One hundred jumping beans are placed along the center line of a gymnasium floor at intervals of six inches (in.). Twelve hours later, the distance of each jumping bean from the line is measured. The sum of the squares of the distances divided by 100 is 36in.2 . Note that no constraint is given on the direction or angle of the jump.

(a) Calculate the diffusion coefficient of the jumping beans.
(b) If the mean jump (or roll) distance of a bean is equal to 0.1in., estimate the mean jump (or roll) frequency of a bean.

1. a net force of 100 newton’s is applied to a wagon for 5 seconds. this causes the wagon to undergo a change in momentum of

Ryan is examining the energy of the particles in a bar of gold. what is ryan most likely studying?

Asilver dollar is dropped from the top of a building that is 1324 feet tall. use the position function below for free-falling objects. s(t) = −16t2 + v0t + s0 (a) determine the position and velocity functions for the coin. s(t) = v(t) = (b) determine the average velocity on the interval [1, 2]. ft/s (c) find the instantaneous velocities when t = 1 second and t = 2 seconds. v(1) = ft/s v(2) = ft/s (d) find the time required for the coin to reach the ground level. (round your answer to three decimal places.) t = s (e) find the velocity of the coin at impact. (round your answer to three decimal places.) ft/s