Bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a dist

Incomplete question: "A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?"

The normalized corrected value at 32.5 mm is 0.1638

Explanation:

The corrected light measurement at 29 mm is equal to:

20.7 - 5.1 = 15.6 mV

The corrected light measurement at 32.5 mm is equal to:

15.6 - 5.1 = 10.5 mV

To normalize the data to its maximum value means that the maximum value must be calculated and the data must be scaled using that value, as in this case the maximum value is 15.6 mm, then the normalized corrected value at 32.5 mm is equal to:

10.5 * 15.6 = 163.8 = 0.1638

what type of charge does it have? negative

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.