Answer:
Parallel universe, or alternate reality, is a hypothetical self-contained plane of existence, co-existing with one's own
Answer: B = 1380T
Explanation: please find the attached file for the solution
Answer:
v = 12.4 [m/s]
Explanation:
With the speed and Area information, we can determine the volumetric flow.
where:
r = radius = 0.0120 [m]
v = 2.88 [m/s]
![A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%280.0120%29%5E%7B2%7D%20%5C%5CA%3D4.523%2A10%5E%7B-4%7D%20%5Bm%5D%5C%5C)
Therefore the flow is:
![V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]](https://tex.z-dn.net/?f=V%3D2.88%2A4.523%2A10%5E%7B-4%7D%20%5C%5CV%3D1.302%2A10%5E%7B-3%7D%20%5Bm%5E%7B3%7D%2Fs%20%5D)
Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.
![v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]](https://tex.z-dn.net/?f=v%3DV%2FA%5C%5Cv%3D1.302%2A10%5E%7B-3%7D%20%2F1.05%2A10%5E%7B-4%7D%20%5C%5Cv%3D12.4%5Bm%2Fs%5D)
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
