Answer:
A generator turns rotary motion into electricity. It is basically the inverse of a motor. Generally a transformer changes one voltage into another based on the number of conductor windings on each side. There are two sets of windings called the “primary” and the “secondary”.
Explanation:
<h2>The distance between students is 2.46 m</h2>
Explanation:
The force of attraction due to Newton's gravitation law is
F = 
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r = 
If we substitute the values in the above equation
r = 
= 2.46 m
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.
We're given two angular speeds, and we need to solve for a time.
Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .
Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.
So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !
At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:
They're in line, SOMEwhere on the circles, when
(a fraction of one orbit) = (the same fraction of the other orbit)
AND
the total elapsed time is a common multiple of their periods.
Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.
The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.
K (2π/ω seconds) = (K+1) (π/ω seconds)
2Kπ/ω = Kπ/ω + π/ω
Subtract Kπ/ω : Kπ/ω = π/ω
Multiply by ω/π : K = 1
(Now I have a feeling that I have just finished re-inventing the wheel.)
And there we have it:
In the time it takes the slower planet to revolve once,
the faster planet revolves twice, and catches up with it.
It will be 2π/ω seconds before the planets line up again.
When they do, they are again in the same position as shown
in the drawing.
To describe it another way . . .
When Kanye has completed its first revolution ...
Bieber has made it halfway around.
Bieber is crawling the rest of the way to the starting point while ...
Kanye is doing another complete revolution.
Kanye laps Bieber just as they both reach the starting point ...
Bieber for the first time, Kanye for the second time.
You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
Answer:
Explanation:
Total length of the wire is 29 m.
Let the length of one piece is d and of another piece is 29 - d.
Let d is used to make a square.
And 29 - d is used to make an equilateral triangle.
(a)
Area of square = d²
Area of equilateral triangle = √3(29 - d)²/4
Total area,
Differentiate both sides with respect to d.
For maxima and minima, dA/dt = 0
d = 8.76 m
Differentiate again we get the
(a) So, the area is maximum when the side of square is 29 m
(b) so, the area is minimum when the side of square is 8.76 m
