Why adhesion is stronger at faster peeling rate? Adhesion is strong at fast peeling rates and weak at slow peeling rate (paper, video; Newtonian adhesive, non-Newtonian adhesive?), i.e. it is rate-sensitive. Why?
(source: Meitl et al.: thanks to rate-sensitive adhesion energy, objects were transferred from one substrate to another -- peel-off at 10 cm/s has higher adhesion than peel-off at 1 cm/s)
 A: The crucial word here is rate. When pulling one gives a dp/dt, momentum to the pulled tape, which is the force that is applied. This can be given as delta(p)/delta(t). For the same transfer of momentum delta(p) to the unstuck tape, a smaller delta(t) (faster) will result in a greater force. This can be exacerbated because one gives a bigger momentum to the unstuck part when in a hurry to do it fast.
A: Adhesion (wiki) can be due to several phenomena. In some, the adhesive is a very viscous liquid which will slowly deform (or "flow") with a small applied force.
A: Adhesion is not stronger at faster peeling rate. If we formally associate adhesion as "the amount of energy needed to separate two parts", then we can describe it physically by the potential joining those two parts together.
In classical mechanics, where potentials are independent of time, like in gravitation or electrostatics, you can still explain why it "feels harder" to separate two parts the faster you do it: it "feels harder" because the force needed is bigger.
But why do we need bigger force to separate two parts attached by a potential which is independent of the external force applied or the time?
Because the effect expected (parts separated to a non-interaction distance) involves using an amount of energy equal to the potential energy bounding the parts. And although this amount of energy is independent of time, the force needed does depend on the time desired. 
In mathematical expression, the work performed by the external force:
$$W_e = \int_0^d F_e(s)ds$$
is a time independent magnitude which does not care about the force involved nor its time dependence. But if we put $ds = v(t)dt$ we get:
$$W_e = \int_0^{t_d} F_e(t) v(t)dt$$
it can be seen how, the shorter is the time desired to achieve full separation $t_d$, the higher values of the integrand is needed. This integrand is a magnitude of power, or energy per unit of time, thus the faster the separation, the higher the power needed.
However, the power is the product of the force and velocity here $F_e(t) v(t)$, so is not clear yet why the force has to be higher. After all, a higher $v(t)$ would also contribute to a higher power so $F_e(t)$ does not have to be necessarily higher for faster separation.
Well actually the speed of separation depends on the balance of the external force to the force related to the bounding potential $F_b$:
$$v(t) = \int_0^{t_d} \frac{F_e - F_b}{m_a}dt$$
which is always smaller than the maximum velocity $v(t) < v_m(t)$ that can be achieved by $F_e(t)$, which is:
$$v_m(t) = \int_0^{t_d} \frac{F_e}{m_a}dt$$
and therefore, whatever the force you apply $F_e(t)$
$$F_e(t)v(t)<F_e(t)v_m(t)$$
it will not be as effective in increasing speed, as if there were no bounding potential (nor the related force $F_b(t)$). 
Thus although $v(t)$ will be higher the faster is the removal, it will not suffice to account for the necessary power, and an increase in $F_e(t)$ will always occur such that the total work performed is constant however small the time employed.
A: The reason is the same as with non-newtonian fluids.
As you can see there, the work done is not independent to the rate.
If you do it with lower velocity, you can get through with less work.
There is a lot of Video's from these "non-newtoniand Fluids" in youtube.
The reason why this is so, lies on the understaning of Turbulence; Turbulent water is less hard than laminar/still water. 
This is not mainstram physics, but Turbulence is a cut/crack in fluid; there is surface's inside the fluid. And to produce these surface's you need surface energy. When your penetration rate is too high, you are producing these surfaces/cuts on bigger volume of the fluid. (conical distribution) If you penetrate with less velocity, you only make a single cut, and thus you need less energy, as less surfaces/cuts is produced.
You can think this even with glass. Normal Glass vs. Safety Glass; you need more force to break the safety glass, cause it needs the force to split it completely when the normal glass just needs the force to cut it in few pieces. 
Answer; 
With the lower rate there is minimum amount of atomic bonds separated, with the higher rate there is much more atomic bonds that are being separated, and thus the adhesion appears much stronger. 
