What we have so far:
Kinetic energy = 0. The reason behind that is because the beam is not moving at a height of 40m.
Gavity, g = 9.8m/s²
Height = 40m
Potential energy = mgh; this is equal to 0 because m, stands for mass and in this problem, we do not have a value for the mass of the beam. Hence, 0 x 9.8m/s² x 40m = 0. Potential energy = 0.
Solution:
We will use the equation of Total energy:
TE = potential energy + kinetic energy
TE = 0 + 0
∴ TE = 0
The answer is: Assuming no air resistance, the total energy of the beam as it hits the ground is 0.
To be honest the numbers up there are like dividing like it say 2 or -2 so i would be 2/-2 i hope these helps i really do
First, find the product (w*r)(x): (w*r)(x) = (x-2)*[2-x^2] = 2x - x^3 - 4 + 2x^2
This is a cubing function. Since the sign of the cube-of-x term is negative, the graph will begin in Quadrant II and pass through Quadrant IV. There are no limits on y. Thus, the range is (-infinity, +infinity).
