You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. part a in one day, by how much does the distance traveled by the tip of the minute hand exceed the distance traveled by the tip of the hour hand?

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

7.4 * 4π = 93 mm

In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is

12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

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force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach