For a flower to be pollinated, pollen from an anther (which is located at the top of the stamen) needs to reach a stigma (at the top of the pistle.) Some plants are genetically capable of pollinating themselves if their own pollen reaches their own stigma; some plants are not capable of self pollination under any circumstances.
For plants that can genetically self pollinate, but would prefer not to, they can avoid this by having their pistil and pollen/stamens mature at different times. If the stamens mature first, the pollen will be dispersed by animals or wind or whatever dispersal mechanism it relies on. Then by the time the pistil is ready to be pollinated, there is no pollen left in that flower to land on the stigma.
To find or discover by investigation?
Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
Answer:
T = 8.55 N
Explanation:
When string makes an angle 40 degree with the vertical then it will have two forces on it
1) gravitational force (mg)
2) Tension force in string (T)
now we know that net force towards the center of the path is known as centripetal force and it is given as
Answer:
See Explanation
Explanation:
The relationship between angle of an incline and the acceleration of an object moving down the incline.
As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object
Parallel force = mgsin, perpendicular = mgcosΘ
With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.
mgsinΘ = ma
Fnet = ma
B.) From Fnet = ma
Fnet = ma
a = Fnet / m
Where Fnet = Net force = mgsinΘ, a = acceleration
