This question is from a physics lab on Gravitational Acceleration. Given that ea

Question

This question is from a physics lab on Gravitational Acceleration.

Given that each tick from a spark timer occurs at one sixtieth of a second, get the slope in units of cm/sec^2. Compare your answer with the standard value of 980 cm/sec^2by computing the percentage difference.

Our lab data for instantaneous velocity gives us a best fit line with and equation of y=.5223 + 2.0692 giving a slope of .5223 cm/ticks^2. This conversion seems as if it should be pretty straight forward using the equation cm/tick^2 = (3600 ticks) (cm/s^2) but this is giving me 1880.28 cm/s^2 and this is about twice as much as it should be. I am not really sure why this is coming out to be twice as much as the experimental data from the position vs time graph suggest it should be.

Explanation / Answer

Hi....the method with which you are doing this is absolutely correct.

The conversion factor for ticks and seconds is 60 ticks / s. Therefore,  [60 ticks / s]² = 3600 ticks² / s² . The answer 1880 which you are getting is due to the fact that your slope value is very high.

Now, I think the problem is with the value of your slope. It should not be 0.5223. This value is too high...I would like you to check again the value and calculate the slope manually. It should be around 0.26 or near to that.

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