bit of math that is needed to calculate/explain it? also you started with 'negligable weight' of the balloon while I took it into account. yes, an acceleration has the same effect as gravity (thanks A. Einstein) it's a good base to come up with the answer but I doubt that you would get your points if that was all you have to say about that question.

The balloon? please tell that you are joking the top would be the bottom if you turned it upside down.

my point is that it isn't because it was 'just' an EM radiation that it won't be a current when it passes a conductor (You). I already said that the spectrum was between 1Khz and 1Mhz The microwave frequency is a lot higher but if you want to be on the safe side. it isn't an ideal filter you know see my link again with diff. dB values. yes you can make more holes just like in the microwave oven, the holes there have the same function. (only they block just one frequency 2.45 GHz, you block a range.)

EM=electric/magnetic radiation they go hand in hand. Don't forget that you ARE a conductor in a strong field (a bad one though) If you want a whole size take the the size of the wholes in the metal of your microwave.

Nice math that isn't helping us. "Your question is rather unfair" I didn't stop you to say that your answers was only valid for certain conditions. but forget about all that lets try to get the formula: only about the horizontal force: -What is the force on the balloon due to the diff. air pressure due to the acceleration? That is the mass of the air volume of the balloon multiplied with the acceleration of car. F1 = RHOair . V . a -Now what should be the force on the balloon to keep it the same speed as the car? That is simple: the total mass of the balloon multiplied with the acceleration of the car. (Mt= total mass balloon) F2 = Mt . a so F2 must be < F1 for the balloon to move to the front. This is the condition for the force on the balloon. like you can see the acceleration doesn't matter or: (Mb=mass empty balloon) Mb + RHOhe . V < RHOair . V But this is the same condition to know if a balloon will go up or not!! So a balloon that goes up in the car will also move forward when the car accelerate. (and that condition was more or less given in the question.) so the weight of the balloon doesn't have to be mentiond only that it goes up. anyone saw any mistakes?

He probably meant the computer you have now Although mine is already out of date

anyone feeling lucky to write a formula down to know what the weight of the ballon can be in a certain situation?

It's about creatng an easier way around you so you won't be hit. about metal bucket etc. keep in mind that there should be a nice clear space between you and the Faraday cage otherwise you will part of the cage. a bad part

so you do realize now that a helium filled balloon will not certainly move to the front of the car? (it will depend on the weight of the ballon in relation to its size)