If so, I need to do the following calculations… If you could provide a formula that I could use to do the calculations that would be very helpful. Also, could you please demonstrate how to use the formula so I can make sure that I’m doing the math right, thanks!

This is the information I have:

I am able to provide a 20mV input voltage at a 1.5mA input current for 1.25 seconds.

If I want to collect all of that energy over the 1.25 seconds in a capacitor, then release it in 0.1 or 0.2 seconds, then what would be my output voltage and current values over the output time of 0.1 or 0.2 be?

Please show applicable formula and calculations! Thank you.
Also, can a capacitor be charged off a variable DC supply?

The input power P= 0.02X 0.0015 =30X10^-6=3X10^-5 (watts)
Energy E=P x t=3X10^-5 x 1.25 =3.75 x 10^-5 (Joules)
From the equation E=0.5 C V^2
we don’t know the Capacitance, but we can see if C is higher the output voltage will be lower.

This is O/T, but is it true that you have to do a total of 4 things in order to have your Q & A hidden?

1. Uncheck box on Y!A that allows you to share updates to Yahoo before clicking Submit
2. In Yahoo! Pulse, hide profile
3. In Yahoo! Pulse, uncheck box allowing you to share updates
4. On Yahoo Answers itself, hide Q & A

Is it true that you have to jump through at least 4 hoops in order to get your basic privacy restored, or am I wrong? (as a noob just discovered Y! Pulse and find it to be a horrid rape of privacy)

And if that fails, will prayer restore this basic right?

===========On-topic question below==================================

Ok, that was seriously OT, so here is a backup question (if you prefer to skip the above):

If you had a choice, would you prefer A) an afterlife of some kind or B) oblivion? (let’s say for the sake of argument that you wouldn’t know what kind of afterlife it is – only that you have to pick between A or B.)

Keeping praying.

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Bible Study@9:30am Saturday . Sabbath Worship@11:00am Saturday.

http://www.webassign.net/rogacalcet1/3-2-figure17.gif

P = 200m^(-1/4 )

(b) Find an equation of the tangent line at the point on the graph that represent goat (m = 32).
(c) Find an equation of the tangent line at the point on the graph that represent man (m = 69).

and can you email me the answer?

b)
P = 200m^(-1/4)

dP/dm = 200 (-1/4) m^(-5/4)
dP/dm = -50m^(-5/4)

At m=32, dP/dm = -50 / 32^(5/4) = -0.65695

At m=32, P= 200/32^(1/4) = 84.0896

Equation of tangent line at (32, 84.0896) is:

P – 84.0896 = -0.65695 ( m – 32)
P = -0.65695m +84.0896+(32)(0.65695)
P= -0.65695m +105.112

c)
m=69
P = 200m^(-1/4)

dP/dm = 200 (-1/4) m^(-5/4)
dP/dm = -50m^(-5/4)

At m=69, dP/dm = -50 / 69^(5/4) = -0.251425

At m=69, P= 200/69^(1/4) = 69.3933
Equation of tangent line at (69, 69.3933) is:

P – 69.3933 = -0.251425 ( m – 69)
P = -0.251425m + (0.251425)(69) + 69.3933
P = -0.251425m + 86.7416

You will the manual for this.

Anyhow, we can plug in the numbers and see what we get. The photon energy Ep at the frequency given is

Ep = h.ν = 6.626 x 10^(-34) x 2.83 x 10^11 = 1.88 x 10^(-22) J

where h is the Planck constant. The energy El in the laser pulse is

El = 65 x 10^3 x 3 x 10^(-3) = 195 J

so the number of ‘photons’ in each pulse is

El/Ep = 195 ÷ 1.88 x 10^(22) = 1.04 x 10^24

for which the closest option is (e) 1.2 x 10^24.

Ion lifers work by having 2 electrodes one at the top and one at the bottom .There is high voltage at the top electrode that ionizes the air and the ions are strongly attracted downwards electrode ions rushing downwards impinges on air molecules causes upward force .

