Rate of applied force versus material failure?

Question: Does the rate at which a force is applied to an object determine the maximum force reached that mechanically fails the object? If so, what are the concepts that need to be understood to understand the force-rate dependence of mechanical failure?

hypothetical scenario that prompted the question: A burst disc is placed between, and isolating, two volumes of gas initially at equal pressures as shown in the image below. The initial pressures of the gases are above the pressure rating of the burst disc. At one end of the chamber there is a valve to let gas contained on the right-hand side (RHS) of the burst disc out. As the gas is let out of the valve, the pressure differential across the burst disc increases and consequently, the resulting force acting on the burst disc (from left to right) increases.

The valve can be opened to a small degree to let the gas out slowly, causing a slow decrease in pressure on the RHS, and a slow increase of resulting force acting on the burst disc from left to right. Given enough time the resulting pressure differential (resulting force) will be enough to burst (fail) the disc.

If the valve is opened to a large degree, letting the gas out rapidly, would the disc fail at a resulting pressure differential (resulting force) less than what was seen for the slow gas scenario?

Edit: The time scale for the rapid pressure release scenario is on the order of milliseconds. I have heard of concepts of Impulse, Jerk, and pressure/load/hydraulic shock. I am not sure if of these concepts would play a role in this situation. If any do, I would like to hear how.

Maybe. My answer wavers between "no" for the idealized case to "possibly" for the real-world case. This is the way I see it: If there is a rapid decompression of the RHS gas by quickly opening the valve fully, the pressure differential on the disc rapidly increases from zero to the pressure of the LHS gas. BUT since there is no overpressure or overshoot of the pressure differential (since the pressure differential can never exceed $P_{LHS}-0=P_{LHS}$), at no time do the stresses on the disc ever exceed the stresses that the disk would experience if the valve were opened slowly.