Recent content by Envy9268

I'm now getting 950 cm/s^2, slightly off from your calculation, but much much more realistic. I wonder where the discrepancy is now, but thank you so much for helping me understand.

Would that not be the length of the glider, 10.15 cm?

That makes sense. According to the lab handout, it says Δx is the object's (vector) displacement.

Hmm, so is (Δt)_avg the average time required for the glider to pass through the photogate from start to finish? That makes more sense, because the photogate couldn't know when the glider was released.

That's good news at least. I'm not sure what I'm missing then.

The time it takes for the glider to reach the photogate

Hi, so this is a lab in which we used an air track at an angle and a glider to gather some data through various trials, ultimately to calculate "g". L_glider = 10.15 cm x (photogate activation point) = 547.5 mm or 54.75 cm x_0 (release point) = 1800.0 mm or 180.00 cm (Δx)_midpoint = | x - x_0 |...