Question

The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an acceleration of 2.5×10⁴ m/s². What is the force exerted on the pollen grains during this ejection?

1) 1.25 N
2) 2.50 N
3) 3.75 N
4) 5.00 N

Answer 1

The force exerted on the pollen grains during the ejection is 0 N.

Step-by-step explanation:

The force exerted on the pollen grains during the ejection can be calculated using Newton's second law of motion, which states that force is equal to mass times acceleration (F = m * a). In this case, the mass of the pollen grains is not given, but we can assume it to be very small compared to the force. So, we can use the equation F = m * a to calculate the force exerted. Given the acceleration of 2.5×10⁴ m/s² and the time of ejection as 0.30 ms, we can calculate the force as follows:

F = m * a

F = m * (Δv / Δt)

F = m * (0 - 0) / (0.30 × 10^-3 s)

F = 0 N

Therefore, the force exerted on the pollen grains during the ejection is 0 N.