A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
1 answer:
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Answer:
t = 1.02 s
Explanation:
The computation of the time required is shown below:
The package speed for belt is
= 3 - 1
= 2 m/s
Moreover, the decelerative force would be acted on the block i.e u.m.g
So, the decelerative produced
= 0.2 × 9.81
= 1.962 m/s^2
And, final velocity = 0
v = u - at
here
V = 0 = final velocity
u = 2 m/s
so,
0 = 2 - 1.962 × t
t = 1.02 s
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
