Answer:

Explanation:



Electron information needed to solve the question:






![E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.11x10%7B-31%7Dkg%2A3.0x10%5E%7B12%7Dm%2Fs%5E2%7D%7B-1.6x10%7B-19%7DC%7D-%5B%2819.0x10%5E3mj%2B18.0x10%5E3m%29xi%28400x10%5E%7B-6%7DT%29%5D)
![E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C](https://tex.z-dn.net/?f=E%3D-i17.08N%2FC-%5B7.6%28-k%29%2B7.2%28j%29%5DN%2FC)

Answer:
I think it's a because it goes thru it and reflects
Answer:
coefficient of static friction of the surface and the normal force
Explanation:
The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR
In fact, entropy of an isolated system never decreases (2nd law of thermodynamics), unless some external energy is provided in order to "restore" order in the system and decrease its entropy.
(note that when external energy is added to the system, it is no longer "isolated").
*This is only true if the question is referring to a certain system within the universe. If we are considering the universe itself as the system, then this option is no longer correct, because no external energy can be provided to the universe, and since the universe is an isolated system, its entropy can never decrease. If we are considering the universe itself as the system, none of the options is true.
Answer:
The length of open-open pipe needed is 6.23 m
The length of open-close pipe needed is 3.11 m
Explanation:
Fundamental frequency for standing wave mode of an open- open pipe is given by

where v is the velocity and L is the length
The length of open-open pipe needed is

Fundamental frequency for standing wave mode of an open- close pipe is given by

The length of open-close pipe needed is
