Due to a faulty tap, water drips from the mixer at N = 53 drops per minute. During the time t = 17 hours

Due to a faulty tap, water drips from the mixer at N = 53 drops per minute. During the time t = 17 hours, the dripping water fills a cylindrical three-liter jar. Assuming all drops to be the same, find the volume of one drop in milliliters, the weight of the drop in millinewtons. What pressure will this water create on the bottom of a can with an area of S = 99 cm2? Give your answer in kilopascals. The density of water is 1 g / cm3. Free fall acceleration g = 9.8 m / s2. Enter the drop volume to the nearest thousandths, weight and pressure – to the hundredths.

Since 1 hour = 60 minutes, then 17 hours will be:

17 * 60 = 1020 minutes.

The number of drops is defined as:

n = N * t = 53 * 1020 = 23460.

Let’s express the volume of a three-liter can in mm:

3 * 10000 = 30,000 mm ^ 3.

Then the volume of one drop will be equal to:

v = V / n = 300000/23460 = 1.279 mm ^ 3.

Its weight is:

T = m * g = 10 ^ (3) * 1.279 * 10 ^ (- 9) * 9.8 = 12.53 * 10 ^ (- 6) = 0.01 mn.

The bank is cylindrical, then its height is:

h = V / S = 0.003 / 99 * 10 ^ (- 4) =