The closed tank of fig. p.2.34 is filled with water and is 5 ft long. the pressure gage on the tank reads 7 psi. determine: (a) the height, h, in the open water column, (b) the gage pressure acting on the bottom tank surface ab, and (c) the absolute pressure of the air in the top of the tank if the local atmospheric pressure is 14.7 ps
the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the box is the same as its force.
hope this !
im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this. hope that you like the video and that it explains more in detail to you.
that is the link.
No because scientist are still changing there mines today so they are not correct to this day
(a) the height in the open water column is: 18.15 (ft), (b) the gauge pressure acting on the bottom tank surface AB is: 8.73 (Psi) and (c) the absolute pressure of the air in the top of the tank with local atmospheric pressure at 14.7 Psi will be: 21.7 (Psi).
Explanation:
We need to apply the Pascal's Law, ([tex]P=Y*h[/tex]) where P is pressure, γ is the specific weight of the fluid and h is the height of the column of fluid, to find the pressure at differents points in the tank (see attached). First we need to assume a property of the water called specific weight as: 62.4 (lbf/ft^3) or 0.03612 (lbf/in^3). Then knowing that 1 feet as the same as 12 inches and appling Pascal's Law, we get: [tex]P_{c}=P_{gauge(air)}+Y_{H2O}*h_{c} =7+0.03612*24=7.87(Psi)[/tex] and the other hand, we have:[tex]P_{d} =Y*h=0.03612*h[/tex] and knowing that the pressure at point C and point D are the same, we can find the h as:[tex]P_{c}=P_{d}[/tex], so [tex]7.87 (Psi)=0.03612*h[/tex] getting h=217.75(in) or 18.15(ft) (a). Now similarly to find the gauge pressure acting on the bottom of the tank surface AB we can apply Pascal's Law as:[tex]P_{gaugeAB}=P_{gauge(air)}+Y_{H2O}*h_{AB} =7+0.03612*48=8.73(Psi)[/tex] (b). Finally we can find the absolute pressure of the air in the top of the tank, assuming a local atmospheric pressure as 14.7 (Psi) so:[tex]P_{abs(air)}=P_{gauge(air)} +P_{atm} =7+14.7=21.7(Psi)[/tex] (c).
[tex]The closed tank of fig. p.2.34 is filled with water and is 5 ft long. the pressure gage on the tank[/tex]