Well a rocket that uses laser the people at the ground shines a laser beam to the craft to get it to go up or the craft shines a laser beam .A craft that runs on pulse laser rides on pulse of the laser going on and off very fast.

My question is all this needs gigawatt power and very costly more so than chemical

Ion lifters do ionize air, however for space travel purpose, you use an ion thruster which is a bit different.

Laser based propulsion generally requires a secondary fixed point and pushes something away from that point, but could possibly be used _on_ the moving object without requiring the secondary fixed point.

The only reason it requires such huge amounts of power is because we’re not IN space when using them (at least right now). We’re fighting against gravity and atmosphere to get TO space, but once you’re there, both ion thrusters and laser propulsion are pretty cost effective.

Ion thrusters don’t produce all that much thrust, but they can keep producing thrust for a very long time, with a relatively low amount of fuel and power input. You can also turn them on and off very easily.

Lasers take a lot of electrical power to produce thrust because they use photons, which don’t mass much, as the opposite force. But, pop a couple of solar arrays up where they are always in direct (very very direct) sunlight, and your power is dang near free.

Without gravity and atmospheric friction, either method means you simply keep going faster and faster, at least until you get to significant fractions of the speed of light.

As for chemical rockets being cheaper, well, no, not really.

It takes over 35Kilowatts hours to produce one kilogram of liquid hydrogen.
The external hydrogen tank on a space shuttle (the one that provides most of the boost/thrust)contains roughly 35,000Kg of hydrogen.
That takes 1,225,000,000 watts to produce, or, about 1.25 gigawatts. That’s not including cooling after production, which is necessary, and doesn’t include the liquid oxygen production and cooling either. Even worse, once it’s burned, it’s burned. We could also talk about the pollution, and let’s not forget thermal pollution either.

What happens when a laser gets too hot? Generally, it melts down, altthough sometimes the tubes do shatter for tube types.
What happens when 35,000kilos of hydrogen and oxygen get too hot? Hmmm..

Now, if you are in space already, where do you get the materials for your hydrogen and oxygen for your chemical rocket? It’s easier to find materials to build solid state equipment, or even tube lasers.

Well, that should be enough for now.

Is this signal AC or DC? If AC, what frequency? What is the current level?

There is a really poor assumption I see on this forum, that the less information you supply the better the answer. That is a totally incorrect assumption.

.

I want to get rid of dragon pulse or dragon rush.

But i can’t decide, because they both have the same accuracy, but dragon pulse has 90 power and dragon rush always deals 40 damage sooooooo what should I get rid of?

You want Dragon Rush, but want to get rid of Dragon Pulse or Dragon Rush? I’m not sure what you’re really wanting to do, but here is some data you could use:

Dragon Rush
Power: 100 Accuracy: 75 PP: 10 Physical Desc: Has a 20% chance of making the target flinch.

Dragon Pulse
Power: 90 Accuracy: 100 PP: 10 Special

Dragon Rage
Power: — Accuracy: 100 PP: 10 Special Desc: Always inflicts 40 HP damage.

It would also help to know what Pokemon you’re trying to make this decision for. Some Dragon type Pokemon have no business running Dragon Pulse and Dragon Rush. Anyway, you should almost never have Dragon Rage on any Pokemon. Get rid of it.

Read through my generator page for a more detailed field flash technique as well as other generator related information.

How it works.

Well a rocket that uses laser the people at the ground shines a laser beam to the craft to get it to go up or the craft shines a laser beam .A craft that runs on pulse laser rides on pulse of the laser going on and off very fast.

The problems are twofold
1) The spacecraft would still need a rocket engine for attainment of earth orbit etc. So the laser equipment would be dead weight that would need to be carried after the initial ascent until reentry, offset by the need for less weight of rocket fuel. Plus, presumably, both the laser and the rocket engine would need to be at the rear of the spacecraft, so a way would need to be found to jettison the rocket assembly and re-attach the laser assembly in mid flight as reentry commences.
2) This is at best an experimental technique. It has never been tried in practice. (or has it?)