1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec
Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s
In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons
Hey there,
Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does

affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).
Hope this really helps you.
Answer:
(A) a heuristic
Explanation:
A heuristic:It is a reasoning strategy to find answers, make judgement about any something."
It is possible to choose between the options given, we have the following interpretation as; Interpretation: "A heuristic: finding answers is a thinking technique, evaluating something." Mechanism: In the heuristic approach the method is to find solutions or answers to a question by choosing the right and optimal compositions.
Answer:

Explanation:
The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

After a quick substitution, the expanded expression is:



The final temperature of the system is:
