Answer:
Explanation:
For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

where the suffix i means initial, and the suffix f means final.
The initial momentum will be:

as the second puck is initially at rest:

Using the unit vector
pointing in the original line of motion:



So:


Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

So, our velocity vectors will be:


We got


So, we got the equations:

and
.
From the last one, we get:




and, for the first one:






so:

and


Answer:
C. software
Explanation:
software, is a collection of data or computer instructions that tell the computer how to work. This is in contrast to physical hardware, from which the system is built and actually performs the work.
<span>Actually in this case heat energy is being transferred. Heat
energy or thermal energy is transferred from the burning of wood to the
sausages for it to be cooked. The sausage is being heated by the fire and is
absorbing the heat or thermal energy.</span>
Because its expose to the wires inside that could electrify you.