The first Stiefel-Whitney class of a trivial line bundle $E=M \times \mathbb{R}$

Question

If the base space $M$ is non-orientable, is the trivial line bundle $E=M \times \mathbb{R}$ also non-orientable? i.e. $w_1(E) \neq 0$. If so, how could it be proved? Could we use $w_k(\xi\times\eta)=\sum_{i=0}^k w_i(\xi)\times w_{k-i}(\eta)$?

(update due to the answer) If the base space is a mobius, all fibers are perpendicular to the base space, like a 'thick mobius'. Is this fiber bundle a line bundle? Is this fiber bundle non-orientable? If so, how could I calculate its first Stiefel-Whitney class $w_1 \neq 0$?

Answer 1

The trivial vector bundle over any manifold has $w = 0$. The tangent bundle $TM$ of a nonorientable (smooth) manifold $M$ has $w(TM)\not = 0$, but that doesn't apply to arbitrary bundles over $M$.

Answer 2

This looks like Proposition 2 on page 38 of this, it is an immediate consequence of the axioms defining characteristic classes.