Answer:
B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.
Explanation:
A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.
In the motion map:
- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration
- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed
In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.
Answer:
nothing interest, i just hate that part i love only the calculations
Answer:
Force is 14.93N along positive y axis.
Explanation:
We know that force 'F' on a current carrying conductor placed in a magnetic field of intensity B is given by
where L is the length of the conductor
Applying values in the equation we have force F =
Thus force is 14.93N along positive y axis.
Answer:
Grow up man, this is completely based on your curriculum, we would need your book to answer, and this has to be done by you.
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.
