Four identical metallic spheres with charges of +8.2 µC, +9.0 µC, −7.8 µC, and −8.8 µC are placed on a piece of paper. The paper
1 answer:
Answer:
a) 0.15 μC b) 9.4*10¹¹ electrons.
Explanation:
As the total charge must be conserved, the total charge on the spheres, after being brought to contact each other, and then separated, must be equal to the total charge present in the spheres prior to be put in contact:
Q = +8.2μC +9.0 μC +(-7.8 μC) + (-8.8 μC) = +0.6 μC
As the spheres are assumed perfect conductors, as they are identical, once in contact each other, the excess charge spreads evenly on each sphere, so the final charge, on each of them, is just the fourth part of the total charge:
Qs = Qt/4 = 0.6 μC / 4 = 0.15 μC.
b) As the charge has a positive sign, this means that each sphere has a defect of electrons.
In order to know how many electrons are absent in each sphere, we can divide the total charge by the charge of one electron, which is the elementary charge e, as follows:
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Answer:
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
Explanation:
The expression for conservation of the angular momentum (L) is
Where
initial moment of inertia and angular velocity
is the final moment of inertia and angular velocity
The expression of moment of inertia of the satellite (a solid sphere) is
Where is the satellite mass
r is the radus of the sphere
Substititute 1900kg for m and 4.6m for r
The final moment of inertia of the satellite about the centre of mass
Where is the antenna's mass and
I is the length of the antenna
So, the Final rotation rate of the satellite is:
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
Answer:
12.1 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement = 120 m
a = Acceleration due to gravity on Moon = 1.67 m/s²
Equation of motion
Time taken by the rock to hit the bottom of the crater is 12.1 seconds
Answer:
electric potential is 3.31 × V
potential energy is 152 MeV
Explanation:
given data
fragment charge Q = 46 protons = 46 × 1.6 × C
to find out
electric potential and potential energy
solution
we know here distance from fragment d = 2 × m
and constant for electric force k that is 9 × N-m²/C²
so that we can find electric potential = kQ/d
electric potential = 9 × / ( 2 ×
)
electric potential = 3.31 × V
and
we know relation between electric potential and potential
that is V = U/q
so U will be = qV
now put all value
we get potential energy U
potential energy = 46 × 3.31 ×
potential energy = 1.52 × eV
so potential energy = 152 MeV
Answer :
Explanation :
Given that
Radius of sphere
The distance between the centers of the two spheres is
The charge of the sphere
The magnitude of the repulsive force between the charges pushing them a part is
Using coulomb law
Hence, this is the required solution.