This is a very interesting problem ... mainly because it's different from the usual questions in the Physics neighborhood.
I can discuss it with you, but maybe not quite give you a final answer with the information you've given in the question.
I agree with all of your calculations so far ... the total energy required, and the power implied if the lift has to happen in 5 seconds.
First of all, let's talk about power. I'm assuming that your battery is a "car" battery, and I'm guessing you measured the battery voltage while the car was running. Turn off the car, and you're likely to read something more like 13 to 13.8 volts. But that's not important right now. What I'm looking for is the CURRENT that your application would require, and then to look around and see whether a car battery would be capable of delivering it.
Power = (volts) x (current)
7,050 W = (14 volts) x (current)
Current = (7,050 watts / 14 volts) = 503 Amperes.
That kind of current knocks the wind out of me. I've never seen that kind of number outside of a power distribution yard. BUT ... I also know that the current demand from a car battery during starting is enormous, so I'd better look around online and try to find out what a car battery is actually capable of.
I picked a manufacturer's name that I'd heard of, then picked their recommended battery for a monster 2003-model car, and looked at the specs for the battery.
The spec I looked at was the 'CCA' ... cold cranking Amps. That's the current the battery is guaranteed to deliver for 30 seconds, at a temperature of 0°F, without dropping below 12 volts.
This battery that I saw is rated 803 Amps CCA !
OK. Let's back up a little bit. I'm pretty sure the battery you have is a nominal "12-volt" battery. Let's say you use to start lifting the lift. As the lift lifts, the battery voltage sags. What is the required current if the battery immediately droops to 12V and stays there, while delivering 7,050 watts continuously ?
Power = (volts) x (current)
7,050 W = (12 V) x (current)
Current = (7,050 W / 12 V) = 588 Amps .
Amazingly, we may be in the ball park. If the battery you have is rated by the manufacturer for 600 Amps CCA (0°F) or CA (32°F), then the battery can deliver the current you need. BUT ... you can't conduct that kind of current through ear-bud wire, or house wiring wire. I'm not even so sure of jumper-cables. You need thick, no-nonsense cable, AND connections with a lot of area ... No alligator clips. Shiny nuts and bolts with no crud on them.
Now ... I still want to check the matter of the total energy. I'm sure you're OK, because the CCA and CA specifications talk about 30 seconds of cranking, and you're only talking about 5 seconds of lifting. But I still want to see the total energy requirement compared to the typical battery specification ... 'AH' ... ampere-hours.
You're talking about 35,000 joules
= 35,000 watt-seconds
= 35,000 volt-amp-seconds.
(35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)
= (35,000 x 1) / (3600 x 12) volt-amp-sec-hour / sec-volt
= 0.81 Amp-Hour .
That's an absurdly small depletion from your car battery. But just because it's only 810 mAh, don't get the idea that you can do it with a few rechargeable AA batteries out of your camera. You still need those 600 cranking amps. That would be a dead short for a stack of camera batteries, and they would shrivel up and die.
We use 1/o + 1/i = 1/f where o is the distance of the object, i as distance of the image and f is the focal length. Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span> <span>i = - 20 cm </span>
<span>For the case of the problem,</span>
<span>o = (20 + 30) = 50 cm </span>
<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>
