Answer:
Explanation:
The acceleration of the ball would be due to the downward force of gravity, 9.8m/s^2. In order to find the displacement given that interval of time, you have to use the corresponding kinematic formula:
The initial velocity was given, the time was given, and the acceleration was given. Therefore:
To find the required time given a desired final velocity, we can use:
Complete Question
The complete question is shown on the first uploaded image
Answer:
the compass direction of the resultant displacement is south of west
Explanation:
Generally using cosine we can obtain the resultant R as follows
=>
=>
We can obtain the direction of the resultant by first using sine rule to obtain angle C as follows
=>
=>
=>
Then the direction is obtained as
=>
=>
Hence the compass direction of the resultant displacement is south of west
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
