-- If the field were inclined to the surface, then it would have
some component parallel to the surface.
-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
field trying to move them.
Answer:
I think the answer is option D ...
bt I m not sure..
Answer:
Matthew should use the safety glasses as his PPE.
Explanation:
Answer:
K = 36 J
Explanation:
In this exercise we must use the conservation of mechanical energy at two points at initial x = 0 and the end point of maximum compression of the spring
Initial x = 0
Em₀ = K = ½ m v²
Final point of maximum compression
= Ke = ½ k x²
Em₀ = 
K = ½ k x²
We put the kinetic energy because it is a data, let's look for the spring constant
k = 2 K / x²
k = 2 9 / d²
k = 18 / d²
Let's calculate the kinetic energy, so that the compression is x = 2d
K = ½ k x²
K = ½ (18 / d²) (2d)²
K = ½ 18/d² 4d²
K = 36 J