What type of energy do you observe from a light bulb
1 answer:
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Answer:
a) q_inner = -Q
, q_outer = -2Q
b) E₁ = k Q / r² r<a
E₂ = 0 a<r<b
E₃ = - k 2Q/r² r>b
d) the charge continues inside the spherical shell, the results do not change
Explanation:
a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign
q_inner = -Q
the outer shell of the shell the load is
q_outer = -3 Q + Q
q_outer = -2Q
b) To find the electric field again, use Gauss's law,
We define as a Gaussian surface a sphere
Ф = E. dA = /ε₀
in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product
E A = q_{int}/ε₀
the area of a sphere is
A = 4 π r²
E = 1 / 4πε₀ Q/ r²
k = 1 / 4πε₀
let's apply this expression to the different radii
i) r <a
in this case the load inside is the point load
= + Q
E₁ = k Q / r²
ii) the field inside the shell
a <r <b
As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.
E₂ = 0
iii) r>b
q_{int} = Q- 3Q = -2Q
E₃ = k (-2Q/r²)
E₃ = - k 2Q/r²
c) see attached
d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface
Answer:
Velocity = 0.4762 m/s
Explanation:
Given the details for the simple harmonic motion from the question as:
Angular frequency, ω = 12 rad/s
Amplitude, A = 0.060 m
Displacement, y = 0.045 m
The initial Energy = U = (1/2) kA²
where A is the amplitude and k is the spring constant.
The final energy is potential and kinetic energy
K + U = (1/2) mv² + (1/2) kx²
where x is the displacement
m is the mass of the object
v is the speed of the object
Since energy is conservative. So, the final and initial energies are equal as:
(1/2) k A² = (1/2) m v² + (1/2) kx²
Using, ω² = k/m, we get:
Velocity:
<u>Velocity = 0.4762 m/s</u>