- #1
eprparadox
- 138
- 2
Homework Statement
A large water tank, open at the top, has a small hole in the bottom. When the water level is
## 30## ##m## above the bottom of the tank, the speed of the water leaking from the hole:
A. is ##2.5## ##m/s##
B. is ##24## ##m/s##
C. is ##44## ##m/s##
D. cannot be calculated unless the area of the hole is given
E. cannot be calculated unless the areas of the hole and tank are given
Homework Equations
## p + \rho g y + \frac{1}{2}\rho v^2 = constant ##
The Attempt at a Solution
## 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 ##
We can set our reference from ## y_1 = 0 ## and because the large opening at the top is so much larger than the small opening, ## v_2 = 0 ##
This gives us
## p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g y_2 ##
Here's where I get confused. What exactly are ## p_1 ## and ## p_2 ##?
I know in solutions to this problem, they're assume to be the same since both openings are open to the atmosphere. And because of this, ## p_1 ## and ## p_2 ## cancel out and we can solve for ## v_1 ##.
But that doesn't make sense to me.
These two holes are 30 meters apart so the pressures should be different.
I'm not sure how to reconcile what ## p_1 ## and ## p_2 ## are.
Any thoughts?