A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. When it is 5 m above the bottom of the hill, how much mechanical energy does the ball have? (Assume the bottom of the hill is 0 height)

How is it possible that phylogenies based on sequences from nuclear genomes and organellar genomes (i.e., chloroplasts and mitochondria) from the same group of species yield phylogenetic trees with different branching orders?   answer :   the genome of the endosymbiont is more closely related to members of the group in which it originally evolved, while the nuclear genome of the engulfing organism has its own evolutionary trajectory.

D=m/v m=d*vd=densitym=massv=volumed(ether)=0.71 gr/cm³=0.71 gr/ mlv=130 mlm=d*vm=0.71 gr/ml*(130 ml)=92.3 gm=92.3 g

Let l be the vertical distance of pendulum when it makes 25.0 degree angle to verticalcos(25.0) = l / 1.20 ml = (1.20m)(cos 25.0)so the height, h, of the pendulum above the equilibrium position is h = 1.20m - (1.20m)(cos 25.0)ep,i + ek, i = ep, f = ek, fwhen pendulum held at 25.0 degrees, initial kinetic energy is 0 jwhen pendulum passes through equilibrium point, final potential energy is 0jep, i = ek fmgh = 1/2 m v^2gh = 1/2 m v^2v = √(2gh)v = √[2 * 9.81 m/s² * ( 1.20m - (1.20m)(cos 25.0) )]v = 1.5 m/s