The tomato is dropped. what is the velocity, v, of the tomato when it hits the ground? assume 85.6 % of the work done in part a is transferred to kinetic energy, e, by the time the tomato hits the ground.

Ok, we know that you drop a tomato of mass M from a height h, because you drop it, it has not initial velocity.

Now, the kinetic energy is:

K = (M/2)*v^2

because at the begginig there is not velocity, the kinetic energy is zero.

the potential energy is:

U = M*g*h

where g = 9.8m/s^2

We know that when the tomato hits the ground, the 85.6% of these potential energy is converted in kinetic; so we have:

K = 0.856*M*g*h = (1/2)*M*v^2

M can be canceled in both sides:

v^2 = 2*0.856*g*h

v = √(1.712*g*h)

is the velocity of the tomato when it hits the ground.

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

the inner core.

good luck: )

klicker