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2 answers:
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Answer:
Choice a. Approximately on average.
Explanation:
The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.
Assume that electric charge of size flowed through a wire cross-section over a period of time
. The average current in that wire would be:
.
For this question:
, whereas
.
Therefore, the average current in this circuit would be:
.
However, the units in the choices are all in (for Amperes.) One Ampere is equal to one
. It will take some unit conversations to change the unit of
(coulombs-per-minute) to coulombs-per-second.
.
Hence, the most accurate choice here would be choice a.
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:
0
Explanation:
Given:
initial velocity,
distance of telescope from the launch pad,
time after which the observation starts, t = 3 min.
Now, from the equation of motion:
Which indicates that the after the given time the rocket would have fallen down on the ground if it did not had an acceleration of its own.