An Open Cylindrical Tank of Acid....

[The-Mad-Lisper]

Homework Statement

An open cylindrical tank of acid rests at the edge of a table 2.20\cdot 10^0\ m above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is 7.00\cdot10^{-1}\ m deep?
Let \rho be the mass density of acid, g be the acceleration of gravity, h_{tank} be the depth of the tank, and h_0 the height of the base of the tank.

Homework Equations

p_1+\rho g y_1+\frac{1}{2}\rho {v_1}^2=p_2+\rho g y_2+\frac{1}{2}\rho {v_2}^2 (Bernoulli's Equation)

The Attempt at a Solution

\Delta x = vt

\frac{1}{2}gt^2=h_0
t^2=\frac{2h_0}{g}
t=\sqrt{\frac{2h_0}{g}}

\rho g h_{tank}=\frac{1}{2}\rho v^2
gh_{tank}=\frac{1}{2}v^2
v^2=2gh_{tank}
v=\sqrt{2gh_{tank}}

\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}
\Delta x=\sqrt{\frac{2gh_{tank}h_0}{g}}
\Delta x=\sqrt{2h_{tank}h_0}
\Delta x=\sqrt{2\cdot 7.00\cdot 10^{-1}\ m\cdot 2.20\cdot 10^0\ m}
\Delta x=\sqrt{3.08\cdot 10^0\ m^2}
\Delta x = 1.75\cdot 10^0\ m

Unfortunately, 1.75\ m is not the correct answer. Perhaps I am not setting up Bernoulli's equation correctly for this problem.

 

[SteamKing]

Homework Statement

An open cylindrical tank of acid rests at the edge of a table 2.20\cdot 10^0\ m above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is 7.00\cdot10^{-1}\ m deep?
Let \rho be the mass density of acid, g be the acceleration of gravity, h_{tank} be the depth of the tank, and h_0 the height of the base of the tank.

Homework Equations

p_1+\rho g y_1+\frac{1}{2}\rho {v_1}^2=p_2+\rho g y_2+\frac{1}{2}\rho {v_2}^2 (Bernoulli's Equation)

The Attempt at a Solution

\Delta x = vt

\frac{1}{2}gt^2=h_0
t^2=\frac{2h_0}{g}
t=\sqrt{\frac{2h_0}{g}}

\rho g h_{tank}=\frac{1}{2}\rho v^2
gh_{tank}=\frac{1}{2}v^2
v^2=2gh_{tank}
v=\sqrt{2gh_{tank}}

\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}
\Delta x=\sqrt{\frac{2gh_{tank}h_0}{g}}
\Delta x=\sqrt{2h_{tank}h_0}
\Delta x=\sqrt{2\cdot 7.00\cdot 10^{-1}\ m\cdot 2.20\cdot 10^0\ m}
\Delta x=\sqrt{3.08\cdot 10^0\ m^2}
\Delta x = 1.75\cdot 10^0\ m

Unfortunately, 1.75\ m is not the correct answer. Perhaps I am not setting up Bernoulli's equation correctly for this problem.

No, the problem is not with the Bernoulli equation.

You have:

t=\sqrt{\frac{2h_0}{g}}

and

v=\sqrt{2gh_{tank}}

when you combine these two equations:

\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}

you dropped one of the factors of 2 under the radical sign for some reason.