Answer:
The total charge Q of the sphere is
.
Explanation:
Given that,
Radius = 5 cm
Charge density 
We need to calculate the total charge Q of the sphere
Using formula of charge

Where,
= charge density
V = volume
Put the value into the formula

Put the value into the formula


Hence, The total charge Q of the sphere is
.
Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
Work needed = 23,520 J
<h3>
Further explanation
</h3>
Given
height = 12 m
mass = 200 kg
Required
work needed by the crane
Solution
Work is the transfer of energy caused by the force acting on a moving object
Work is the product of force with the displacement of objects.
Can be formulated
W = F x d
W = Work, J, Nm
F = Force, N
d = distance, m
F = m x g
Input the value :
W = mgd
W = 200 kg x 9.8 m/s²x12 m
W = 23520 J
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)