Prove or disprove that DTIME(n^2)=NL
Let's say that a class $\cal C$ is closed under "reverse quadratic padding" if for any language $L$, if $Q(L) = \{\langle |x|, 0^{|x|^2} \rangle : x \in L \}$ belongs to $\cal C$ then $L$ also belongs to $\cal C$.

The class $\mathsf{NL}$ is closed under reverse quadratic padding: if $Q(L) \in \mathsf{NL}$, then we can decide $L$ itself in $\mathsf{NL}$ by simulating the decider for $Q(L)$. For that we use our ability to count up to $n^2$ (here $n$ is the input length), noticing that $O(\log(n^2)) = O(\log n)$.

In contrast, the class $\mathsf{DTIME}(n^2)$ isn't closed under reverse quadratic padding. Indeed, use the time hierarchy theorem to find a language $L \in \mathsf{DTIME}(n^3) \setminus \mathsf{DTIME}(n^2)$. Then $Q(L) \in \mathsf{DTIME}(n^2)$, whereas $L \notin \mathsf{DTIME}(n^2)$.