Respuesta :
As we know that pollens are ejected out from rest
So its initial speed is
[tex]v_i = 0[/tex]
acceleration of the pollen that comes out is
[tex]a = 2.5 \times 10^4 m/s^2[/tex]
now it took time
[tex]t = 0.30 ms[/tex]
so we will use kinematics to find out the speed with which it will ejected out
[tex]v_f = v_i + at[/tex]
[tex]v_f = 0 + (2.5 \times 10^4)(0.30 \times 10^{-3})[/tex]
[tex]v_f = 7.5 m/s[/tex]
so final speed will be 7.5 m/s
Pollen grains speed : 7.5 m/s
Further explanation
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration
• An equation of uniformly accelerated motion
[tex] \large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}} [/tex]
V = vo + at
Vt² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
Acceleration is a change in speed within a certain time interval
a = Δv / Δ t
[tex] \displaystyle a = \frac {v2-v1} {t2-t1} [/tex]
The velocity graph with respect to time (v-t) in this motion will be in the form of sloping straight lines with a gradient of tan θ = a
For acceleration that is positive (a> 0) is called accelerated motion, the graph is sloping upward, while for acceleration that is negative (a <0) is called slowed down motion, the graph is sloping downward
The pollen grains to be ejected out of the flower in 0.30 ms at an acceleration of 2.5×10⁴m/s², then :
Δt = 0.3 ms = 3.10⁻⁴ s
a = 2.5 × 10⁴ m/s²
so :
[tex]\rm \Delta v=a\times \Delta t\\\\\Delta v=2.5\times 10^4\times 3\times 10^{-4}\\\\\Delta v=7.5\\\\\Delta v=vf-vi(vi=0),vf=\boxed{\bold{7.5\:m/s}}[/tex]
Learn more
The distance of the elevator
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resultant velocity
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the average velocity
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