Answer:
![[\psi]= [Length^{-3/2}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%3D%20%5BLength%5E%7B-3%2F2%7D%5D)
- This means that the integral of the square modulus over the space is dimensionless.
Explanation:
We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral
must be dimensionless, as represents a probability.
As the differentials has units of length
![[dx]=[dy]=[dz]=[Length]](https://tex.z-dn.net/?f=%5Bdx%5D%3D%5Bdy%5D%3D%5Bdz%5D%3D%5BLength%5D)
for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:
![[\psi]^2 = [Length^{-3}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%5E2%20%3D%20%5BLength%5E%7B-3%7D%5D)
taking the square root this gives us :
![[\psi] = [Length^{-3/2}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%20%3D%20%5BLength%5E%7B-3%2F2%7D%5D)
Answer:
resistor R₂ has the lowest current density
Explanation:
The current density is
j = I / A
now let's analyze each case
a) R₂ has an area 2A₀ and a length L₀ that R₁
b) R₃ has an area Ao and a length 3L₀ what R₁
we can see that all the area is given in relation to the resistance R₁
the current density in R₁ is
j₁ = I / A₀
the current density in R₂
j₂ = I / 2A₀
j₂ 2 = ½ I/A₀
the current density in R₃
j₃ = I / A₀
j₂ < j₁ = j₃
therefore resistor R₂ has the lowest current density
The center of the ghole is the singularity. or a point where extremely large amounts of matter are crushed into an infinitely small amount of space. hope it helps :)
The independent variable would be the fertilizer as it is what the dependent variable (the growth of grass) depends on.
