A cylindrical container with a diameter of 7.62 cm is resting on the bottom of y

Question

A cylindrical container with a diameter of 7.62 cm is resting on the bottom of your friend's swimming pool in the orientation shown in the figure. the swimming pool is 9.14 m deep. the density of the chlorinated water in the pool is 1.15 x 10^3 kg/m^3

a) what is the magnitude of the force that hte water is exerting n a circular side of the cylinder?

b) the cylinder's weight in air, before it was dropped into the pool, was 22.2 N. it's weight at the bottom of the pool is 14.0 N. what is the horizontal length of the cylinder, in cm?

to get full points please include detailed equations and steps used, thanks!

Explanation / Answer

Part A)

P = pgh + Patm

P = (1000)(9.8)(9.14) + 1.013 X 10^5

P = 190872 Pa

P = F/A

F = PA

F = (190872)(pi)(.0381)^2

F = 870 N

Part B)
Fb = 22.2 - 14 = 8.2 N

Fb = Vpg and V of a cylinder is (pi)r^2(h)

8.2 = (pi)(.0381^2)(h)(1000)(9.8)

h = .183 m (18.3 cm)

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