The apparent weight of a 1.1 g drop of water is 4.24084 N.
Solution:
N = Speed of rotation = 1250 rpm
D = Diameter = 45 cm
r = Radius = 22.5 cm
M = Mass of drop = 1.1 g
Angular speed of the water = [tex]\omega = \frac{2\pi N}{60}[/tex]
[tex]\omega = \frac{2\pi \times 1250}{60}[/tex]
[tex]\omega = 130.89 rad/s[/tex]
Apparent weight is given by
[tex]W _a = M\omega^{2}R[/tex]
[tex]W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225[/tex]
[tex]W_a[/tex] = 4.24084 N
Know more about Apparent weight https://brainly.com/question/14323035
#SPJ4
Question:
The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?