Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
Answer: vf = 51 m/s
d = 112 m
Explanation: Solution attached:
To find vf we use acceleration equation:
a = vf - vi / t
Derive to find vf
vf = at + vi
Substitute the values
vf = 3.5 m/s² ( 8.0 s) + 23 m/s
= 51 m/s
To solve for distance we use
d = (∆v)² / 2a
= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)
= (28 m/s)² / 7 m/s²
= 784 m/s / 7 m/s²
= 112 m
For these question, it has two separate equations: 2f(a) and f(2a) .
For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation
For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.