Efficiency is calculated through dividing the actual mechanical advantage by the hypothetical mechanical advantage:
- the actual mechanical advantage is 9J because that's how much work the light bulb doing
- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.
change the decimal to a percentage:
the light bulb had 9% efficiency
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
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Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
Preserved fossil<span> (like a fossil in amber, ice or tar.</span>
Explanation:
It is given that,
Voltage of the battery, V = 12 V
Current, I = 100 ampere-hours
Energy stored is given by the product of power and time taken. So,
P is the power, 
P = 1200 watts
This power can be used for 1 hour or 3600 seconds
Energy, 
E = 4320000 J
So, the energy stored in this battery is 4320000 J. Hence, this is the required solution.
