What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.

26467.21 N

Explanation:

Initial height of water ( h1 )  = 5 m

diameter of container ( d1 )= 100 cm

pressure inside the container ( p1 )= 0.850 atm

Diameter of faucet ( d2 )= 1 inch

Calculate the value of the net force on the side of container

lets assume ; pressure outside the container ( p2 ) = 1 atm

Fnet = ( P1A1  + mg ) - ( P2A2 )

       = [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]

       = [ 16902.2766 + pvg ] - [ 51.3161 ]

where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25

       = [ 16902.2766 + 9616.25 ] - [ 51.3161 ]   = 26467.21 N ( downwards )

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the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

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