C++11 move sematincs, am I understanding this right?

I'm approaching C++ for the first time, and I'm reading The C++ Programming Language as a guide.

I understand the move semantics in principle, I want to know if I understand when they are used, but I don't really know how to test it by myself.

If I have this:

vector<int> a{1,2,3,4,5};
vector<int> b = a;

this is a (slow) copy. a as ha name and it's an lvalue (it can be modified), so the compiler copies it's content into b by allocating a new vector and copying each element one by one.

In this case:

vector<int> a{1,2,3,4,5};
vector<int> b = std::move(a);

I'm forcing a "to be" a rvalue (can't be modified) and it's content it's moved to b without overhead. (By the way, is std::move an expensive operation to perform?)

If I have a function that returns a vector (foo()) and I do this

b = foo();

When foo() returns the vector, I'm getting a move, because the returned value is an rvalue. Memory is allocated only once in foo(), when I declared the vector that will be returned.

When I'm using a constructor like this:

vector<Object> x{new Object(),new Object(),new Object()};

I'm still getting moves, since the results of those three new operations are rvalues, right?

x = new Object();

This would result in a move too, assuming whatever x type is, it has a move assignment.

In the case of a container that supports push_back() (or other similar operations), if I pass to push_back() an rvalue, there will be a move, otherwise a copy. Like:

vector<int> a{1,2,3,4,5};
b.push_back(a);
b.push_back(std::move(a));

in the second case, a will be appended to b with a move and not a copy. In the first case it will be a copy.

Another thing:

vector<int> a{1,2,3,4,5};
vector<int> b = std::move(a);

In this case a is "static" memory, not dynamically allocated. It's content is moved to b. Is a taking up any more memory that I would need to free somehow?

Thank you very much for your help!