The bowling ball is at rest, so it only has gravitational potential energy.
Ug = mgy
Ug = (2)(9.8)(40) = 784 J
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Answer:
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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
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)
Since it was dropped, it should be the speed of gravity which is 9.8 meters/second
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