Asealed tank containing seawater to a height of 10.7 m also contains air above the water at a gauge pressure of 3.20 atm. water flows out from the bottom through a small hole. how fast is this water moving?

28.97 m/s

Explanation:

height of water (H1) = 10.7 m

Gauge pressure (Pg) = 3.2 x 1.01325 x 10^{5} = 324240 N/ m^[2}

pressure at the top (P1) = gauge pressure (Pg) + atmospheric pressure(Pa)  (Pg + Pa)

acceleration due to gravity (g) = 9.8 m/s^{2}

height at the bottom (H2) = 0

pressure at the bottom (P2) = atmospheric pressure = (Pa)

density of sea water (ρ) = 1030 kg/m^{3}

speed of water at the top (V1) = 0 (assuming the tank is big, the speed of water at any point on the surface is almost 0 and also the hole is very small compares to the size of the tank)

speed of water at the top (V2) = ?

applying Bernoulli's equation we can find the value of V2.

(take note that represents ρ whcih is the density of sea water)

recall that P1 = Pg + Pa and P2 = Pa

)

)

V2 =

V2 =

V2 = 28.97 m/s

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever