Some plants disperse their seeds by having the fruit split and contract, propelling the seeds through the air. the trajectory of these seeds can be determined by using a high-speed camera. in one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.7 m/s . the launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight upward and −90∘ straight downward.

23000 frames/s

Explanation:

The rest of the statement of the question is presented below:

The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?

We know the maximum initial speed at which the seeds are dispersed is:

In addition, we know the maximum distance at which the seeds move between photographic frames is:

And we need to find the minimum frame rate of the camera with these given conditions. This can be found by finding the time for each frame and then the frame rate:

Finding the time:

This  is the time for each frame

Now we need to find the frame rate, which is the frequency at which the photos are taken.

In this sense, frequency is defined as:

Finally:

Hence, the minimum frame rate is 23000 frames per second.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

answer: a. same unit

explanation: