Answer :
Answer:
[tex]v_h=2.8m/s[/tex]
Explanation:
The Bernoulli equation tells us that in a fluid [tex]P+\frac{\rho v^2}{2}+\rho gh=constant[/tex], so we will apply this to a point on the surface and a point on the hole.
For a point on the surface (at height [tex]h_s[/tex]) the velocity will be approximated to 0m/s since we assume it barely moves (unless other information is given), so we have:
[tex]P_{atm}+\rho gh_s=constant[/tex]
For a point on the hole (at height [tex]h_h[/tex] and velocity of the fluid [tex]v_h[/tex]) we have:
[tex]P_{atm}+\frac{\rho v_h^2}{2}+\rho gh_h=constant[/tex]
Since both constant are the same, cancelling [tex]P_{atm}[/tex] we have:
[tex]\rho gh_s=\frac{\rho v_h^2}{2}+\rho gh_h[/tex]
Which means:
[tex]\frac{\rho v_h^2}{2}=\rho gh_s-\rho gh_h=\rho g(h_s-h_h)[/tex]
So the velocity of the fluid flowing through the hole will be:
[tex]v_h=\sqrt{2g(h_s-h_h)}=\sqrt{2(9.8m/s^2)((0.5m)-(0.1m))}=2.8m/s[/tex]