Are you asking about independent and dependent variables?
The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.
Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:
w * t = E
w: Watts
t: Time
E: Energy required
(Watts times time is equal to the energy required)
<u>Input our values:</u>
400 * t = 3000
(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)
<u>Divide both sides by 400 to isolate t:</u>
<u />
= 
t = 7.5 (s)
<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>
<u></u>
If you have any questions on how I got to the answer, just ask!
- breezyツ
I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³.
I'm not sure whether that is the answer you were looking for, but that's what I found.
We can solve the problem by using the first law of thermodynamics, which states that:

where

is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
Answer:
Thin, aluminium and buried underground.
Explanation:
When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.
i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.
ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.
iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.