Answer:
Explanation:
First of all, we need to find the pressure exerted on the sphere, which is given by:
where
is the atmospheric pressure
is the water density
is the gravitational acceleration
is the depth
Substituting,
The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m
So the total area of the sphere is
And so, the inward force exerted on it is
Answer:
Explanation:
First, we are going to calculate the electrical potential in the point middle between the two charges
Remember that the electrical potential can be calculated as:
Where 
and it is satisfy the superposition principle, thus
The electrical potential at 10 cm from charge 1 is:
Since the work - energy theorem, we have:
where q is the electron's charge and m is the electron's mass
Therefore:
Judge a source's reliability
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
