Question

The flowers of the bunchberry plant open with astonishing force and speed, with pollen grains accelerating from rest to 7.5 m/s. a. What impulse is delivered to a 1.0 * 10-10 kg pollen grain? b. What is the average force if this impulse is delivered in 0.30 ms? c. How does this force compare to the weight of the pollen grain?

Answer 1

Answer:

a) J=$7.5*10^{-10}$

b) F = $2.5*10^{-6}N$

c) The force is 2551.02 times bigger than the weight of the pollen grain.

Explanation:

a) we know that the impulse J is equal to:

J =$MV_f-MV_i$

where M is the mass, $V_f$ is the final velocity and $V_i$ the initial velocity. So, replacing values, we get:

J =$(10^{-10})(7.5m/s)-(10^{-10}(0)$

J =$(10^{-10})(7.5m/s)$

J=$7.5*10^{-10}$

b) We know that the force is equal to:

F = J/t

where J is the impulse and t the time. So, replacing values, we get:

F = $\frac{7.5*10^{-10}}{0.0003}$

F = $2.5*10^{-6}N$

c) The weight W of the pollen grain is calculated as:

W = $10^{-10}*9.8$

W = $9.8*10^{-10}N$

So, if we compare the force and the weight of the pollen grain, we get:

$\frac{2.5*10^{-6} }{9.8*10^{-10} } =2551.02$

it means that the force is 2551.02 times bigger than the weight of the pollen grain.

Answer 2

Answer:

a)  I = 7.5 10⁻¹⁰ N s , b)  F = 25 10⁻¹⁰ N , c)  F / W = 2.55

Explanation:

a) The momentum is related to the moment by the equation

       I = F t = Δp

       I = m v - m v₀

In our case the initial speed is zero, let's calculate}

       I = m v -0

       I = 1.0 10⁻¹⁰ 7.5

       I = 7.5 10⁻¹⁰ N s

b) the average force

       F t = mv

       F = mv / t

Let's calculate

       F = 1.0 10⁻¹⁰ 7.5 / 0.30

       F = 25 10⁻¹⁰ N

The weight of a pollen grain is

      W =m g

      W = 1.0 10⁻¹⁰ 9.8

      W = 9.8 10⁻¹⁰

The relationship between the two is

      F / W = 25 / 9.8

      F / W = 2.55