Weeds are very important agronomically because they reduce the yield of crops in three ways: by competing with the crop for water, light and nutrients, by interfering with crop harvest, and by contaminating harvested products with weed seeds and toxins. Weeds can reduce yield up to 50% and are responsible for millions of dollars in crop losses each year.
It’s biotic factor would be competition as Competition can be defined as an interaction between organisms or species, in which the fitness of one is lowered by the presence of another.
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.
Answer:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem = 
fem = 
- d (B l y) / dt = - B lv
B = 
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West
Answer:
F₁ = 4 F₀
Explanation:
The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:
F₀ = mv₀²/r --------------- equation (1)
where,
F₀ = Force on string at t₀
m = mass of ball
v₀ = speed of ball at t₀
r = radius of circular path
Now, at time t₁:
v₁ = 2v₀
F₁ = mv₁²/r
F₁ = m(2v₀)²/r
F₁ = 4 mv₀²/r
using equation (1):
<u>F₁ = 4 F₀</u>
Answer:
a) this stone was first called Schorl stone
b)the first modern name for this mineral was Black tourmaline
Explanation:
hope this helps you!
:)
