Answer:
<h2><em><u>0</u></em></h2>
Step-by-step explanation:
<em><u>The rational number 0 is the additive identity for rational numbers</u></em><em><u>.</u></em>
<em><u>As</u></em><em><u>,</u></em>
<em><u>If</u></em><em><u> </u></em><em><u>we</u></em><em><u> </u></em><em><u>add</u></em><em><u> </u></em><em><u>any</u></em><em><u> </u></em><em><u>rational</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>zero</u></em><em><u> </u></em><em><u>we</u></em><em><u> </u></em><em><u>get</u></em><em><u> </u></em><em><u>that</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>itself</u></em><em><u>.</u></em>
<em>E.g</em><em>.</em>

and

Answer:
the answer is c
Step-by-step explanation:
Answer:
Is this even possible
Step-by-step explanation:
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.