Refer to the diagram shown below.
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s
Thew energy stored in a capacitor of capacitance
and voltage between the plates
is
.
Substituting numerical value

I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.
The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.
You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.
Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.
<span>The answers are as follows:
(a) how many meters are there in 11.0 light-years?
11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m
(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?
1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au
(c) what is the speed of light in au/h? au/h
</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h